//M2//QN1//SUB//DL0

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

//X

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.

//M0//QN2//SUB//DL0

What transition elements? Explain with examples.

//X

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] 3d 1 4s 2
42Mo: [Kr] 4d 5 5s 1
25Mn: [Ar] 3d 5 4s 2

//M2//QN3//SUB//DL0

What is ƒ-block element?

//X

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.

//M2//QN4//SUB//DL0

Explain position of d-block in periodic table.

//X

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.

//M0//QN5//SUB//DL0

Electronic configuration of first transition metal.

//X

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] 3 d 1 4 s 2 [Ar] 3 d 2 4 s 2 [Ar] 3 d 3 4 s 2 [Ar] 3 d 5 4 s 1 [Ar] 3 d 5 4 s 2 [Ar] 3 d 6 4 s 2 [Ar] 3 d 7 4 s 2 [Ar] 3 d 8 4 s 2 [Ar] 3 d 10 4 s 1 [Ar] 3 d 10 4 s 2

//M0//QN6//SUB//DL0

Electronic configuration of second transition metal.

//X

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] 4 d 1 5 s 2 [Kr] 4 d 2 5 s 2 [Kr] 4 d 4 5 s 1 [Kr] 4 d 5 5 s 1 [Kr] 4 d 5 5 s 2 [Kr] 4 d 7 5 s 1 [Kr] 4 d 8 5 s 1 [Kr] 4 d 10 5 s 0 [Kr] 4 d 10 5 s 1 [Kr] 4 d 10 5 s 2

//M0//QN7//SUB//DL0

Electronic configuration of third transition metal.

//X

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] 5 d 1 6 s 2 [Xe] 4 f 14 5 d 2 6 s 2 [Xe] 4 f 14 5 d 3 6 s 2 [Xe] 4 f 14 5 d 4 6 s 2 [Xe] 4 f 14 5 d 5 6 s 2 [Xe] 4 f 14 5 d 6 6 s 2 [Xe] 4 f 14 5 d 7 6 s 2 [Xe] 4 f 14 5 d 9 6s 1 [Xe] 4 f 14 5 d 10 6 s 1 [Xe] 4 f 14 5 d 10 6 s 2

//M0//QN8//SUB//DL0

Why the chromium and copper have exceptional electronic configuration?

//X

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] 3d 5 4s 1 instead of [Ar] 3d 4 4s 2
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] 3d 10 4s 1 instead of [Ar] 3d 9 4s 2.

//M3//QN9//SUB//DL0

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?

//X

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 d 10 configuration in their ground state as well as in their common oxidation states and hence, are not regarded as transition metals
Examples: 30Zn: [Ar] 3d 10 4s 2
48Cd: [Kr] 4d 10 5s 2
80Hg: [Xe] 4ƒ 14 5d 10 6s2

//M2//QN10//SUB//DL0

In what way is the electronic configuration of the transition elements different from that of the non-transition elements?[Topic 4.0] [2 Marks]

//X

Transition metals have a partially filled d-orbital. Therefore, the electronic configuration of transition elements is (n - 1)d 1-10 ns 0-2.
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 ns 1-2 or ns 2 np 1-6.

//M0//QN11//SUB//DL0

Explain physical properties of transition elements.

//X

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.

//M2//QN12//SUB//DL0

Aqueous solution of TiCl 3 is coloured where as aqueous solution of TiCl 4 is colourless. Give reason.

//X

In TiCl3 oxidation number of Ti is +3.
Ti3+: [Ar] 3d 1.
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] 3d 0.
Here in 3d-orbital there is no unpaired electron so it is colourless.

//M2//QN13//SUB//DL0//EQ

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

//X

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 d 5 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.

//M0//QN14//SUB//DL0//EQ

Write note on enthalpy of atomization of transition metals.

//X

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.

//M0//QN15//SUB//DL0//EQ

What is lanthanoid contraction? What are the consequences of lanthanoid contraction? or Explain atomic radii of transition element along the period.

