Chapter 4 · The d -and f -Block Elements

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Question

What are d-block elements? Explain classification of d-block elements.

Answer

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.

The name transition metals is often used to refer to the elements of d-block.

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).

The fourth 6d-series which begins with Ac is still incomplete.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

What are d-block elements? Explain classification of d-block elements.

Answer

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.

The name transition metals is often used to refer to the elements of d-block.

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).

The fourth 6d-series which begins with Ac is still incomplete.

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Question

What transition elements? Explain with examples.

Answer

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.

Examples: 21Sc: [Ar] 3d1 4s2

42Mo: [Kr] 4d5 5s1

25Mn: [Ar] 3d5 4s2

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

What transition elements? Explain with examples.

Answer

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.

Examples: 21Sc: [Ar] 3d1 4s2

42Mo: [Kr] 4d5 5s1

25Mn: [Ar] 3d5 4s2

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Question

What is ƒ-block element?

Answer

The element in which last electron fill in ƒ-orbital is known as ƒ-block element.

ƒ-block element is known as inner transition element.

The elements constituting the ƒ-block are those in which the 4ƒ – and 5ƒ – orbitals progressively filled.

The two series of the inner transition metals,
4ƒ and 5ƒ are known as lanthanoids and actinoids respectively.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

What is ƒ-block element?

Answer

The element in which last electron fill in ƒ-orbital is known as ƒ-block element.

ƒ-block element is known as inner transition element.

The elements constituting the ƒ-block are those in which the 4ƒ – and 5ƒ – orbitals progressively filled.

The two series of the inner transition metals,
4ƒ and 5ƒ are known as lanthanoids and actinoids respectively.

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Question

Explain position of d-block in periodic table.

Answer

The d-block occupies the large middle section flanked by s- and p-blocks in the periodic table.

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.

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.

The fourth row of 6d is still incomplete.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Explain position of d-block in periodic table.

Answer

The d-block occupies the large middle section flanked by s- and p-blocks in the periodic table.

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.

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.

The fourth row of 6d is still incomplete.

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Question

Electronic configuration of first transition metal.

Answer

Element	Atomic Number	Electronic Configuration
Sc Ti V Cr Mn Fe Co Ni Cu Zn	21 22 23 24 25 26 27 28 29 30	[Ar] 3d1 4s2 [Ar] 3d2 4s2 [Ar] 3d3 4s2 [Ar] 3d5 4s1 [Ar] 3d5 4s2 [Ar] 3d6 4s2 [Ar] 3d7 4s2 [Ar] 3d8 4s2 [Ar] 3d10 4s1 [Ar] 3d10 4s2

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Electronic configuration of first transition metal.

Answer

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Question

Electronic configuration of second transition metal.

Answer

Element	Atomic Number	Electronic configuration
Y Zr Nb Mo Tc Ru Rh Pd Ag Cd	39 40 41 42 43 44 45 46 47 48	[Kr] 4d1 5s2 [Kr] 4d2 5s2 [Kr] 4d4 5s1 [Kr] 4d5 5s1 [Kr] 4d5 5s2 [Kr] 4d7 5s1 [Kr] 4d8 5s1 [Kr] 4d10 5s0 [Kr] 4d10 5s1 [Kr] 4d10 5s2

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Electronic configuration of second transition metal.

Answer

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Question

Electronic configuration of third transition metal.

Answer

Element	Atomic Number	Electronic configuration
La Hf Ta W Re Os Ir Pt Au Hg	57 72 73 74 75 76 77 78 79 80	[Xe] 5d1 6s2 [Xe] 4f14 5d2 6s2 [Xe] 4f14 5d3 6s2 [Xe] 4f14 5d4 6s2 [Xe] 4f14 5d5 6s2 [Xe] 4f14 5d6 6s2 [Xe] 4f14 5d7 6s2 [Xe] 4f14 5d9 6s1 [Xe] 4f14 5d10 6s1 [Xe] 4f14 5d10 6s2

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Electronic configuration of third transition metal.

Answer

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Question

Why the chromium and copper have exceptional electronic configuration?

Answer

There is little energy difference between (n-1)d and ns-orbitals.

Furthermore, half and completely filled sets of orbitals are relatively more stable.

A consequence of this factor is reflected in the electronic configurations of Cr and Cu in the 3d-series.

In case of Cr

24Cr: [Ar] 3d5 4s1 instead of [Ar] 3d4 4s2

The energy gap between the two sets (3d and 4s) of orbitals is small enough to prevent electron entering the 3d orbitals.

Similarly in case of Cu,

29Cu: [Ar] 3d10 4s1 instead of [Ar] 3d9 4s2.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Why the chromium and copper have exceptional electronic configuration?

Answer

There is little energy difference between (n-1)d and ns-orbitals.

Furthermore, half and completely filled sets of orbitals are relatively more stable.

A consequence of this factor is reflected in the electronic configurations of Cr and Cu in the 3d-series.

In case of Cr

24Cr: [Ar] 3d5 4s1 instead of [Ar] 3d4 4s2

The energy gap between the two sets (3d and 4s) of orbitals is small enough to prevent electron entering the 3d orbitals.

Similarly in case of Cu,

29Cu: [Ar] 3d10 4s1 instead of [Ar] 3d9 4s2.

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Question

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

Characteristics of transition element:

All the transition elements are metallic elements.

These elements are hard and strong.

Their melting points are high.

These elements can form alloys with each other.

Most of these elements dissolve in acid, but acid has no effect on certain noble elements.

These elements possess various valencies.

They possess property of malleability and ductility.

They are good conductors of electricity and heat.

Some of their ions possess paramagnetic property.

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.

Elements such as Zn, Cd, and Hg cannot be classified as transition elements because these have completely filled d-subshell.

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

Examples: 30Zn: [Ar] 3d10 4s2

48Cd: [Kr] 4d10 5s2

80Hg: [Xe] 4ƒ14 5d10 6s2

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

Characteristics of transition element:

All the transition elements are metallic elements.

These elements are hard and strong.

Their melting points are high.

These elements can form alloys with each other.

Most of these elements dissolve in acid, but acid has no effect on certain noble elements.

These elements possess various valencies.

They possess property of malleability and ductility.

They are good conductors of electricity and heat.

