(a) (I) (i) Because $\mathrm{Mn}^{3+}$ has the outer electronic configuration of $3 d^4 \& \mathrm{Mn}^{2+}$ has the outer electron configuration of $3 d^5$. Thus the conversion is favourable. However in case of $\mathrm{Cr}^{3+}/\mathrm{Cr}^{2+}$ undergoes a change in outer configuration from $3 d^3$ to $3 d^4$ which is not stable. (ii) Due to absence of unpaired electron $\mathrm{Sc}^{3+}$ is colourless and due to presence of one unpaired electron d-d transition takes place making $\mathrm{Ti}^{3+}$ coloured in nature. (iii) It is due to comparable energies of $5 f, 6 d$ and $7 s$ orbitals. (II) $2\mathrm{MnO}_2 + 4\mathrm{KOH} + \mathrm{O}_2 \xrightarrow{\Delta}$ $2\mathrm{K}_2\mathrm{MnO}_4 + 2\mathrm{H}_2\mathrm{O}$ $3\mathrm{K}_2\mathrm{MnO}_4 \xrightarrow{4\mathrm{HCl}} 2\mathrm{KMnO}_4$ $+\mathrm{MnO}_2 + 2\mathrm{H}_2\mathrm{O} + 4\mathrm{KCl}$ 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) $\mathrm{Ce}^{4+}$ has the tendency to accept one electron to get the +3 oxidation state, hence $\mathrm{Ce}^{+4}$ is a good oxidising agent. (II) In case of lanthonoids, differentiating electron enters in $4 f$ orbital whereas in case of Actinoids it enters in $5 f$ orbital. They both have $(3+)$ as their most common oxidation state. (III) $\mathrm{Cr}_2\mathrm{O}_7^{2-} + 2\mathrm{OH}^- \rightarrow 2\mathrm{CrO}_4^{2-} + \mathrm{H}_2\mathrm{O}$