{ "chapter_id": 4, "chapter_title": "The d -and f -Block Elements", "uploaded_filename": "Chapter4.zip", "uploaded_sha256": "31268c46f8684e8ff1cc758166d8d4d5b06fe941dcc1344509415ce7fc9d2bef", "status": "confirmed", "sections": [ { "section_id": "4.2", "section_title": "Topic-wise Q & A", "heading_text": "Topic-wise Questions and Answers", "detected_type": "SUB", "match_confidence": 0.609, "questions": [ { "sequence_no": 1, "question_text": "
What are d-block elements? Explain classification of d-block elements.
", "answer_text": "
The d-block of the periodic table contains the elements of the groups 3-12 in which the
d-orbitals are progressively filled in each of the four long periods.
\n
The name transition metals is often used to refer to the elements of d-block.
\n
There are mainly three series of the transition metals, 3d-series (Sc to Zn), 4d-series (Y to Cd) and 5d-series (La and Hf to Hg).
\n
The fourth 6d-series which begins with Ac is still incomplete.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-3", "marks": 2, "source_number": "1", "original_raw_text": " What are d-block elements? Explain classification of d-block elements.", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 2, "question_text": "
What transition elements? Explain with examples.
", "answer_text": "
Transition element is defined as the one which has incompletely filled d-orbitals in its ground state or in any one of its oxidation states.
\n
Examples:\t21Sc: [Ar] 3d1 4s2
\n
42Mo: [Kr] 4d5 5s1
\n
25Mn: [Ar] 3d5 4s2
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "source_number": "2", "original_raw_text": " What transition elements? Explain with examples. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 3, "question_text": "
What is ƒ-block element?
", "answer_text": "
The element in which last electron fill in ƒ-orbital is known as ƒ-block element.
\n
ƒ-block element is known as inner transition element.
\n
The elements constituting the ƒ-block are those in which the 4ƒ – and 5ƒ – orbitals progressively filled.
\n
The two series of the inner transition metals,
4ƒ and 5ƒ are known as lanthanoids and actinoids respectively.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "marks": 2, "source_number": "3", "original_raw_text": " What is ƒ-block element? ", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 4, "question_text": "
Explain position of d-block in periodic table.
", "answer_text": "
The d-block occupies the large middle section flanked by s- and p-blocks in the periodic table.
\n
The name 'transition' given to the elements of d-block is only because of their position between s- and p-block elements, and their properties is transition between s-block and p-block elements.
\n
The d-orbitals of the penultimate energy level in their atoms receive electrons giving rise to the three rows of the transition metals, i.e., 3d, 4d and 5d.
\n
The fourth row of 6d is still incomplete.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-6", "marks": 2, "source_number": "4", "original_raw_text": " Explain position of d-block in periodic table. ", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 5, "question_text": "
Electronic configuration of first transition metal.
", "answer_text": "", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [ { "id": "tbl-1", "rows": 2, "columns": 3, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Element

\n		\n

Atomic Number

\n		\n

Electronic Configuration

\n
\n

Sc

\n

Ti

\n

V

\n

Cr

\n

Mn

\n

Fe

\n

Co

\n

Ni

\n

Cu

\n

Zn

\n		\n

21

\n

22

\n

23

\n

24

\n

25

\n

26

\n

27

\n

28

\n

29

\n

30

\n		\n

[Ar] 3d1 4s2

\n

[Ar] 3d2 4s2

\n

[Ar] 3d3 4s2

\n

[Ar] 3d5 4s1

\n

[Ar] 3d5 4s2

\n

[Ar] 3d6 4s2

\n

[Ar] 3d7 4s2

\n

[Ar] 3d8 4s2

\n

[Ar] 3d10 4s1

\n

[Ar] 3d10 4s2

\n
", "position": "answer", "context": "Element Atomic Number Electronic Configuration Sc Ti V Cr Mn…" } ], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-3", "source_number": "5", "original_raw_text": " Electronic configuration of first transition metal. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 6, "question_text": "
Electronic configuration of second transition metal.
", "answer_text": "", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [ { "id": "tbl-1", "rows": 2, "columns": 3, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Element

\n		\n

Atomic Number

\n		\n

Electronic configuration

\n
\n

Y

\n

Zr

\n

Nb

\n

Mo

\n

Tc

\n

Ru

\n

Rh

\n

Pd

\n

Ag

\n

Cd

\n		\n

39

\n

40

\n

41

\n

42

\n

43

\n

44

\n

45

\n

46

\n

47

\n

48

\n		\n

[Kr] 4d1 5s2

\n

[Kr] 4d2 5s2

\n

[Kr] 4d4 5s1

\n

[Kr] 4d5 5s1

\n

[Kr] 4d5 5s2

\n

[Kr] 4d7 5s1

\n

[Kr] 4d8 5s1

\n

[Kr] 4d10 5s0

\n

[Kr] 4d10 5s1

\n

[Kr] 4d10 5s2

\n
", "position": "answer", "context": "Element Atomic Number Electronic configuration Y Zr Nb Mo Tc…" } ], "metadata": { "source_class": "Quest", "source_number": "6", "original_raw_text": " Electronic configuration of second transition metal. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 7, "question_text": "
Electronic configuration of third transition metal.
", "answer_text": "", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [ { "id": "tbl-1", "rows": 2, "columns": 3, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Element

\n		\n

Atomic Number

\n		\n

Electronic configuration

\n
\n

La

\n

Hf

\n

Ta

\n

W

\n

Re

\n

Os

\n

Ir

\n

Pt

\n

Au

\n

Hg

\n		\n

57

\n

72

\n

73

\n

74

\n

75

\n

76

\n

77

\n

78

\n

79

\n

80

\n		\n

[Xe] 5d1 6s2

\n

[Xe] 4f14 5d2 6s2

\n

[Xe] 4f14 5d3 6s2

\n

[Xe] 4f14 5d4 6s2

\n

[Xe] 4f14 5d5 6s2

\n

[Xe] 4f14 5d6 6s2

\n

[Xe] 4f14 5d7 6s2

\n

[Xe] 4f14 5d9 6s1

\n

[Xe] 4f14 5d10 6s1

\n

[Xe] 4f14 5d10 6s2

\n
", "position": "answer", "context": "Element Atomic Number Electronic configuration La Hf Ta W Re…" } ], "metadata": { "source_class": "Quest", "source_number": "7", "original_raw_text": " Electronic configuration of third transition metal.", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 8, "question_text": "
Why the chromium and copper have exceptional electronic configuration?
", "answer_text": "
There is little energy difference between (n-1)d and ns-orbitals.
\n
Furthermore, half and completely filled sets of orbitals are relatively more stable.
\n
A consequence of this factor is reflected in the electronic configurations of Cr and Cu in the 3d-series.
\n
In case of Cr
\n
24Cr: [Ar] 3d5 4s1 instead of [Ar] 3d4 4s2
\n
The energy gap between the two sets (3d and 4s) of orbitals is small enough to prevent electron entering the 3d orbitals.
\n
Similarly in case of Cu,
\n
29Cu: [Ar] 3d10 4s1 instead of [Ar] 3d9 4s2.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "source_number": "8", "original_raw_text": " Why the chromium and copper have exceptional electronic configuration?", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 9, "question_text": "
What are the characteristics of the transition elements and why are they called transition elements? Which d-block elements may not be regarded as the transition elements?
", "answer_text": "
Characteristics of transition element:
\n
All the transition elements are metallic elements.
\n
These elements are hard and strong.
\n
Their melting points are high.
\n
These elements can form alloys with each other.
\n
Most of these elements dissolve in acid, but acid has no effect on certain noble elements.
\n
These elements possess various valencies.
\n
They possess property of malleability and ductility.
\n
They are good conductors of electricity and heat.
\n
Some of their ions possess paramagnetic property.
\n
Transition elements are those elements in which the atoms or ions (in stable oxidation state) contain partially filled d-orbital. These elements lie in the d-block and show a transition of properties between s-block and p-block. Therefore, these are called transition elements.
\n
Elements such as Zn, Cd, and Hg cannot be classified as transition elements because these have completely filled d-subshell.
\n
Zinc, cadmium and mercury of group 12 have full d10 configuration in their ground state as well as in their common oxidation states and hence, are not regarded as transition metals
\n
Examples:\t30Zn: [Ar] 3d10 4s2
\n
48Cd: [Kr] 4d10 5s2
\n
80Hg: [Xe] 4ƒ14 5d10 6s2
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-3", "marks": 3, "source_number": "9", "original_raw_text": " What are the characteristics of the transition elements and why are they called transition elements? Which d-block elements may not be regarded as the transition elements? ", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 10, "question_text": "
In what way is the electronic configuration of the transition elements different from that of the non-transition elements?[Topic 4.0] [2 Marks]
", "answer_text": "
Transition metals have a partially filled d-orbital. Therefore, the electronic configuration of transition elements is (n - 1)d1-10 ns0-2.
\n
The non-transition elements either do not have a d-orbital or have a fully filled d-orbital. Therefore, the electronic configuration of non-transition elements is ns1-2 or ns2 np1-6.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "marks": 2, "source_number": "10", "original_raw_text": " In what way is the electronic configuration of the transition elements different from that of the non-transition elements?[Topic 4.0] [2 Marks]", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 11, "question_text": "
Explain physical properties of transition elements.
", "answer_text": "
All the transition elements are metallic element.
\n
All the transition metals have high tensile strength, ductility, malleability, high thermal and electrical conductivity and metallic lustre.
\n
With the exceptions of Zn, Cd, Hg and Mn, they have one or more typical metallic structures at normal temperatures.
\n
The transition metals (with the exception of Zn, Cd and Hg) are very much hard and have low volatility.
\n
Their melting and boiling points are high.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-17", "source_number": "11", "original_raw_text": " Explain physical properties of transition elements. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 12, "question_text": "
Aqueous solution of TiCl3 is coloured where as aqueous solution of TiCl4 is colourless. Give reason.
", "answer_text": "
In TiCl3 oxidation number of Ti is +3.
Ti3+: [Ar] 3d1.
\n
Here in 3d-orbital one unpaired electron so d-d transition is possible so it is coloured.
\n
In TiCl4 oxidation number of Ti is. +4.
Ti4+: [Ar] 3d0.
\n
Here in 3d-orbital there is no unpaired electron so it is colourless.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-17", "marks": 2, "source_number": "12", "original_raw_text": " Aqueous solution of TiCl3 is coloured where as aqueous solution of TiCl4 is colourless. Give reason. ", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 13, "question_text": "
Why the tansition elements are hard and have high melting and boiling points?
", "answer_text": "
The high melting points of these metals are attributed to the involvement of greater number of electrons from (n-1)d in addition to the ns electrons in the interatomic metallic bonding.
\n
In any row the melting points of these metals rise to a maximum at d5 except for anomalous values of Mn and Tc and fall regularly as the atomic number increases.
\n
\"\"
\n
Zn, Cd, and Hg are soft and volatile and they have low melting point because all these elements have no unpaired electrons.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/20.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest", "marks": 2, "source_number": "13", "original_raw_text": " Why the tansition elements are hard and have high melting and boiling points? ", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 14, "question_text": "
Write note on enthalpy of atomization of transition metals.
", "answer_text": "
The transition elements have high enthalpy of atomization because of presence of strong bonds between the atoms.
\n
The enthalpy of atomization increases with the increase in number of unpaired electrons which results in formation of strong bonds due to strong interatomic interactions.
\n
The metals of the second and third series have greater enthalpies of atomization than the corresponding elements of the first series; this is an important factor in accounting for the occurrence of much more frequent metal–metal bonding in compounds of the heavy transition metals.
\n
\"\"
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/22.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-20", "source_number": "14", "original_raw_text": " Write note on enthalpy of atomization of transition metals. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 15, "question_text": "
What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
\n
or
\n
\tExplain atomic radii of transition element along the period.
", "answer_text": "
In a given series decrease in radius with increasing atomic number.
\n
This is because the new electron enters a
d-orbital each time the nuclear charge increases by unity. Hence, the net electrostatic attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases.
\n
The same trend is observed in the atomic radii of a given series. However, the variation within a series is quite small.
\n
The curves in Fig. show an increase from the first (3d) to the second (4d) series of the elements but the radii of the third (5d) series are virtually the same as those of the corresponding members of the second series.
\n
This phenomenon is associated with the intervention of the 4ƒ-orbitals which must be filled before the 5d series of elements begin.
\n
The filling of 4f before 5d-orbital results in a regular decrease in atomic radii called lanthanoid contraction.
\n
The net result of the lanthanoid contraction is that the second and the third d series exhibit similar radii (e.g. Zr-160 pm, Hf-159 pm) and have very similar physical and chemical properties much more than that expected on the basis of usual family relationship.
\n
\"\"
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/25.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-22", "source_number": "15", "boundary_log": [ "or_alt→stem" ], "original_raw_text": " What is lanthanoid contraction? What are the consequences of lanthanoid contraction? \nor\n\tExplain atomic radii of transition element along the period. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 16, "question_text": "
Explain the variation in ionization enthalpies of transition element in 3d-series.
", "answer_text": "
Due to an increase in nuclear charge which accompanies the filling of the inner d-orbitals, there is an increase in ionization enthalpy along each series of the transition elements from left to right.
\n
The irregular trend in the first ionization enthalpy of the 3d metals, though of little chemical significance, can be accounted for by considering that the removal of one electron alters the relative energies of 4s and 3d-orbitals.
\n
So the unipositive ions have dn configurations with no 4s electrons. There is a reorganization energy accompanying ionization with some gains in exchange energy as the number of electrons increases and from the transference of s electrons into d-orbitals.
\n
There is generally expected increasing trend in the values as the effective nuclear charge increases. However, the value of Cr is lower because of the absence of any change in the d configuration and the value for Zn higher because it represent an ionization from the 4s level.
\n
The lowest common oxidation state of these metals is +2. To form the M2+ ions from the gaseous atoms, the sum of the first and second ionization energies is required in addition to the enthalpy of atomization for each element.
\n
The dominant term is the second ionization enthalpy which shows unusually high values for Cr and Cu where the d5 and d10 configurations of the M+ ions are disrupted, with considerable loss of exchange energy. The value for Zn is correspondingly low as the ionization consists of the removal of an electron which allows the production of the stable d10 configuration.
\n
The trend in the third ionization enthalpies is not complicated by the 4s orbital factor and shows the greater difficulty of removing and electron from the d5 (Mn2+) and d10 (Zn2+) ions superimposed upon the general increasing trend.
\n
In general, the third ionization enthalpies are quite high and there is a marked break between the values for Mn2+ and Fe2+.
\n
Also, the high values for copper, nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "source_number": "16", "original_raw_text": " Explain the variation in ionization enthalpies of transition element in 3d-series. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 17, "question_text": "
Write about oxidation states of transition series?
", "answer_text": "
One of the notable features of a transition element is the great variety of oxidation states it may show in its compounds.
\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Oxidation States of the first row Transition Metals

\n

(the most common ones are in bold types)

\n
\n

Sc

\n		\n

Ti

\n		\n

V

\n		\n

Cr

\n		\n

Mn

\n		\n

Fe

\n		\n

Co

\n		\n

Ni

\n		\n

Cu

\n		\n

Zn

\n
\n

+3

\n		\n

+2

\n

+3

\n

+4

\n		\n

+2

\n

+3

\n

+4

\n

+5

\n		\n

+2

\n

+3

\n

+4

\n

+5

\n

+6

\n		\n

+2

\n

+3

\n

+4

\n

+5

\n

+6

\n

+7

\n		\n

+2

\n

+3

\n

+4

\n

+6

\n		\n

+2

\n

+3

\n

+4

\n		\n

+2

\n

+3

\n

+4

\n		\n

+1

\n

+2

\n		\n

+2

\n
\n
The elements which give the greatest number of oxidation states occur in or near the middle of the series. Manganese, for example, exhibits all the oxidation states from +2 to +7.
\n
In the starting of series, very less number of d-electrons are available for chemical bonding. Hence, less number of oxidation states are shown by elements present at the starting of series.
\n
\te.g.: Sc+3, Ti+2, Ti+3, Ti+4
\n
At the end of the series there are too many d-electrons and d-orbitals are completely occupied. Hence, these elements show very less number of oxidation states.
\n
Down the group the stability of elements in higher oxidation states increases because removal of electrons from d-orbitals become easy.
\n
For example, in group 6, Mo(VI) and W(VI) are found to be more stable than Cr(VI). Thus, Cr(VI) in the form of dichromate in acidic medium is a strong oxidising agent, whereas MoO3 and WO3 are not.
\n
Low oxidation states are found when a complex compound has ligands capable of p-acceptor character in addition to the s-bonding. For example, in Ni(CO)4 and Fe(CO)5, the oxidation state of nickel and iron is zero.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "marks": 3, "source_number": "17", "original_raw_text": " Write about oxidation states of transition series? ", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2", "needs_review": true, "needs_figure_review": true, "figure_review_reason": "leaked-figure signature in answer: 42 consecutive short label lines (shredded vector figure?)" } }, { "sequence_no": 18, "question_text": "
Explain trends in M2+/M standard electrode potentials.
", "answer_text": "
An element in M2+ state in aqueous medium is more stable if the electrode potential (M2+/M) value is more negative.
\n
The general trend towards less negative Eo values across the series is related to the general increase in the sum of the first and second ionization enthalpies.
\n
The unique behaviour of Cu, having a positive E°/V, accounts for its inability to liberate H2 from acids. Only oxidising acids (nitric and hot concentrated sulphuric) react with Cu, the acids being reduced.
\n
\"\"
\n
It is interesting to note that the value of E°/V for Mn, Ni and Zn are more negative than expected from the trend.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/31.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest", "source_number": "18", "original_raw_text": " Explain trends in M2+/M standard electrode potentials. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 19, "question_text": "
Write a note on halide compounds of Transition elements.
", "answer_text": "
The Transition elements form Ionic halides with fluorine and co-valent halides with chlorine, bromine and iodine.
\n
The highest oxidation numbers are achieved in TiX4 (tetrahalides), VF5 and CrF6.
\n
Mn is not known in +7 oxidation state with fluorine i.e. MnF7 is not known. But MnO3F is known because oxygen stabilizes the compound due to its tendency to form a multiple bonds.
\n
The ability of fluorine to stabilise the highest oxidation state is due to either higher lattice energy as in the case of CoF3, or higher bond enthalpy terms for the higher covalent compounds, e.g., VF5 and CrF6.
\n
Although V5+ is represented only by VF5, the other halides, however, undergo hydrolysis to give oxohalides, VOX3. Another feature of fluorides.
\n
\tVF5 + H2O → VOF3 + 2HF
\n
The solution of metal halides are acidic as they produce acid on hydrolysis.
\n
All halides of copper such as CuF2, CuCl2 and CuBr2 are known. However CuI2 is not known because Cu2+ is good oxidising agent while I– is good reducing agent. So they form Cu2I2 on reaction with each other.
\n
\t2Cu2+ + 4I– → Cu2I2(s) + I2
\n
However, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation.
\n
\t2Cu+ → Cu2+ + Cu
\n
The stability of Cu2+(aq) rather than Cu+(aq) is due to the much more negative DhydH° of Cu2+(aq) than Cu+, which more than compensates for the second ionization enthalpy of Cu.
\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Formulas of Halides of 3d Metals

\n
\n

Oxidation Number

\n
\n

+6

\n

+5

\n

+4

\n

+3

\n

+2

\n

+1

\n		\n

TiX4

\n

TiX3

\n

TiX2III

\n		\n

VF5

\n

VX14

\n

VX3

\n

VX2

\n		\n

CrF6

\n

CrF5

\n

CrX4

\n

CrX3

\n

CrX2

\n		\n

MnF4

\n

MnF3

\n

MnF2

\n		\n

FeX13

\n

FeX2

\n		\n

CoF3

\n

CoX2

\n		\n

NiX2

\n		\n

CuX211

\n

CuX111

\n		\n

ZnX2

\n
\n
Key: X = F → I; X 1 = F → Br; X 11 = F, Cl; X 111 = Cl → I
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "source_number": "19", "original_raw_text": " Write a note on halide compounds of Transition elements. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2", "needs_review": true, "needs_figure_review": true, "figure_review_reason": "leaked-figure signature in answer: 6 consecutive short label lines (shredded vector figure?)" } }, { "sequence_no": 20, "question_text": "
Explain chemical reactivity of transition metals.
", "answer_text": "
Transition metals vary widely in their chemical reactivity.
\n
Many of them are sufficiently electropositive to dissolve in mineral acids, although a few are 'noble' – that is, they are unaffected by simple acids.
\n
The metals of the first series with the exception of copper are relatively more reactive and are oxidised by 1 M H+, though the actual rate at which these metals react with oxidising agents like hydrogen ion (H+) is sometimes slow.
\n
The Eo values for M2+/M indicate a decreasing tendency to form divalent cations across the series.
\n
This general trend towards less negative E° values is related to the increase in the sum of the first and second ionization enthalpies.
\n
It is interesting to note that the E° values for Mn, Ni and Zn are more negative than expected from the general trend.
\n
An examination of the EK values for the redox couple M3+/M2+ shows that Mn3+ and Co3+ ions are the strongest oxidising agents in aqueous solutions. The ions Ti2+, V2+ and Cr2+ are strong reducing agents and will liberate hydrogen from a dilute acid,
e.g., 2Cr2+(aq) + 2H+(aq) → 2Cr3+(aq) + H2(g)
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "marks": 2, "source_number": "20", "original_raw_text": " Explain chemical reactivity of transition metals. ", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 21, "question_text": "
Explain magnetic properties of transition elements.
", "answer_text": "
When a magnetic field is applied to substances, mainly two types of properties magnetic behaviour are observed: diamagnetism and paramagnetism.
\n
Diamagnetic substances are repelled by the applied field while the paramagnetic substances are attracted.
\n
Substances which are attracted very strongly are said to be ferromagnetic.
\n
In fact, ferromagnetism is an extreme form of paramagnetism. Many of the transition metal ions are paramagnetic.
\n
Paramagnetism arises from the presence of unpaired electrons, each such electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum.
\n
\tFor these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the 'spin-only' formula,
i.e., m = \"\"
\n
Where, n = The number of unpaired electrons µ = Magnetic moment in units of Bohr
magneton (BM).
\n
The magnetic moment increases with the increasing number of unpaired electrons.

Calculated and Observed Magnetic Moments (BM)

Ion

Configu-ration

Unpaired

electron(s)

Magnetic moment

Calculated

Observed

Sc3+

Ti3+

Ti2+

V2+

Cr2+

Mn2+

Fe2+

Co2+

Ni2+

Cu2+

Zn2+

3d 0

3d 1

3d 2

3d 3

3d 4

3d 5

3d 6

3d 7

3d 8

3d 9

3d 10

0

1

2

3

4

5

4

3

2

1

0

0

1.73

2.84

3.87

4.90

5.92

4.90

3.87

2.84

1.73

0

0

1.75

2.76

3.86

4.80

5.96

5.3 - 5.5

4.4 - 5.2

2.9 - 3, 4

1.8 - 2.2

0

", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/EQ-0318-143240.png", "position": "answer", "context": "For these, the magnetic moment is determined by the number o…" } ], "tables": [], "metadata": { "source_class": "Quest", "marks": 3, "source_number": "21", "original_raw_text": " Explain magnetic properties of transition elements. ", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2", "needs_review": true, "needs_figure_review": true, "figure_review_reason": "leaked-figure signature in answer: 12 consecutive short label lines (shredded vector figure?)" } }, { "sequence_no": 22, "question_text": "
Calculate the magnetic moment of a tivalent ion in aqueous solution if its atomic number
is 26.[May 2021] [3 Marks]
", "answer_text": "
26M: [Ar] 3d5 4s2
\n
M3+: [Ar] 3d5
\n
\"\"
\n
Unpaired electrons = 5
\n
∴ Magnetic moment (µ)= \"\"
\n
\t\t\t= \"\"
\n
\t\t\t= \"\"
\n
\t\t\t= 5.92 B.M
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/39.png", "position": "answer", "context": "" }, { "id": "img-1", "path": "Chapter4/EQ-1222-153945.png", "position": "answer", "context": "∴ Magnetic moment ( µ ) =" }, { "id": "img-2", "path": "Chapter4/EQ-0908-095623.png", "position": "answer", "context": "=" }, { "id": "img-3", "path": "Chapter4/EQ-0908-095712.png", "position": "answer", "context": "=" } ], "tables": [], "metadata": { "source_class": "Quest", "marks": 3, "pyq": true, "source_number": "22", "original_raw_text": " Calculate the magnetic moment of a tivalent ion in aqueous solution if its atomic numberis 26.[May 2021] [3 Marks] ", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 23, "question_text": "
Why transition elements form a coloured ions?
", "answer_text": "
Most of the ionic and covalent compounds of transition elements are coloured. It is due to the presence of incompletely filled d-orbitals.
\n
When an electron from a lower energy d-orbital is excited to a higher energy d-orbital, the energy of excitation corresponds to the frequency of light absorbed.
\n
This frequency generally lies in the visible region. The colour observed corresponds to the complementary colour of the light absorbed.
\n
The frequency of the light absorbed is determined by the nature of the ligand.
\n
In aqueous solutions where water molecules are the ligands, the colours of the ions observed are listed in Table.