//X

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.

//M0//QN16//SUB//DL0

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

//X

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 d n 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 d 5 and d 10 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 d 10 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 d 5 (Mn2+) and d 10 (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.

//M3//QN17//SUB//DL0

Write about oxidation states of transition series?

//X

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.

//M0//QN18//SUB//DL0//EQ

Explain trends in M2+/M standard electrode potentials.

//X

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.

//M0//QN19//SUB//DL0

Write a note on halide compounds of Transition elements.

//X

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 D hydH° of Cu2+(aq) than Cu+, which more than compensates for the second ionization enthalpy of Cu.

Formulas of Halides of 3 d Metals
Oxidation Number
+6 +5 +4 +3 +2 +1 | TiX 4 TiX 3 TiX 2 III | VF 5 VX 1 4 VX 3 VX 2 | CrF 6 CrF 5 CrX 4 CrX 3 CrX 2 | MnF 4 MnF 3 MnF 2 | FeX 1 3 FeX 2 | CoF 3 CoX 2 | NiX 2 | CuX 2 11 CuX 111 | ZnX 2

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

//M2//QN20//SUB//DL0

Explain chemical reactivity of transition metals.

//X

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., 2Cr 2+(aq) + 2H+(aq) → 2Cr 3+(aq) + H 2(g)

//M3//QN21//SUB//DL0//EQ

Explain magnetic properties of transition elements.

//X

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
Sc 3+ Ti 3+ Ti 2+ V 2+ Cr 2+ Mn 2+ Fe 2+ Co 2+ Ni 2+ Cu 2+ Zn 2+ | 3 d 0 3 d 1 3 d 2 3 d 3 3 d 4 3 d 5 3 d 6 3 d 7 3 d 8 3 d 9 3 d 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

//M3//QN22//SUB//DL0//EQ//PYQ

Calculate the magnetic moment of a tivalent ion in aqueous solution if its atomic number

is 26.[May 2021] [3 Marks]

//X

26M: [Ar] 3d 5 4 s 2
M3+: [Ar] 3d 5
Unpaired electrons = 5
∴ Magnetic moment (µ)=
=
=
= 5.92 B.M

//M3//QN23//SUB//DL0

Why transition elements form a coloured ions?

//X

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
3 d 0 | Sc 3+ | Colourless
3 d 0 | Ti 4+ | Colourless
3 d 1 | Ti 3+ | Purple
3 d 1 | V 4+ | Blue
3 d 2 | V 3+ | Green
3 d 3 | V 2+ | Violet
3 d 3 | Cr 3+ | Violet
3 d 4 | Mn 3+ | Violet
3 d 4 | Cr 2+ | Blue
3 d 5 | Mn 2+ | Pink
3 d 5 | Fe 3+ | Yellow
3 d 6 | Fe 2+ | Green
3 d 6 | Co 3+ | Blue
3 d 7 | Co 2+ | Pink
3 d 8 | Ni 2+ | Green
3 d 9 | Cu 2+ | Blue
3 d 10 | Zn 2+ | Colourless

//M2//QN24//SUB//DL0

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

//X

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.

//M2//QN25//SUB//DL0

Explain Catalytic property of transition elements.

//X

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 → I 2 + 2SO 2–4
An explanation of this catalytic action can be given as:
2Fe 3+ + 2I – → 2Fe 2+ + I 2
2Fe 2+ + S 2 O 2–8 → 2Fe 3+ + 2SO 2–4

//M3//QN26//SUB//DL0

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.

//X

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.

//M0//QN27//SN//DL0

Write short note on alloys.

//X

“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.

