A. Account for the following:

(i) Transition metals form a large number of complex compounds.

(ii) The lowest oxide of transition metal is basic whereas the highest oxide is amphoteric or acidic.

(iii) E° value for the Mn³⁺/Mn²⁺ couple is highly positive (+1.57 V) as compare to Cr³⁺/Cr²⁺.

B. Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements.

OR

A. (i) How is the variability in oxidation states of transition metals different from that of the p-block elements?

(ii) Out of Cu⁺ and Cu²⁺, which ion is unstable in aqueous solution and why?

(iii) Orange colour of Cr₂O₇^2– ion changes to yellow when treated with an alkali. Why?

B. Chemistry of actinoids is complicated as compared to lanthanoids. Give two reasons.

(A) (i) 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.

(ii) In the case of a lower oxide of a transition metal, the metal atom has a low oxidation state. This means that some of the valence electrons of the metal atom are not involved in bonding. As a result, it can donate electrons and behave as a base.

On the other hand, in the case of a higher oxide of a transition metal, the metal atom has a high oxidation state. This means that the valence electrons are involved in bonding and so, they are unavailable. There is also a high effective nuclear charge.

As a result, it can accept electrons and behave as an acid.

(iii) +1.57 – Mn³⁺/Mn²⁺

–0.41 – Cr³⁺/Cr²⁺.

Mn²⁺ exists in 3d⁵ configuration which is half filled configuration and such half filled configurations provide extra stability to the Metal Ion. While Mn³⁺ will exist in 3d⁴ configuration which is less stable than 3d⁵.

Hence the conversion from 3+ to 2+ Oxidation state is very feasible. Hence E° value is +, depicting that ∆G° being -ve. And we know when ∆G° becomes -ve the reaction is feasible.

On the other hand, Cr²⁺ has 3d⁴ configuration , which can become 3d³ which is more stable, which is obtained by losing an electron by Cr⁺².

The reverse reaction is not that feasible, giving negative E° and positive G.

(B) Similarity: They are involved in the filling of (n-2) f orbitals. n = outermost orbital.

Difference: Lanthanoids are involved in the filling of 4f- orbitals whereas actinoids are involved in the filling of 5f-orbitals.

OR

(i) The variability of oxidation states arises out of incomplete filling of d orbitals in such a way that their oxidation states differ from each other by unity. In the case of non transition elements the oxidation states differ normally differ by a unit of 2.

(ii) copper (I) compounds are unstable in aqueous solution and undergo disproportionation.

2Cu⁺→ Cu⁺² + Cu

The stability of Cu⁺² (aq) rather than Cu⁺(aq) is due to the much more negative Δ_hydH^- of Cu2+ (aq) than Cu+, which more than compensates for the second ionisation enthalpy of Cu.

(iii) On treatment of Cr₂O₇^2- ion which is orange in colour, with alkali forms CrO₄^2- which is yellow in colour.

Cr₂O₇^2- + 2OH^- 2 CrO₄^2- + H₂O

(B) The complication arises due to the occurrence of a wide range of oxidation states in these elements and because their radioactivity creates special problems in their study.