(i) ${\mathrm{KMnO}}_4$ can be prepared from pyrolusite $\left({\mathrm{MnO}}_2\right)\cdot {\mathrm{MnO}}_2$ is ignited with $\mathrm{KOH}$ in the presence of catalysts agents, such as oxygen from the air or ${\mathrm{KNO}}_3$ or ${\mathrm{KClO}}_4$ to give ${\mathrm{KMnO}}_4$.
$+2{\mathrm{H}}_2\mathrm{O}$
(ii) For the preparation of ${\mathrm{Na}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$, the yellow solution of sodium chromate $\left({\mathrm{Na}}_2{\mathrm{CrO}}_4\right)$ is acidified with sulphuric acid to give a solution from which orange sodium dichromate, ${\mathrm{Na}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\cdot 2{\mathrm{H}}_2\mathrm{O}$ can be crystallized.

(b) (i) Electronic configuration of ${\mathrm{Mn}}^{2+}$ is $\left[\mathrm{Ar}\right]3{\mathit{d}}^5$
Electronic configuration of ${\mathrm{Fe}}^{2+}$ is $\left[\mathrm{Ar}\right]3{\mathit{d}}^6$
It is known that the half-filled orbitals are more stable.
Therefore, $\mathrm{Mn}$ in +2 state has a stable ${\mathit{d}}^5$ configuration, shows resistance to the oxidation to ${\mathrm{Mn}}^{3+}$.
${\mathrm{Fe}}^{2+}$ has $3{\mathit{d}}^6$ configuration and by losing one electron, its configuration changes to a more stable $3{\mathit{d}}^5$ configuration and it gets oxidised to ${\mathrm{Fe}}^{3+}$ easily.
(ii) In all transition metals (except $\mathrm{Zn}$, electronic configuration: $3{\mathit{d}}^{10}4{\mathit{s}}^2$ ), there are some unpaired electrons that account for their stronger metallic bonding.
$\mathrm{Zn}$ has the least enthalpy of atomization because it lacks these unpaired electrons, which makes its inter-atomic electronic bonding the weakest.
(iii) Actinides exhibit larger oxidation states because of very small energy gap between $5\mathit{f},6\mathit{d}$ and $7\mathit{s}$ sub-shells.
Since, all these sub-shells have similar $\left(\mathit{n}+\mathit{l}\right)$ value, therefore all can be involved in bonding resulting in a larger oxidation number for actinoids.
OR
(i) Manganese ([Ar] $3{\mathit{d}}^{5}4{\mathit{s}}^2)$ shows maximum number of oxidation state as its atoms have five unpaired electrons in $3\mathit{d}$ orbitals. It shows all the oxidation state from +2 to +7 .
(ii) $\mathrm{Cu}$ has positive ${\mathrm{E}}_{\mathrm{O}}\frac{{\mathrm{M}}^{2+}}{\mathrm{M}}$ value, because the sum of enthalpies of sublimation and ionization is not balanced by hydration enthalpy.
(iii) ${\mathrm{Mn}}^{3+}$ is stronger oxidising agent as the charge from ${\mathrm{Mn}}^{3+}$ to ${\mathrm{Mn}}^{2+}$ results in half filled, ${\mathit{d}}^5$ configuration which has extra stability.
(iv) Europium, $\left({\mathrm{Eu}}^{2+}\right)$ is formed by losing the two $5\mathrm{}\mathit{s}$ electrons and its electronic configuration becomes [Xe $]4{\mathit{f}}^7$ which is quite stable configuration.
(v) ${\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+8{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}$ $+4{\mathrm{H}}_2\mathrm{O}$