(a) (I) (i) Because ${\mathit{M}\mathit{n}}^{3+}$ has the outer electronic configuration of has the outer electron configuration of $3{\mathit{d}}^5$. Thus the conversion is favourable.
However in case of ${\mathit{C}\mathit{r}}^{3+}∕{\mathit{C}\mathit{r}}^{2+}$ undergoes a change in outer configuration from $3{\mathit{d}}^3$ to $3{\mathit{d}}^4$ which is not stable.
(ii) Due to absence of unpaired electron ${\mathit{S}\mathit{c}}^{3+}$ is colourless and due to presence of one unpaired electron d-d transition takes place making ${\mathit{T}\mathit{i}}^{3+}$ coloured in nature.
(iii) It is due to comparable energies of $5\mathit{f},6\mathit{d}$ and $7\mathit{s}$ orbitals.
(II) $2{\mathit{M}\mathit{n}\mathit{O}}_2+4\mathit{K}\mathit{O}\mathit{H}+{\mathit{O}}_2\stackrel{\mathrm{\Delta }}{\mathit{─}\mathit{─}\mathit{─}\mathit{─}▸}$ $2{\mathit{K}}_2{\mathit{M}\mathit{n}\mathit{O}}_4+2{\mathit{H}}_2\mathit{O}$
$3{\mathit{K}}_2{\mathit{M}\mathit{n}\mathit{O}}_4\stackrel{4\mathit{H}\mathit{C}\mathit{l}}{\mathit{─}\mathit{─}\mathit{─}\mathit{─}▸}2{\mathit{K}\mathit{M}\mathit{n}\mathit{O}}_4$ $+{\mathit{M}\mathit{n}\mathit{O}}_2+2{\mathit{H}}_2\mathit{O}+4\mathit{K}\mathit{C}\mathit{l}$
OR (b) (I) (i) The atomic radii of the transition elements in any series are not much different from each other.
As a result they can very easily replace each other in the lattice and form alloys.
(ii) ${\mathit{C}\mathit{e}}^{4+}$ has the tendency to accept one electron to get the +3 oxidation state, hence ${\mathit{C}\mathit{e}}^{+4}$ is a good oxidising agent.
(II) In case of lanthonoids, differentiating electron enters in $4\mathit{f}$ orbital whereas in case of Actinoids it enters in $5\mathit{f}$ orbital.
They both have $\left(3+\right)$ as their most common oxidation state.
(III) ${\mathit{C}\mathit{r}}_2{\mathit{O}}_7^{2-}+2{\mathit{O}\mathit{H}}^{-}\mathrm{}\mathit{─}\mathit{─}\mathit{─}\mathit{─}▸\mathrm{}2\mathrm{}{\mathit{C}\mathit{r}\mathit{O}}_4^{2-}+\mathrm{}{\mathit{H}}_2\mathrm{}\mathit{O}$