Configuration

Example

Colour

3d0

Sc3+

Colourless

3d0

Ti4+

Colourless

3d1

Ti3+

Purple

3d1

V4+

Blue

3d2

V3+

Green

3d3

V2+

Violet

3d3

Cr3+

Violet

3d4

Mn3+

Violet

3d4

Cr2+

Blue

3d5

Mn2+

Pink

3d5

Fe3+

Yellow

3d6

Fe2+

Green

3d6

Co3+

Blue

3d7

Co2+

Pink

3d8

Ni2+

Green

3d9

Cu2+

Blue

3d10

Zn2+

Colourless

", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "marks": 3, "source_number": "23", "original_raw_text": " Why transition elements form a coloured ions?", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 24, "question_text": "
Why the transition element form a large number of complex compounds?
", "answer_text": "
Complex compounds are those in which the metal ions bind a number of anions or neutral molecules giving complex species with characteristic properties.
\n
A few examples are: [Fe(CN)6]3–, [Fe(CN)6]4–, [Cu(NH3)4]2+ and [PtCl4]2–.
\n
The transition metals form a large number of complex compounds.
\n
This is due to the comparatively smaller sizes of the metal ions. Their high ionic charges and the availability of d-orbitals for bond formation.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "marks": 2, "source_number": "24", "original_raw_text": " Why the transition element form a large number of complex compounds? ", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 25, "question_text": "
Explain Catalytic property of transition elements.
", "answer_text": "
The transition metals and their compounds are known for their catalytic activity.
\n
This activity is ascribed to their ability to adopt multiple oxidation states and to form complexes.
\n
Vanadium (V) oxide (in Contact Process), finely divided iron (in Haber's Process), and nickel (in Catalytic Hydrogenation) are some of the examples.
\n
Catalysts at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst (first row transition metals utilise 3d and 4s electrons for bonding).
\n
This has the effect of increasing the concentration of the reactants at the catalyst surface and also weakening of the bonds in the reacting molecules (the activation energy is lowering).
\n
The Transition metal ions can change their oxidation states, they become more effective as catalysts.
\n
For example, iron (III) catalyses the reaction between iodide and persulphate ions.
\n
\t2I – + S2O2–8 → I2 + 2SO2–4
\n
An explanation of this catalytic action can be given as:
\n
\t2Fe3+ + 2I – → 2Fe2+ + I2
\n
\t2Fe2+ + S2O2–8 → 2Fe3+ + 2SO2–4
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "marks": 2, "source_number": "25", "original_raw_text": "Explain Catalytic property of transition elements. ", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 26, "question_text": "
What are interstitial compounds? Why are such compounds well known for transition metals?
\n
or
\n
\tWhat are interstitial compounds. Write its any two characteristics.
\n
or
\n
\tWrite note on interstitial compounds.
", "answer_text": "
Transition metals are large in size and contain lots of interstitial sites. Transition elements can trap atoms of other elements (that have small atomic size), such as H, C, N, in the interstitial sites of their crystal lattices. The resulting compounds are called interstitial compounds.
\n
They are usually non-stoichiometric and are neither typically ionic nor covalent,
\n
For example, TiC, Mn4N, Fe3H, VH0.56 and TiH1.7, etc.
\n
These compounds are referred to as interstitial compounds.
\n
The principal physical and chemical characteristics of these compounds are as follows:
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-22", "source_number": "26", "marks": 3, "boundary_log": [ "or_alt→stem" ], "original_raw_text": " What are interstitial compounds? Why are such compounds well known for transition metals?\nor\n\tWhat are interstitial compounds. Write its any two characteristics. \nor\n\tWrite note on interstitial compounds.", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 27, "question_text": "
Write short note on alloys.
", "answer_text": "
“An alloy is a blend of metals prepared by mixing the components.”
\n
Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other.
\n
The atomic size of two metals forming the alloy must be the same. There must not be more than 15% difference in their atomic radii.
\n
The chemical properties of the metals used for preparation of alloys must be same, that is, their electronic configurations of valence shell must be the same.
\n
The crystal structures of pure metallic elements used for alloys must be similar.
\n
The alloys so formed are hard and have often high melting points.
\n
The best known are ferrous alloys: chromium, vanadium, tungsten, molybdenum and manganese are used for the production of a variety of steels and stainless steel.
\n
Alloys of transition metals with non transition metals such as brass (copper-zinc) and bronze (copper-tin), are also of considerable industrial importance.
", "type": "SN", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-40", "source_number": "27", "original_raw_text": " Write short note on alloys.", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 28, "question_text": "
Write note on oxides and oxoanions of 3d transition metals.
", "answer_text": "
The ability of oxygen to stabilize the high oxidation states of metal than fluorine is higher because it forms a multiple bonds with a metal.
\n
The metal oxide in lower oxidation state is basic while in higher oxidation state are acidic.
\n
The metal oxides having metals in intermediate oxidation states are amphoteric.
\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Oxides of 3d Metals

\n
\n

Acidic

\n

oxide

\n		\n

Basic

\n

oxide

\n		\n

Amphoteric

\n

oxide

\n
\n

Mn2O7

\n		\n

MnO

\n		\n

Mn3O4, Mn2O3, MnO2

\n
\n

CrO

\n		\n

Cr2O3

\n
\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Oxidation Number

\n
\n

3

\n		\n

4

\n		\n

5

\n		\n

6

\n		\n

7

\n		\n

8

\n		\n

9

\n		\n

10

\n		\n

11

\n		\n

12

\n
\n

+7

\n

+6

\n

+5

\n

+4

\n

+3

\n

+2

\n

+1

\n		\n

Sc2O3

\n		\n

TiO2

\n

Ti2O3

\n

TiO

\n		\n

V2O5

\n

V2O4 V2O3

\n

VO

\n		\n

CrO3

\n

CrO2 Cr2O3

\n

(CrO)

\n		\n

Mn2O7

\n

MnO2 Mn2O3 Mn3O4*

\n

MnO

\n		\n

Fe2O3

\n

Fe3O4*

\n

FeO

\n		\n

Co3O*4

\n

CoO

\n		\n

NiO

\n		\n

CuO

\n

Cu2O

\n		\n

ZnO

\n
\n

Note: *Denotes mixed oxides

\n
\n
In the covalent oxide Mn2O7, each Mn is tetrahedrally surrounded by O's including a Mn–O–Mn bridge. The tetrahedral [MO4]n- ions are known for VV, CrVI, MnV, MnVI and MnVII.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "source_number": "28", "original_raw_text": " Write note on oxides and oxoanions of 3d transition metals. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2", "needs_review": true, "needs_figure_review": true, "figure_review_reason": "leaked-figure signature in answer: 17 consecutive short label lines (shredded vector figure?)" } }, { "sequence_no": 29, "question_text": "
Discuss structures and magnetic properties of manganate and permanganate ions.
", "answer_text": "
The manganate and permanganate ions are tetrahedral.
\n
\"\"
\n
The green manganate is paramagnetic with one unpaired electron but the permanganate is diamagnetic.
\n
The π-bonding takes place by overlap of
p-orbitals of oxygen with d-orbitals of manganese.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/54.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-40", "marks": 2, "source_number": "29", "original_raw_text": " Discuss structures and magnetic properties of manganate and permanganate ions. ", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 30, "question_text": "
Give the structure of chromate and dichromate ions, and explain interconvertible property of these ions.
", "answer_text": "
The structures of chromate ion, CrO2–4 and the dichromate ion, Cr2O2–7 are shown below.
\n
\"\"
\n
\"\"
\n
The chromate ion is tetrahedral whereas the dichromate ion consists of two tetrahedral sharing one corner with Cr–O–Cr bond angle of 126°.
\n
The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. The oxidation state of chromium in chromate and dichromate is the same.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/56.png", "position": "answer", "context": "" }, { "id": "img-1", "path": "Chapter4/57.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-40", "marks": 2, "source_number": "30", "original_raw_text": " Give the structure of chromate and dichromate ions, and explain interconvertible property of these ions. ", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 31, "question_text": "
Explain the chemical properties of KMnO4.
", "answer_text": "
KMnO4 act as strong oxidising agent in neutral, alkaline and in acidic medium.
\n
(I)\tIn Acidic Solution:
\n
(a)\tIodine is liberated from potassium iodide:
\n
\"\"
\n
(b)\tFe2+ ion (green) is converted to Fe3+ (yellow):
\n
\"\"
\n
(c)\tOxalate ion or oxalic acid is oxidised at 333K :
\n
\"\"
\n
(d)\tHydrogen sulphide is oxidised, sulphur being precipitated:
\n
\tH2S → 2H+ + S2–
\n
\t5S2– + 2MnO–4 + 16H+ → 2Mn2+ + 8H2O + 5S
\n
(e)\tSulphurous acid or sulphite is oxidised to a sulphate or sulphuric acid:
\n
\t5SO2–3 + 2MnO–4 + 6H+ → 2Mn2+ + 3H2O + 5SO2–4
\n
(f)\tNitrite is oxidised to nitrate:
\n
\t5NO–2 + 2MnO–4 + 6H+ → 2Mn2+ + 5NO–3 + 3H2O
\n
(2)\tIn neutral or faintly alkaline solutions:
\n
(a)\tA notable reaction is the oxidation of iodide to iodate:
\n
\t2MnO–4 + H2O + I– → 2MnO2 + 2OH– + IO–3
\n
(b)\tThiosulphate is oxidised almost quantitatively to sulphate:
\n
\t8MnO–4 + 3S2O2–3 + H2O → 8MnO2 + 6SO2–4 + 2OH–
\n
(c)\tManganous salt is oxidised to MnO2; the presence of zinc sulphate or zinc oxide catalyses the oxidation:
\n
\t2MnO–4 + 3Mn2+ + 2H2O → 5MnO2 + 4H+
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/59.png", "position": "answer", "context": "" }, { "id": "img-1", "path": "Chapter4/60.png", "position": "answer", "context": "" }, { "id": "img-2", "path": "Chapter4/61.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest", "source_number": "31", "original_raw_text": " Explain the chemical properties of KMnO4.", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 32, "question_text": "
Write down the uses of KMnO4.
", "answer_text": "
Potassium permanganate is used as a favourite oxidant in preparative organic chemistry.
\n
Its uses are for the bleaching of wool, cotton, silk and other textile fibres and for the decolourisation of oils are also dependent on its strong oxidising power.
\n
An aqueous solution of potassium permanganate is used for gargling to keep mouth germfree as it is antiseptic.
\n
It is useful as titrant in redox titrations to know the proportion of metal ions like iron and organic compounds like oxalic acid.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-40", "marks": 2, "source_number": "32", "original_raw_text": " Write down the uses of KMnO4.", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 33, "question_text": "
Complete the reactions and balance them.
\n
\t(i) MnO2 + KOH + O2 →
\n
\t(ii) MnO–4 + H2O + I– \"\"
\n
\t(iii) MnO–2 + S2O2–3 + H2O \"\"
", "answer_text": "
(i) MnO2 + 4KOH + O2 → 2K2 MnO4 + 2H2O
\n
\t(ii) 2MnO–4 + H2O + I– → 2MnO2 + 2OH– + IO–3
\n
\t(iii) 8MnO–4 + 2S2O2–2 + H2O → 8MnO2 + 6SO2–4 + 2OH–
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/64.png", "position": "stem", "context": "(ii) MnO – 4 + H 2 O + I –" }, { "id": "img-1", "path": "Chapter4/65.png", "position": "stem", "context": "(iii) MnO – 2 + S 2 O 2 – 3 + H 2 O" } ], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-20", "source_number": "33", "marks": 3, "original_raw_text": " Complete the reactions and balance them. \n\t(i) MnO2 + KOH + O2 →\n\t(ii) MnO–4 + H2O + I– \n\t(iii) MnO–2 + S2O2–3 + H2O ", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 34, "question_text": "
Write note on inner transition elements or ƒ-block elements.
", "answer_text": "
The ƒ-block consists of the two series, lanthanoids (the fourteen elements following lanthanum) and actinoids (the fourteen elements following actinium).
\n
Because lanthanum closely resembles the lanthanoids, it is usually included in any discussion of the lanthanoids for which the general symbol Ln is often used.
\n
Similarly, a discussion of the actinides includes actinium besides the fourteen elements constituting the series.
\n
The lanthanoids resemble one another more closely than do the members of ordinary transition elements in any series.
\n
They have only one stable oxidation state and their chemistry provides an excellent opportunity to examine the effect of small changes in size and nuclear charge along a series of otherwise similar elements.
\n
The chemistry of the actinoids is, on the other hand, much more complicated. The complication arises partly owing to the occurrence of a wide range of oxidation states in these elements and partly because their radioactivity creates special problems in their study.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-17", "source_number": "34", "original_raw_text": " Write note on inner transition elements or ƒ-block elements.", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 35, "question_text": "
Write note on physical properties of lanthanoids.
", "answer_text": "
All the lanthanoids are silvery white soft metals and tarnish rapidly in air.
\n
The hardness increases with increasing atomic number, samarium being steel hard.
\n
Their melting points range between 1000 to 1200K but samarium melts at 1623K.
\n
They have typical metallic structure and are good conductors of heat and electricity.
\n
Density and other properties change smoothly except for Eu and Yb and occasionally for Sm and Tm.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-20", "source_number": "35", "original_raw_text": " Write note on physical properties of lanthanoids.", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 36, "question_text": "
Write down uses of lanthanoids and its compounds.
", "answer_text": "
Use of the lanthanoids is for the production of alloy steels for plates and pipes.
\n
A well known alloy is mischmetal which consists of a lanthanoid metal (∼ 95%) and iron (∼ 5%) and traces of S, C, Ca and Al.
\n
A good deal of mischmetal is used in Mg-based alloy to produce bullets, shell and lighter flint.
\n
Mixed oxides of lanthanoids are employed as catalysts in petroleum cracking.
\n
Some individual Ln oxides are used as phosphors in television screens and similar fluorescing surfaces.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "source_number": "36", "original_raw_text": " Write down uses of lanthanoids and its compounds.", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 37, "question_text": "
Discuss chemical properties of the lanthanoids.
", "answer_text": "
In their chemical behaviour, in general, the earlier members of the series are quite reactive similar to calcium but, with increasing atomic number, they behave more like aluminium.
\n
The value of E° of half reaction
\n
\"\" + 3e– \"\" Ln(s) range from – 2.2 V to 2.4 V except for Eu that has – 2.0 V.
\n
\"\"
\n
Lanthanoids heated with carbon forms carbides Ln3C, Ln2C3 and LnC2.
\n
Lanthanoids form trihalides LnX3 with halogen and liberates hydrogen gas when treated with dilute acids.
\n
These metals combine with oxygen to form oxides of type Ln2O3.
\n
Oxides are basic in nature and when dissolved in water they form hydroxide Ln(OH)3.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/EQ-0910-100339.png", "position": "answer", "context": "+ 3e – Ln (s) range from – 2.2 V to 2.4 V except for Eu that…" }, { "id": "img-1", "path": "Chapter4/71.png", "position": "answer", "context": "+ 3e – Ln (s) range from – 2.2 V to 2.4 V except for Eu that…" }, { "id": "img-2", "path": "Chapter4/72.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-6", "marks": 3, "source_number": "37", "original_raw_text": " Discuss chemical properties of the lanthanoids.", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 38, "question_text": "
Write down electronic configuration of all lanthanoids elements. [3 Marks]
", "answer_text": "
General electronic configuration - [xe] 4ƒ1-14 5d0-1
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-34", "marks": 3, "source_number": "38", "original_raw_text": " Write down electronic configuration of all lanthanoids elements. [3 Marks]", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 39, "question_text": "
Explain the lanthanoid contraction.
", "answer_text": "
As we move along the lanthanoid series, the atomic number increases gradually by one. This means that the number of electrons and protons present in an atom also increases by one.
\n
As electrons are being added to the same shell, the effective nuclear charge increases.
\n
This happens because the increase in nuclear attraction due to the addition of proton is more pronounced than the increase in the interelectronic repulsions due to the addition of electron.
\n
Also, with the increase in atomic number, the number of electrons in the 4ƒ-orbital also increases. The 4ƒ electrons have poor shielding effect.
\n
Therefore, the effective nuclear charge experienced by the outer electrons increases. Consequently, the attraction of the nucleus for the outermost electrons increases.
\n
This results in a steady decrease in the size of lanthanoids with the increase in the atomic number. This is termed as lanthanoid contraction.
\n
\"\"
\n
Consequences of lanthanoid contraction:
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/75.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest", "marks": 3, "source_number": "39", "original_raw_text": " Explain the lanthanoid contraction.", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 40, "question_text": "
Write down general electronic configuration of elements of actinoids. Why there are irregularities in the electronic configurations?
", "answer_text": "
The general electronic configuration of actinoids is [Rn]5ƒ0-14 6d0-2 7s2
\n
The irregularities in the electronic configuration of actinoid series is due to very less difference in the energies of 5ƒ-orbitals and 6d-orbitals.
\n
All the actinoids are believed to have the electronic configuration of 7s2 and variable occupancy of the 5f and 6d-subshells.
\n
The irregularities in the electronic configurations of the actinoids, like those in the lanthanoids are related to the stabilities of the ƒ 0, ƒ 7 and ƒ 14 occupancies of the 5ƒ-orbitals.
\n
Thus, the configurations of Am and Cm are [Rn] 5f 77s2 and [Rn] 5f 7. 6d 1. 7s2.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-20", "source_number": "40", "original_raw_text": " Write down general electronic configuration of elements of actinoids. Why there are irregularities in the electronic configurations? ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 41, "question_text": "
Discuss comparison between lathanoids and actinoids.
", "answer_text": "
The actinoid metals are all silvery in appearance but display a variety of structures. The structural variability is obtained due to irregularities in metallic radii which are far greater than in lanthanoids.
\n
The magnetic properties of the actinoids are more complex than those of the lanthanoids.
\n
It is evident from the behaviour of the actinoids that the ionization enthalpies of the early actinoids, though not accurately known, but are lower than for the early lanthanoids.
\n
The stable oxidation state of all the lanthanoids is (+3). In actinoids, oxidation states (+2) to (+6) are seen.
\n
In lanthanoids only promethium is radioactive but all the actinoids are radioactive.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "source_number": "41", "original_raw_text": " Discuss comparison between lathanoids and actinoids. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 42, "question_text": "
Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
\n
\t(i)\tElectronic configuration,
\n
\t(ii)\tOxidation state,
\n
\t(iii)\tAtomic and ionic sizes and
\n
\t(iv)\tChemical reactivity.
", "answer_text": "
(i)\tElectronic configuration: The general electronic configuration for lanthanoids is [Xe]54 4f0-14 5d0-1 6s2 and that for actinoids is [Rn]86 5f1-14 6d0-1 7s2. Unlike 4f-orbitals, 5f-orbitals are not deeply buried and participate in bonding to a greater extent.
\n
(ii)\tOxidation states: The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of +2 and +4. This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d and 7s levels are of comparable energies. Again, (+3) is the principle oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in +3 state than in +4 state.
\n
(iii)\tAtomic and Ionic sizes: Similar to lanthanoids, actinoids also exhibit actinoid contraction (overall decrease in atomic and ionic radii). The contraction is greater due to the poor shielding effect of
5f-orbitals.
\n
(iv)\tChemical reactivity: In the lanthanoid series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In case of acids, they are slightly affected by nitric acid (because of the formation of a protective oxide layer).
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-50", "source_number": "42", "original_raw_text": " Compare the chemistry of actinoids with that of the lanthanoids with special reference to:\n\t(i)\tElectronic configuration,\n\t(ii)\tOxidation state,\n\t(iii)\tAtomic and ionic sizes and\n\t(iv)\tChemical reactivity. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 43, "question_text": "
Discuss applications of d and ƒ-block elements.
", "answer_text": "
Iron and steels are the most important construction materials.
\n
Their production is based on the reduction of iron oxides, the removal of impurities and the addition of carbon and alloying metals such as Cr, Mn and Ni.
\n
Some compounds are manufactured for special purposes such as TiO for the pigment industry and MnO2 for use in dry battery cells.
\n
The battery industry also requires Zn and Ni/Cd.
\n
UK 'copper' coins are copper-coated steel. The 'silver' UK coins are a Cu/Ni alloy. Many of the metals and/or their compounds are essential catalysts in the chemical industry.
\n
V2O5 catalyses the oxidation of SO2 in the manufacture of sulphuric acid.
\n
TiCl4 with Al(CH3)3 forms the basis of the Ziegler catalysts used to manufacture polyethylene (polythene).
\n
Iron catalysts are used in the Haber process for the production of ammonia from N2/H2 mixtures. Nickel catalysts enable the hydrogenation of fats to proceed.
\n
In the Wacker process the oxidation of ethyne to ethanal is catalysed by PdCl2. Nickel complexes are useful in the polymerisation of alkynes and other organic compounds such as benzene. The photographic industry relies on the special light-sensitive properties of AgBr.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-20", "source_number": "43", "original_raw_text": " Discuss applications of d and ƒ-block elements. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2" } }, { "sequence_no": 44, "question_text": "
Give reason:[March 2023] [3 Marks]
", "answer_text": "
(i)\tTransition elements exhibit higher enthalpies of atomisation.
\n
(ii)\tIn aqueous solution, Cr2+ is stronger reducing agent than Fe2+.
\n
(iii)\tThe second ionisation enthalpy of Cu is higher then Zn.
\n
Ans. (i)\tBecause transition elements have more unpaired electrons, they have stronger interatomic interactions and hence stronger bonding.
\n
(ii) In Cr2+ to Cr3+ d4→ d3 occurs and in Fe2+ to Fe3+ d6 → d5 occurs. In aq. medium d3
is more stable than d5.
\n
(iii) In 2nd ionisation enthalpy of Cu,
e– is removed from 3d10 and in Zn from 4s1. 3d10 is more stable configuration than 4s1.
\n
Class 12 Chemistry (Part 1) 017
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-53", "marks": 3, "pyq": true, "source_number": "44", "original_raw_text": " Give reason:[March 2023] [3 Marks]", "deduction_level": 0, "source_html": "Chapter4", "section": "Topic-wise Q & A", "section_id": "4.2", "needs_review": true, "needs_boundary_review": true, "boundary_review_reason": "answer has 3 content line(s) before its answer marker ('Ans. (i) Because transition el') — pre-marker content belongs to the question" } } ] }, { "section_id": "", "section_title": "NCERT Intext Questions and Answers", "heading_text": "NCERT Intext Questions and Answers", "detected_type": "SUB", "match_confidence": 0.542, "questions": [ { "sequence_no": 45, "question_text": "
Silver atom has completely filled d-orbitals (4d10) in its ground state. How can you say that it is a transition element?
", "answer_text": "
Silver (Z = 47) can exhibit +2 oxidation state where in it will have incompletely filled d-orbitals (4d), hence a transition element.
\n
47Ag\t: [Kr] 4d10 5s1
\n
Ag2+: [Kr] 4d9
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-3", "marks": 2, "source_number": "1", "original_raw_text": " Silver atom has completely filled d-orbitals (4d10) in its ground state. How can you say that it is a transition element? ", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Intext Questions and Answers", "section_id": "" } }, { "sequence_no": 46, "question_text": "
In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomization of zinc is the lowest, i.e., 126 kJ mol-1. Why?
", "answer_text": "
In the formation of metallic bonds, no electrons from 3d-orbitals are involved in case of zinc, while in all other metals of the 3d series, electrons from the d-orbitals are always involved in the formation of metallic bonds.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-17", "marks": 2, "source_number": "2", "original_raw_text": " In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomization of zinc is the lowest, i.e., 126 kJ mol-1. Why? ", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Intext Questions and Answers", "section_id": "" } }, { "sequence_no": 47, "question_text": "
Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?
", "answer_text": "
Manganese (25Mn) shows largest number of oxidation states because it has maximum number of unpaired electrons.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-20", "source_number": "3", "original_raw_text": " Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why? ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Intext Questions and Answers", "section_id": "" } }, { "sequence_no": 48, "question_text": "
The E°(M2+/M) value for copper is positive (+0.34 V). What is possibly the reason for this? (Hint: consider its high DaH° and low DhydH°)
", "answer_text": "
The E°(M2+/M) value of a metal depends on the energy changes involved in the following:
\n
(1) Sublimation: The energy required for converting one mole of an atom from the solid state to the gaseous state.
\n
\tCu(s) → Cu(g) ∆sH° or ∆aH° = 339 kJ Mol–1
\n
(2)\tIonization: The energy required to take out electrons from one mole of atoms in the gaseous state.
\n
\tCu(g) → \"\" + 2e– ∆iH° = 2703 kJ Mol–1
\n
(3)\tHydration: The energy released when one mole of ions are hydrated.
\n
\"\" \"\" \"\" ∆hydH° = –2121 kJ Mol–1
\n
E°Cu2+/Cu\t= ∆aH° + ∆iH° + ∆hydH°
\n
\t\t= 339 + 2703 – 2121
\n
= 1921 kJ Mol–1
\n
Total value of all the enthalpies is positive so reaction is endothermic so E°Cu+2/Cu is positive.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/EQ-0206-122430.png", "position": "answer", "context": "Cu (g) → + 2e – ∆ i H ° = 2703 kJ Mol –1" }, { "id": "img-1", "path": "Chapter4/EQ-0206-122430.png", "position": "answer", "context": "∆ hyd H ° = –2121 kJ Mol –1" }, { "id": "img-2", "path": "Chapter4/EQ-0907-165214.png", "position": "answer", "context": "∆ hyd H ° = –2121 kJ Mol –1" }, { "id": "img-3", "path": "Chapter4/EQ-0213-100023.png", "position": "answer", "context": "∆ hyd H ° = –2121 kJ Mol –1" } ], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-44", "marks": 3, "source_number": "4", "original_raw_text": " The E°(M2+/M) value for copper is positive (+0.34 V). What is possibly the reason for this? (Hint: consider its high DaH° and low DhydH°)", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Intext Questions and Answers", "section_id": "" } }, { "sequence_no": 49, "question_text": "
How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements?
", "answer_text": "
Irregular variation of ionization enthalpies is mainly attributed to varying degree of stability of different 3d-configurations (e.g., d0, d5, d10 are exceptionally stable).
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "marks": 2, "source_number": "5", "original_raw_text": " How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements? ", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Intext Questions and Answers", "section_id": "" } }, { "sequence_no": 50, "question_text": "
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?\t
", "answer_text": "
Because of small size and high electronegativity oxygen or fluorine can oxidise the metal to its highest oxidation state.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-3", "marks": 2, "source_number": "6", "original_raw_text": " Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?\t ", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Intext Questions and Answers", "section_id": "" } }, { "sequence_no": 51, "question_text": "
Which is a stronger reducing agent Cr2+ or Fe2+ and why?
", "answer_text": "
Cr2+ is stronger reducing agent than Fe2+.
\n
Reason: d4 → d3 occurs in case of Cr2+ to Cr3+.
\n
\tBut d6 → d5 occurs in case of Fe2+ to Fe3+.
\n
In a medium (like water) d3 is more stable as compared to d5 (see CFSE).
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-3", "marks": 2, "source_number": "7", "original_raw_text": " Which is a stronger reducing agent Cr2+ or Fe2+ and why?", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Intext Questions and Answers", "section_id": "" } }, { "sequence_no": 52, "question_text": "
Calculate the 'spin only' magnetic moment of M2+(aq) ion (Z = 27).
", "answer_text": "
Z = 27
\n
µ\t= \"\"
\n
\t= \"\"
\n
\t= \"\"
\n
\t= 3.87 B.M
\n
\t[Ar] 3d7 4s2
\n
\\ M2+ = [Ar] 3d7
\n
\t3d7 = \"\"
\n
\ti.e., 3 unpaired electrons
\n
\\ n = 3
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/EQ-0215-150741.png", "position": "answer", "context": "µ =" }, { "id": "img-1", "path": "Chapter4/EQ-0215-150803.png", "position": "answer", "context": "=" }, { "id": "img-2", "path": "Chapter4/EQ-0215-150819.png", "position": "answer", "context": "=" }, { "id": "img-3", "path": "Chapter4/94.png", "position": "answer", "context": "3 d 7 =" } ], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-59", "marks": 2, "source_number": "8", "original_raw_text": " Calculate the 'spin only' magnetic moment of M2+(aq) ion (Z = 27).", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Intext Questions and Answers", "section_id": "" } }, { "sequence_no": 53, "question_text": "
Explain why Cu+ ion is not stable in aqueous solutions?
", "answer_text": "
Cu+ in aqueous solution undergoes disproportionation, i.e., 2Cu+(aq) → Cu2+(aq) + Cu(s)
\n
The E° value for this is favourable.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-20", "marks": 2, "source_number": "9", "original_raw_text": " Explain why Cu+ ion is not stable in aqueous solutions?", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Intext Questions and Answers", "section_id": "" } }, { "sequence_no": 54, "question_text": "
Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
", "answer_text": "
The 5ƒ electrons are more effectively shielded from nuclear charge. In other words the 5ƒ electrons themselves provide poor shielding from element to element in the series.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-3", "marks": 2, "source_number": "10", "original_raw_text": " Actinoid contraction is greater from element to element than lanthanoid contraction. Why?", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Intext Questions and Answers", "section_id": "" } } ] }, { "section_id": "4.5", "section_title": "NCERT Textbook Exercise Q & A", "heading_text": "NCERT Textbook Exercise Questions And Answers", "detected_type": "SUB", "match_confidence": 0.75, "questions": [ { "sequence_no": 55, "question_text": "
Write down the electronic configuration of:
\n
\t(i) Cr3+\t\t(ii) Pm3+
\n
\t(iii) Cu+\t\t(iv) Ce4+
\n
\t(v) Co2+\t\t(vi) Lu2+
\n
\t(vii) Mn2+\t(viii) Th4+
", "answer_text": "
(i)\tCr3+\t:\t1s2 2s2 2p6 3s2 3p6 3d3
\n
(ii)\tPm3+ :\t1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 \t\t\t\t\t4d10 5s2 5p6 4f4
\n
: [Xe]54 4f4
\n
(iii)\tCu+ :\t1s2 2s2 2p6 3s2 3p6 3d10
\n
: [Ar]18 3d10
\n
(iv)\tCe4+\t:\t1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 \t\t\t\t\t4d10 5s2 5p6
\n
: [Xe]54
\n
(v)\tCo2+\t:\t1s2 2s2 2p6 3s2 3p6 3d7
\n
(vi)\tLu2+\t:\t1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 \t\t\t\t\t4d10 5s2 5p6 4f14 5d1
\n
\t\t\t:\t[Xe]54 4f14 5d1
\n
(vii)\tMn2+\t:\t1s2 2s2 2p6 3s2 3p6 3d5
\n
(viii)\tTh4+\t:\t1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 \t\t\t\t\t4d10 4f14 5s2 5p6 5d10 6s2 6s6
\n
\t\t\t:\t[Rn]86
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "source_number": "1", "marks": 1, "original_raw_text": " Write down the electronic configuration of:\n\t(i) Cr3+\t\t(ii) Pm3+ \n\t(iii) Cu+\t\t(iv) Ce4+\n\t(v) Co2+\t\t(vi) Lu2+ \n\t(vii) Mn2+\t(viii) Th4+", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 56, "question_text": "
Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state?
", "answer_text": "
Electronic configuration of Mn2+ is [Ar]183d5.
\n
Electronic configuration of Fe2+ is [Ar]183d6.
\n
It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in (+2) state has a stable d5 configuration.
\n
This is the reason Mn2+ shows resistance to oxidation to Mn3+. Also, Fe2+ has 3d6 configuration and by losing one electron, its configuration changes to a more stable 3d5 configuration. Therefore, Fe2+ easily gets oxidized to Fe3+ oxidation state.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-3", "marks": 2, "source_number": "2", "original_raw_text": " Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state? ", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 57, "question_text": "
Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
", "answer_text": "
It can be easily observed that except Sc, all others metals display +2 oxidation state. Also, on moving from Sc to Mn, the atomic number increases from 21 to 25. This means the number of electrons in the 3d-orbital also increases from 1 to 5.
\n
Sc\t(+2)\t=\td1
\n
\tTi\t(+2)\t=\td2
\n
\tV\t(+2)\t=\td3
\n
\tCr\t(+2)\t=\td4
\n
\tMn (+2)\t=\td5
\n
+2 oxidation state is attained by the loss of the two 4s electrons by these metals. Since the number of d electrons in (+2) state also increases from Ti(+2) to Mn(+2), the stability of +2 state increases (as d-orbital is becoming more and more half-filled). Mn(+2) has d5 electrons (that is half-filled d-shell, which is highly stable).
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-59", "marks": 2, "source_number": "3", "original_raw_text": "Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 58, "question_text": "
To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.
", "answer_text": "
If an element contain d0, d5, d10 electronic configuration it become more stable.
\n
Example,
\n
Sc3+ is more stable than Sc+ because of d0 configuration.
\n
V5+ is more stable than V3+ because of d0 configuration.
\n
Mn2+ is more stable than Mn3+ because of
d5 configuration.
\n
Fe3+ is more stable than Fe2+ because of
d5 configuration.
\n
Zn2+ is more stable than Zn+ because of
d10 configuration.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-59", "marks": 2, "source_number": "4", "original_raw_text": " To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 59, "question_text": "
What may be the stable oxidation state of the transition element with the following
d electron configurations in the ground state of their atoms: 3d3, 3d5, 3d8 and 3d4?
", "answer_text": "", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-40", "marks": 1, "source_number": "5", "original_raw_text": " What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: 3d3, 3d5, 3d8 and 3d4?", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5", "needs_review": true, "missing_answer": true } }, { "sequence_no": 60, "question_text": "
Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
", "answer_text": "
(1)\tVanadate ion (VO–3 )
\n
Oxidation state of vanadium is +5. which is equal to its group number 5.
\n
(2)\tDichromate (Cr2O2– 7 ) and chromate (CrO2–4 )
\n
Oxidation state of chromium in both oxoanions is +6. which is equal to its group number 6.
\n
(3)\tPermanganate ion (MnO4– )
\n
Oxidation state of manganese is +7. which is equal to its group number 7.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-59", "source_number": "6", "original_raw_text": " Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 61, "question_text": "
What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
", "answer_text": "
Refer Que. 15 (Page No.240)
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-22", "source_number": "7", "original_raw_text": " What is lanthanoid contraction? What are the consequences of lanthanoid contraction? ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 62, "question_text": "
What are the characteristics of the transition elements and why are they called transition elements? Which d-block elements may not be regarded as the transition elements?
", "answer_text": "
Refer Que. 9 (Page No.238)
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-3", "marks": 3, "source_number": "8", "original_raw_text": " What are the characteristics of the transition elements and why are they called transition elements? Which d-block elements may not be regarded as the transition elements? ", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 63, "question_text": "
In what way is the electronic configuration of the transition elements different from that of the non-transition elements?
[Topic 4.0] [2 Marks]
", "answer_text": "
Refer Que. 10 (Page No.238)
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "marks": 2, "source_number": "9", "original_raw_text": " In what way is the electronic configuration of the transition elements different from that of the non-transition elements? [Topic 4.0] [2 Marks]", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 64, "question_text": "
What are the different oxidation states exhibited by the lanthanoids?
", "answer_text": "
In the lanthanide series, +3 oxidation state is most common i.e., Ln(III) compounds are predominant. However, +2 and +4 oxidation states can also be found in the solution or in solid compounds.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "source_number": "10", "original_raw_text": " What are the different oxidation states exhibited by the lanthanoids?", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 65, "question_text": "
Explain giving reasons:
\n
\t(i)\tTransition metals and many of their compounds show paramagnetic behaviour.
\n
\t(ii)\tThe enthalpies of atomization of the transition metals are high.
\n
\t(iii)\tThe transition metals generally form coloured compounds.
\n
\t(iv)\tTransition metals and their many compounds act as good catalyst.
", "answer_text": "
(i)\tTransition metals show paramagnetic behaviour. Paramagnetism arises due to the presence of unpaired electrons with each electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum.
\n
However, in the first transition series, the orbital angular momentum is quenched. Therefore, the resulting paramagnetism is only because of the unpaired electron.
\n
Magnetic moment is calculated by following formula.
\n
µ = \"\" B.M.
\n
\t\twhere, n = No. of unpaired electron
\n
(ii) Transition elements have high effective nuclear charge and a large number of valence electrons. Therefore, they form very strong metallic bonds. As a result, the enthalpy of atomization of transition metals in high.
\n
(iii) Most of the complexes of transition metals are coloured. This is because of the absorption of radiation from visible light region to promote an electron from one of the d-orbitals to another.
\n
In the presence of ligands, the d-orbitals split up into two sets of orbitals having different energies. Therefore, the transition of electrons can take place from one set to another.
\n
The energy required for these transitions is quite small and falls in the visible region of radiation. The ions of transition metals absorb the radiation of a particular wavelength and the rest is reflected, imparting colour to the solution.
\n
(iv)\tThe catalytic activity of the transition elements can be explained by two basic facts.
\n
(a)\tOwing to their ability to show variable oxidation states and form complexes, transition metals form unstable intermediate compounds. Thus, they provide a new path with lower activation energy, Ea, for the reaction.
\n
(b)\tTransition metals also provide a suitable surface for the reactions to occur.
\n
Example, V2O5 in contact process, Fe in Haber's process.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/EQ-0121-121708.png", "position": "answer", "context": "µ = B.M." } ], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-6", "source_number": "11", "marks": 2, "original_raw_text": " Explain giving reasons:\n\t(i)\tTransition metals and many of their compounds show paramagnetic behaviour.\n\t(ii)\tThe enthalpies of atomization of the transition metals are high.\n\t(iii)\tThe transition metals generally form coloured compounds.\n\t(iv)\tTransition metals and their many compounds act as good catalyst. ", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 66, "question_text": "
What are interstitial compounds? Why are such compounds well known for transition metals?
", "answer_text": "
Refer Que. 26 (Page No.244)
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-22", "source_number": "12", "original_raw_text": " What are interstitial compounds? Why are such compounds well known for transition metals?", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 67, "question_text": "
How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.
", "answer_text": "
In transition elements, the oxidation state can vary from +1 to the highest oxidation state by removing all its valence electrons. Also, in transition elements, the oxidation states differ by 1 (Fe2+ and Fe3+; Cu+ and Cu2+). In non-transition elements, the oxidation states differ by 2, for example, +2 and +4 or +3 and +5, etc.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "source_number": "13", "original_raw_text": " How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 68, "question_text": "
Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate? [Topic 4.4]
", "answer_text": "
Dichromates are generally prepared from chromate. which in turn are obtained by the fusion of chromite ore (FeCr2O4) with sodium or potassium carbonate in free access of air.
\n
The reaction with sodium carbonate occurs as follows:
\n
\t4FeCr2O4 + 8Na2CO3 + 7O2 →
8Na2CrO4 + 2Fe2O3 + 8CO2
\n
The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na2Cr2O7 . 2H2O can be crystallised.
\n
\t2Na2CrO4 + 2 H+ → Na2Cr2O7 + 2Na+ + H2O
\n
Sodium dichromate is more soluble than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride.
\n
\tNa2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl
\n
Orange crystals of potassium dichromate crystallise out.
\n
The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution.
\n
\t2CrO2–4 + 2H+ → Cr2O2–7 + H2O
\n
\tCr2O2–7 + 2OH– → 2CrO2–4 + H2O
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-22", "marks": 4, "source_number": "14", "original_raw_text": " Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate? [Topic 4.4]", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 69, "question_text": "
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with ; (i) iodide, (ii) iron(II) solution and (iii) H2S.
", "answer_text": "
Potassium dichromates are strong oxidising agents
\n
In acidic solution, its oxidising action can be represented as follows
\n
\tCr2O2–7 + 14H+ + 6e– → 2Cr3+ + 7H2O
\n
(1)\tAcidified potassium dichromate will oxidise iodides to iodine,
\n
\"\"
\n
(2)\tAcidified potassium dichromate will oxidise iron (II) solution to iron (III) solution.
\n
\"\"
\n
(3)\tAcidified potassium dichromate will oxidise H2S to sulphur.
\n
\"\"
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/112.png", "position": "answer", "context": "" }, { "id": "img-1", "path": "Chapter4/113.png", "position": "answer", "context": "" }, { "id": "img-2", "path": "Chapter4/114.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-3", "source_number": "15", "original_raw_text": " Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with ; (i) iodide, (ii) iron(II) solution and (iii) H2S.", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 70, "question_text": "
Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) SO2 and (iii) oxalic acid? Write the ionic equations for the reactions.
\n
OR
\n
\tState the balance reaction equations of acidic permanganate ion with
\n
(i) Fe2+ (ii) C2 O2–4 (iii) SO2–3 [June 2024] [3 Marks]
", "answer_text": "
Preparation:
\n
Potassium permanganate is prepared by fusion of MnO2 with an alkali metal hydroxide and an oxidising agent like KNO3. This produces the dark green K2MnO4 which disproportionates in a neutral or acidic solution to give permanganate.
\n
\t2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
\n
\t3MnO2–4 + 4H+ → 2MnO–4 + MnO2 + 2H2O
\n
Commercially it is prepared by the alkaline oxidative fusion of MnO2 followed by the electrolytic oxidation of manganate (VI).
\n
\"\"
\n
\"\"
\n
In the laboratory, a manganese (II) ion salt is oxidised by peroxodisulphate to permanganate.
\n
\t2Mn2+ + 5S2O2–8 + 8H2O → 2MnO–4 + 10SO2–4 + 16H+
\n
In Acidic Medium oxidising action of KMnO4
\n
\"\" + 8H+ + 5e– → Mn2+ + 4H2O
\n
(i)\tAcidified KMnO4 solution oxidizes Fe (II) ions to Fe (III) ions i.e., ferrous ions to ferric ions.
\n
\"\"
\n
(ii)\tAcidified potassium permanganate oxidizes SO2 to sulphuric acid.
\n
\"\"
\n
(iii)\tAcidified potassium permanganate oxidizes oxalic acid to carbon dioxide.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/EQ-0224-155150.png", "position": "answer", "context": "" }, { "id": "img-1", "path": "Chapter4/EQ-0224-155211.png", "position": "answer", "context": "" }, { "id": "img-2", "path": "Chapter4/EQ-0206-133705.png", "position": "answer", "context": "+ 8H + + 5e – → Mn 2+ + 4H 2 O" }, { "id": "img-3", "path": "Chapter4/116.png", "position": "answer", "context": "" }, { "id": "img-4", "path": "Chapter4/117.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-3", "marks": 3, "source_number": "16", "boundary_log": [ "or_alt→stem" ], "original_raw_text": " Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) SO2 and (iii) oxalic acid? Write the ionic equations for the reactions.\nOR\n\tState the balance reaction equations of acidic permanganate ion with\n (i) Fe2+ (ii) C2 O2–4 (iii) SO2–3 [June 2024] [3 Marks]", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 71, "question_text": "
For M2+/M and M3+/M2+ systems the Eo values for some metals are as follows:
\n
\tCr2+/Cr\t–0.9 V\tCr3/Cr2+:\t–0.4 V
\n
\tMn2+/Mn\t–1.2 V\tMn3+/Mn2+\t+1.5 V
\n
\tFe2+/Fe\t–0.4 V\tFe3+/Fe2\t+0.8 V
\n
\tUse this data to comment upon:
\n
\t(i)\tThe stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+.
\n
\t(ii)\tThe ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.
", "answer_text": "
(i)\tThe E° value for Fe3+/Fe2+ is higher than that for Cr3+/Cr2+ and lower than that for
Mn3+/Mn2+. So, the reduction of Fe3+ to Fe2+ is easier than the reduction of Mn3+ to Mn2+, but not as easy as the reduction of Cr3+ to Cr2+. Hence, Fe3+ is more stable than Mn3+, but less stable than Cr3+. These metal ions can be arranged in the increasing order of their stability as: Mn3+ < Fe3+ < Cr3+.
\n
(ii)\tThe reduction potentials for the given pairs increase in the following order:
\n
\t\tMn2+ / Mn < Cr2+ / Cr < Fe2+ / Fe
\n
So, the oxidation of Fe to Fe2+ is not as easy as the oxidation of Cr to Cr2+ and the oxidation of Mn to Mn2+. Thus, these metals can be arranged in the increasing order of their ability to get oxidised as: Fe < Cr < Mn.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-6", "source_number": "17", "original_raw_text": " For M2+/M and M3+/M2+ systems the Eo values for some metals are as follows:\n\tCr2+/Cr\t–0.9 V\tCr3/Cr2+:\t–0.4 V\n\tMn2+/Mn\t–1.2 V\tMn3+/Mn2+\t+1.5 V\n\tFe2+/Fe\t–0.4 V\tFe3+/Fe2\t+0.8 V\n\tUse this data to comment upon:\n\t(i)\tThe stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+.\n\t(ii)\tThe ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 72, "question_text": "
Predict which of the following will be coloured in aqueous solution? Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+ and Co2+. Give reasons for each.
", "answer_text": "
Only the ions that have electrons in d-orbital and in which d-d transition is possible will be coloured.
\n
The ions in which d-orbitals are empty or completely filled will be colourless as no d-d transition is possible in those configurations.