Some of their ions possess paramagnetic property.

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.

Elements such as Zn, Cd, and Hg cannot be classified as transition elements because these have completely filled d-subshell.

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

Examples: 30Zn: [Ar] 3d10 4s2

48Cd: [Kr] 4d10 5s2

80Hg: [Xe] 4ƒ14 5d10 6s2

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Question

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

Transition metals have a partially filled d-orbital. Therefore, the electronic configuration of transition elements is (n - 1)d1-10 ns0-2.

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

Transition metals have a partially filled d-orbital. Therefore, the electronic configuration of transition elements is (n - 1)d1-10 ns0-2.

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.

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Question

Explain physical properties of transition elements.

Answer

All the transition elements are metallic element.

All the transition metals have high tensile strength, ductility, malleability, high thermal and electrical conductivity and metallic lustre.

With the exceptions of Zn, Cd, Hg and Mn, they have one or more typical metallic structures at normal temperatures.

The transition metals (with the exception of Zn, Cd and Hg) are very much hard and have low volatility.

Their melting and boiling points are high.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Explain physical properties of transition elements.

Answer

All the transition elements are metallic element.

All the transition metals have high tensile strength, ductility, malleability, high thermal and electrical conductivity and metallic lustre.

With the exceptions of Zn, Cd, Hg and Mn, they have one or more typical metallic structures at normal temperatures.

The transition metals (with the exception of Zn, Cd and Hg) are very much hard and have low volatility.

Their melting and boiling points are high.

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Question

Aqueous solution of TiCl3 is coloured where as aqueous solution of TiCl4 is colourless. Give reason.

Answer

In TiCl3 oxidation number of Ti is +3.
Ti3+: [Ar] 3d1.

Here in 3d-orbital one unpaired electron so d-d transition is possible so it is coloured.

In TiCl4 oxidation number of Ti is. +4.
Ti4+: [Ar] 3d0.

Here in 3d-orbital there is no unpaired electron so it is colourless.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Aqueous solution of TiCl3 is coloured where as aqueous solution of TiCl4 is colourless. Give reason.

Answer

In TiCl3 oxidation number of Ti is +3.
Ti3+: [Ar] 3d1.

Here in 3d-orbital one unpaired electron so d-d transition is possible so it is coloured.

In TiCl4 oxidation number of Ti is. +4.
Ti4+: [Ar] 3d0.

Here in 3d-orbital there is no unpaired electron so it is colourless.

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Question

Why the tansition elements are hard and have high melting and boiling points?

Answer

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.

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.

Zn, Cd, and Hg are soft and volatile and they have low melting point because all these elements have no unpaired electrons.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Why the tansition elements are hard and have high melting and boiling points?

Answer

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.

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.

Zn, Cd, and Hg are soft and volatile and they have low melting point because all these elements have no unpaired electrons.

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Question

Write note on enthalpy of atomization of transition metals.

Answer

The transition elements have high enthalpy of atomization because of presence of strong bonds between the atoms.

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.

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.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Write note on enthalpy of atomization of transition metals.

Answer

The transition elements have high enthalpy of atomization because of presence of strong bonds between the atoms.

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.

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.

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Question

What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

or

Explain atomic radii of transition element along the period.

Answer

In a given series decrease in radius with increasing atomic number.

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.

The same trend is observed in the atomic radii of a given series. However, the variation within a series is quite small.

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.

This phenomenon is associated with the intervention of the 4ƒ-orbitals which must be filled before the 5d series of elements begin.

The filling of 4f before 5d-orbital results in a regular decrease in atomic radii called lanthanoid contraction.

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.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

or

Explain atomic radii of transition element along the period.

Answer

In a given series decrease in radius with increasing atomic number.

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.

The same trend is observed in the atomic radii of a given series. However, the variation within a series is quite small.

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.

This phenomenon is associated with the intervention of the 4ƒ-orbitals which must be filled before the 5d series of elements begin.

The filling of 4f before 5d-orbital results in a regular decrease in atomic radii called lanthanoid contraction.

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.

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#16 SUB 0M ✎ Edit ⚠ Log Error

Question

Explain the variation in ionization enthalpies of transition element in 3d-series.

Answer

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.

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.

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.

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.

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.

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.

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.

In general, the third ionization enthalpies are quite high and there is a marked break between the values for Mn2+ and Fe2+.

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Explain the variation in ionization enthalpies of transition element in 3d-series.

Answer

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.

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.

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.

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.

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.

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.

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.

In general, the third ionization enthalpies are quite high and there is a marked break between the values for Mn2+ and Fe2+.

Also, the high values for copper, nickel and zinc indicate why it is difficult to obtain oxidation state greater than two for these elements.

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#17 SUB 3M ✎ Edit ⚠ Log Error

Question

Write about oxidation states of transition series?

Answer

One of the notable features of a transition element is the great variety of oxidation states it may show in its compounds.

Oxidation States of the first row Transition Metals (the most common ones are in bold types)
Sc	Ti	V	Cr	Mn	Fe	Co	Ni	Cu	Zn
+3	+2 +3 +4	+2 +3 +4 +5	+2 +3 +4 +5 +6	+2 +3 +4 +5 +6 +7	+2 +3 +4 +6	+2 +3 +4	+2 +3 +4	+1 +2	+2

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.

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.

e.g.: Sc+3, Ti+2, Ti+3, Ti+4

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.

Down the group the stability of elements in higher oxidation states increases because removal of electrons from d-orbitals become easy.

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.

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Write about oxidation states of transition series?

Answer

One of the notable features of a transition element is the great variety of oxidation states it may show in its compounds.

Oxidation States of the first row Transition Metals (the most common ones are in bold types)
Sc	Ti	V	Cr	Mn	Fe	Co	Ni	Cu	Zn
+3	+2 +3 +4	+2 +3 +4 +5	+2 +3 +4 +5 +6	+2 +3 +4 +5 +6 +7	+2 +3 +4 +6	+2 +3 +4	+2 +3 +4	+1 +2	+2

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.

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.

e.g.: Sc+3, Ti+2, Ti+3, Ti+4

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.

Down the group the stability of elements in higher oxidation states increases because removal of electrons from d-orbitals become easy.

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.

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.