//M0//QN28//SUB//DL0

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

//X

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 3 d Metals
Acidic oxide | Basic oxide | Amphoteric oxide
Mn 2 O 7 | MnO | Mn 3 O 4, Mn 2 O 3, MnO 2
| CrO | Cr 2 O 3

Oxidation Number
| 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12
+7 +6 +5 +4 +3 +2 +1 | Sc 2 O 3 | TiO 2 Ti 2 O 3 TiO | V 2 O 5 V 2 O 4 V 2 O 3 VO | CrO 3 CrO 2 Cr 2 O 3 (CrO) | Mn 2 O 7 MnO 2 Mn 2 O 3 Mn 3 O 4* MnO | Fe 2 O 3 Fe 3 O 4* FeO | Co 3 O*4 CoO | NiO | CuO Cu 2 O | ZnO
Note: *Denotes mixed oxides

In the covalent oxide Mn 2 O 7, each Mn is tetrahedrally surrounded by O's including a Mn–O–Mn bridge. The tetrahedral [MO 4]n- ions are known for V V, Cr VI, Mn V, Mn VI and Mn VII.

//M2//QN29//SUB//DL0//EQ

Discuss structures and magnetic properties of manganate and permanganate ions.

//X

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.

//M2//QN30//SUB//DL0//EQ

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

//X

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.

//M0//QN31//SUB//DL0//EQ

Explain the chemical properties of KMnO4.

//X

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+

//M2//QN32//SUB//DL0

Write down the uses of KMnO4.

//X

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.

//M3//QN33//SUB//DL0//EQ

Complete the reactions and balance them. (i) MnO 2 + KOH + O 2 → (ii) MnO–4 + H 2 O + I– (iii) MnO–2 + S 2 O 2–3 + H 2 O

//X

(i) MnO 2 + 4KOH + O 2 → 2K 2 MnO 4 + 2H 2 O
(ii) 2MnO–4 + H 2 O + I– → 2MnO 2 + 2OH– + IO–3
(iii) 8MnO–4 + 2S 2 O 2–2 + H 2 O → 8MnO 2 + 6SO 2–4 + 2OH–

//M0//QN34//SUB//DL0

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

//X

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.

//M0//QN35//SUB//DL0

Write note on physical properties of lanthanoids.

//X

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.

//M0//QN36//SUB//DL0

Write down uses of lanthanoids and its compounds.

//X

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.

//M3//QN37//SUB//DL0//EQ

Discuss chemical properties of the lanthanoids.

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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.

//M3//QN38//SUB//DL0

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

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General electronic configuration - [xe] 4ƒ 1-14 5d 0-1

//M3//QN39//SUB//DL0//EQ

Explain the lanthanoid contraction.

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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.)

//M0//QN40//SUB//DL0

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

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The general electronic configuration of actinoids is [Rn]5ƒ 0-14 6d 0-2 7s 2
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 7s 2 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 77s 2 and [Rn] 5f 7. 6d 1. 7s 2.

//M0//QN41//SUB//DL0

Discuss comparison between lathanoids and actinoids.

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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.

//M0//QN42//SUB//DL0

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.

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(i) Electronic configuration: The general electronic configuration for lanthanoids is [Xe]54 4 f 0-14 5 d 0-1 6 s 2 and that for actinoids is [Rn]86 5 f 1-14 6 d 0-1 7 s 2. Unlike 4 f-orbitals, 5 f-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
5 f-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).

//M0//QN43//SUB//DL0

Discuss applications of d and ƒ-block elements.

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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.

//M3//QN44//SUB//DL0//PYQ

Give reason:[March 2023] [3 Marks]

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(i) Transition elements exhibit higher enthalpies of atomisation.
(ii) In aqueous solution, Cr 2+ is stronger reducing agent than Fe 2+.
(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 Cr 2+ to Cr 3+ d 4→ d 3 occurs and in Fe 2+ to Fe 3+ d 6 → d 5 occurs. In aq. medium d 3
is more stable than d 5.
(iii) In 2 nd ionisation enthalpy of Cu,
e– is removed from 3 d 10 and in Zn from 4 s 1. 3 d 10 is more stable configuration than 4 s 1.
Class 12 Chemistry (Part 1) 017