Element

Atomic number

Ionic state

Electronic configuration in ionic state

Number of unpaired electrons

Ti

22

T13+

[Ar] 3d1

1

V

23

V3+

[Ar] 3d2

2

Cu

29

Cu+

[Ar] 3d10

0

Sc

21

Sc3+

[Ar]

0

Mn

25

Mn2+

[Ar] 3d5

5

Fe

26

Fe3+

[Ar] 3d5

5

Co

27

Co2+

[Ar] 3d7

3

\n
From the above table, it can be easily observed that only Sc3+ has an empty d-orbital and Cu+ has completely filled d-orbitals. All other ions, except Sc3+ and Cu+, will be coloured in aqueous solution because of d-d transitions.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-59", "marks": 2, "source_number": "18", "original_raw_text": " Predict which of the following will be coloured in aqueous solution? Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+ and Co2+. Give reasons for each.", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 73, "question_text": "
Compare the stability of +2 oxidation state for the elements of the first transition series.
", "answer_text": "
The number of oxidation states increases on moving from Sc to Mn. On moving from Mn to Zn, the number of oxidation states decreases due to a decrease in the number of available unpaired electrons.
\n
The relative stability of the +2 oxidation state increases on moving from top to bottom. This is because on moving from top to bottom, it becomes more and more difficult to remove the third electron from the d-orbital.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "source_number": "19", "original_raw_text": " Compare the stability of +2 oxidation state for the elements of the first transition series. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 74, "question_text": "
Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
\n
\t(i)\tElectronic configuration,
\n
\t(ii)\tOxidation state,
\n
\t(iii)\tAtomic and ionic sizes and
\n
\t(iv)\tChemical reactivity.
", "answer_text": "
Refer Que. 42 (Page No.250)
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-50", "source_number": "20", "original_raw_text": " Compare the chemistry of actinoids with that of the lanthanoids with special reference to:\n\t(i)\tElectronic configuration,\n\t(ii)\tOxidation state,\n\t(iii)\tAtomic and ionic sizes and\n\t(iv)\tChemical reactivity. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 75, "question_text": "
How would you account for the following:
\n
(i)\tCr2+ is strongly reducing while manganese (III )is strongly oxidising.
\n
\t(ii)\tCobalt (II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.
\n
\t(iii)\tThe d 1 configuration is very unstable in ions.
", "answer_text": "
(i)\tCr2+ is strongly reducing in nature. It has a d 4 configuration. While acting as a reducing agent, it gets oxidized to Cr3+ (electronic configuration, d 3). This d 3 configuration can be written as \"\" configuration, which is a more stable configuration. In the case of Mn3+ (d 4), it acts as an oxidizing agent and gets reduced to Mn2+ (d 5). This has an exactly half-filled d-orbital and is highly stable.
\n
(ii)\tCo (II) is stable in aqueous solutions. However, in the presence of strong field complexing reagents, it is oxidized to Co (III). Although the 3rd ionization energy for Co is high, but the higher amount of crystal field stabilization energy (CFSE) released in the presence of strong field ligands overcomes this ionization energy.
\n
(iii)\tThe ions in d 1 configuration tend to lose one more electron to get into stable d 0 configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the d-orbital of these ions. Therefore, they act as reducing agents.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/EQ-0318-140506.png", "position": "answer", "context": "Ans. (i)\tCr 2+ is strongly reducing in nature. It has a d 4 …" } ], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-6", "source_number": "21", "original_raw_text": " How would you account for the following:\n (i)\tCr2+ is strongly reducing while manganese (III )is strongly oxidising.\n\t(ii)\tCobalt (II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.\n\t(iii)\tThe d 1 configuration is very unstable in ions. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 76, "question_text": "
What is meant by 'disproportionation'? Give an example of disproportionation reaction in aqueous solution.
", "answer_text": "
It is found that sometimes a relatively less stable oxidation state undergoes an oxidation – reduction reaction in which it is simultaneously oxidised and reduced. This is called disproportionation.
\n
\te.g. 3\"\" + 4H+ → 2\"\" + MnO2 + 2H2O
\n
\"\"
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/EQ-0206-134806.png", "position": "answer", "context": "e.g. 3 + 4H+ → 2 + MnO 2 + 2H 2 O" }, { "id": "img-1", "path": "Chapter4/EQ-0206-134749.png", "position": "answer", "context": "e.g. 3 + 4H+ → 2 + MnO 2 + 2H 2 O" }, { "id": "img-2", "path": "Chapter4/125.png", "position": "answer", "context": "" } ], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-77", "marks": 2, "source_number": "22", "original_raw_text": " What is meant by 'disproportionation'? Give an example of disproportionation reaction in aqueous solution.", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 77, "question_text": "
Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?
", "answer_text": "
In the first transition series, Cu exhibits
+1 oxidation state very frequently. It is because Cu (+1) has an electronic configuration of [Ar] 3d10. The completely filled d-orbital makes
it highly stable.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "source_number": "23", "original_raw_text": " Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 78, "question_text": "
Calculate the number of unpaired electrons in the following gaseous ions: Mn3+, Cr3+, V3+ and Ti3+. Which one of these is the most stable in aqueous solution?
", "answer_text": "
Cr3+ is the most stable in aqueous solutions owing to a \"\" configuration.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/EQ-0318-140506.png", "position": "answer", "context": "Cr 3+ is the most stable in aqueous solutions owing to a con…" } ], "tables": [ { "id": "tbl-1", "rows": 5, "columns": 3, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Gaseous ions

\n		\n

Number of unpaired electrons

\n
\n

(i)

\n		\n

Mn3+: [Ar] 3d4

\n		\n

4

\n
\n

(ii)

\n		\n

Cr3+: [Ar] 3d3

\n		\n

3

\n
\n

(iii)

\n		\n

V3+: [Ar] 3d2

\n		\n

2

\n
\n

(iv)