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#18 SUB 0M 🖼 1 ✎ Edit ⚠ Log Error

Question

Explain trends in M2+/M standard electrode potentials.

Answer

An element in M2+ state in aqueous medium is more stable if the electrode potential (M2+/M) value is more negative.

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.

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.

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Explain trends in M2+/M standard electrode potentials.

Answer

An element in M2+ state in aqueous medium is more stable if the electrode potential (M2+/M) value is more negative.

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.

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.

It is interesting to note that the value of E°/V for Mn, Ni and Zn are more negative than expected from the trend.

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#19 SUB 0M ✎ Edit ⚠ Log Error

Question

Write a note on halide compounds of Transition elements.

Answer

The Transition elements form Ionic halides with fluorine and co-valent halides with chlorine, bromine and iodine.

The highest oxidation numbers are achieved in TiX4 (tetrahalides), VF5 and CrF6.

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.

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.

Although V5+ is represented only by VF5, the other halides, however, undergo hydrolysis to give oxohalides, VOX3. Another feature of fluorides.

VF5 + H2O → VOF3 + 2HF

The solution of metal halides are acidic as they produce acid on hydrolysis.

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.

2Cu2+ + 4I– → Cu2I2(s) + I2

However, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation.

2Cu+ → Cu2+ + Cu

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.

Formulas of Halides of 3d Metals
Oxidation Number
+6 +5 +4 +3 +2 +1	TiX4 TiX3 TiX2III	VF5 VX14 VX3 VX2	CrF6 CrF5 CrX4 CrX3 CrX2	MnF4 MnF3 MnF2	FeX13 FeX2	CoF3 CoX2	NiX2	CuX211 CuX111	ZnX2

Key: X = F → I; X 1 = F → Br; X 11 = F, Cl; X 111 = Cl → I

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Write a note on halide compounds of Transition elements.

Answer

The Transition elements form Ionic halides with fluorine and co-valent halides with chlorine, bromine and iodine.

The highest oxidation numbers are achieved in TiX4 (tetrahalides), VF5 and CrF6.

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.

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.

Although V5+ is represented only by VF5, the other halides, however, undergo hydrolysis to give oxohalides, VOX3. Another feature of fluorides.

VF5 + H2O → VOF3 + 2HF

The solution of metal halides are acidic as they produce acid on hydrolysis.

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.

2Cu2+ + 4I– → Cu2I2(s) + I2

However, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation.

2Cu+ → Cu2+ + Cu

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.

Formulas of Halides of 3d Metals
Oxidation Number
+6 +5 +4 +3 +2 +1	TiX4 TiX3 TiX2III	VF5 VX14 VX3 VX2	CrF6 CrF5 CrX4 CrX3 CrX2	MnF4 MnF3 MnF2	FeX13 FeX2	CoF3 CoX2	NiX2	CuX211 CuX111	ZnX2

Key: X = F → I; X 1 = F → Br; X 11 = F, Cl; X 111 = Cl → I

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#20 SUB 2M ✎ Edit ⚠ Log Error

Question

Explain chemical reactivity of transition metals.

Answer

Transition metals vary widely in their chemical reactivity.

Many of them are sufficiently electropositive to dissolve in mineral acids, although a few are 'noble' – that is, they are unaffected by simple acids.

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.

The Eo values for M2+/M indicate a decreasing tendency to form divalent cations across the series.

This general trend towards less negative E° values is related to the increase in the sum of the first and second ionization enthalpies.

It is interesting to note that the E° values for Mn, Ni and Zn are more negative than expected from the general trend.

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Explain chemical reactivity of transition metals.

Answer

Transition metals vary widely in their chemical reactivity.

Many of them are sufficiently electropositive to dissolve in mineral acids, although a few are 'noble' – that is, they are unaffected by simple acids.

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.

The Eo values for M2+/M indicate a decreasing tendency to form divalent cations across the series.

This general trend towards less negative E° values is related to the increase in the sum of the first and second ionization enthalpies.

It is interesting to note that the E° values for Mn, Ni and Zn are more negative than expected from the general trend.

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)

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#21 SUB 3M 🖼 1 ✎ Edit ⚠ Log Error

Question

Explain magnetic properties of transition elements.

Answer

When a magnetic field is applied to substances, mainly two types of properties magnetic behaviour are observed: diamagnetism and paramagnetism.

Diamagnetic substances are repelled by the applied field while the paramagnetic substances are attracted.

Substances which are attracted very strongly are said to be ferromagnetic.

In fact, ferromagnetism is an extreme form of paramagnetism. Many of the transition metal ions are paramagnetic.

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.

For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the 'spin-only' formula,
i.e., m =

Where, n = The number of unpaired electrons µ = Magnetic moment in units of Bohr
magneton (BM).

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Explain magnetic properties of transition elements.

Answer

When a magnetic field is applied to substances, mainly two types of properties magnetic behaviour are observed: diamagnetism and paramagnetism.

Diamagnetic substances are repelled by the applied field while the paramagnetic substances are attracted.

Substances which are attracted very strongly are said to be ferromagnetic.

In fact, ferromagnetism is an extreme form of paramagnetism. Many of the transition metal ions are paramagnetic.

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.

For these, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the 'spin-only' formula,
i.e., m =

Where, n = The number of unpaired electrons µ = Magnetic moment in units of Bohr
magneton (BM).

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

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#22 SUB 3M PYQ 🖼 4 ✎ Edit ⚠ Log Error

Question

Calculate the magnetic moment of a tivalent ion in aqueous solution if its atomic number
is 26.[May 2021] [3 Marks]

Answer

26M: [Ar] 3d5 4s2

M3+: [Ar] 3d5

Unpaired electrons = 5

∴ Magnetic moment (µ)=

=

=

= 5.92 B.M

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Calculate the magnetic moment of a tivalent ion in aqueous solution if its atomic number
is 26.[May 2021] [3 Marks]

Answer

26M: [Ar] 3d5 4s2

M3+: [Ar] 3d5

Unpaired electrons = 5

∴ Magnetic moment (µ)=

=

=

= 5.92 B.M

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#23 SUB 3M ✎ Edit ⚠ Log Error

Question

Why transition elements form a coloured ions?

Answer

Most of the ionic and covalent compounds of transition elements are coloured. It is due to the presence of incompletely filled d-orbitals.

