(a) When a chromite ore (A) is fused with an aqueous solution of sodium carbonate in free excess of air, a yellow solution of compound (B) is obtained. This solution is filtered and acidified with sulphuric acid to form compound (C). Compound (C) on treatment with solution of KCl gives orange crystals of compound (D). Write the chemical formulae of compounds A to D.

(b) Describe the cause of the following variations with respect to lanthanoids and actinoids:

(i) Greater range of oxidation states of actinoids as compared to lanthanoids.

(ii) Greater actinoid contraction as compared to lanthanoid contraction.

(iii) Lower ionisation enthalpy of early actinoids as compared to the early lanthanoids.

OR

(a) What happens when:

(i) Manganate ions (MnO₄^2-) undergoes disproportionation reaction in acidic medium?

(ii) Lanthanum is heated with Sulphur?

(b) Explain the following trends in the properties of the members of the First series of transition elements:

(i) E⁰(M²⁺|M) value for copper is positive(+0.34 V) in contrast to the other members of the series.

(ii) Cr²⁺ is reducing while Mn³⁺ is oxidising, though both have d⁴ configuration.

(iii) The oxidising power in the series increases in the order

VO₂⁺< Cr₂O₇^2-< MnO₄^-.

(a)Chromite ore FeCr₂O₄ compound A is fused with an aqueous solution of sodium carbonate in excess of air, to form compound B that is sodium chromate that is yellow in colour. Sodium chromate is filtered and acidified with sulphuric acid to form compound C that is sodium dichromate, which is an orange crystalline compound. Sodium dichromate reacts with KCl to form potassium dichromate which is less soluble than sodium dichromate. This is compound D.

4FeCr₂O₄ + 8Na₂CO₃ + 7O₂→ 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂

(A)

B (i) The actinoids have a greater range of oxidation states than lanthanoids, which are both members of the f-block group of elements. This is because the the 5f, 6d and 7s levels are of comparable energies. The first half of the series frequently exhibit higher oxidation states but it decreases in the succeeding elements. The elements in general show +3 state and then increases to +4 in Th to +5, +6 and +7 respectively in Pa, U and Np.

(ii) Actinoids show a gradual decrease in the size of atoms or M³⁺ ions across the series. This is actinoid contraction, just like lanthanoid contraction, but there is a greater contraction from element to element in this series because of poor shielding by 5f electrons.

(iii) The ionization enthalpies of the early actinoids, though not accurately known, are lower compared to early lanthanoids. This is understood because it is to be expected that when 5f orbitals are beginning to be occupied, they will penetrate less into the inner core of electrons. The 5f electrons will be more effectively shielded from the nuclear charge than the 4f electrons of the corresponding lanthanoids. Because the outer electrons are less firmly held, they are available for bonding in the actinoids.

OR

A (i) Disproportionation is a reaction where a particular oxidation state becomes less stable relative to other oxidation states. Disproportionation is a type of redox reaction in which a species is simultaneously reduced and oxidised forming two different products. Manganate ions (MnO₄^2-) has an oxidation number of +6. In acidic medium, it undergoes disproportionation reaction to give MnO₂ and MnO₄^-.

3MnO₄^2- + 4H⁺→ MnO₂ + 2MnO₄^- + 2H₂O

(ii) When Lanthanum is heated with Sulphur, it forms Ln₂S₃.

2Ln + 3S → Ln₂S₃

B (i) The E⁰(M²⁺/M) value of a metal depends on the energy changes involved in the following reactions:

1. Sublimation energy which is the energy needed to convert one mole of atoms from a solid state to gaseous state. 2. Ionization energy which is the energy supplied to remove electrons from one mole of atoms, which are in the gaseous state. 3. Hydration energy which is the energy emitted to hydrate one mole of ions. Copper has a high ionisation energy and low hydration energy. Copper also has high atomization energy Δ_aH°. Hence, the E⁰(M²⁺/M) value for copper is positive.

(ii) Cr²⁺ is reducing in nature while Mn³⁺ is oxidizing. Both of them have the same d⁴ configuration. When Cr²⁺ acts as a reducing agent, it gets oxidized to Cr³⁺, with configuration d³. It can be written as 3t_2g configuration, which is the stable configuration. In the case for Mn³⁺, when it oxidises, it gets reduced to Mn²⁺ with a d⁵ configuration. This configuration has a half-filled orbital and has extra stability.

Cr³⁺ configuration Mn²⁺ configuration

(iii)VO₂⁺ is an oxocation and Cr₂O₇^2- and MnO₄^- are both oxoanions of the transition metals. The ions in which the central metal atom is has the highest oxidation state will have the highest oxidising power. In VO₂⁺, vanadium is present in the +5 oxidation state, while in Cr₂O₇^2- ion, Cr is present in the +6 oxidation state. Similarly in the MnO₄^-, Mn is present in the +7 oxidation state. Thus as the oxidation state of the central metal atom increases in the given order.