\n		\n

Ti3+: [Ar] 3d1

\n		\n

1

\n
", "position": "stem", "context": "Gaseous ions Number of unpaired electrons (i) Mn 3+ : [Ar] 3…" } ], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-59", "source_number": "24", "original_raw_text": " Calculate the number of unpaired electrons in the following gaseous ions: Mn3+, Cr3+, V3+ and Ti3+. Which one of these is the most stable in aqueous solution? ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 79, "question_text": "
Give examples and suggest reasons for the following features of the transition metal chemistry:
\n
\t(i)\tThe lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
\n
\t(ii)\tA transition metal exhibits highest oxidation state in oxides and fluorides.
\n
\t(iii)\tThe highest oxidation state is exhibited in oxoanions of a metal.
", "answer_text": "
(i)\tIn the case of a lower oxide of a transition metal, the metal atom has a low oxidation state. This means that some of the valence electrons of the metal atom are not involved in bonding. As a result, it can donate electrons and behave as a base.
\n
On the other hand, in the case of a higher oxide of a transition metal, the metal atom has a high oxidation state. This means that the valence electrons are involved in bonding and so, they are unavailable. There is also a high effective nuclear charge. As a result, it can accept electrons and behave as an acid.
\n
\te.g.: MnO = Basic, MnO2 = Amphoteric,
Mn2O7 = Acidic
\n
(ii)\tOxygen and fluorine act as strong oxidising agents because of their high electronegativities and small sizes. Hence, they bring out the highest oxidation states from the transition metals. In other words, a transition metal exhibits higher oxidation states in oxides and fluorides. For example, in OsF6 and V2O5, the oxidation states of Os and V are
+6 and +5 respectively.
\n
(iii)\tOxygen is a strong oxidising agent due to its high electronegativity and small size. So, oxo-anions of a metal have the highest oxidation state. For example, in MnO–4, the oxidation state of Mn is +7.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-50", "source_number": "25", "marks": 3, "original_raw_text": " Give examples and suggest reasons for the following features of the transition metal chemistry:\n\t(i)\tThe lowest oxide of transition metal is basic, the highest is amphoteric/acidic.\n\t(ii)\tA transition metal exhibits highest oxidation state in oxides and fluorides.\n\t(iii)\tThe highest oxidation state is exhibited in oxoanions of a metal.", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 80, "question_text": "
Indicate the steps in the preparation of:
\n
\t(i) K2Cr2O7 from chromite ore.
\n
\t(ii) KMnO4 from pyrolusite ore.
", "answer_text": "
(i) K2Cr2O7 from chromite ore
\n
4FeCr2O4 + 8Na2CO3 + 7O2
\n
→ 8Na2CrO4 + 2Fe2O3 + 8CO2
\n
2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O
\n
Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl
\n
\t(ii) KMnO4 from pyrolusite ore
\n
Potassium permanganate is prepared by fusion of MnO2 with an alkali metal hydroxide and an oxidising agent like KNO3.
\n
This produces the dark green K2MnO4 which disproportionates in a neutral or acidic solution to give permanganate.
\n
2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
\n
3MnO2–4 + 4H+ → 2MnO–4 + MnO2 + 2H2O
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-3", "source_number": "26", "marks": 3, "original_raw_text": " Indicate the steps in the preparation of: \n\t(i) K2Cr2O7 from chromite ore. \n\t(ii) KMnO4 from pyrolusite ore. ", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 81, "question_text": "
What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
", "answer_text": "
“An alloy is a solid solution of two or more elements in a metallic matrix. It can either be a partial solid solution or a complete solid solution.”
\n
Alloys are usually found to possess different physical properties than those of the component elements.
\n
An important alloy of lanthanoids is Mischmetal. It contains lanthanoids (94-95%), iron (5%), and traces of S, C, Si, Ca, and Al.
\n
Uses:
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "source_number": "27", "original_raw_text": " What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 82, "question_text": "
What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29, 59, 74, 95, 102, 104.
", "answer_text": "
Inner transition metals are those elements in which the last electron enters the ƒ-orbital.
\n
The elements in which the 4ƒ-orbitals and the 5ƒ-orbitals are progressively filled are called ƒ-block elements.
\n
Among the given atomic numbers, the atomic numbers of the inner transition elements are
59, 95, and 102.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-50", "source_number": "28", "original_raw_text": " What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29, 59, 74, 95, 102, 104.", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 83, "question_text": "
The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.
", "answer_text": "
Lanthanoids primarily show three oxidation states (+2, +3, +4).
\n
Among these oxidation states, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4ƒ, 5d, and 6s-orbitals is quite large.
\n
On the other hand, the energy difference between 5ƒ, 6d, and 7s-orbitals is very less. Hence, actinoids display a large number of oxidation states.
\n
For example, uranium and plutonium display +3, +4, +5, and +6 oxidation states while neptunium displays +3, +4, +5, +7. The most common oxidation state in case of actinoids is also +3.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-77", "marks": 2, "source_number": "29", "original_raw_text": " The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements. ", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 84, "question_text": "
Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.
", "answer_text": "
The last element in the actinoid series is lawrencium, Lr. Its atomic number is 103 and its electronic configuration is [Rn]5ƒ14. 6d1. 7s2. The most common oxidation state displayed by it is +3; because after losing 3 electrons it attains stable f14 configuration.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-3", "marks": 1, "source_number": "30", "original_raw_text": " Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 85, "question_text": "
Use Hund's rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of 'spin-only' formula.
", "answer_text": "
Electronic configuration of Ce: [xe] 4f 1 5d1 6s2
\n
\t\tCe3+: [xe] 4ƒ 1\t(n = 1)
\n
µ = \"\" B.M.
\n
\t\t= \"\"
\n
\t\t= \"\"
\n
\t\t= 1.73 B.M.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/EQ-0121-121708.png", "position": "answer", "context": "µ = B.M." }, { "id": "img-1", "path": "Chapter4/EQ-0206-121833.png", "position": "answer", "context": "=" }, { "id": "img-2", "path": "Chapter4/EQ-0206-121847.png", "position": "answer", "context": "=" } ], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-3", "source_number": "31", "original_raw_text": " Use Hund's rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of 'spin-only' formula.", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 86, "question_text": "
Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.
", "answer_text": "
The lanthanoids that exhibit +2 and +4 states are shown in the given table. The atomic numbers of the elements are given in the parenthesis.
\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

+2

\n		\n

+4 Oxidation State

\n
\n

Nd (60)

\n		\n

Ce (58)

\n
\n

Sm (62)

\n		\n

Pr (59)

\n
\n

Eu (63)

\n		\n

Nd (60)

\n
\n

Tm (69)

\n		\n

Tb (65)

\n
\n

Yb (70)

\n		\n

Dy (66)

\n
\n
Ce after forming Ce4+ attains a stable electronic configuration of [Xe].
\n
Tb after forming Tb4+ attains a stable electronic configuration of [Xe] 4f 7.
\n
Eu after forming Eu2+ attains a stable electronic configuration of [Xe] 4f 7.
\n
Yb after forming Yb2+ attains a stable electronic configuration of [Xe] 4f 14.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-3", "source_number": "32", "original_raw_text": " Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 87, "question_text": "
Compare the chemistry of the actinoids with that of lanthanoids with reference to:
\n
\t(i) Electronic configuration
\n
\t(ii) Oxidation states and
\n
\t(iii) Chemical reactivity.
", "answer_text": "
Refer Que. 42 (Page No.250)
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-51", "source_number": "33", "original_raw_text": "Compare the chemistry of the actinoids with that of lanthanoids with reference to:\n\t(i) Electronic configuration\n\t(ii) Oxidation states and\n\t(iii) Chemical reactivity.", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 88, "question_text": "
Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.
", "answer_text": "
61Pm (Promethium)\t= [xe] 4ƒ5 5d0 6s2
\n
91Pa (Protactinium)\t= [Rn] 5ƒ2 6d1 7s2
\n
\t101Md (Mendelevium)\t= [Rn] 5ƒ13 6d0 7s2
\n
109Mt (Meitnerium)\t= [Rn] 5ƒ14 6d7 7s2
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "source_number": "34", "original_raw_text": " Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 89, "question_text": "
Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:
\n
\t(i)\tElectronic configurations,
\n
\t(ii)\tOxidation states,
\n
\t(iii)\tIonisation enthalpies and
\n
\t(iv)\tAtomic sizes.
", "answer_text": "
(i)\tIn the 1st, 2nd and 3rd transition series, the 3d, 4d and 5d-orbitals are respectively filled.
\n
(ii)\tIn each of the three transition series the number of oxidation states shown by the elements is the maximum in the middle and the minimum at the extreme ends.
\n
However, +2 and +3 oxidation states are quite stable for all elements present in the first transition series. All metals present in the first transition series form stable compounds in the +2 and +3 oxidation states.
\n
The stability of the +2 and +3 oxidation states decreases in the second and the third transition series, wherein higher oxidation states are more important.
\n
(iii)\tIn each of the three transition series, the first ionization enthalpy increases from left to right. However, there are some exceptions.
\n
The first ionization enthalpies of the third transition series are higher than those of the first and second transition series. This occurs due to the poor shielding effect of
4f electrons in the third transition series.
\n
(iv)\tAtomic size generally decreases from left to right across a period. Now, among the three transition series, atomic sizes of the elements in the second transition series are greater than those of the elements corresponding to the same vertical column in the first transition series.
\n
However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-34", "source_number": "35", "original_raw_text": " Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:\n\t(i)\tElectronic configurations,\n\t(ii)\tOxidation states,\n\t(iii)\tIonisation enthalpies and\n\t(iv)\tAtomic sizes.", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 90, "question_text": "
Write down the number of 3d electrons in each of the following ions: Ti2+, V2+, Cr3+, Mn2+, Fe2+, Fe3+, Co2+, Ni2+ and Cu2+. Indicate how would you expect the five 3d orbitals to the occupied for these hydrated ions (octahedral).
", "answer_text": "", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [ { "id": "tbl-1", "rows": 10, "columns": 4, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Ions

\n		\n

Electronic

\n

Configuration

\n		\n

Number of

\n

d-electron

\n		\n

Filling of

\n

d-electron

\n
\n

Ti2+

\n		\n

[Ar] 3d2

\n		\n

2

\n		\n

t2g2 eg0

\n
\n

V2+

\n		\n

[Ar] 3d3

\n		\n

3

\n		\n

t2g3 eg0

\n
\n

Cr3+

\n		\n

[Ar] 3d3

\n		\n

3

\n		\n

t2g3 eg0

\n
\n

Mn2+

\n		\n

[Ar] 3d5

\n		\n

5

\n		\n

t2g3 eg2

\n
\n

Fe2+

\n		\n

[Ar] 3d6

\n		\n

6

\n		\n

t2g4 eg2

\n
\n

Fe3+

\n		\n

[Ar] 3d5

\n		\n

5

\n		\n

t2g3 eg2

\n
\n

Co2+

\n		\n

[Ar] 3d7

\n		\n

7

\n		\n

t2g5 eg2

\n
\n

Ni2+

\n		\n

[Ar] 3d8

\n		\n

8

\n		\n

t2g6 eg2

\n
\n

Cu2+

\n		\n

[Ar] 3d9

\n		\n

9

\n		\n

t2g6 eg3

\n
", "position": "answer", "context": "Ions Electronic Configuration Number of d -electron Filling …" } ], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-40", "marks": 3, "source_number": "36", "original_raw_text": " Write down the number of 3d electrons in each of the following ions: Ti2+, V2+, Cr3+, Mn2+, Fe2+, Fe3+, Co2+, Ni2+ and Cu2+. Indicate how would you expect the five 3d orbitals to the occupied for these hydrated ions (octahedral). ", "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 91, "question_text": "
Elements of the first transition series possess many properties different from those of heavier transition elements.
", "answer_text": "
The properties of the elements of the first transition series differ from those of the heavier transition element in many ways.
\n
(i)\tThe atomic sizes of the elements of the first transition series are smaller than those of the heavier elements (elements of 2nd and 3rd transition series).
\n
However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.
\n
(ii)\t+2 and +3 oxidation states are more common for elements in the first transition series, while higher oxidation states are more common for the heavier elements.
\n
(iii)\tThe enthalpies of atomionization of the elements in the first transition series are lower than those of the corresponding elements in the second and third transition series.
\n
(iv)\tThe melting and boiling points of the first transition series are lower than those of the heavier transition elements. This is because of the occurrence of stronger metallic bonding
(M-M bonding).
\n
(v)\tThe elements of the first transition series form
low-spin or high-spin complexes depending upon the strength of the ligand field. However, the heavier transition elements form only low-spin complexes, irrespective of the strength of the ligand field.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "source_number": "37", "original_raw_text": " Elements of the first transition series possess many properties different from those of heavier transition elements. ", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } }, { "sequence_no": 92, "question_text": "
What can be inferred from the magnetic moment values of the following complex species?
", "answer_text": "
(i)\tK4[Mn(CN)6]
\n
For in transition metals, the magnetic moment is calculated from the spin-only formula.
\n
Therefore,
\n
µ = \"\" = 2.2 B.M.
\n
We can see from the above calculation that the given value is closest to n = 1. Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital.
\n
Hence, we can say that CN– is a strong field ligand that causes the pairing of electrons.
\n
(ii)\t[Fe(H2O)6]2+
\n
µ = \"\" = 5.3 B.M.
\n
(iii)\tK2[MnCl4]
\n
µ = \"\" = 5.9 B.M.
\n
We can see from the above calculation that the given value is closest to n = 5. Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital.
\n
Hence, we can say that Cl– is a weak field ligand and does not cause the pairing of electrons.
", "type": "SUB", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/EQ-0318-154305.png", "position": "answer", "context": "µ = = 2.2 B.M." }, { "id": "img-1", "path": "Chapter4/EQ-0318-154305.png", "position": "answer", "context": "µ = = 5.3 B.M. We can see from the above calculation that th…" }, { "id": "img-2", "path": "Chapter4/EQ-0318-143240.png", "position": "answer", "context": "µ = = 5.9 B.M." } ], "tables": [ { "id": "tbl-1", "rows": 4, "columns": 3, "raw_html": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Example

\n		\n

Magnetic moment (B.M.)

\n
\n

(i)

\n		\n

K4[Mn(CN)6]

\n		\n

2.2

\n
\n

(ii)

\n		\n

[Fe(H2O)6]2+

\n		\n

5.3

\n
\n

(iii)

\n		\n

K2[MnCl4]