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.

This frequency generally lies in the visible region. The colour observed corresponds to the complementary colour of the light absorbed.

The frequency of the light absorbed is determined by the nature of the ligand.

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Why transition elements form a coloured ions?

Answer

Most of the ionic and covalent compounds of transition elements are coloured. It is due to the presence of incompletely filled d-orbitals.

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.

This frequency generally lies in the visible region. The colour observed corresponds to the complementary colour of the light absorbed.

The frequency of the light absorbed is determined by the nature of the ligand.

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

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#24 SUB 2M ✎ Edit ⚠ Log Error

Question

Why the transition element form a large number of complex compounds?

Answer

Complex compounds are those in which the metal ions bind a number of anions or neutral molecules giving complex species with characteristic properties.

A few examples are: [Fe(CN)6]3–, [Fe(CN)6]4–, [Cu(NH3)4]2+ and [PtCl4]2–.

The transition metals form a large number of complex compounds.

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Why the transition element form a large number of complex compounds?

Answer

Complex compounds are those in which the metal ions bind a number of anions or neutral molecules giving complex species with characteristic properties.

A few examples are: [Fe(CN)6]3–, [Fe(CN)6]4–, [Cu(NH3)4]2+ and [PtCl4]2–.

The transition metals form a large number of complex compounds.

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.

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#25 SUB 2M ✎ Edit ⚠ Log Error

Question

Explain Catalytic property of transition elements.

Answer

The transition metals and their compounds are known for their catalytic activity.

This activity is ascribed to their ability to adopt multiple oxidation states and to form complexes.

Vanadium (V) oxide (in Contact Process), finely divided iron (in Haber's Process), and nickel (in Catalytic Hydrogenation) are some of the examples.

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).

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).

The Transition metal ions can change their oxidation states, they become more effective as catalysts.

For example, iron (III) catalyses the reaction between iodide and persulphate ions.

2I – + S2O2–8 → I2 + 2SO2–4

An explanation of this catalytic action can be given as:

2Fe3+ + 2I – → 2Fe2+ + I2

2Fe2+ + S2O2–8 → 2Fe3+ + 2SO2–4

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Explain Catalytic property of transition elements.

Answer

The transition metals and their compounds are known for their catalytic activity.

This activity is ascribed to their ability to adopt multiple oxidation states and to form complexes.

Vanadium (V) oxide (in Contact Process), finely divided iron (in Haber's Process), and nickel (in Catalytic Hydrogenation) are some of the examples.

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).

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).

The Transition metal ions can change their oxidation states, they become more effective as catalysts.

For example, iron (III) catalyses the reaction between iodide and persulphate ions.

2I – + S2O2–8 → I2 + 2SO2–4

An explanation of this catalytic action can be given as:

2Fe3+ + 2I – → 2Fe2+ + I2

2Fe2+ + S2O2–8 → 2Fe3+ + 2SO2–4

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#26 SUB 3M ✎ Edit ⚠ Log Error

Question

What are interstitial compounds? Why are such compounds well known for transition metals?

or

What are interstitial compounds. Write its any two characteristics.

or

Write note on interstitial compounds.

Answer

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.

They are usually non-stoichiometric and are neither typically ionic nor covalent,

For example, TiC, Mn4N, Fe3H, VH0.56 and TiH1.7, etc.

These compounds are referred to as interstitial compounds.

The principal physical and chemical characteristics of these compounds are as follows:

• They have high melting points, higher than those of pure metals.
• They are very hard, some borides approach diamond in hardness.
• They retain metallic conductivity.
• They are chemically inert.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

What are interstitial compounds? Why are such compounds well known for transition metals?

or

What are interstitial compounds. Write its any two characteristics.

or

Write note on interstitial compounds.

Answer

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.

They are usually non-stoichiometric and are neither typically ionic nor covalent,

For example, TiC, Mn4N, Fe3H, VH0.56 and TiH1.7, etc.

These compounds are referred to as interstitial compounds.

The principal physical and chemical characteristics of these compounds are as follows:

• They have high melting points, higher than those of pure metals.
• They are very hard, some borides approach diamond in hardness.
• They retain metallic conductivity.
• They are chemically inert.

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#27 SN 0M ✎ Edit ⚠ Log Error

Question

Write short note on alloys.

Answer

“An alloy is a blend of metals prepared by mixing the components.”

Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other.

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.

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.

The crystal structures of pure metallic elements used for alloys must be similar.

The alloys so formed are hard and have often high melting points.

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.

Alloys of transition metals with non transition metals such as brass (copper-zinc) and bronze (copper-tin), are also of considerable industrial importance.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Write short note on alloys.

Answer

“An alloy is a blend of metals prepared by mixing the components.”

Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other.

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.

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.

The crystal structures of pure metallic elements used for alloys must be similar.

The alloys so formed are hard and have often high melting points.

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.

Alloys of transition metals with non transition metals such as brass (copper-zinc) and bronze (copper-tin), are also of considerable industrial importance.

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#28 SUB 0M ✎ Edit ⚠ Log Error

Question

Write note on oxides and oxoanions of 3d transition metals.

Answer

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.

The metal oxide in lower oxidation state is basic while in higher oxidation state are acidic.

The metal oxides having metals in intermediate oxidation states are amphoteric.

Oxides of 3d Metals
Acidic oxide	Basic oxide	Amphoteric oxide
Mn2O7	MnO	Mn3O4, Mn2O3, MnO2
	CrO	Cr2O3

Oxidation Number
	3	4	5	6	7	8	9	10	11	12
+7 +6 +5 +4 +3 +2 +1	Sc2O3	TiO2 Ti2O3 TiO	V2O5 V2O4 V2O3 VO	CrO3 CrO2 Cr2O3 (CrO)	Mn2O7 MnO2 Mn2O3 Mn3O4* MnO	Fe2O3 Fe3O4* FeO	Co3O*4 CoO	NiO	CuO Cu2O	ZnO
Note: *Denotes mixed oxides

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Write note on oxides and oxoanions of 3d transition metals.

Answer

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.

The metal oxide in lower oxidation state is basic while in higher oxidation state are acidic.

The metal oxides having metals in intermediate oxidation states are amphoteric.