\n		\n

5.9

\n
", "position": "stem", "context": "Example Magnetic moment (B.M.) (i) K 4 [Mn(CN) 6 ] 2.2 (ii) …" } ], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-3", "source_number": "38", "original_raw_text": " What can be inferred from the magnetic moment values of the following complex species?", "marks": 0, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Exercise Q & A", "section_id": "4.5" } } ] }, { "section_id": "4.7", "section_title": "Multiple Choice Questions (MCQs)", "heading_text": "Multiple Choice Questions (MCQs)", "detected_type": "MCQ", "match_confidence": 1.0, "questions": [ { "sequence_no": 93, "question_text": "
\tMn2O7, CrO3, Cr2O3, CrO, V2O5, V2O4
\n
13. Gadolinium belongs to 4f series. It's atomic number is 64. Which of the following is the correct electronic configuration of gadolinium?
\n
14. The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of Cr3+ ion is ______ .
\n
15. Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Which of the following is not the characteristic property of interstitial compounds ?
\n
16. KMnO4 acts as an oxidising agent in alkaline medium. When alkaline KMnO4 is treated with KI, iodide ion is oxidised to ______ .
\n
17. Which of the following statements is not correct?
\n
18. When acidified K2Cr2O7 solution is added to Sn2+ salts then Sn2+ changes to ______ .
\n
19. Highest oxidation state of manganese in fluoride (MnF4) is (+4) but highest oxidation state in oxides (Mn2O7) is (+7) because _____.
\n
20. Although zirconium belongs to 4d transition series and hafnium to 5d transition series even then they show similar physical and chemical properties because ______ .
\n
21. Why is HCl not used to make the medium acidic in oxidation reactions of KMnO4 in acidic medium?
\n
1. The elements whose ground state or any of its oxidation states is incompletely filled by
4d- orbital electrons, are called elements of which transition series?
\n
2. Which element of third transition elements is not considered as transition element?
\n
3. How many groups are included in d-block?
\n
4. The elements which in their ground-state or any one of its oxidation state, have incompletely filled d-orbitals with electrons are called:
\n
5. Which of the following is not considered transition element?
\n
6. What is the general electronic configuration of transition elements?
\n
7. Which element of the following not considered as transition element?
\n
8. Elements of which groups are called d-block element?
\n
9. Which sets are the transitions elements?
\n
10. Among the following series of transition metal ions, the one where all metal ions have 3d2 electronic configuration is:
\n
11. Which of the following are d-block elements but not regarded as transition elements?
\n
12. Zn and Cd do not show variable valency, because:
\n
13. What is the name of element X by the following electronic configuration X = [Ar] 3d104s1?
\n
14. Electronic structure of ________ is 3d3 4s0.
\n
15. Which ion has 5.93 BM magnetic moment?
\n
16. Which of the following is not an interstitial compound?
\n
17. Transition metals and their compounds contain catalytic property, because________ .
\n
18. What is responsible from the following for inner transition elements having +3 stable oxidation state?
\n
19. Which element shows minimum paramagnetism?
\n
20. Which ion of the following contains minimum magnetic moment?
\n
21. What is the general characteristic property of transition elements?
\n
22. What is the change in ionization energy while going from left to right in first transition elements?
\n
23. Which ion of 3d-orbital in transition series contains maximum magnetic moment?
\n
24. Which ion from the following contain magnetic moment same as Co3+ ion?
\n
25. Laws of Hume and Rothery are associated with formation of which of the following?
\n
26. Which catalyst is used in Haber's process for industrial production of NH3?
\n
27. Which of the following metal containing liquid form?
\n
28. Which non-metallic elements does not make interstitial compound?
\n
29. Which statement is incorrect for transition metallic elements?
\n
30. Transition elements containing higher oxidation number possess ________ property.
\n
31. What is the formula to calculate magnetic moment?
\n
32. The voids of interstitial compounds present
in ________ .
\n
33. Which ion from the following contains diamagnetic property?
\n
34. Which metal does not contains more than one oxidation state?
\n
35. The difference between two atomic radii should not be more than ________ to form an alloy.
\n
36. The value of magnetic moment increases as ________ number increases.
\n
37. Which of the following compounds is diamagnetic ?
\n
38. Whose aqueous solution is colourless?
\n
39. Transition metallic elements behaves as reducing agent because ________ .
\n
40. Which of the following descending order of second ionization enthalpy?
\n
41. What will be the correct order of theoretical magnetic moment (in BM unit) of Mn2+, Cr2+, and V2+?
\n
42. In contact process manufacture of H2SO4 which catalyst is used to form SO3 from SO2?
\n
43. Which ionic pair from the following is coloured in aqueous solution?
\n
44. Mention the number of d-electrons in Fe2+ (Z = 26).
\n
45. Which is the third element except Ni and Zn in German silver alloy?
\n
46. Which alloy does not contain Ni metal?
\n
47. Which of the following reason is not correct for formation of complex compounds of transition metal ions?
\n
48. What will be the correct order theoretical magnetic moment of Cr3+, Mn2+ and Fe3+ having magnetic moment x, y and z in BM unit respectively?
\n
49. Which of the following statements about the interstitial compounds is incorrect?
\n
50. In which of the following ion d-d transition is not possible?
\n
51. The theoretical magnetic moment of 27Co is 3.87 BM. Which one is the correct compound from the following?
\n
52. What is the experimental value of magnetic moment of metal ion on [Fe(CN)6]4–?
\n
53. For the tetrahedral complex [MnBr4]2– the spin only magnetic moment value is ______ .
[Atomic no. of Mn = 25]
\n
54. Which of the following is colourless?
\n
55. Highest oxidation state (+7) is shown by:
\n
56. Transition elements are coloured because:
\n
57. Which of the following set has all the coloured ion?
\n
58. The catalytic activity of the transition metal and their compound is ascribed to their _____.
\n
59. Which one of the following sets correctly represents the increase in the paramagnetic property of the ions?
\n
60. Transition metals in their compounds show:
\n
61. Which of the following metal ions is not coloured?
\n
62. Which of the following forms interstitial compounds?
\n
63. The properties of Zr and Hf are similar because;
\n
64. Number of electrons transferred in each case when KMnO4 acts as an oxidising agent to give MnO2, Mn2+, Mn(OH)3 and MnO2–4, are respectively.
\n
65. Which shows a jump in second ionization potential?
\n
66. Number of electrons in 3d-orbital V2+, Cr2+, Mn2+ and Fe2+ are 3, 4, 5 and 6 respectively. Which of the following ions will have largest value of magnetic moment (m)?
\n
67. Which metal has the highest melting point?
\n
68. Non-stoichiometric compounds are formed by:
\n
69. d-block elements generally form:
\n
70. Which has the lowest melting point?
\n
71. The most stable ion is:
\n
72. Identify the correct order of ionic radii.
\n
73. Which elements in first transition series has maximum third ionization enthalpy?
\n
74. Which element in 3d series does not exhibit variable oxidation states?
\n
75. The element with positive reduction potential in 3d series is ________ .
\n
76. Which of the following is diamagnetic?
\n
77. Which of the following has highest magnetic moment?
\n
78. What is the oxidation number of Ti, whose magnetic moment is 1.73 BM?
\n
79. Which metal does not contain more than one oxidation state?
\n
80. Which is least stable in aqueous medium?
\n
81. Which of the following ion has the maximum theoretical magnetic moment?
\n
82. The correct order of catalytic properties of Cr, V, Fe and Mn metals in increasing order is ______ .
\n
83. What is the colour of potassium per manganate (KMnO4)?
\n
84. What is the colour of potassium manganate (K2MnO4)?
\n
85. What will be the product and its colour when MnO2 reacts with KOH in presence of air?
\n
86. In which compound Mn possesses maximum oxidation state?
\n
87. Which compound is used as an oxidizing agent in acidic, basic and in neutral medium?
\n
88. What is the colour of crystals of potassium dichromate?
\n
89. What is the formula of chromite?
\n
90. Which of the following properties is possessed by KMnO4?
\n
91. Which ion converts ferrous ion into ferric ion in acidic medium?
\n
92. The colour of KMnO4 is due to ________.
\n
93. Which one of the following ions exhibits d-d transition and paramagnetism as well?
\n
94. What is obtained by the oxidation I– with MnO4– in alkaline solution?
\n
95. The basic character of the transition metal oxides follow the order ________.
\n
96. Find basic oxides from the followings:
\n
\t(1) Mn2O7 (2) V2O3 (3) V2O5 (4) CrO (5) Cr2O3
\n
97. Which statement is not suitable for interstitial compound?
\n
98. Which of the following statement is incorrect for KMnO4?
\n
99. Identify the correct order of oxidizing strength:
\n
100. What is equivalent weight of K2Cr2O7 in acidic medium, if its molecular weight is taken as \"M\"?
\n
101. Which of the following undergoes disproportionation reaction?
\n
102. What is the structure of chromate ion?
\n
103. What is chemical formula of pyrolusite?
\n
104. What is oxidation number of Mn in KMnO4?
\n
105. Which of the following compounds has colour but no unpaired electrons?
\n
106. In chromate ore, the oxidation number of iron and chromium are respectively:
\n
107. When KMnO4 reacts with acidified FeSO4:
\n
108. The pair of amphoteric oxides is ________ .
\n
109. The number of unpaired electrons in 28Ni is:
\n
110. Why KMnO4 is of purple colour?
\n
111. In presence of acidic medium KMnO4 converts H2S into ________.
\n
112. In acidic medium, which of the following becomes colourless?
\n
113. Which block elements are called inner
transition?
\n
114. Which of the following elements are included in lanthanide series?
\n
115. What valency of all lanthanide series elements?
\n
116. Which of the following statements is incorrect?
\n
117. Which type of hydroxides are formed by lanthanoids?
\n
118. Which oxides are formed by lanthanoids?
\n
119. Which elements is radioactive among lanthanoid elements?
\n
120. Inner transition elements are of which periods?
\n
121. Zr and Hf have almost equal atomic size because ________ .
\n
122. Identify perfect order of ionic radii.
\n
123. Which of the following is most basic?
\n
124. Which of the following is strongest base?
\n
125. Which of the following lanthanoid has smallest atomic radius?
\n
126. Which of the following is lanthanoid element?
\n
127. Which of the following trivalent ions is colourless?
\n
128. Which of the following element have half filled f-orbital?
\n
129. Which of the following statement is incorrect?
\n
130. Across the lanthanide series, the basicity of lanthanoid hydroxides.
\n
131. Lanthanoids are:
\n
132. Lanthanoid contraction occurs because;
\n
133. Which among the following elements is radioactive?
\n
134. Which of the following statement is correct?
\n
135. Which maximum oxidation state possesses by actinoides?
\n
136. Lanthanoid and actinoids resembles in _____ .
\n
137. What is used in dry cell?
\n
138. What is the percentage of iron in pyrophoric alloy?
\n
139. Which is used in the stone of gas lighter?
\n
140. Which compounds of lanthanoids are used in pigments?
\n
141. What in used in thermometer?
\n
142. Which of the following substance is used in measurement of Chemical Oxygen Demand (COD) in polluted water?
\n
143. Assertion\t:\tCo (IV) is known, but Ni (IV) is not.
\n
\tReason \t:\tNi (IV) has a d4 electronic configuration.
\n
144. Assertion\t:\tTransition metals are strong reducing agents.
\n
Reason\t: \tTransition metals form intermetallic alloys with other elements.
\n
145. Assertion\t:\tKMnO4 is stored in deep purple bottles.
\n
Reason \t:\tHeating KMnO4 with alkali converts it into manganate.
\n
146. Assertion\t:\tEu2+ is more stable than Ce2+.
\n
Reason\t:\tEu2+ has a half-filled 4f-orbital and an empty 5d-orbital, which increases its stability.
\n
147. Assertion\t:\tCe4+ is a good analytical reagent.
\n
Reason\t:\tCe4+ has a tendency to change into Ce3+.
\n
148. Assertion\t:\tIn acidic medium, Cr+6 exists as a strong oxidising agent in the form of chromate, whereas MoO3 and WO3 are not strong oxidising agent.
\n
Reason\t:\tHeavy elements in the d-block tend to show higher oxidation states.
\n
149. Assertion\t:\tZn, Hg, and Cd are not considered transition elements.
\n
Reason\t:\tThe electron configuration of Zn, Hg, and Cd is represented by the formula (n–1)d10 ns2.
\n
150. Assertion\t:\tCr2+ undergoes reduction, and Mn3+ undergoes oxidation.
\n
Reason\t:\tCr2+ and Mn3+ have d4 and d5 electron configuration respectively.
\n
151. Assertion\t:\tCu dissolves in dilute HNO3 but not in dilute HCl.
\n
Reason\t:\tThe standard electrode potential (E°) of Cu is positive.
\n
152. Assertion\t:\tFe+3 catalyzes the reaction between iodide and persulfate ions.
\n
\tReason\t:\tTransition metal act as catalyst.
\n
153. Assertion\t:\tIn transition elements, the radii of the 5d series are almost equal to the corresponding members of the 4d series.
\n
Reason\t:\tThe filling of 4f-orbitals before
5d-orbitals causes a regular decrease in atomic radii.
\n
154. Assertion\t:\tAmong the elements from Sc to Zn, Zn has the lowest atomic ionization enthalpy.
\n
Reason\t:\tZinc has a greater number of unpaired electrons in its electronic configuration.
\n
155. Assertion\t:\tZr and Hf are similar in nature and difficult to separate.
\n
Reason\t:\tDue to lanthanide contraction, Zr and Hf have similar radii.
\n
156. Assertion\t:\tCopper (II) iodide is not known.
\n
Reason\t:\tCu2+ oxidizes I– to I2.
\n
157. Assertion\t:\tThe magnetic moment of Mn2+ is lower than that of Cr2+.
\n
\tReason\t:\tThe higher the atomic number, the lower the magnetic moment.
\n
158. Assertion\t:\tIn chromium compounds, the maximum oxidation state of chromium is +6.
\n
\tReason\t:\tChromium has only six electrons in the ns and (n–1)d-orbitals.
\n
159. Assertion\t:\tIn an acidic medium, K2Cr2O7 is orange in colour, while in a basic medium it converts CrO2−4 (yellow).
\n
\tReason\t:\tK2Cr2O7 is hygroscopic in nature and reacts with water to change its colour.
\n
160. Assertion\t:\tThe ability of oxygen to stabilize higher oxidation states exceeds that of fluorine.
\n
\tReason\t:\tThe highest oxidation state in oxides corresponds to the group number.
\n
161. Assertion\t:\tKMnO4 acts as an oxidizing agent in acidic, basic, or neutral media.
\n
\tReason\t:\tKMnO4 oxidizes ferrous sulfate to ferric sulfate.
\n
162. Assertion\t:\tThe atomic radii of Zr and Hf are nearly the same.
\n
Reason\t:\tZr and Hf belong to the same group.
\n
163. Assertion\t:\tCe4+ acts as an oxidizing agent in aqueous medium.
\n
\tReason\t:\tThe +4 oxidation state is common in lanthanoids.
\n
164. Assertion\t:\tHg is the only metal that is liquid at 0°C.
\n
\tReason\t:\tThis is due to its very high ionization energy and weak metallic bonding.
\n
165. Assertion\t:\tThe solution of Na2CrO4 in water is coloured.
\n
\tReason\t:\tIn Na2CrO4, the oxidation state of Cr is +6.
\n
166. Assertion\t:\tThe melting point of Mn is lower than that of Fe.
\n
\tReason\t:\tMn has fewer unpaired electrons compared to Fe, making its atomic bonding weaker.
\n
167. Assertion\t:\tThe coordination number of transition elements is variable.
\n
\tReason\t:\tThe energy of the ns and (n–1) d-orbitals is similar.
\n
168. Assertion\t:\tLa2O3 is basic in nature.
\n
\tReason\t:\tLa forms La(OH)3 in an aqueous solution.
\n
169. Assertion\t:\tAll compounds containing lanthanoid elements typically exhibit +3 oxidation state.
\n
\tReason\t:\tLanthanoid elements have three electrons in their outermost shell.
\n
170. Assertion\t:\tEu2+ and Yb2+ ions act as reducing agents.
\n
\tReason\t:\tBoth ions possess a stable half-filled electron configuration.
\n
171. Assertion\t:\tNeptunium is a transuranic element.
\n
\tReason\t:\tIt is heavier than uranium.
\n
172. Assertion\t:\tThe second ionization enthalpy of chromium is higher than that of its neighbouring elements.
\n
\tReason \t:\tAfter removing one electron from chromium, it attains the stable electron configuration [Ar]3d5, making the removal of the second electron require more energy.
\n
173. Assertion\t:\tWhen ions are formed from the first transition series elements, the electrons from the 4s-orbital are removed before those from the 3d-orbital.
\n
\tReason\t:\tThe attractive force towards the nucleus is stronger for the electrons in the 4s-orbital than for those in the 3d-orbital.
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": { "A": "H2SO4", "B": "KMnO4", "C": "(D)", "D": "K2Cr2O7" }, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/EQ-0121-123942.png", "position": "option:A", "context": "(A) (B)" }, { "id": "img-1", "path": "Chapter4/EQ-0121-123951.png", "position": "option:A", "context": "(A) (B)" }, { "id": "img-2", "path": "Chapter4/EQ-0121-124000.png", "position": "option:C", "context": "(C) (D)" }, { "id": "img-3", "path": "Chapter4/EQ-0121-124009.png", "position": "option:C", "context": "(C) (D)" }, { "id": "img-4", "path": "Chapter4/EQ-1206-143657.png", "position": "option:A", "context": "(A) M\t(B) (C) (D)" }, { "id": "img-5", "path": "Chapter4/EQ-1206-143708.png", "position": "option:A", "context": "(A) M\t(B) (C) (D)" }, { "id": "img-6", "path": "Chapter4/EQ-1206-143717.png", "position": "option:A", "context": "(A) M\t(B) (C) (D)" } ], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-17", "question_subtype": "assertion-reason", "original_raw_text": "\tMn2O7, CrO3, Cr2O3, CrO, V2O5, V2O4\n13. Gadolinium belongs to 4f series. It's atomic number is 64. Which of the following is the correct electronic configuration of gadolinium?\n14. The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of Cr3+ ion is ______ .\n15. Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Which of the following is not the characteristic property of interstitial compounds ?\n16. KMnO4 acts as an oxidising agent in alkaline medium. When alkaline KMnO4", "marks": 1, "deduction_level": 0, "source_html": "Chapter4", "boundary_log": [ "match/mcq table→stem" ], "section": "Multiple Choice Questions (MCQs)", "section_id": "4.7", "needs_review": true, "needs_ar_review": true, "ar_review_reason": "Assertion-Reason MCQ has an empty-text option marker", "needs_layout_review": true, "layout_review_reason": "item body has 31 'Assertion :' lines — an Assertion-Reason run merged into one item; split per number", "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "correct option not detected" } }, { "sequence_no": 94, "question_text": "
Choose the correct option for the statements with T (True) and F (False) indicators:
\n
(i) Mercury is not considered a transition element because it is liquid.
\n
(ii) K2Cr2O7 is a coloured compound because d-d transitions are possible in it.
\n
(iii) CuCl2–4 is known, but CuI2–4 is not known.
\n
\t(iv) Mn2O3 is acidic, while Mn2O7 is basic.
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": { "A": "FTTF", "B": "FFFF", "C": "TTTT", "D": "FTFF" }, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-17", "source_number": "174", "original_raw_text": " Choose the correct option for the statements with T (True) and F (False) indicators:\n (i) Mercury is not considered a transition element because it is liquid.\n (ii) K2Cr2O7 is a coloured compound because d-d transitions are possible in it.\n (iii) CuCl2–4 is known, but CuI2–4 is not known.\n\t(iv) Mn2O3 is acidic, while Mn2O7 is basic.", "marks": 1, "deduction_level": 0, "source_html": "Chapter4", "section": "Multiple Choice Questions (MCQs)", "section_id": "4.7", "needs_review": true, "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "correct option not detected" } }, { "sequence_no": 95, "question_text": "
Choose the correct option for the statements with T (True) and F (False) indicators:
\n
(i) In octahedral complexes, the stability of Co3+ is higher.
\n
(ii) Zinc forms coloured complexes.
\n
(iii) Most of the d-block elements and their compounds are ferromagnetic.
\n
(iv) In acidic medium, Cr+6 exists as dichromate and is a strong oxidizing agent.
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": { "A": "TFTF", "B": "TFFT", "C": "FTTF", "D": "TFFF" }, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest idfb425c3e0-ParaOverride-20", "source_number": "175", "original_raw_text": " Choose the correct option for the statements with T (True) and F (False) indicators:\n (i) In octahedral complexes, the stability of Co3+ is higher.\n (ii) Zinc forms coloured complexes.\n (iii) Most of the d-block elements and their compounds are ferromagnetic.\n (iv) In acidic medium, Cr+6 exists as dichromate and is a strong oxidizing agent.", "marks": 1, "deduction_level": 0, "source_html": "Chapter4", "section": "Multiple Choice Questions (MCQs)", "section_id": "4.7", "needs_review": true, "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "correct option not detected" } }, { "sequence_no": 96, "question_text": "
Choose the correct T (True) or F (False) for the following statements:
\n
(i) All Cr–O bonds in Cr2O2–7 are identical.
\n
(ii) CrO2–4 has a tetrahedral structure.
\n
(iii) Fe, Co, and Ni are collectively known as ferrous metals.
\n
(iv) In acidic K2Cr2O7, reaction with Sn2+ forms Sn4+.
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": { "A": "FTTT", "B": "TFFT", "C": "TFTF", "D": "FTFT" }, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "source_number": "176", "original_raw_text": " Choose the correct T (True) or F (False) for the following statements:\n (i) All Cr–O bonds in Cr2O2–7 are identical.\n (ii) CrO2–4 has a tetrahedral structure.\n (iii) Fe, Co, and Ni are collectively known as ferrous metals.\n (iv) In acidic K2Cr2O7, reaction with Sn2+ forms Sn4+.", "marks": 1, "deduction_level": 0, "source_html": "Chapter4", "section": "Multiple Choice Questions (MCQs)", "section_id": "4.7", "needs_review": true, "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "correct option not detected" } }, { "sequence_no": 97, "question_text": "
Choose the correct T (True) or F (False) for the following statements:
\n
(i) The stability of transition metal ions is low in aqueous medium.
\n
(ii) The magnetic moment of CoCl3 is 4.90 BM.
\n
(iii) Samarium is as hard as steel.
\n
(iv) Gadolinium has maximum paramagnetism.
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": { "A": "FTTF", "B": "FTTT", "C": "TFTF", "D": "FTFT" }, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Quest", "source_number": "177", "original_raw_text": " Choose the correct T (True) or F (False) for the following statements:\n (i) The stability of transition metal ions is low in aqueous medium.\n (ii) The magnetic moment of CoCl3 is 4.90 BM.\n (iii) Samarium is as hard as steel.\n (iv) Gadolinium has maximum paramagnetism.", "marks": 1, "deduction_level": 0, "source_html": "Chapter4", "section": "Multiple Choice Questions (MCQs)", "section_id": "4.7", "needs_review": true, "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "correct option not detected" } } ] }, { "section_id": "4.8", "section_title": "NCERT Exemplar MCQs", "heading_text": "NCERT Exemplar MCQs", "detected_type": "MCQ", "match_confidence": 1.0, "questions": [ { "sequence_no": 98, "question_text": "
(B)\t2.\t(A)\t3.\t(D)\t4.\t(B)\t5.\t(A)\t6.\t(B)\t7.\t(B)\t8.\t(A)\t9.\t(D)\t10.\t(C)
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Normal idfb425c3e0-ParaOverride-97", "source_number": "1", "original_raw_text": "(B)\t2.\t(A)\t3.\t(D)\t4.\t(B)\t5.\t(A)\t6.\t(B)\t7.\t(B)\t8.\t(A)\t9.\t(D)\t10.\t(C)", "marks": 1, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Exemplar MCQs", "section_id": "4.8", "needs_review": true, "needs_layout_review": true, "layout_review_reason": "question body is a consolidated answer key (NUMBER.(LETTER) pairs) — parse it into a number→letter map, not a question", "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "options + correct option not detected" } }, { "sequence_no": 99, "question_text": "
(A)\t12.\t(A)\t13.\t(A)\t14.\t(D) \t15.\t(B)\t16.\t(C)\t17.\t(A)\t18.\t(C)\t19.\t(D)\t20.\t(C)
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Normal idfb425c3e0-ParaOverride-90", "source_number": "11", "original_raw_text": "(A)\t12.\t(A)\t13.\t(A)\t14.\t(D) \t15.\t(B)\t16.\t(C)\t17.\t(A)\t18.\t(C)\t19.\t(D)\t20.\t(C)", "marks": 1, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Exemplar MCQs", "section_id": "4.8", "needs_review": true, "needs_layout_review": true, "layout_review_reason": "question body is a consolidated answer key (NUMBER.(LETTER) pairs) — parse it into a number→letter map, not a question", "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "options + correct option not detected" } }, { "sequence_no": 100, "question_text": "
(B)
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Normal idfb425c3e0-ParaOverride-24", "answer_key_continuation": true, "source_number": "21", "original_raw_text": "(B)", "marks": 1, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Exemplar MCQs", "section_id": "4.8", "needs_review": true, "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "options + correct option not detected" } } ] }, { "section_id": "4.9", "section_title": "NCERT Textbook Based MCQs", "heading_text": "NCERT Textbook Based MCQs", "detected_type": "MCQ", "match_confidence": 1.0, "questions": [ { "sequence_no": 101, "question_text": "
(B)\t2. (A)\t3.\t(B)\t4.\t(C)\t5.\t(A)\t6.\t(A)\t7.\t(D)\t8.\t(B)\t9.\t(A)\t10.\t(D)
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Normal idfb425c3e0-ParaOverride-24", "source_number": "1", "original_raw_text": "(B)\t2. (A)\t3.\t(B)\t4.\t(C)\t5.\t(A)\t6.\t(A)\t7.\t(D)\t8.\t(B)\t9.\t(A)\t10.\t(D)", "marks": 1, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Based MCQs", "section_id": "4.9", "needs_review": true, "needs_layout_review": true, "layout_review_reason": "question body is a consolidated answer key (NUMBER.(LETTER) pairs) — parse it into a number→letter map, not a question", "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "options + correct option not detected" } }, { "sequence_no": 102, "question_text": "
(B)\t12. (B)\t13.\t(B)\t14.\t(A) 15.\t(B)\t16.\t(D)\t17.\t(D)\t18.\t(C)\t19.\t(C)\t20.\t(B)
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Normal idfb425c3e0-ParaOverride-24", "source_number": "11", "original_raw_text": "(B)\t12. (B)\t13.\t(B)\t14.\t(A) 15.\t(B)\t16.\t(D)\t17.\t(D)\t18.\t(C)\t19.\t(C)\t20.\t(B)", "marks": 1, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Based MCQs", "section_id": "4.9", "needs_review": true, "needs_layout_review": true, "layout_review_reason": "question body is a consolidated answer key (NUMBER.(LETTER) pairs) — parse it into a number→letter map, not a question", "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "options + correct option not detected" } }, { "sequence_no": 103, "question_text": "
(A)\t22.\t(A)\t23.\t(D)\t24.\t(C)\t25.\t(A)\t26.\t(C)\t27.\t(B)\t28.\t(D)\t29.\t(D)\t30.\t(B)
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Normal idfb425c3e0-ParaOverride-24", "source_number": "21", "original_raw_text": "(A)\t22.\t(A)\t23.\t(D)\t24.\t(C)\t25.\t(A)\t26.\t(C)\t27.\t(B)\t28.\t(D)\t29.\t(D)\t30.\t(B)", "marks": 1, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Based MCQs", "section_id": "4.9", "needs_review": true, "needs_layout_review": true, "layout_review_reason": "question body is a consolidated answer key (NUMBER.(LETTER) pairs) — parse it into a number→letter map, not a question", "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "options + correct option not detected" } }, { "sequence_no": 104, "question_text": "
(B)\t32.\t(B)\t33.\t(B)\t34.\t(B)\t35.\t(A)\t36.\t(B)\t37.\t(A)\t38.\t(A)\t39.\t(B)\t40.\t(D)
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Normal idfb425c3e0-ParaOverride-24", "source_number": "31", "original_raw_text": "(B)\t32.\t(B)\t33.\t(B)\t34.\t(B)\t35.\t(A)\t36.\t(B)\t37.\t(A)\t38.\t(A)\t39.\t(B)\t40.\t(D)", "marks": 1, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Based MCQs", "section_id": "4.9", "needs_review": true, "needs_layout_review": true, "layout_review_reason": "question body is a consolidated answer key (NUMBER.(LETTER) pairs) — parse it into a number→letter map, not a question", "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "options + correct option not detected" } }, { "sequence_no": 105, "question_text": "
(C)\t42.\t(D)\t43.\t(D)\t44.\t(C)\t45. (A)\t46.\t(B)\t47.\t(C)\t48.\t(C)\t49. (D)\t50.\t(C)
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Normal idfb425c3e0-ParaOverride-24", "source_number": "41", "original_raw_text": "(C)\t42.\t(D)\t43.\t(D)\t44.\t(C)\t45. (A)\t46.\t(B)\t47.\t(C)\t48.\t(C)\t49. (D)\t50.\t(C)", "marks": 1, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Based MCQs", "section_id": "4.9", "needs_review": true, "needs_layout_review": true, "layout_review_reason": "question body is a consolidated answer key (NUMBER.(LETTER) pairs) — parse it into a number→letter map, not a question", "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "options + correct option not detected" } }, { "sequence_no": 106, "question_text": "
(A)\t52.\t(C)\t53.\t(B)\t54.\t(A)\t55.\t(D)\t56.\t(D)\t57.\t(C)\t58.\t(D)\t59.\t(C)\t60.\t(C)
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Normal idfb425c3e0-ParaOverride-24", "source_number": "51", "original_raw_text": "(A)\t52.\t(C)\t53.\t(B)\t54.\t(A)\t55.\t(D)\t56.\t(D)\t57.\t(C)\t58.\t(D)\t59.\t(C)\t60.\t(C)", "marks": 1, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Based MCQs", "section_id": "4.9", "needs_review": true, "needs_layout_review": true, "layout_review_reason": "question body is a consolidated answer key (NUMBER.(LETTER) pairs) — parse it into a number→letter map, not a question", "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "options + correct option not detected" } }, { "sequence_no": 107, "question_text": "
(D) 62. (D)\t63.\t(C)\t64.\t(A)\t65.\t(D)\t66.\t(C)\t67.\t(B)\t68.\t(B)\t69. (C)\t70.\t(C)\t\t71. (A)\t72.\t(A)\t73.\t(A)\t74.\t(B)\t75.\t(B)\t76.\t(B)\t77.\t(B)\t78.\t(A)\t79.\t(B)\t80.\t(B)\t\t81. (D)\t82.\t(D)\t83.\t(A)\t84.\t(B)\t85. (B)\t86.\t(C)\t87.\t(A)\t88.\t(A)\t89.\t(B)\t90.\t(B)\t\t91. (B)\t92.\t(A)\t93.\t(A)\t94.\t(B)\t95.\t(D)\t96.\t(D)\t97.\t(B)\t98. (D)\t99.\t(D)\t100.\t(D)\t\t101. (B)\t102.\t(A)\t103.\t(C)\t104.\t(D)\t105.\t(A)\t106.\t(D)\t107.\t(C)\t108.\t(D)\t109.\t(A)\t110.\t(A)\t\t111. (B)\t112.\t(C)\t113.\t(D)\t114.\t(B)\t115.\t(B)\t116.\t(C)\t117.\t(A)\t118.\t(C)\t119.\t(B)\t120.\t(B)\t\t121. (B)\t122.\t(A)\t123.\t(D)\t124.\t(A)\t125.\t(C)\t126.\t(D)\t127.\t(A)\t128.\t(C)\t129.\t(D)\t130.\t(B)\t\t131. (C)\t132.\t(D)\t133. (A)\t134.\t(D)\t135.\t(C)\t136.\t(C)\t137.\t(A)\t138.\t(D)\t139. (C)\t140.\t(B)\t\t141. (B)\t142.\t(D)\t143. (D)\t144. (B)\t145. (B)\t146. (A)\t147. (A)\t148. (B)\t149. (A)\t150. (D)\t\t151. (A)\t152. (A)\t153. (A)\t154. (D)\t155. (A)\t156. (A)\t157. (D)\t158. (A)\t159. (C)\t160. (B)\t\t161. (B)\t162. (B)\t163. (C)\t164. (A)\t165. (B)\t166. (C)\t167. (A)\t168. (A)\t169. (C)\t170. (C)\t\t171. (A)\t172. (A)\t173. (C)\t174. (D)\t175. (B)\t176. (D)\t177. (B)
", "answer_text": "", "type": "MCQ", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Normal idfb425c3e0-ParaOverride-24", "source_number": "61", "original_raw_text": "(D) 62. (D)\t63.\t(C)\t64.\t(A)\t65.\t(D)\t66.\t(C)\t67.\t(B)\t68.\t(B)\t69. (C)\t70.\t(C)\t\t71. (A)\t72.\t(A)\t73.\t(A)\t74.\t(B)\t75.\t(B)\t76.\t(B)\t77.\t(B)\t78.\t(A)\t79.\t(B)\t80.\t(B)\t\t81. (D)\t82.\t(D)\t83.\t(A)\t84.\t(B)\t85. (B)\t86.\t(C)\t87.\t(A)\t88.\t(A)\t89.\t(B)\t90.\t(B)\t\t91. (B)\t92.\t(A)\t93.\t(A)\t94.\t(B)\t95.\t(D)\t96.\t(D)\t97.\t(B)\t98. (D)\t99.\t(D)\t100.\t(D)\t\t101. (B)\t102.\t(A)\t103.\t(C)\t104.\t(D)\t105.\t(A)\t106.\t(D)\t107.\t(C)\t108.\t(D)\t109.\t(A)\t110.\t(A)\t\t111. (B)\t112.\t(C)\t113.\t(D)\t114.\t(B)\t115.\t(B)\t116.\t(C)\t117.\t(A)\t118.\t(C)\t119.\t(B)\t120.\t(B)\t\t121. (B)\t122.\t(A)\t123.\t(D)\t124.\t(A)\t125.\t(C)\t126.\t(D)\t127.\t(A)\t128.\t(C)\t129.\t(D)\t130.\t(B)\t\t131. (", "marks": 1, "deduction_level": 0, "source_html": "Chapter4", "section": "NCERT Textbook Based MCQs", "section_id": "4.9", "needs_review": true, "needs_layout_review": true, "layout_review_reason": "question body is a consolidated answer key (NUMBER.(LETTER) pairs) — parse it into a number→letter map, not a question", "missing_answer": true, "needs_answer_review": true, "answer_review_reason": "options + correct option not detected" } } ] }, { "section_id": "4.12", "section_title": "Case Study Based Q & A", "heading_text": "Case Study Based Questions and Answers", "detected_type": "CS", "match_confidence": 0.69, "questions": [ { "sequence_no": 108, "question_text": "
\tLanthanoid elements are classified by their 4f energy levels. They are extremely similar in properties. As the atomic number increases, the atomic and ionic radii decrease due to lanthanoid contraction from lanthanum to lutetium, and basicity also decreases.
\n
The sum of the three ionization enthalpies for each element is low, resulting in a predominant
+3 oxidation state. Ln3+ dominates the chemistry of these elements. Ln2+ and Ln4+ ions are always less stable than Ln3+. Inner orbital 4f electrons do not participate in bonding; they are neither removed nor significantly contribute to crystal field stabilization in complexes.
\n
(a) Which of the following is the most basic?
\n
\t(A) Ce(OH)3\t(B) Lu(OH)3
\n
\t(C) Yb(OH)3\t(D) Tb(OH)3
\n
(b) In which of the following lanthanoids is the
+2 oxidation state the most stable?
\n
\t(A) Ce\t(B) Eu\t(C) Tb\t(D) Dy
\n
(c) The colour of the trivalent ions of lanthanoids is due to:
\n
(A)\tLanthanoid contraction
\n
(B)\tTheir fluorescent properties
\n
(C)\tThe number of unpaired electrons in the
4f orbital
\n
(D)\tThe similar +3 oxidation state
\n
(d) Lanthanoids are ________.
\n
(A)\tthe 14 elements of the sixth period (atomic numbers 90 to 103) that fill the 4f subshell.
\n
(B)\tthe 14 elements of the seventh period (atomic numbers 90 to 103) that fill the 5f subshell.
\n
(C)\tthe 14 elements of the sixth period (atomic numbers 58 to 71) that fill the 4f subshell.
\n
(D)\tthe 14 elements of the seventh period (atomic numbers 58 to 71) that fill the 4f subshell.
", "answer_text": "
(a) (A) Ce(OH)3
\n
The atomic number increases, the ionic radii decrease due to lanthanoid contraction, and basicity also decreases. Therefore, Ce(OH)3 is the most ionic and the most basic.
\n
(b) (B) Eu
\n
While +3 is the predominant state, Europium (Eu) can achieve a stable +2 state. This is because the electronic configuration of Eu2+ is [Xe] 4f 7.
\n
(c) (C) The number of unpaired electrons in the
4f orbital
\n
The color of lanthanoid ions (Ln3+) is primarily due to f−f transitions. Because the 4f subshell is partially filled with unpaired electrons, these electrons can absorb specific wavelengths of visible light to jump between different 4f energy levels.
\n
(d) (C) the 14 elements of the sixth period (atomic numbers 58 to 71) that fill the 4f subshell.
\n
Lanthanoids are the 14 elements of the sixth period (atomic numbers 58 to 71) that fill the 4f subshell.
", "type": "CS", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Question_S2-to-S4_Que_Casestudy_01", "marks": 0, "deduction_level": 0, "original_raw_text": "\tLanthanoid elements are classified by their 4f energy levels. They are extremely similar in properties. As the atomic number increases, the atomic and ionic radii decrease due to lanthanoid contraction from lanthanum to lutetium, and basicity also decreases.\n The sum of the three ionization enthalpies for each element is low, resulting in a predominant+3 oxidation state. Ln3+ dominates the chemistry of these elements. Ln2+ and Ln4+ ions are always less stable than Ln3+. Inner orbital 4f electrons do not participate in bonding; they are neither removed nor significantly contribute to crystal f", "source_html": "Chapter4", "sub_questions": [ { "label": "(a)", "question": "Which of the following is the most basic?", "answer": "(A) Ce(OH) 3\nThe atomic number increases, the ionic radii decrease due to lanthanoid contraction, and basicity also decreases. Therefore, Ce(OH) 3 is the most ionic and the most basic." }, { "label": "(A)", "question": "the 14 elements of the sixth period (atomic numbers 90 to 103) that fill the 4 f subshell.", "answer": "" }, { "label": "(C)", "question": "the 14 elements of the sixth period (atomic numbers 58 to 71) that fill the 4 f subshell.", "answer": "" }, { "label": "(b)", "question": "In which of the following lanthanoids is the\n+ 2 oxidation state the most stable?", "answer": "(B) Eu\nWhile +3 is the predominant state, Europium (Eu) can achieve a stable +2 state. This is because the electronic configuration of Eu 2+ is [Xe] 4 f 7 ." }, { "label": "(c)", "question": "The colour of the trivalent ions of lanthanoids is due to:", "answer": "(C) The number of unpaired electrons in the\n4 f orbital\nThe color of lanthanoid ions (Ln 3+ ) is primarily due to f − f transitions. Because the 4 f subshell is partially filled with unpaired electrons, these electrons can absorb specific wavelengths of visible light to jump between different 4 f energy levels." }, { "label": "(B)", "question": "the 14 elements of the seventh period (atomic numbers 90 to 103) that fill the 5 f subshell.", "answer": "" }, { "label": "(D)", "question": "the 14 elements of the seventh period (atomic numbers 58 to 71) that fill the 4 f subshell.", "answer": "" }, { "label": "(d)", "question": "Lanthanoids are ________.", "answer": "(C) the 14 elements of the sixth period (atomic numbers 58 to 71) that fill the 4 f subshell.\nLanthanoids are the 14 elements of the sixth period (atomic numbers 58 to 71) that fill the 4 f subshell." } ], "section": "Case Study Based Q & A", "section_id": "4.12" } }, { "sequence_no": 109, "question_text": "
All transition elements are metals and are therefore good conductors of electricity and heat. The melting and boiling points of transition elements are generally very high. Most transition elements melt above 1000°C, with a few exceptions.
\n
Transition metals have high enthalpy of atomization, which increases with an increase in the number of d- electrons and then decreases. This behaviour can also be explained based on increasing interatomic interactions with the increasing number of electrons.
\n
(a) Among the first transition series (atomic numbers 21 to 30), which element has the lowest enthalpy of atomization?
\n
\t(A) Sc\t(B) Mn\t(C) Cu\t(D) Zn
\n
(b) Which electronic configuration show the highest magnetic moment?
\n
\t(A) 3d2\t(B) 3d5\t(C) 3d7\t(D) 3d9
\n
(c) The metallic character of transition elements is:
\n
\t(A) more than alkali metals
\n
\t(B) less than alkali metals
\n
\t(C) same as alkali metals
\n
\t(D) not fixed
\n
(d) Which group of transition elements has nearly the same atomic size?
\n
\t(A) Sc, Ti, V\t(B) Ni, Cu, Zn
\n
\t(C) Fe, Co, Ni\t(D) V, Ni, Cu
", "answer_text": "
(a) (D) Zn
\n
Zinc (Zn) has an electronic configuration of [Ar]3d104s2. Since all its d-orbitals are completely filled, there are no unpaired d-electrons to participate in metallic bonding.
\n
(b) (B) 3d5
\n
μ = \"\", where n is the number of unpaired electrons.
\n
\t3d5: 5 unpaired electrons (n)
\n
μ = \"\"
\n
μ ≈ 5.92 BM
\n
(c) (B) Less than alkali metals
\n
Due to the presence of d-electrons and stronger effective nuclear charge, transition metals generally show less metallic character than alkali metals.
\n
(d) (C) Fe, Co, Ni
\n
In a transition series, the atomic radius initially decreases but then becomes almost constant in the middle of the series. For the elements Fe, Co, and Ni, the increasing nuclear charge is almost perfectly balanced by the increasing screening effect of the
d-electrons.
", "type": "CS", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/EQ-0206-192409.png", "position": "answer", "context": "μ = , where n is the number of unpaired electrons." }, { "id": "img-1", "path": "Chapter4/EQ-0206-192533.png", "position": "answer", "context": "μ =" } ], "tables": [], "metadata": { "source_class": "Question_S2-to-S4_Que_Casestudy_01", "marks": 0, "deduction_level": 0, "original_raw_text": " All transition elements are metals and are therefore good conductors of electricity and heat. The melting and boiling points of transition elements are generally very high. Most transition elements melt above 1000°C, with a few exceptions.\n Transition metals have high enthalpy of atomization, which increases with an increase in the number of d- electrons and then decreases. This behaviour can also be explained based on increasing interatomic interactions with the increasing number of electrons.\n(a) Among the first transition series (atomic numbers 21 to 30), which element has the lowest entha", "source_html": "Chapter4", "sub_questions": [ { "label": "(a)", "question": "Among the first transition series (atomic numbers 21 to 30), which element has the lowest enthalpy of atomization?", "answer": "(D) Zn\nZinc (Zn) has an electronic configuration of [Ar]3d 10 4s 2 . Since all its d -orbitals are completely filled, there are no unpaired d -electrons to participate in metallic bonding." }, { "label": "(A)", "question": "Sc, Ti, V (B) Ni, Cu, Zn", "answer": "" }, { "label": "(b)", "question": "Which electronic configuration show the highest magnetic moment?", "answer": "(B) 3 d 5\nμ = , where n is the number of unpaired electrons.\n3d 5 : 5 unpaired electrons (n)\nμ =\nμ ≈ 5.92 BM" }, { "label": "(c)", "question": "The metallic character of transition elements is :", "answer": "(B) Less than alkali metals\nDue to the presence of d-electrons and stronger effective nuclear charge, transition metals generally show less metallic character than alkali metals." }, { "label": "(B)", "question": "less than alkali metals", "answer": "" }, { "label": "(C)", "question": "Fe, Co, Ni (D) V, Ni, Cu", "answer": "" }, { "label": "(D)", "question": "not fixed", "answer": "" }, { "label": "(d)", "question": "Which group of transition elements has nearly the same atomic size?", "answer": "(C) Fe, Co, Ni\nIn a transition series, the atomic radius initially decreases but then becomes almost constant in the middle of the series. For the elements Fe, Co, and Ni, the increasing nuclear charge is almost perfectly balanced by the increasing screening effect of the\nd -electrons." } ], "section": "Case Study Based Q & A", "section_id": "4.12" } }, { "sequence_no": 110, "question_text": "
A metallurgical research institute is developing high strength alloy material for aircraft engines. For this purpose, they search transition metals such as Cr, Mn, Fe, Co, Ni and their ions.
\n
\tDuring analysis they observe:
\n
Many transition metal show multiple oxidation states.
\n
Ions like MnO–4, Cr2O2–7 exhibit strong oxidizing properties
\n
Coloured compounds form due to d-d transitions.
\n
Complex formation tendency varies across the series.
\n
To select the best alloying element, scientists measure the following oxidation states in real samples:
\n
\tElement\tObserved oxin states
\n
\tChromium (Cr)\t+2, +3, +6
\n
\tManganese (Mn)\t+2, +4, +7
\n
\tIron (Fe)\t+2, +3
\n
\tCobalt (Co)\t+2, +3
\n
(a) Which element in the table show the maximum number of oxidation states? Explain the reason.
\n
(b) Which metal ion from the list will form the strongest coordination complexes and why?
\n
(c) Between MnO4– and Cr2O2–7, Which is the strongest oxidizing agent? Justify answer using oxidation states.
\n
(d) Explain why transition metal compounds are often colored, using Fe2+/Fe3+ as an example.
", "answer_text": "
(a) Mn shows the highest variation: +2, +4, +7
\n
\tReason: It has half-filled 3d5 configuration, allowing oxidation from +2 to +7 by losing both 3d and 4s electrons.
\n
(b) Cobalt (Co3+) forms strongest complexes.
\n
\tReason: High charge density → strong ligand attraction → stable octahedral complexes.
\n
\tEx. [Co(NH3)6]3+
\n
(c) MnO4– is the strongest oxidizing agent
\n
\tReason: Mn is in +7 oxidation state (higher oxidation state than Cr6+ in Cr2O72–).
\n
\tHigher oxidation state → greater ability to accept electrons.
\n
(d) Transition metals show colour due to d–d electronic transitions.
\n
Ex. Fe2+ (d6) and Fe3+ (d5) absorb different wavelengths → appear green/yellow → characteristic colours.
", "type": "CS", "type_uncertain": false, "options": {}, "correct_option": null, "images": [], "tables": [], "metadata": { "source_class": "Question_S2-to-S4_Que_Casestudy_01", "marks": 0, "deduction_level": 0, "original_raw_text": " A metallurgical research institute is developing high strength alloy material for aircraft engines. For this purpose, they search transition metals such as Cr, Mn, Fe, Co, Ni and their ions.\n\tDuring analysis they observe:\nMany transition metal show multiple oxidation states.\nIons like MnO–4, Cr2O2–7 exhibit strong oxidizing properties\nColoured compounds form due to d-d transitions.\nComplex formation tendency varies across the series.\n To select the best alloying element, scientists measure the following oxidation states in real samples:\n\tElement\tObserved oxin states\n\tChromium (Cr)\t+2, +3, +6", "source_html": "Chapter4", "sub_questions": [ { "label": "(a)", "question": "Which element in the table show the maximum number of oxidation states? Explain the reason.", "answer": "Mn shows the highest variation: +2, +4, +7\nReason: It has h alf-filled 3 d 5 configuration, allowing oxidation from +2 to + 7 by losing both 3 d and 4 s electrons." }, { "label": "(b)", "question": "Which metal ion from the list will form the strongest coordination complexes and why?", "answer": "Cobalt (Co 3+ ) forms strongest complexes.\nReason: High charge density → strong ligand attraction → stable octahedral complexes.\nEx. [Co(NH 3 ) 6 ] 3+" }, { "label": "(c)", "question": "Between MnO 4 – and Cr 2 O 2 – 7 , Which is the strongest oxidizing agent? Justify answer using oxidation states.", "answer": "MnO 4 – is the strongest oxidizing agent\nReason: Mn is in + 7 oxidation state (higher oxidation state than Cr 6+ in Cr 2 O 7 2– ).\nHigher oxidation state → greater ability to accept electrons." }, { "label": "(d)", "question": "Explain why transition metal compounds are often colored, using Fe 2+ /Fe 3+ as an example.", "answer": "Transition metals show colour due to d–d electronic transitions.\nEx. Fe 2+ ( d 6 ) and Fe 3+ ( d 5 ) absorb different wavelengths → appear green/yellow → characteristic colours." } ], "section": "Case Study Based Q & A", "section_id": "4.12" } }, { "sequence_no": 111, "question_text": "
A company producing permanent magnets studies the magnetics behavior of transition metals. They analyze ions of Fe, Co, Ni and Cu to understand which metal ions have:
\n
Maximum unpaired electrons
\n
Highest paramagnetic character
\n
Strong orbital contribution to magnetism
\n
Stable metallic bonding for magnet development
\n
\tMagnetic moment: µ = \"\" = BM
\n
\tElectronic Configurations:
\n
\tFe2+ → 3d6\tNi2+ → 3d8
\n
\tCo2+ → 3d7\tCu2+ → 3d9
\n
(a) Calculate the magnetic moment of Fe2+, Co2+, Ni2+, Cu2+.
\n
(b) Which metal ion is the most paramagnetic and
why?
\n
(c) Why do transition metals form strong metallic bond?
\n
(d) Explain why Cu2+ is coloured but Cu+ is colourless.
", "answer_text": "
(a) Fe2+ → 3d 6 → 4 unpaired e–
\n
µ = \"\" = \"\" = 4.90 BM
\n
Co2+ → 3d 7 → 3 unpaired e–
\n
µ = \"\" = \"\" = 3.87 BM
\n
Ni2+ → 3d 8 → 2 unpaired e–
\n
µ = \"\" = \"\" = 2.83 BM
\n
Cu2+ → 3d 9 → 1 unpaired e–
\n
µ = \"\" = \"\" = 1.73 BM
\n
(b) Fe2+ is most paramagnetic because of 4 unpaired electrons.
\n
(c) Due to presence of delocalized d-electrons, giving : High cohesive energy, strong metallic bonding, high melting point.
\n
(d) Cu2+ (d 9) → has one unpaired electron → d - d transitions → coloured.
", "type": "CS", "type_uncertain": false, "options": {}, "correct_option": null, "images": [ { "id": "img-0", "path": "Chapter4/EQ-1213-231439.png", "position": "stem", "context": "Magnetic moment: µ = = BM" }, { "id": "img-1", "path": "Chapter4/EQ-1213-232224.png", "position": "answer", "context": "µ = = = 4.90 BM" }, { "id": "img-2", "path": "Chapter4/EQ-1213-232231.png", "position": "answer", "context": "µ = = = 4.90 BM" }, { "id": "img-3", "path": "Chapter4/EQ-1213-232408.png", "position": "answer", "context": "µ = = = 3.87 BM" }, { "id": "img-4", "path": "Chapter4/EQ-1213-232416.png", "position": "answer", "context": "µ = = = 3.87 BM" }, { "id": "img-5", "path": "Chapter4/EQ-1213-232516.png", "position": "answer", "context": "µ = = = 2.83 BM" }, { "id": "img-6", "path": "Chapter4/EQ-1213-232522.png", "position": "answer", "context": "µ = = = 2.83 BM" }, { "id": "img-7", "path": "Chapter4/EQ-1213-232554.png", "position": "answer", "context": "µ = = = 1.73 BM" }, { "id": "img-8", "path": "Chapter4/EQ-0206-121847.png", "position": "answer", "context": "µ = = = 1.73 BM" } ], "tables": [], "metadata": { "source_class": "Question_S2-to-S4_Que_Casestudy_01", "marks": 0, "deduction_level": 0, "original_raw_text": " A company producing permanent magnets studies the magnetics behavior of transition metals. They analyze ions of Fe, Co, Ni and Cu to understand which metal ions have:\nMaximum unpaired electrons\nHighest paramagnetic character\nStrong orbital contribution to magnetism\nStable metallic bonding for magnet development\n\tMagnetic moment: µ = = BM\n\tElectronic Configurations:\n\tFe2+ → 3d6\tNi2+ → 3d8\n\tCo2+ → 3d7\tCu2+ → 3d9\n(a) Calculate the magnetic moment of Fe2+, Co2+, Ni2+, Cu2+.\n(b) Which metal ion is the most paramagnetic andwhy?\n(c) Why do transition metals form strong metallic bond?\n(d) Explain wh", "source_html": "Chapter4", "sub_questions": [ { "label": "(a)", "question": "Calculate the magnetic moment of Fe 2+ , Co 2+ , Ni 2+ , Cu 2+ .", "answer": "Fe 2+ → 3 d 6 → 4 unpaired e –\nµ = = = 4.90 BM\nCo 2+ → 3 d 7 → 3 unpaired e –\nµ = = = 3.87 BM\nNi 2+ → 3 d 8 → 2 unpaired e –\nµ = = = 2.83 BM\nCu 2+ → 3 d 9 → 1 unpaired e –\nµ = = = 1.73 BM" }, { "label": "(b)", "question": "Which metal ion is the most paramagnetic and\nwhy?", "answer": "Fe 2+ is most paramagnetic because of 4 unpaired electrons." }, { "label": "(c)", "question": "Why do transition metals form strong metallic bond?", "answer": "Due to presence of delocalized d-electrons, giving : High cohesive energy, strong metallic bonding, high melting point." }, { "label": "(d)", "question": "Explain why Cu 2+ is coloured but Cu + is colourless.", "answer": "Cu 2+ ( d 9 ) → has one unpaired electron → d - d transitions → coloured." } ], "section": "Case Study Based Q & A", "section_id": "4.12", "needs_review": true, "needs_optsplit_review": true, "optsplit_review_reason": "answer body is an option list (A/B/C/D) — options mis-routed into the answer; they belong to the question" } }, { "sequence_no": 112, "question_text": "
Cu+ (d 10) → No unpaired electron → No d - d transition → colourless.
\n
Class 12 Chemistry (Part 1) 019
\n
d-Block Elements: Periodic table elements where (n−1)d orbitals are progressively filled (Groups 3–12). Known as transition elements.
\n
f-Block Elements: Elements in which 4f or 5f orbitals are progressively filled. Divided into lanthanoids and actinoids.
\n
Transition Metals: Metals having incomplete d-subshells in atoms or common ions. Show characteristics like coloured ions, variable oxidation states, catalysis, etc.
\n
Inner Transition Metals: f-block elements (lanthanoids + actinoids).
\n
Electronic Configuration: Arrangement of electrons in orbitals; basis for classifying d- and f-block elements.
\n
3d/4d/5d/6d Series: Four horizontal rows of transition metals based on the filling of d-orbitals.
\n
Exceptional Configurations: Cases like Cr (3d54s1), Cu (3d104s1), Pd (4d105s0) due to stability of half-filled or fully filled orbitals.
\n
Lanthanoids (Ln): Series from Ce to Lu, involving progressive filling of 4f orbitals. Show lanthanoid contraction.
\n
Actinoids: Series from Th to Lr, involving progressive filling of 5f orbitals. Show multiple oxidation states and radioactivity.
\n
Lanthanoid Contraction: Steady decrease in atomic/ionic radii from La to Lu due to poor shielding of 4f electrons.
\n
Actinoid Contraction: Similar decrease across the actinoid series due to poor shielding in
5f orbitals.
\n
Oxidation State: Charge of an atom in a compound; transition metals show a wide range due to variable d-electron participation.
\n
Variable Oxidation States: Ability of transition metals to show many oxidation states (e.g., Mn: +2 to +7).
\n
Standard Electrode Potential (E°): Measure of tendency of an element to get reduced. Governs redox behaviour of transition metals.
\n
M2+/M Potential: Standard reduction potential for M2+ → M; used to compare reactivities across transition series.
\n
Ionisation Enthalpy: Energy required to remove electrons; increases across a transition series but less steeply than in main groups.
\n
Atomic Radius: Distance from nucleus to valence shell; decreases gradually across the d-block due to poor d-electron shielding.
\n
Ionic Radius: Radius of the ion; decreases with increasing charge and atomic number in a series.
\n
Enthalpy of Atomisation: Energy required to convert atoms of a metal into gaseous state; high for transition metals due to strong metallic bonding.
\n
Metal–Metal Bonding: Formation of bonds between metal atoms; stronger and more common in heavier transition metals (4d, 5d).
\n
Coloured Ions: Transition ions show colour due to d–d transitions when electrons absorb visible light.
\n
Paramagnetism: Magnetic behaviour due to unpaired electrons; measured using spin only formula \"\".
\n
Diamagnetism: Weak repulsion from magnetic fields due to absence of unpaired electrons.
\n
Ferromagnetism: Strong attraction to magnetic fields; pronounced in elements like Fe, Co, Ni.
\n
d–d Transitions: Electron jumps from lower to higher d-orbitals in complexes; cause colours in transition metal ions.
\n
Crystal Field Theory (CFT): Theory explaining splitting of d-orbitals in complexes (connected to colours and magnetism).
\n
Complex Compounds: Species where metal ions bind ligands to form coordinate complexes (e.g., [Fe(CN)6]3–).
\n
Ligands: Ions/molecules that donate lone pairs to metals in complexes.
\n
Coordination Number: Number of ligand donor atoms attached to central metal.
\n
Catalytic Activity: Transition metals catalyse due to variable oxidation states and ability to form complexes.
\n
Interstitial Compounds: Compounds where small atoms (H, C, N) occupy spaces in metal lattices (e.g., TiC, Fe₃H).
\n
Alloys: Mixtures of metals forming solid solutions (e.g., brass, bronze, stainless steel).
\n
Chromate Ion (CrO42–): Tetrahedral ion; yellow; forms in alkaline medium.
\n
Dichromate Ion (Cr2O72–): Orange ion; dominant in acidic medium; strong oxidising agent.
\n
Potassium Dichromate (K2Cr2O7): Important oxidising agent, used in volumetric analysis and organic chemistry.
\n
Permanganate Ion (MnO4–): Purple ion; strong oxidising agent in acidic medium, weaker in neutral/alkaline media.
\n
Potassium Permanganate (KMnO4): Strong oxidant used in analysis, organic synthesis, and bleaching.
\n
Manganate Ion (MnO42–): Green ion; paramagnetic; disproportionates in neutral/acidic medium.
\n
Oxide Formation: Transition metals form oxides across various oxidation states from +2 to +7 depending on element.
\n
Oxocations: Metal cations with oxygen (VO2+, VO2+, TiO2+), stabilising higher oxidation states.
\n
Disproportionation: Process where the same element is simultaneously oxidised and reduced. Example: Mn(VI) → Mn(VII) + Mn(IV).
\n
Lanthanide Contraction Effect: Causes elements like Zr and Hf to have nearly identical radii and similar properties.
\n
Radioactivity (Actinoids): Many actinoids are radioactive, forming multiple oxidation states and complex ions.
\n
Nuclear Fuels: Actinoids like U and Pu used due to their fission properties.
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{\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.25;\n\tmargin-bottom:0;\n\tmargin-left:40px;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-40px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Answer---MCQ_MCQ-4 {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.25;\n\tmargin-bottom:0;\n\tmargin-left:0;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\np.Answer---MCQ_Prashnottar-Heding {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:13px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:bold;\n\tline-height:1.25;\n\tmargin-bottom:3px;\n\tmargin-left:23px;\n\tmargin-right:11px;\n\tmargin-top:6px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:center;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\np.Answer-1 {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.5;\n\tmargin-bottom:0;\n\tmargin-left:26px;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-26px;\n\ttext-transform:none;\n\twidows:1;\n}\nli.Body-text-for-Q---A, p.Body-text-for-Q---A {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.5;\n\tlist-style-position:outside;\n\tmargin-bottom:0;\n\tmargin-left:28px;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\nli.Body-text-for-Q---A-copy-2 {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.5;\n\tlist-style-position:outside;\n\tmargin-bottom:0;\n\tmargin-left:48px;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\np.Box-Text {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.25;\n\tmargin-bottom:1px;\n\tmargin-left:0;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\np.Center-Heading {\n\t-epub-hyphens:none;\n\tcolor:#ec008c;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:13px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:bold;\n\tline-height:1.25;\n\tmargin-bottom:3px;\n\tmargin-left:23px;\n\tmargin-right:11px;\n\tmargin-top:6px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:center;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\np.Heding {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"B Bharati TitleTwo\", sans-serif;\n\tfont-size:15px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.25;\n\tmargin-bottom:3px;\n\tmargin-left:0;\n\tmargin-right:0;\n\tmargin-top:6px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:center;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\np.Hint-Solution {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.3;\n\tmargin-bottom:0;\n\tmargin-left:28px;\n\tmargin-right:0;\n\tmargin-top:6px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-28px;\n\ttext-transform:none;\n\twidows:1;\n}\np.MCQ-2 {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.4;\n\tmargin-bottom:0;\n\tmargin-left:48px;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-20px;\n\ttext-transform:none;\n\twidows:1;\n}\np.MCQ-3 {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.4;\n\tmargin-bottom:0;\n\tmargin-left:28px;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-28px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Mind-Map {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:10px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.5;\n\tmargin-bottom:0;\n\tmargin-left:0;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\nli.Mind-Map-Body- {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:10px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.5;\n\tlist-style-position:outside;\n\tmargin-bottom:0;\n\tmargin-left:14px;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\np.Normal {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.5;\n\tmargin-bottom:0;\n\tmargin-left:28px;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-28px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Normal-2 {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.3;\n\tmargin-bottom:0;\n\tmargin-left:48px;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-20px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Normal-copy {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.3;\n\tmargin-bottom:0;\n\tmargin-left:28px;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-28px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Basic-Paragraph {\n\tcolor:#000000;\n\tfont-family:\"Minion Pro\", serif;\n\tfont-size:12px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.2;\n\tmargin-bottom:0;\n\tmargin-left:0;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:left;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\np.Quest {\n\t-epub-hyphens:none;\n\tcolor:#008bbf;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:bold;\n\tline-height:1.5;\n\tmargin-bottom:3px;\n\tmargin-left:28px;\n\tmargin-right:0;\n\tmargin-top:9px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-28px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Quest_Black {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:bold;\n\tline-height:1.31;\n\tmargin-bottom:3px;\n\tmargin-left:28px;\n\tmargin-right:0;\n\tmargin-top:6px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-28px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Question_S1-to-S4_Questions_S1-to-S4 {\n\t-epub-hyphens:none;\n\tcolor:#ef4d89;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:bold;\n\tline-height:1.25;\n\tmargin-bottom:0;\n\tmargin-left:23px;\n\tmargin-right:0;\n\tmargin-top:6px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-23px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Question_S1_Questions_S1 {\n\t-epub-hyphens:none;\n\tcolor:#4171b8;\n\tfont-family:\"B Bharati AlokTwo\", sans-serif;\n\tfont-size:12px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.25;\n\tmargin-bottom:0;\n\tmargin-left:23px;\n\tmargin-right:0;\n\tmargin-top:6px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-23px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Question_S2-to-S4_Que_Casestudy_01 {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.25;\n\tmargin-bottom:0;\n\tmargin-left:23px;\n\tmargin-right:0;\n\tmargin-top:6px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-23px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Question_S2-to-S4_Questions_S2-to-S4 {\n\t-epub-hyphens:none;\n\tcolor:#00aeef;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:bold;\n\tline-height:1.25;\n\tmargin-bottom:0;\n\tmargin-left:23px;\n\tmargin-right:0;\n\tmargin-top:6px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:-23px;\n\ttext-transform:none;\n\twidows:1;\n}\np.Spacing {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.5;\n\tmargin-bottom:0;\n\tmargin-left:0;\n\tmargin-right:0;\n\tmargin-top:4px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\np.Summary-Style_Akruti-Font {\n\tcolor:#000000;\n\tfont-family:\"B Bharati