Oxides of 3d Metals
Acidic oxide	Basic oxide	Amphoteric oxide
Mn2O7	MnO	Mn3O4, Mn2O3, MnO2
	CrO	Cr2O3

Oxidation Number
	3	4	5	6	7	8	9	10	11	12
+7 +6 +5 +4 +3 +2 +1	Sc2O3	TiO2 Ti2O3 TiO	V2O5 V2O4 V2O3 VO	CrO3 CrO2 Cr2O3 (CrO)	Mn2O7 MnO2 Mn2O3 Mn3O4* MnO	Fe2O3 Fe3O4* FeO	Co3O*4 CoO	NiO	CuO Cu2O	ZnO
Note: *Denotes mixed oxides

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.

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#29 SUB 2M 🖼 1 ✎ Edit ⚠ Log Error

Question

Discuss structures and magnetic properties of manganate and permanganate ions.

Answer

The manganate and permanganate ions are tetrahedral.

The green manganate is paramagnetic with one unpaired electron but the permanganate is diamagnetic.

The π-bonding takes place by overlap of
p-orbitals of oxygen with d-orbitals of manganese.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Discuss structures and magnetic properties of manganate and permanganate ions.

Answer

The manganate and permanganate ions are tetrahedral.

The green manganate is paramagnetic with one unpaired electron but the permanganate is diamagnetic.

The π-bonding takes place by overlap of
p-orbitals of oxygen with d-orbitals of manganese.

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#30 SUB 2M 🖼 2 ✎ Edit ⚠ Log Error

Question

Give the structure of chromate and dichromate ions, and explain interconvertible property of these ions.

Answer

The structures of chromate ion, CrO2–4 and the dichromate ion, Cr2O2–7 are shown below.

The chromate ion is tetrahedral whereas the dichromate ion consists of two tetrahedral sharing one corner with Cr–O–Cr bond angle of 126°.

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Give the structure of chromate and dichromate ions, and explain interconvertible property of these ions.

Answer

The structures of chromate ion, CrO2–4 and the dichromate ion, Cr2O2–7 are shown below.

The chromate ion is tetrahedral whereas the dichromate ion consists of two tetrahedral sharing one corner with Cr–O–Cr bond angle of 126°.

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.

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#31 SUB 0M 🖼 3 ✎ Edit ⚠ Log Error

Question

Explain the chemical properties of KMnO4.

Answer

KMnO4 act as strong oxidising agent in neutral, alkaline and in acidic medium.

(I) In Acidic Solution:

(a) Iodine is liberated from potassium iodide:

(b) Fe2+ ion (green) is converted to Fe3+ (yellow):

(c) Oxalate ion or oxalic acid is oxidised at 333K :

(d) Hydrogen sulphide is oxidised, sulphur being precipitated:

H2S → 2H+ + S2–

5S2– + 2MnO–4 + 16H+ → 2Mn2+ + 8H2O + 5S

(e) Sulphurous acid or sulphite is oxidised to a sulphate or sulphuric acid:

5SO2–3 + 2MnO–4 + 6H+ → 2Mn2+ + 3H2O + 5SO2–4

(f) Nitrite is oxidised to nitrate:

5NO–2 + 2MnO–4 + 6H+ → 2Mn2+ + 5NO–3 + 3H2O

(2) In neutral or faintly alkaline solutions:

(a) A notable reaction is the oxidation of iodide to iodate:

2MnO–4 + H2O + I– → 2MnO2 + 2OH– + IO–3

(b) Thiosulphate is oxidised almost quantitatively to sulphate:

8MnO–4 + 3S2O2–3 + H2O → 8MnO2 + 6SO2–4 + 2OH–

(c) Manganous salt is oxidised to MnO2; the presence of zinc sulphate or zinc oxide catalyses the oxidation:

2MnO–4 + 3Mn2+ + 2H2O → 5MnO2 + 4H+

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Explain the chemical properties of KMnO4.

Answer

KMnO4 act as strong oxidising agent in neutral, alkaline and in acidic medium.

(I) In Acidic Solution:

(a) Iodine is liberated from potassium iodide:

(b) Fe2+ ion (green) is converted to Fe3+ (yellow):

(c) Oxalate ion or oxalic acid is oxidised at 333K :

(d) Hydrogen sulphide is oxidised, sulphur being precipitated:

H2S → 2H+ + S2–

5S2– + 2MnO–4 + 16H+ → 2Mn2+ + 8H2O + 5S

(e) Sulphurous acid or sulphite is oxidised to a sulphate or sulphuric acid:

5SO2–3 + 2MnO–4 + 6H+ → 2Mn2+ + 3H2O + 5SO2–4

(f) Nitrite is oxidised to nitrate:

5NO–2 + 2MnO–4 + 6H+ → 2Mn2+ + 5NO–3 + 3H2O

(2) In neutral or faintly alkaline solutions:

(a) A notable reaction is the oxidation of iodide to iodate:

2MnO–4 + H2O + I– → 2MnO2 + 2OH– + IO–3

(b) Thiosulphate is oxidised almost quantitatively to sulphate:

8MnO–4 + 3S2O2–3 + H2O → 8MnO2 + 6SO2–4 + 2OH–

(c) Manganous salt is oxidised to MnO2; the presence of zinc sulphate or zinc oxide catalyses the oxidation:

2MnO–4 + 3Mn2+ + 2H2O → 5MnO2 + 4H+

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#32 SUB 2M ✎ Edit ⚠ Log Error

Question

Write down the uses of KMnO4.

Answer

Potassium permanganate is used as a favourite oxidant in preparative organic chemistry.

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.

An aqueous solution of potassium permanganate is used for gargling to keep mouth germfree as it is antiseptic.

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Write down the uses of KMnO4.

Answer

Potassium permanganate is used as a favourite oxidant in preparative organic chemistry.

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.

An aqueous solution of potassium permanganate is used for gargling to keep mouth germfree as it is antiseptic.

It is useful as titrant in redox titrations to know the proportion of metal ions like iron and organic compounds like oxalic acid.

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#33 SUB 3M 🖼 2 ✎ Edit ⚠ Log Error

Question

Complete the reactions and balance them.

(i) MnO2 + KOH + O2 →

(ii) MnO–4 + H2O + I–

(iii) MnO–2 + S2O2–3 + H2O

Answer

(i) MnO2 + 4KOH + O2 → 2K2 MnO4 + 2H2O

(ii) 2MnO–4 + H2O + I– → 2MnO2 + 2OH– + IO–3

(iii) 8MnO–4 + 2S2O2–2 + H2O → 8MnO2 + 6SO2–4 + 2OH–

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Complete the reactions and balance them.

(i) MnO2 + KOH + O2 →

(ii) MnO–4 + H2O + I–

(iii) MnO–2 + S2O2–3 + H2O

Answer

(i) MnO2 + 4KOH + O2 → 2K2 MnO4 + 2H2O

(ii) 2MnO–4 + H2O + I– → 2MnO2 + 2OH– + IO–3

(iii) 8MnO–4 + 2S2O2–2 + H2O → 8MnO2 + 6SO2–4 + 2OH–

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#34 SUB 0M ✎ Edit ⚠ Log Error

Question

Write note on inner transition elements or ƒ-block elements.

Answer

The ƒ-block consists of the two series, lanthanoids (the fourteen elements following lanthanum) and actinoids (the fourteen elements following actinium).

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.

Similarly, a discussion of the actinides includes actinium besides the fourteen elements constituting the series.

The lanthanoids resemble one another more closely than do the members of ordinary transition elements in any series.

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.

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Write note on inner transition elements or ƒ-block elements.

Answer

The ƒ-block consists of the two series, lanthanoids (the fourteen elements following lanthanum) and actinoids (the fourteen elements following actinium).

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.

Similarly, a discussion of the actinides includes actinium besides the fourteen elements constituting the series.

The lanthanoids resemble one another more closely than do the members of ordinary transition elements in any series.

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.

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.

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#35 SUB 0M ✎ Edit ⚠ Log Error

Question

Write note on physical properties of lanthanoids.

Answer

All the lanthanoids are silvery white soft metals and tarnish rapidly in air.

The hardness increases with increasing atomic number, samarium being steel hard.

Their melting points range between 1000 to 1200K but samarium melts at 1623K.

They have typical metallic structure and are good conductors of heat and electricity.

Density and other properties change smoothly except for Eu and Yb and occasionally for Sm and Tm.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Write note on physical properties of lanthanoids.

Answer

All the lanthanoids are silvery white soft metals and tarnish rapidly in air.

The hardness increases with increasing atomic number, samarium being steel hard.

Their melting points range between 1000 to 1200K but samarium melts at 1623K.

They have typical metallic structure and are good conductors of heat and electricity.

Density and other properties change smoothly except for Eu and Yb and occasionally for Sm and Tm.

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#36 SUB 0M ✎ Edit ⚠ Log Error

Question

Write down uses of lanthanoids and its compounds.

Answer

Use of the lanthanoids is for the production of alloy steels for plates and pipes.

A well known alloy is mischmetal which consists of a lanthanoid metal (∼ 95%) and iron (∼ 5%) and traces of S, C, Ca and Al.

A good deal of mischmetal is used in Mg-based alloy to produce bullets, shell and lighter flint.

Mixed oxides of lanthanoids are employed as catalysts in petroleum cracking.

Some individual Ln oxides are used as phosphors in television screens and similar fluorescing surfaces.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Write down uses of lanthanoids and its compounds.

Answer

Use of the lanthanoids is for the production of alloy steels for plates and pipes.

A well known alloy is mischmetal which consists of a lanthanoid metal (∼ 95%) and iron (∼ 5%) and traces of S, C, Ca and Al.

A good deal of mischmetal is used in Mg-based alloy to produce bullets, shell and lighter flint.

Mixed oxides of lanthanoids are employed as catalysts in petroleum cracking.

Some individual Ln oxides are used as phosphors in television screens and similar fluorescing surfaces.

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#37 SUB 3M 🖼 3 ✎ Edit ⚠ Log Error

Question

Discuss chemical properties of the lanthanoids.

Answer

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.

The value of E° of half reaction

+ 3e– Ln(s) range from – 2.2 V to 2.4 V except for Eu that has – 2.0 V.

Lanthanoids heated with carbon forms carbides Ln3C, Ln2C3 and LnC2.

Lanthanoids form trihalides LnX3 with halogen and liberates hydrogen gas when treated with dilute acids.

These metals combine with oxygen to form oxides of type Ln2O3.

Oxides are basic in nature and when dissolved in water they form hydroxide Ln(OH)3.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Discuss chemical properties of the lanthanoids.

Answer

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.

The value of E° of half reaction

+ 3e– Ln(s) range from – 2.2 V to 2.4 V except for Eu that has – 2.0 V.

Lanthanoids heated with carbon forms carbides Ln3C, Ln2C3 and LnC2.

Lanthanoids form trihalides LnX3 with halogen and liberates hydrogen gas when treated with dilute acids.

These metals combine with oxygen to form oxides of type Ln2O3.

Oxides are basic in nature and when dissolved in water they form hydroxide Ln(OH)3.

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#38 SUB 3M ✎ Edit ⚠ Log Error

Question

Write down electronic configuration of all lanthanoids elements. [3 Marks]

Answer

General electronic configuration - [xe] 4ƒ1-14 5d0-1

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Write down electronic configuration of all lanthanoids elements. [3 Marks]

Answer

General electronic configuration - [xe] 4ƒ1-14 5d0-1

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#39 SUB 3M 🖼 1 ✎ Edit ⚠ Log Error

Question

Explain the lanthanoid contraction.

Answer

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.

As electrons are being added to the same shell, the effective nuclear charge increases.

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.

Also, with the increase in atomic number, the number of electrons in the 4ƒ-orbital also increases. The 4ƒ electrons have poor shielding effect.

Therefore, the effective nuclear charge experienced by the outer electrons increases. Consequently, the attraction of the nucleus for the outermost electrons increases.

This results in a steady decrease in the size of lanthanoids with the increase in the atomic number. This is termed as lanthanoid contraction.

Consequences of lanthanoid contraction:

• There is similarity in the properties of second and third transition series.
• Separation of lanthanoids is possible due to lanthanoid contraction.
• It is due to lanthanoid contraction that there is variation in the basic strength of lanthanoid hydroxides. (Basic strength decreases from La(OH)3 to Lu(OH)3.)