AlokTwo\", sans-serif;\n\tfont-size:10px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1;\n\tmargin-bottom:0;\n\tmargin-left:0;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:left;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\np.Summary-Style_Normal {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.25;\n\tmargin-bottom:0;\n\tmargin-left:0;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\np.Summary-Style_spacing {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"B Bharati TitleLightTwo\", sans-serif;\n\tfont-size:12px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.25;\n\tmargin-bottom:6px;\n\tmargin-left:0;\n\tmargin-right:0;\n\tmargin-top:6px;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\nli.body-text-copy {\n\t-epub-hyphens:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\tline-height:1.5;\n\tlist-style-position:outside;\n\tmargin-bottom:0;\n\tmargin-left:28px;\n\tmargin-right:0;\n\tmargin-top:0;\n\torphans:1;\n\tpage-break-after:auto;\n\tpage-break-before:auto;\n\ttext-align:justify;\n\ttext-align-last:left;\n\ttext-decoration:none;\n\ttext-indent:0;\n\ttext-transform:none;\n\twidows:1;\n}\nspan.Biology-Character-Style-1_Alok-Two {\n\tcolor:#000000;\n\tfont-family:\"B Bharati AlokTwo\", sans-serif;\n\tfont-size:12px;\n\tfont-style:normal;\n\tfont-weight:normal;\n}\nspan.Biology-Character-Style-1_CaseStudy-Number {\n\tcolor:#ec008c;\n\tfont-family:\"Myriad Pro\", sans-serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Biology-Character-Style-1_English {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:10px;\n\tfont-style:normal;\n\tfont-weight:normal;\n}\nspan.Biology-Character-Style-1_Enlighs-Bold {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:10px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Biology-Character-Style-1_Ques-Number {\n\tcolor:#11aea7;\n\tfont-family:\"Myriad Pro\", sans-serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Character-Style-13-copy {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:italic;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\ttext-transform:none;\n}\nspan.Character-Style-15 {\n\tcolor:#ec008c;\n\ttext-decoration:underline;\n}\nspan.Character-Style-2 {\n\tcolor:#00aeef;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:8px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\ttext-transform:none;\n}\nspan.Chemistry-Character-Style-1_English {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-weight:normal;\n}\nspan.Chemistry-Character-Style-1_Enlighs-Bold {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Chemistry-Character-Style-1_Myriad-Pro-BOLD-Hint-Solution {\n\tcolor:#ec008c;\n\tfont-family:\"Myriad Pro\", sans-serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Chemistry-Character-Style-1_Ques-Number {\n\tcolor:#008bbf;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Chemistry-Character-Style-1_Ques-Number-2 {\n\tcolor:#ec008c;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.English {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:10px;\n\tfont-style:normal;\n\tfont-weight:normal;\n}\nspan.English-bold {\n\tfont-family:Calibri, sans-serif;\n\tfont-size:12px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Marks-Magenta {\n\tcolor:#ec008c;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:8px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Marks-Magenta----Without-Skew {\n\tcolor:#ec008c;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:8px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Mind-Map-English-Bold {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:10px;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Myriad-Pro {\n\tfont-family:\"Myriad Pro\", sans-serif;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.Normal-English {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\ttext-transform:none;\n}\nspan.Normal-English-Bold {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:bold;\n\ttext-transform:none;\n}\nspan.Normal-Italic {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:10px;\n\tfont-style:italic;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\ttext-transform:none;\n}\nspan.Physics-Character-Style-1_Title-Two {\n\tcolor:#000000;\n\tfont-family:\"B Bharati TitleTwo\", sans-serif;\n\tfont-size:12px;\n\tfont-style:normal;\n\tfont-weight:normal;\n}\nspan.Symbol {\n\tfont-family:Symbol, sans-serif;\n\tfont-size:10px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\ttext-transform:none;\n}\ntable.TableOverride-1 {\n\tborder-collapse:collapse;\n\tmargin-left:28px;\n}\ntable.TableOverride-2 {\n\tborder-collapse:collapse;\n\tmargin-left:auto;\n\tmargin-right:auto;\n}\ntable.TableOverride-3 {\n\tborder-collapse:collapse;\n}\ntable.TableOverride-4 {\n\tborder-collapse:collapse;\n\tborder-color:#00aeef;\n\tmargin-left:auto;\n\tmargin-right:auto;\n}\ntable.TableOverride-5 {\n\tborder-collapse:collapse;\n\tborder-color:#00aeef;\n\tborder-width:1px;\n\tmargin-left:28px;\n}\ntable.TableOverride-6 {\n\tborder-collapse:collapse;\n\tborder-color:#00aeef;\n\tborder-width:1px;\n}\ntable.TableOverride-7 {\n\tborder-collapse:collapse;\n\tborder-color:#00aeef;\n\tborder-width:1px;\n\tmargin-left:auto;\n\tmargin-right:auto;\n}\ntable.TableOverride-8 {\n\tborder-collapse:collapse;\n\tborder-color:#00aeef;\n\tmargin-left:28px;\n}\ntable.TableOverride-9 {\n\tborder-collapse:collapse;\n\tborder-width:1px;\n\tmargin-bottom:-4px;\n\tmargin-top:4px;\n}\ntd.CellOverride-1 {\n\tbackground-color:#d8e3d6;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:4px;\n\tpadding-top:4px;\n}\ntd.CellOverride-2 {\n\tbackground-color:#eff3ee;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:4px;\n\tpadding-top:4px;\n}\ntd.CellOverride-3 {\n\tbackground-color:#d8e3d6;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:4px;\n\tpadding-top:4px;\n\tvertical-align:middle;\n}\ntd.CellOverride-4 {\n\tbackground-color:#d8e3d6;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:3px;\n\tpadding-top:3px;\n\tvertical-align:middle;\n}\ntd.CellOverride-5 {\n\tbackground-color:#eff3ee;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n}\ntd.CellOverride-6 {\n\tbackground-color:#5fa070;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n}\ntd.CellOverride-7 {\n\tbackground-color:#d8e3d6;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n}\ntd.CellOverride-8 {\n\tbackground-color:#5fa070;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n}\ntd.CellOverride-9 {\n\tbackground-color:#d8e3d6;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n}\ntd.CellOverride-10 {\n\tbackground-color:#eff3ee;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-left:0px;\n\tpadding-right:0px;\n\tpadding-top:6px;\n}\ntd.CellOverride-11 {\n\tbackground-color:#5fa070;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:3px;\n\tpadding-top:3px;\n\tvertical-align:middle;\n}\ntd.CellOverride-12 {\n\tbackground-color:#eff3ee;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:3px;\n\tpadding-top:3px;\n}\ntd.CellOverride-13 {\n\tbackground-color:#d8e3d6;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:7px;\n\tpadding-top:6px;\n}\ntd.CellOverride-14 {\n\tbackground-color:#eff3ee;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:7px;\n\tpadding-top:6px;\n}\ntd.CellOverride-15 {\n\tbackground-color:#d8e3d6;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:0px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-left:0px;\n\tpadding-right:0px;\n\tpadding-top:6px;\n\tvertical-align:middle;\n}\ntd.CellOverride-16 {\n\tbackground-color:#d8e3d6;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:0px;\n\tpadding-bottom:6px;\n\tpadding-left:0px;\n\tpadding-right:0px;\n\tpadding-top:6px;\n\tvertical-align:middle;\n}\ntd.CellOverride-17 {\n\tbackground-color:#d8e3d6;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-left:0px;\n\tpadding-right:0px;\n\tpadding-top:6px;\n\tvertical-align:middle;\n}\ntd.CellOverride-18 {\n\tbackground-color:#eff3ee;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:14px;\n\tpadding-left:0px;\n\tpadding-right:0px;\n\tpadding-top:14px;\n}\ntd.CellOverride-19 {\n\tbackground-color:#eff3ee;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:4px;\n\tpadding-left:0px;\n\tpadding-right:0px;\n\tpadding-top:4px;\n}\ntd.CellOverride-20 {\n\tbackground-color:#ebf0f9;\n\tborder-bottom-color:#6ba7db;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#6ba7db;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#aac8e9;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#6ba7db;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n}\ntd.CellOverride-21 {\n\tbackground-color:#ebf0f9;\n\tborder-bottom-color:#6ba7db;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#aac8e9;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#aac8e9;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#6ba7db;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n}\ntd.CellOverride-22 {\n\tbackground-color:#ebf0f9;\n\tborder-bottom-color:#6ba7db;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#aac8e9;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#6ba7db;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#6ba7db;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n}\ntd.CellOverride-23 {\n\tbackground-color:#ffffff;\n\tborder-bottom-color:#6ba7db;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#6ba7db;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#aac8e9;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#6ba7db;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tvertical-align:middle;\n}\ntd.CellOverride-24 {\n\tbackground-color:#ffffff;\n\tborder-bottom-color:#6ba7db;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#aac8e9;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#6ba7db;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#6ba7db;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tvertical-align:middle;\n}\ntd.CellOverride-25 {\n\tbackground-color:#ffffff;\n\tborder-bottom-color:#6ba7db;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#aac8e9;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#aac8e9;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#6ba7db;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tvertical-align:middle;\n}\ntd.CellOverride-26 {\n\tbackground-color:#d8e3d6;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:5px;\n}\ntd.CellOverride-27 {\n\tbackground-color:#eff3ee;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:5px;\n}\ntd.CellOverride-28 {\n\tbackground-color:#d8e3d6;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n\tvertical-align:middle;\n}\ntd.CellOverride-29 {\n\tbackground-color:#d8e3d6;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-width:1px;\n\tpadding-bottom:8px;\n\tpadding-top:8px;\n}\ntd.CellOverride-30 {\n\tbackground-color:#eff3ee;\n\tborder-bottom-color:#8bb490;\n\tborder-bottom-width:1px;\n\tborder-left-color:#8bb490;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-width:1px;\n\tborder-top-color:#8bb490;\n\tborder-top-width:1px;\n\tpadding-bottom:8px;\n\tpadding-top:8px;\n}\ntd.CellOverride-31 {\n\tborder-bottom-color:#3cafd6;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#3cafd6;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#8bb490;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#3cafd6;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n\tvertical-align:middle;\n}\ntd.CellOverride-32 {\n\tborder-bottom-color:#3cafd6;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#3cafd6;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#3cafd6;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#3cafd6;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n\tvertical-align:middle;\n}\ntd.CellOverride-33 {\n\tborder-bottom-color:#3cafd6;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#3cafd6;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#3cafd6;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#3cafd6;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n}\ntd.CellOverride-34 {\n\tbackground-color:#5aa0d7;\n\tborder-bottom-color:#9bbfe5;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#9bbfe5;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#9bbfe5;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#9bbfe5;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n\tvertical-align:middle;\n}\ntd.CellOverride-35 {\n\tbackground-color:#f0f4fb;\n\tborder-bottom-color:#9bbfe5;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#9bbfe5;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#9bbfe5;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#9bbfe5;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:5px;\n\tpadding-top:5px;\n\tvertical-align:middle;\n}\ntd.CellOverride-36 {\n\tborder-bottom-color:#9bbfe5;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#9bbfe5;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#9bbfe5;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#9bbfe5;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n}\ntd.CellOverride-37 {\n\tborder-bottom-color:#808285;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#808285;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#808285;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#808285;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n}\ntd.CellOverride-38 {\n\tbackground-color:#eff9fe;\n\tborder-bottom-color:#9bbfe5;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#9bbfe5;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#9bbfe5;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#9bbfe5;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n}\ntd.CellOverride-39 {\n\tborder-bottom-width:1px;\n\tborder-left-width:1px;\n\tborder-right-width:1px;\n\tborder-top-width:1px;\n\tpadding-bottom:6px;\n\tpadding-top:6px;\n}\ntd.CellOverride-40 {\n\tborder-bottom-color:#000000;\n\tborder-bottom-style:solid;\n\tborder-bottom-width:1px;\n\tborder-left-color:#000000;\n\tborder-left-style:solid;\n\tborder-left-width:1px;\n\tborder-right-color:#000000;\n\tborder-right-style:solid;\n\tborder-right-width:1px;\n\tborder-top-color:#000000;\n\tborder-top-style:solid;\n\tborder-top-width:1px;\n\tpadding-bottom:5px;\n\tpadding-top:5px;\n}\np.idfb425c3e0-ParaOverride-1 {\n\tmargin-left:31px;\n\tmargin-right:3px;\n\ttext-align:left;\n}\np.idfb425c3e0-ParaOverride-2 {\n\tmargin-top:9px;\n\ttext-align:center;\n}\np.idfb425c3e0-ParaOverride-3 {\n\tmargin-top:3px;\n}\nli.idfb425c3e0-ParaOverride-4 {\n\ttext-indent:0px;\n}\np.idfb425c3e0-ParaOverride-5 {\n\tmargin-bottom:3px;\n}\np.idfb425c3e0-ParaOverride-6 {\n\tmargin-bottom:0px;\n}\nli.idfb425c3e0-ParaOverride-7, 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{\n\t-webkit-text-combine:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:bold;\n\ttext-decoration:none;\n\ttext-transform:none;\n}\nspan.idfb425c3e0-CharOverride-72 {\n\tcolor:#ec008c;\n\tfont-family:\"Myriad Pro\", sans-serif;\n\tfont-style:normal;\n\tfont-weight:normal;\n}\nspan.idfb425c3e0-CharOverride-73 {\n\tcolor:#000000;\n\tfont-size:8px;\n\tvertical-align:super;\n}\nspan.idfb425c3e0-CharOverride-74 {\n\tcolor:#000000;\n\tfont-size:70%;\n\tvertical-align:super;\n}\nspan.idfb425c3e0-CharOverride-75 {\n\tcolor:#ec008c;\n\tfont-style:normal;\n\tfont-weight:bold;\n}\nspan.idfb425c3e0-CharOverride-76 {\n\tcolor:#ffffff;\n\tfont-family:\"Hind Vadodara\", sans-serif;\n\tfont-style:normal;\n\tfont-weight:normal;\n}\nspan.idfb425c3e0-CharOverride-77 {\n\tfont-family:\"Times New Roman\", serif;\n\tfont-style:normal;\n\tfont-weight:normal;\n}\nspan.idfb425c3e0-CharOverride-78 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serif;\n\tfont-size:11px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\ttext-decoration:none;\n\ttext-transform:none;\n}\nspan.idfb425c3e0-CharOverride-85 {\n\tcolor:#000000;\n\ttext-decoration:underline;\n}\nspan.idfb425c3e0-CharOverride-86 {\n\t-webkit-text-combine:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:8px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\ttext-decoration:none;\n\ttext-transform:none;\n\tvertical-align:sub;\n}\nspan.idfb425c3e0-CharOverride-87 {\n\t-webkit-text-combine:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", serif;\n\tfont-size:8px;\n\tfont-style:normal;\n\tfont-variant:normal;\n\tfont-weight:normal;\n\ttext-decoration:none;\n\ttext-transform:none;\n\tvertical-align:super;\n}\nspan.idfb425c3e0-CharOverride-88 {\n\t-webkit-text-combine:none;\n\tcolor:#000000;\n\tfont-family:\"Times New Roman\", 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{\n\twidth:30px;\n}\ncol._idGenTableRowColumn-42 {\n\twidth:27px;\n}\ncol._idGenTableRowColumn-43 {\n\twidth:20px;\n}\ntr._idGenTableRowColumn-44 {\n\tmin-height:151px;\n}\ntr._idGenTableRowColumn-45 {\n\tmin-height:21px;\n}\ncol._idGenTableRowColumn-46 {\n\twidth:193px;\n}\ncol._idGenTableRowColumn-47 {\n\twidth:45px;\n}\ncol._idGenTableRowColumn-48 {\n\twidth:40px;\n}\ncol._idGenTableRowColumn-49 {\n\twidth:34px;\n}\ncol._idGenTableRowColumn-50 {\n\twidth:66px;\n}\ntr._idGenTableRowColumn-51 {\n\tbackground-color:#e4ebf7;\n\tmin-height:38px;\n}\ntr._idGenTableRowColumn-52 {\n\tbackground-color:#fff6ef;\n\tmin-height:18px;\n}\ntr._idGenTableRowColumn-53 {\n\tbackground-color:#e4ebf7;\n\tmin-height:18px;\n}\ncol._idGenTableRowColumn-54 {\n\twidth:80px;\n}\ncol._idGenTableRowColumn-55 {\n\twidth:115px;\n}\ntr._idGenTableRowColumn-56 {\n\tbackground-color:#e4ebf7;\n\tmin-height:35px;\n}\ntr._idGenTableRowColumn-57 {\n\tbackground-color:#fff6ef;\n\tmin-height:19px;\n}\ntr._idGenTableRowColumn-58 {\n\tbackground-color:#e4ebf7;\n\tmin-height:19px;\n}\ncol._idGenTableRowColumn-59 {\n\twidth:91px;\n}\ncol._idGenTableRowColumn-60 {\n\twidth:112px;\n}\ntr._idGenTableRowColumn-61 {\n\tmin-height:24px;\n}\ncol._idGenTableRowColumn-62 {\n\twidth:77px;\n}\ncol._idGenTableRowColumn-63 {\n\twidth:63px;\n}\ncol._idGenTableRowColumn-64 {\n\twidth:59px;\n}\ntr._idGenTableRowColumn-65 {\n\tbackground-color:#e4ebf7;\n\tmin-height:36px;\n}\ntr._idGenTableRowColumn-66 {\n\tbackground-color:#e4ebf7;\n\tmin-height:33px;\n}\ntr._idGenTableRowColumn-67 {\n\tbackground-color:#fff6ef;\n\tmin-height:23px;\n}\ntr._idGenTableRowColumn-68 {\n\tbackground-color:#e4ebf7;\n\tmin-height:23px;\n}\ntr._idGenTableRowColumn-69 {\n\tbackground-color:#fff6ef;\n\tmin-height:22px;\n}\ncol._idGenTableRowColumn-70 {\n\twidth:33px;\n}\ncol._idGenTableRowColumn-71 {\n\twidth:400px;\n}\ncol._idGenTableRowColumn-72 {\n\twidth:44px;\n}\ntr._idGenTableRowColumn-73 {\n\tmin-height:35px;\n}\ntr._idGenTableRowColumn-74 {\n\tmin-height:19px;\n}\ntr._idGenTableRowColumn-75 {\n\tmin-height:33px;\n}\ntr._idGenTableRowColumn-76 {\n\tmin-height:40px;\n}\ntr._idGenTableRowColumn-77 {\n\tmin-height:31px;\n}\ntr._idGenTableRowColumn-78 {\n\tmin-height:85px;\n}\ntr._idGenTableRowColumn-79 {\n\tmin-height:46px;\n}\ntr._idGenTableRowColumn-80 {\n\tmin-height:60px;\n}\ncol._idGenTableRowColumn-81 {\n\twidth:88px;\n}\ncol._idGenTableRowColumn-82 {\n\twidth:31px;\n}\ncol._idGenTableRowColumn-83 {\n\twidth:220px;\n}\ntr._idGenTableRowColumn-84 {\n\tmin-height:98px;\n}\ntr._idGenTableRowColumn-85 {\n\tmin-height:29px;\n}\ncol._idGenTableRowColumn-86 {\n\twidth:93px;\n}\ncol._idGenTableRowColumn-87 {\n\twidth:518px;\n}\ntr._idGenTableRowColumn-88 {\n\tmin-height:22px;\n}\n#_idContainer000, #_idContainer101, #_idContainer130, #_idContainer205, #_idContainer305 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:37.98px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:518.74px;\n\tz-index:0;\n}\n#_idContainer001, #_idContainer102, #_idContainer131, #_idContainer206, #_idContainer306 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:26.68px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:26.68px;\n\tz-index:0;\n}\n#_idContainer002, #_idContainer103, #_idContainer132, #_idContainer207, #_idContainer307 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.46px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:17.58px;\n\tz-index:0;\n}\n#_idContainer003, #_idContainer104, #_idContainer133, #_idContainer208, #_idContainer308 {\n\t-ms-transform:translate(3.244px,10.458px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(3.244px,10.458px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:9.41px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(3.244px,10.458px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(3.244px,10.458px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:11.09px;\n\tz-index:1;\n}\n#_idContainer004, #_idContainer105, #_idContainer134, #_idContainer209, #_idContainer309 {\n\t-ms-transform:translate(6.062px,3.121px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(6.062px,3.121px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.46px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(6.062px,3.121px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(6.062px,3.121px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:17.58px;\n\tz-index:1;\n}\n#_idContainer005, #_idContainer106, #_idContainer135, #_idContainer210, #_idContainer310 {\n\t-ms-transform:translate(0.000px,-0.001px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,-0.001px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:29.70px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,-0.001px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,-0.001px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:29.70px;\n\tz-index:1;\n}\n#_idContainer006 {\n\t-ms-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.95px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:267.58px;\n\tz-index:0;\n}\n#_idContainer007 {\n\t-ms-transform:translate(122.054px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(122.054px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:29.70px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(122.054px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(122.054px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:274.63px;\n\tz-index:1;\n}\n#_idContainer008, #_idContainer109, #_idContainer138, #_idContainer213, #_idContainer313 {\n\tdisplay:inline-block;\n\theight:38px;\n\tposition:relative;\n\twidth:519px;\n}\n#_idContainer107 {\n\t-ms-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.95px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:284.30px;\n\tz-index:0;\n}\n#_idContainer108 {\n\t-ms-transform:translate(113.692px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(113.692px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:29.70px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(113.692px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(113.692px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:291.35px;\n\tz-index:1;\n}\n#_idContainer125 {\n\tdisplay:inline-block;\n\twidth:75px;\n}\n#_idContainer136 {\n\t-ms-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.95px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:366.13px;\n\tz-index:0;\n}\n#_idContainer137 {\n\t-ms-transform:translate(72.779px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(72.779px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:29.70px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(72.779px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(72.779px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:373.18px;\n\tz-index:1;\n}\n#_idContainer211 {\n\t-ms-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.95px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:264.10px;\n\tz-index:0;\n}\n#_idContainer212 {\n\t-ms-transform:translate(123.792px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(123.792px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:29.70px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(123.792px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(123.792px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:271.16px;\n\tz-index:1;\n}\n#_idContainer237 {\n\t-ms-transform:translate(0.000px,9.609px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,9.609px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:324.78px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,9.609px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,9.609px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:512.74px;\n\tz-index:0;\n}\n#_idContainer238 {\n\t-ms-transform:translate(7.370px,24.094px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(7.370px,24.094px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:51.69px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(7.370px,24.094px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(7.370px,24.094px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:504.00px;\n\tz-index:2;\n}\n#_idContainer239 {\n\t-ms-transform:translate(7.370px,74.683px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(7.370px,74.683px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:259.84px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(7.370px,74.683px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(7.370px,74.683px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:504.00px;\n\tz-index:3;\n}\n#_idContainer240 {\n\t-ms-transform:translate(229.731px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(229.731px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:20.30px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(229.731px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(229.731px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:59.53px;\n\tz-index:1;\n}\n#_idContainer241 {\n\tdisplay:inline-block;\n\theight:340px;\n\tposition:relative;\n\twidth:519px;\n}\n#_idContainer242 {\n\t-ms-transform:translate(0.000px,11.175px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,11.175px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:0.00px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,11.175px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,11.175px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:517.74px;\n\tz-index:0;\n}\n#_idContainer243 {\n\t-ms-transform:translate(202.677px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(202.677px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:22.85px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(202.677px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(202.677px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:113.39px;\n\tz-index:1;\n}\n#_idContainer244 {\n\tdisplay:inline-block;\n\theight:23px;\n\tposition:relative;\n\twidth:519px;\n}\n#_idContainer284 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:675.06px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:0;\n}\n#_idContainer285 {\n\t-ms-transform:translate(177.165px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(177.165px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:675.06px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(177.165px,0.000px) rotate(0.000deg) skew(0.000deg) 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{\n\t-ms-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.95px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(7.051px,2.771px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:313.22px;\n\tz-index:0;\n}\n#_idContainer312 {\n\t-ms-transform:translate(99.236px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(99.236px,2.898px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 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{\n\t-ms-transform:translate(4.018px,2.069px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(4.018px,2.069px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:15.55px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(4.018px,2.069px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(4.018px,2.069px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:11.65px;\n\tz-index:1;\n}\n#_idContainer339 {\n\t-ms-transform:translate(4.655px,2.334px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(4.655px,2.334px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:19.69px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(4.655px,2.334px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(4.655px,2.334px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:19.69px;\n\tz-index:1;\n}\n#_idContainer340 {\n\tdisplay:inline-block;\n\theight:24px;\n\tposition:relative;\n\twidth:94px;\n}\n#_idContainer341 {\n\tdisplay:inline-block;\n\theight:277px;\n\tposition:relative;\n\twidth:0px;\n}\n#_idContainer342 {\n\tdisplay:inline-block;\n\theight:695px;\n\tposition:relative;\n\twidth:0px;\n}\n#_idContainer343 {\n\tdisplay:inline-block;\n\theight:708px;\n\twidth:518px;\n}\n#_idContainer344 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 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{\n\t-ms-transform:translate(11.049px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(11.049px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.26px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(11.049px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(11.049px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:10.15px;\n\tz-index:2;\n}\n#_idContainer347 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.26px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:10.15px;\n\tz-index:0;\n}\n#_idContainer348 {\n\t-ms-transform:translate(10.693px,3.197px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(10.693px,3.197px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:16.30px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(10.693px,3.197px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(10.693px,3.197px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:8.25px;\n\tz-index:3;\n}\n#_idContainer349 {\n\t-ms-transform:translate(0.753px,3.197px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.753px,3.197px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:16.30px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.753px,3.197px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.753px,3.197px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:8.25px;\n\tz-index:1;\n}\n#_idContainer350 {\n\t-ms-transform:translate(3.023px,2.091px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(3.023px,2.091px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.26px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(3.023px,2.091px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(3.023px,2.091px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:21.20px;\n\tz-index:1;\n}\n#_idContainer351 {\n\t-ms-transform:translate(6.218px,5.984px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(6.218px,5.984px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:27.44px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(6.218px,5.984px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(6.218px,5.984px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:27.25px;\n\tz-index:1;\n}\n#_idContainer352 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) 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0%;\n\theight:30.05px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(19.842px,9.637px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(19.842px,9.637px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:117.82px;\n\tz-index:0;\n}\n#_idContainer356 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:40px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:138px;\n}\n#_idContainer361 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:9px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer362, #_idContainer381 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:8px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer363 {\n\t-ms-transform:translate(-0.001px,126.269px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(-0.001px,126.269px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:49.33px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(-0.001px,126.269px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(-0.001px,126.269px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:1.00px;\n\tz-index:0;\n}\n#_idContainer364 {\n\t-ms-transform:translate(0.000px,127.461px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,127.461px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:126.46px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,127.461px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,127.461px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:1;\n}\n#_idContainer365 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:128px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:50px;\n}\n#_idContainer366 {\n\t-ms-transform:translate(0.000px,2.