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Explain the lanthanoid contraction.

Answer

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.

As electrons are being added to the same shell, the effective nuclear charge increases.

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.

Also, with the increase in atomic number, the number of electrons in the 4ƒ-orbital also increases. The 4ƒ electrons have poor shielding effect.

Therefore, the effective nuclear charge experienced by the outer electrons increases. Consequently, the attraction of the nucleus for the outermost electrons increases.

This results in a steady decrease in the size of lanthanoids with the increase in the atomic number. This is termed as lanthanoid contraction.

Consequences of lanthanoid contraction:

• There is similarity in the properties of second and third transition series.
• Separation of lanthanoids is possible due to lanthanoid contraction.
• It is due to lanthanoid contraction that there is variation in the basic strength of lanthanoid hydroxides. (Basic strength decreases from La(OH)3 to Lu(OH)3.)

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#40 SUB 0M ✎ Edit ⚠ Log Error

Question

Write down general electronic configuration of elements of actinoids. Why there are irregularities in the electronic configurations?

Answer

The general electronic configuration of actinoids is [Rn]5ƒ0-14 6d0-2 7s2

The irregularities in the electronic configuration of actinoid series is due to very less difference in the energies of 5ƒ-orbitals and 6d-orbitals.

All the actinoids are believed to have the electronic configuration of 7s2 and variable occupancy of the 5f and 6d-subshells.

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.

Thus, the configurations of Am and Cm are [Rn] 5f 77s2 and [Rn] 5f 7. 6d 1. 7s2.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Write down general electronic configuration of elements of actinoids. Why there are irregularities in the electronic configurations?

Answer

The general electronic configuration of actinoids is [Rn]5ƒ0-14 6d0-2 7s2

The irregularities in the electronic configuration of actinoid series is due to very less difference in the energies of 5ƒ-orbitals and 6d-orbitals.

All the actinoids are believed to have the electronic configuration of 7s2 and variable occupancy of the 5f and 6d-subshells.

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.

Thus, the configurations of Am and Cm are [Rn] 5f 77s2 and [Rn] 5f 7. 6d 1. 7s2.

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#41 SUB 0M ✎ Edit ⚠ Log Error

Question

Discuss comparison between lathanoids and actinoids.

Answer

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.

The magnetic properties of the actinoids are more complex than those of the lanthanoids.

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.

The stable oxidation state of all the lanthanoids is (+3). In actinoids, oxidation states (+2) to (+6) are seen.

In lanthanoids only promethium is radioactive but all the actinoids are radioactive.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Discuss comparison between lathanoids and actinoids.

Answer

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.

The magnetic properties of the actinoids are more complex than those of the lanthanoids.

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.

The stable oxidation state of all the lanthanoids is (+3). In actinoids, oxidation states (+2) to (+6) are seen.

In lanthanoids only promethium is radioactive but all the actinoids are radioactive.

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#42 SUB 0M ✎ Edit ⚠ Log Error

Question

Compare the chemistry of actinoids with that of the lanthanoids with special reference to:

(i) Electronic configuration,

(ii) Oxidation state,

(iii) Atomic and ionic sizes and

(iv) Chemical reactivity.

Answer

(i) Electronic 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.

(ii) Oxidation 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.

(iii) Atomic 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.

(iv) Chemical 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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Compare the chemistry of actinoids with that of the lanthanoids with special reference to:

(i) Electronic configuration,

(ii) Oxidation state,

(iii) Atomic and ionic sizes and

(iv) Chemical reactivity.

Answer

(i) Electronic 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.

(ii) Oxidation 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.

(iii) Atomic 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.

(iv) Chemical 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).

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#43 SUB 0M ✎ Edit ⚠ Log Error

Question

Discuss applications of d and ƒ-block elements.

Answer

Iron and steels are the most important construction materials.

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.

Some compounds are manufactured for special purposes such as TiO for the pigment industry and MnO2 for use in dry battery cells.

The battery industry also requires Zn and Ni/Cd.

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.

V2O5 catalyses the oxidation of SO2 in the manufacture of sulphuric acid.

TiCl4 with Al(CH3)3 forms the basis of the Ziegler catalysts used to manufacture polyethylene (polythene).

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.

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Discuss applications of d and ƒ-block elements.

Answer

Iron and steels are the most important construction materials.

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.

Some compounds are manufactured for special purposes such as TiO for the pigment industry and MnO2 for use in dry battery cells.

The battery industry also requires Zn and Ni/Cd.

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.

V2O5 catalyses the oxidation of SO2 in the manufacture of sulphuric acid.

TiCl4 with Al(CH3)3 forms the basis of the Ziegler catalysts used to manufacture polyethylene (polythene).

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.

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.

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#44 SUB 3M PYQ ✎ Edit ⚠ Log Error

Question

Give reason:[March 2023] [3 Marks]

Answer

(i) Transition elements exhibit higher enthalpies of atomisation.

(ii) In aqueous solution, Cr2+ is stronger reducing agent than Fe2+.

(iii) The second ionisation enthalpy of Cu is higher then Zn.

Ans. (i) Because transition elements have more unpaired electrons, they have stronger interatomic interactions and hence stronger bonding.

(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.

(iii) In 2nd ionisation enthalpy of Cu,
e– is removed from 3d10 and in Zn from 4s1. 3d10 is more stable configuration than 4s1.

Class 12 Chemistry (Part 1) 017

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Give reason:[March 2023] [3 Marks]

Answer

(i) Transition elements exhibit higher enthalpies of atomisation.

(ii) In aqueous solution, Cr2+ is stronger reducing agent than Fe2+.

(iii) The second ionisation enthalpy of Cu is higher then Zn.

Ans. (i) Because transition elements have more unpaired electrons, they have stronger interatomic interactions and hence stronger bonding.

(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.

(iii) In 2nd ionisation enthalpy of Cu,
e– is removed from 3d10 and in Zn from 4s1. 3d10 is more stable configuration than 4s1.

Class 12 Chemistry (Part 1) 017

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#45 SUB 2M ✎ Edit ⚠ Log Error

Question

Silver atom has completely filled d-orbitals (4d10) in its ground state. How can you say that it is a transition element?

Answer

Silver (Z = 47) can exhibit +2 oxidation state where in it will have incompletely filled d-orbitals (4d), hence a transition element.

47Ag : [Kr] 4d10 5s1

Ag2+: [Kr] 4d9

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Silver atom has completely filled d-orbitals (4d10) in its ground state. How can you say that it is a transition element?

Answer

Silver (Z = 47) can exhibit +2 oxidation state where in it will have incompletely filled d-orbitals (4d), hence a transition element.

47Ag : [Kr] 4d10 5s1

Ag2+: [Kr] 4d9

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#46 SUB 2M ✎ Edit ⚠ Log Error

Question

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

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

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.

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#47 SUB 0M ✎ Edit ⚠ Log Error

Question

Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?

Answer

Manganese (25Mn) shows largest number of oxidation states because it has maximum number of unpaired electrons.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?

Answer

Manganese (25Mn) shows largest number of oxidation states because it has maximum number of unpaired electrons.

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#48 SUB 3M 🖼 4 ✎ Edit ⚠ Log Error

Question

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

The E°(M2+/M) value of a metal depends on the energy changes involved in the following:

(1) Sublimation: The energy required for converting one mole of an atom from the solid state to the gaseous state.

Cu(s) → Cu(g) ∆sH° or ∆aH° = 339 kJ Mol–1

(2) Ionization: The energy required to take out electrons from one mole of atoms in the gaseous state.

Cu(g) → + 2e– ∆iH° = 2703 kJ Mol–1

(3) Hydration: The energy released when one mole of ions are hydrated.

∆hydH° = –2121 kJ Mol–1

E°Cu2+/Cu = ∆aH° + ∆iH° + ∆hydH°

= 339 + 2703 – 2121

= 1921 kJ Mol–1

Total value of all the enthalpies is positive so reaction is endothermic so E°Cu+2/Cu is positive.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

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

The E°(M2+/M) value of a metal depends on the energy changes involved in the following:

(1) Sublimation: The energy required for converting one mole of an atom from the solid state to the gaseous state.

Cu(s) → Cu(g) ∆sH° or ∆aH° = 339 kJ Mol–1

(2) Ionization: The energy required to take out electrons from one mole of atoms in the gaseous state.

Cu(g) → + 2e– ∆iH° = 2703 kJ Mol–1

(3) Hydration: The energy released when one mole of ions are hydrated.

∆hydH° = –2121 kJ Mol–1

E°Cu2+/Cu = ∆aH° + ∆iH° + ∆hydH°

= 339 + 2703 – 2121

= 1921 kJ Mol–1

Total value of all the enthalpies is positive so reaction is endothermic so E°Cu+2/Cu is positive.

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#49 SUB 2M ✎ Edit ⚠ Log Error

Question

How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements?

Answer

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements?

Answer

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).

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#50 SUB 2M ✎ Edit ⚠ Log Error

Question

Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Answer

Because of small size and high electronegativity oxygen or fluorine can oxidise the metal to its highest oxidation state.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Answer

Because of small size and high electronegativity oxygen or fluorine can oxidise the metal to its highest oxidation state.

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#51 SUB 2M ✎ Edit ⚠ Log Error

Question

Which is a stronger reducing agent Cr2+ or Fe2+ and why?

Answer

Cr2+ is stronger reducing agent than Fe2+.

Reason: d4 → d3 occurs in case of Cr2+ to Cr3+.

But d6 → d5 occurs in case of Fe2+ to Fe3+.

In a medium (like water) d3 is more stable as compared to d5 (see CFSE).

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Which is a stronger reducing agent Cr2+ or Fe2+ and why?

Answer

Cr2+ is stronger reducing agent than Fe2+.

Reason: d4 → d3 occurs in case of Cr2+ to Cr3+.

But d6 → d5 occurs in case of Fe2+ to Fe3+.

In a medium (like water) d3 is more stable as compared to d5 (see CFSE).

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#52 SUB 2M 🖼 4 ✎ Edit ⚠ Log Error

Question

Calculate the 'spin only' magnetic moment of M2+(aq) ion (Z = 27).

Answer

Z = 27

µ =

=

=

= 3.87 B.M

[Ar] 3d7 4s2

\ M2+ = [Ar] 3d7

3d7 =

i.e., 3 unpaired electrons

\ n = 3

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Calculate the 'spin only' magnetic moment of M2+(aq) ion (Z = 27).

Answer

Z = 27

µ =

=

=

= 3.87 B.M

[Ar] 3d7 4s2

\ M2+ = [Ar] 3d7

3d7 =

i.e., 3 unpaired electrons

\ n = 3

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#53 SUB 2M ✎ Edit ⚠ Log Error

Question

Explain why Cu+ ion is not stable in aqueous solutions?

Answer

Cu+ in aqueous solution undergoes disproportionation, i.e., 2Cu+(aq) → Cu2+(aq) + Cu(s)

The E° value for this is favourable.

Type SUB — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Explain why Cu+ ion is not stable in aqueous solutions?

Answer

Cu+ in aqueous solution undergoes disproportionation, i.e., 2Cu+(aq) → Cu2+(aq) + Cu(s)

The E° value for this is favourable.

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#54 SUB 2M ✎ Edit ⚠ Log Error

Question

Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Answer

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 — Subjective MCQ — Multiple Choice TF — True / False YN — Yes / No IA — Interactive Activity FB — Fill in the Blanks OW — One Word MT — Match the Following GLO — Glossary RCEMT — Reading Comp. – Extracts – MT SN — Short Notes CBTFQ — Content-Based T/F RCMTP — Reading Comp. – MT – Poem RCSR — Reading Comp. – Supp. Reader RCNV — Reading Comp. – Nonverbal RCUDI — Reading Comp. – Unseen D/I IFMC — Identify Function – Match Cols CTD — Complete the Dialogue CTS — Complete the Sentences VCPB — Vocab: Complete Para – Blanks VFNM — Vocab: Nearest Meaning – SR INDS — Indirect Speech CONJ — Conjunction EDTNG — Editing DASD — Do as Directed STCQ — Select the Correct PARAW — Paragraph Writing EFRW — Email / Report Writing PICD — Picture Description PARA — Paragraph CS — Case Study Marks

Question

Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Answer

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.

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