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,2.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:56.82px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,2.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,2.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:1.00px;\n\tz-index:1;\n}\n#_idContainer367 {\n\t-ms-transform:translate(-0.001px,0.839px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(-0.001px,0.839px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:20.16px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(-0.001px,0.839px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(-0.001px,0.839px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:0;\n}\n#_idContainer368 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:22px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:58px;\n}\n#_idContainer369 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:33px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer370 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:85px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer374 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:232px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:44px;\n}\n#_idContainer375 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:43px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer378 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:16px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer382 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:153px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:22px;\n}\n#_idContainer386, #_idContainer395 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:13px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer387, #_idContainer404, #_idContainer405, #_idContainer406, #_idContainer408, #_idContainer409 {\n\tdisplay:inline-block;\n\theight:3px;\n\twidth:1px;\n}\n#_idContainer389 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:136px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:44px;\n}\n#_idContainer392 {\n\t-ms-transform:translate(0.000px,78.142px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,78.142px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:77.14px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,78.142px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,78.142px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:1;\n}\n#_idContainer393 {\n\t-ms-transform:translate(-0.001px,76.811px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(-0.001px,76.811px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:23.07px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(-0.001px,76.811px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(-0.001px,76.811px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:1.00px;\n\tz-index:0;\n}\n#_idContainer394 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:79px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:24px;\n}\n#_idContainer396 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:10px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer397 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:127px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:22px;\n}\n#_idContainer398 {\n\t-ms-transform:translate(0.000px,0.499px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.499px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:215.51px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.499px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.499px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:0;\n}\n#_idContainer399 {\n\t-ms-transform:translate(215.119px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(215.119px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:26.10px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(215.119px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(215.119px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:1.00px;\n\tz-index:1;\n}\n#_idContainer400 {\n\tdisplay:inline-block;\n\theight:27px;\n\tposition:relative;\n\twidth:217px;\n}\n#_idContainer401 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:112px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:0px;\n}\n#_idContainer402 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:257px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:0px;\n}\n#_idContainer403 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:47px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer407 {\n\tdisplay:inline-block;\n\twidth:8px;\n}\n#_idContainer412 {\n\t-ms-transform:translate(0.000px,128.269px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,128.269px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:86.43px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,128.269px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,128.269px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:128.27px;\n\tz-index:0;\n}\n#_idContainer413 {\n\t-ms-transform:translate(21.464px,120.585px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(21.464px,120.585px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:49.51px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(21.464px,120.585px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(21.464px,120.585px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:113.39px;\n\tz-index:1;\n}\n#_idContainer414 {\n\tdisplay:inline-block;\n\theight:128px;\n\tposition:relative;\n\twidth:86px;\n}\n#_idContainer415 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:20px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer418 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:6px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(180.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer419 {\n\tdisplay:inline-block;\n\theight:15px;\n\tposition:relative;\n\twidth:1px;\n}\n#_idContainer420 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:134px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:0px;\n}\n#_idContainer421 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:44px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:19px;\n}\n#_idContainer422 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:124px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:0px;\n}\n#_idContainer425 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:40px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer426 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:14px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer429 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:47px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer431 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:8px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer432, #_idContainer433 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:62px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:0px;\n}\n#_idContainer434 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:61px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:0px;\n}\n#_idContainer436, #_idContainer463, #_idContainer487, #_idContainer491, #_idContainer496, #_idContainer502 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:9px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer437 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:45px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer438 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:94px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:0px;\n}\n#_idContainer439 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:123px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:0px;\n}\n#_idContainer440 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:436px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(90.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:14px;\n}\n#_idContainer441 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:468px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:0px;\n}\n#_idContainer445 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:6px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer446, #_idContainer501 {\n\tdisplay:inline-block;\n\theight:25px;\n\tposition:relative;\n\twidth:1px;\n}\n#_idContainer447 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:7px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer452 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:9px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer456 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:30px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer457 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:27px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer461 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:17px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer462, #_idContainer486, #_idContainer490, #_idContainer495 {\n\tdisplay:inline-block;\n\theight:24px;\n\tposition:relative;\n\twidth:1px;\n}\n#_idContainer464 {\n\t-ms-transform:translate(143.993px,363.625px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(143.993px,363.625px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:142.99px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(143.993px,363.625px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(143.993px,363.625px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:1.00px;\n\tz-index:2;\n}\n#_idContainer465 {\n\t-ms-transform:translate(143.993px,267.690px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(143.993px,267.690px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:40.77px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(143.993px,267.690px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(143.993px,267.690px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:266.69px;\n\tz-index:0;\n}\n#_idContainer466 {\n\t-ms-transform:translate(143.993px,369.588px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(143.993px,369.588px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:40.73px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(143.993px,369.588px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(143.993px,369.588px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:78.02px;\n\tz-index:1;\n}\n#_idContainer467 {\n\t-ms-transform:translate(163.859px,290.571px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(163.859px,290.571px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:21.88px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(163.859px,290.571px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(163.859px,290.571px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:1.00px;\n\tz-index:3;\n}\n#_idContainer468 {\n\tdisplay:inline-block;\n\theight:370px;\n\tposition:relative;\n\twidth:186px;\n}\n#_idContainer469 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:5px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer470 {\n\t-ms-transform:translate(129.968px,354.781px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(129.968px,354.781px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:128.97px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(129.968px,354.781px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(129.968px,354.781px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:3;\n}\n#_idContainer471 {\n\t-ms-transform:translate(122.699px,325.197px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(122.699px,325.197px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:13.02px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(122.699px,325.197px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(122.699px,325.197px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:78.02px;\n\tz-index:0;\n}\n#_idContainer472 {\n\t-ms-transform:translate(123.199px,227.014px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(123.199px,227.014px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:13.02px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(123.199px,227.014px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(123.199px,227.014px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:226.01px;\n\tz-index:4;\n}\n#_idContainer473 {\n\t-ms-transform:translate(128.706px,246.240px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(128.706px,246.240px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:18.23px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(128.706px,246.240px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(128.706px,246.240px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:1.00px;\n\tz-index:1;\n}\n#_idContainer474 {\n\t-ms-transform:translate(128.706px,354.781px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(128.706px,354.781px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:28.58px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(128.706px,354.781px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(128.706px,354.781px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:1.00px;\n\tz-index:2;\n}\n#_idContainer475 {\n\tdisplay:inline-block;\n\theight:355px;\n\tposition:relative;\n\twidth:137px;\n}\n#_idContainer476 {\n\t-ms-transform:translate(58.942px,292.434px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(58.942px,292.434px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:57.94px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(58.942px,292.434px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(58.942px,292.434px) rotate(270.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:5;\n}\n#_idContainer477 {\n\t-ms-transform:translate(52.184px,270.179px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(52.184px,270.179px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:13.02px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(52.184px,270.179px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(52.184px,270.179px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:78.02px;\n\tz-index:0;\n}\n#_idContainer478 {\n\t-ms-transform:translate(58.192px,292.434px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(58.192px,292.434px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:21.25px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(58.192px,292.434px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(58.192px,292.434px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:1.00px;\n\tz-index:4;\n}\n#_idContainer479 {\n\t-ms-transform:translate(58.942px,191.222px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(58.942px,191.222px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:37.45px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(58.942px,191.222px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(58.942px,191.222px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:0.00px;\n\tz-index:3;\n}\n#_idContainer480 {\n\t-ms-transform:translate(59.192px,152.008px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(59.192px,152.008px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:13.76px;\n\tleft:0px;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(59.192px,152.008px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(59.192px,152.008px) rotate(90.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:1.00px;\n\tz-index:2;\n}\n#_idContainer481 {\n\t-ms-transform:translate(73.957px,280.091px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(73.957px,280.091px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:40.73px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(73.957px,280.091px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(73.957px,280.091px) rotate(270.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:279.09px;\n\tz-index:1;\n}\n#_idContainer482 {\n\tdisplay:inline-block;\n\theight:292px;\n\tposition:relative;\n\twidth:116px;\n}\n#_idContainer499 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-ms-transform-origin:50% 50%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\t-webkit-transform-origin:50% 50%;\n\tdisplay:inline-block;\n\theight:22px;\n\tposition:relative;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,-1.000);\n\ttransform-origin:50% 50%;\n\ttransform-origin:50% 50%;\n\twidth:1px;\n}\n#_idContainer504 {\n\t-ms-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:27.63px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,0.000px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:517.61px;\n\tz-index:0;\n}\n#_idContainer505 {\n\t-ms-transform:translate(0.000px,38.534px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-ms-transform-origin:0% 0%;\n\t-webkit-transform:translate(0.000px,38.534px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\t-webkit-transform-origin:0% 0%;\n\theight:670.13px;\n\tleft:0px;\n\toverflow:hidden;\n\tposition:absolute;\n\ttop:0px;\n\ttransform:translate(0.000px,38.534px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform:translate(0.000px,38.534px) rotate(0.000deg) skew(0.000deg) scale(1.000,1.000);\n\ttransform-origin:0% 0%;\n\ttransform-origin:0% 0%;\n\twidth:518.74px;\n\tz-index:1;\n}\n#_idContainer506 {\n\tdisplay:inline-block;\n\theight:709px;\n\tposition:relative;\n\twidth:519px;\n}\nimg._idGenObjectAttribute-1 {\n\theight:100.00%;\n\tmin-width:100%;\n\twidth:100.00%;\n}\nimg._idGenObjectAttribute-2 {\n\theight:24px;\n\twidth:253px;\n}\nimg._idGenObjectAttribute-3 {\n\theight:13px;\n\twidth:35px;\n}\nimg._idGenObjectAttribute-4 {\n\theight:17px;\n\twidth:35px;\n}\nimg._idGenObjectAttribute-5 {\n\theight:91px;\n\twidth:252px;\n}\nimg._idGenObjectAttribute-6 {\n\theight:37px;\n\twidth:252px;\n}\nimg._idGenObjectAttribute-7 {\n\theight:17px;\n\twidth:64px;\n}\nimg._idGenObjectAttribute-8 {\n\theight:17px;\n\twidth:34px;\n}\nimg._idGenObjectAttribute-9 {\n\theight:14px;\n\twidth:75px;\n}\nimg._idGenObjectAttribute-10 {\n\theight:253px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-11 {\n\theight:163px;\n\twidth:252px;\n}\nimg._idGenObjectAttribute-12 {\n\theight:13px;\n\twidth:79px;\n}\nimg._idGenObjectAttribute-13 {\n\theight:214px;\n\twidth:216px;\n}\nimg._idGenObjectAttribute-14 {\n\theight:129px;\n\twidth:175px;\n}\nimg._idGenObjectAttribute-15 {\n\theight:18px;\n\twidth:75px;\n}\nimg._idGenObjectAttribute-16 {\n\theight:219px;\n\twidth:242px;\n}\nimg._idGenObjectAttribute-17 {\n\theight:120px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-18 {\n\theight:92px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-19 {\n\theight:192px;\n\twidth:181px;\n}\nimg._idGenObjectAttribute-20 {\n\theight:155px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-21 {\n\theight:143px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-22 {\n\theight:184px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-23 {\n\theight:135px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-24 {\n\theight:14px;\n\twidth:48px;\n}\nimg._idGenObjectAttribute-25 {\n\theight:17px;\n\twidth:44px;\n}\nimg._idGenObjectAttribute-26 {\n\theight:12px;\n\twidth:41px;\n}\nimg._idGenObjectAttribute-27 {\n\theight:12px;\n\twidth:40px;\n}\nimg._idGenObjectAttribute-28 {\n\theight:12px;\n\twidth:19px;\n}\nimg._idGenObjectAttribute-29 {\n\theight:221px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-30 {\n\theight:18px;\n\twidth:76px;\n}\nimg._idGenObjectAttribute-31 {\n\theight:14px;\n\twidth:107px;\n}\nimg._idGenObjectAttribute-32 {\n\theight:13px;\n\twidth:124px;\n}\nimg._idGenObjectAttribute-33 {\n\theight:186px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-34 {\n\theight:153px;\n\twidth:252px;\n}\nimg._idGenObjectAttribute-35 {\n\theight:186px;\n\twidth:252px;\n}\nimg._idGenObjectAttribute-36 {\n\theight:130px;\n\twidth:234px;\n}\nimg._idGenObjectAttribute-37 {\n\theight:108px;\n\twidth:121px;\n}\nimg._idGenObjectAttribute-38 {\n\theight:88px;\n\twidth:164px;\n}\nimg._idGenObjectAttribute-39 {\n\theight:17px;\n\twidth:76px;\n}\nimg._idGenObjectAttribute-40 {\n\theight:74px;\n\twidth:217px;\n}\nimg._idGenObjectAttribute-41 {\n\theight:75px;\n\twidth:236px;\n}\nimg._idGenObjectAttribute-42 {\n\theight:20px;\n\twidth:42px;\n}\nimg._idGenObjectAttribute-43 {\n\theight:13px;\n\twidth:77px;\n}\nimg._idGenObjectAttribute-44 {\n\theight:16px;\n\twidth:25px;\n}\nimg._idGenObjectAttribute-45 {\n\theight:5px;\n\twidth:32px;\n}\nimg._idGenObjectAttribute-46 {\n\theight:183px;\n\twidth:225px;\n}\nimg._idGenObjectAttribute-47 {\n\theight:299px;\n\twidth:491px;\n}\nimg._idGenObjectAttribute-48 {\n\theight:13px;\n\twidth:134px;\n}\nimg._idGenObjectAttribute-49 {\n\theight:259px;\n\twidth:216px;\n}\nimg._idGenObjectAttribute-50 {\n\theight:129px;\n\twidth:252px;\n}\nimg._idGenObjectAttribute-51 {\n\theight:24px;\n\twidth:519px;\n}\nimg._idGenObjectAttribute-52 {\n\theight:17px;\n\twidth:49px;\n}\nimg._idGenObjectAttribute-53 {\n\theight:345px;\n\twidth:491px;\n}\nimg._idGenObjectAttribute-54 {\n\theight:76px;\n\twidth:256px;\n}\nimg._idGenObjectAttribute-55 {\n\theight:54px;\n\twidth:252px;\n}\nimg._idGenObjectAttribute-56 {\n\theight:18px;\n\twidth:72px;\n}\nimg._idGenObjectAttribute-57 {\n\theight:14px;\n\twidth:76px;\n}\nimg._idGenObjectAttribute-58 {\n\theight:17px;\n\twidth:24px;\n}\nimg._idGenObjectAttribute-59 {\n\theight:11px;\n\twidth:37px;\n}\nimg._idGenObjectAttribute-60 {\n\theight:17px;\n\twidth:27px;\n}\nimg._idGenObjectAttribute-61 {\n\theight:12px;\n\twidth:43px;\n}\nimg._idGenObjectAttribute-62 {\n\theight:23px;\n\twidth:102px;\n}\nimg._idGenObjectAttribute-63 {\n\theight:14px;\n\twidth:72px;\n}\nimg._idGenObjectAttribute-64 {\n\theight:14px;\n\twidth:167px;\n}\nimg._idGenObjectAttribute-65 {\n\theight:172px;\n\twidth:224px;\n}\nimg._idGenObjectAttribute-66 {\n\theight:13px;\n\twidth:119px;\n}\nimg._idGenObjectAttribute-67 {\n\theight:13px;\n\twidth:46px;\n}\nimg._idGenObjectAttribute-68 {\n\theight:18px;\n\twidth:73px;\n}\nimg._idGenObjectAttribute-69 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{\n\theight:17px;\n\twidth:71px;\n}\nimg._idGenObjectAttribute-114 {\n\theight:13px;\n\twidth:43px;\n}\nimg._idGenObjectAttribute-115 {\n\theight:13px;\n\twidth:20px;\n}\nimg._idGenObjectAttribute-116 {\n\theight:13px;\n\twidth:42px;\n}\nimg._idGenObjectAttribute-117 {\n\theight:41px;\n\twidth:14px;\n}\nimg._idGenObjectAttribute-118 {\n\theight:164px;\n\twidth:122px;\n}\ndiv._idGenObjectStyleOverride-1 {\n\tbackground-color:#dae6f5;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-2 {\n\tbackground-color:#5aa0d7;\n\tborder-color:#ffffff;\n\tborder-style:solid;\n\tborder-width:2px;\n}\ndiv._idGenObjectStyleOverride-3 {\n\tbackground-color:#ffffff;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-4 {\n\tbackground-color:#cadbf1;\n\tborder-bottom-left-radius:4px;\n\tborder-bottom-right-radius:4px;\n\tborder-top-left-radius:4px;\n\tborder-top-right-radius:4px;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-5 {\n\tbackground-color:#f3f6fc;\n\tborder-color:#bad2ed;\n\tborder-style:solid;\n\tborder-width:3px;\n}\ndiv._idGenObjectStyleOverride-6 {\n\tbackground-color:#5aa0d7;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n}\ndiv._idGenObjectStyleOverride-7 {\n\tbackground-color:#ffffff;\n\tborder-color:#5aa0d7;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-8 {\n\tbackground-color:#5aa0d7;\n\tborder-bottom-left-radius:4px;\n\tborder-bottom-right-radius:4px;\n\tborder-top-left-radius:4px;\n\tborder-top-right-radius:4px;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-9 {\n\tborder-style:dotted;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-10 {\n\tbackground-color:#f3f6fc;\n\tborder-color:#5aa0d7;\n\tborder-style:solid;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-11 {\n\tbackground-color:#5aa0d7;\n\tborder-color:#ffffff;\n\tborder-style:solid;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-12 {\n\tborder-color:#bb9d6e;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-13 {\n\tbackground-color:#58595b;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-14 {\n\tbackground-color:#ffffff;\n\tborder-bottom-right-radius:2px;\n\tborder-top-right-radius:2px;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-15 {\n\tbackground-color:#4ec8eb;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-16 {\n\tbackground-color:#bd202e;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-17 {\n\tborder-bottom-left-radius:12px;\n\tborder-bottom-right-radius:12px;\n\tborder-color:#ffffff;\n\tborder-style:solid;\n\tborder-top-left-radius:12px;\n\tborder-top-right-radius:12px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-18 {\n\tbackground-color:#f0e4f1;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#b055a0;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-19 {\n\tbackground-color:#fde9f1;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#ef5ba1;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-20 {\n\tbackground-color:#fad5e5;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#ef5ba1;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-21 {\n\tbackground-color:#ededee;\n\tborder-color:#ec008c;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-22 {\n\tbackground-color:#ededee;\n\tborder-color:#b055a0;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-23 {\n\tbackground-color:#a3238e;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#a3238e;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-24 {\n\tborder-color:#a3238e;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-25 {\n\tbackground-color:#0080c6;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#0080c6;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-26 {\n\tbackground-color:#e4ebf7;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#3f92cf;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-27 {\n\tbackground-color:#ededee;\n\tborder-color:#006cb7;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-28 {\n\tborder-color:#006cb7;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-29 {\n\tbackground-color:#ec0085;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#ec0085;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-30 {\n\tbackground-color:#39b54a;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#39b54a;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-31 {\n\tbackground-color:#eaf4e7;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#6dc067;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-32 {\n\tbackground-color:#ededee;\n\tborder-color:#39b54a;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-33 {\n\tborder-color:#ec008c;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-34 {\n\tborder-color:#39b54a;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-35 {\n\tbackground-color:#ed1c24;\n\tborder-width:0px;\n}\ndiv._idGenObjectStyleOverride-36 {\n\tborder-bottom-left-radius:8px;\n\tborder-bottom-right-radius:8px;\n\tborder-top-left-radius:8px;\n\tborder-top-right-radius:8px;\n}\ndiv._idGenObjectStyleOverride-37 {\n\tborder-color:#f78f37;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-38 {\n\tbackground-color:#ededee;\n\tborder-color:#00aeef;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-39 {\n\tbackground-color:#fee5ce;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#f9a35d;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-40 {\n\tbackground-color:#ffefe3;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#f89958;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-41 {\n\tbackground-color:#ededee;\n\tborder-color:#f89958;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-42 {\n\tbackground-color:#f58232;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#f58232;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-43 {\n\tbackground-color:#ededee;\n\tborder-color:#f78f37;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-44 {\n\tbackground-color:#d8ecd4;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#39b54a;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-45 {\n\tbackground-color:#00aeef;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#00aeef;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-46 {\n\tbackground-color:#c7eafb;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#00b9f2;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-47 {\n\tbackground-color:#ededee;\n\tborder-color:#00b9f2;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-48 {\n\tbackground-color:#e1f4fd;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#00b9f2;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-49 {\n\tbackground-color:#f4efe7;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-color:#c7ad85;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyleOverride-50 {\n\tbackground-color:#bb9d6e;\n\tborder-bottom-left-radius:3px;\n\tborder-bottom-right-radius:3px;\n\tborder-top-left-radius:3px;\n\tborder-top-right-radius:3px;\n}\ndiv._idGenObjectStyleOverride-51 {\n\tbackground-color:#ededee;\n\tborder-color:#c7ad85;\n\tborder-width:1px;\n}\ndiv._idGenObjectStyle-Disabled {\n\tbackground-color:transparent;\n\tborder-width:0px;\n}\ndiv._idGenObjectLayout-1 {\n\ttext-align:center;\n}\ndiv._idGenObjectLayout-2 {\n\ttext-align:left;\n}\ndiv._idGenObjectLayout-3 {\n\tmargin:49px -49px 49px -49px;\n}\ndiv._idGenObjectLayout-4 {\n\tmargin:-39px 39px -39px 39px;\n}\ndiv._idGenObjectLayout-5 {\n\tmargin:18px -18px 18px -18px;\n}\ndiv._idGenObjectLayout-6 {\n\tmargin:-16px 16px -16px 16px;\n}\ndiv._idGenObjectLayout-7 {\n\tmargin:-42px 42px -42px 42px;\n}\ndiv._idGenObjectLayout-8 {\n\tmargin:-94px 94px -94px 94px;\n}\ndiv._idGenObjectLayout-9 {\n\tmargin:-65px 65px -65px 65px;\n}\ndiv._idGenObjectLayout-10 {\n\tmargin:-46px 46px -46px 46px;\n}\ndiv._idGenObjectLayout-11 {\n\tmargin:-27px 27px -27px 27px;\n}\ndiv._idGenObjectLayout-12 {\n\tmargin:-53px 53px -53px 53px;\n}\ndiv._idGenObjectLayout-13 {\n\tmargin:-56px 56px -56px 56px;\n}\ndiv._idGenObjectLayout-14 {\n\tmargin:-129px 129px -129px 129px;\n}\ndiv._idGenObjectLayout-15 {\n\tmargin:-23px 23px -23px 23px;\n}\ndiv._idGenObjectLayout-16 {\n\tmargin:-9px 9px -9px 9px;\n}\ndiv._idGenObjectLayout-17 {\n\tmargin:-67px 67px -67px 67px;\n}\ndiv._idGenObjectLayout-18 {\n\tmargin:-13px 13px -13px 13px;\n}\ndiv._idGenObjectLayout-19 {\n\tmargin:-62px 62px -62px 62px;\n}\ndiv._idGenObjectLayout-20 {\n\tmargin:-7px 7px -7px 7px;\n}\ndiv._idGenObjectLayout-21 {\n\tmargin:-3px 3px -3px 3px;\n}\ndiv._idGenObjectLayout-22 {\n\tmargin:-31px 31px -31px 31px;\n}\ndiv._idGenObjectLayout-23 {\n\tmargin:-4px 4px -4px 4px;\n}\ndiv._idGenObjectLayout-24 {\n\tmargin:-22px 22px -22px 22px;\n}\ndiv._idGenObjectLayout-25 {\n\tmargin:-47px 47px -47px 47px;\n}\ndiv._idGenObjectLayout-26 {\n\tmargin:-211px 211px -211px 211px;\n}\ndiv._idGenObjectLayout-27 {\n\tmargin:-234px 234px -234px 234px;\n}\ndiv._idGenObjectLayout-28 {\n\tmargin:-2px 2px -2px 2px;\n}\n" }