(a) (i) Mn exhibits highest oxidation states (upto +7 ) in the oxides, for example $\mathrm{Mn}_2\mathrm{O}_7$ in comparison to $\mathrm{Mn}$ fluorides, $\mathrm{MnF}_4$ as oxygen is able to form multiple bond with metal. And also fluorine does not have $d$-orbital. (ii) $\mathrm{Cr}^{2+}$ is a strong reducing agent as its electrode potential is more negative. Also, it can loose on electron to become $\mathrm{Cr}^{3+}$ which has half-filled $d$-orbital which is its stable state. (iii) Electronic configuration : $\,{}_{29}\mathrm{Cu}=[\mathrm{Ar}]\,3d^{10}4s^1\,\mathrm{Cu}^{2+}$ $=[\mathrm{Ar}]\,3d^9$ $\,{}_{30}\mathrm{Zn}=[\mathrm{Ar}]\,3d^{10}4s^2\,\mathrm{Zn}^{2+}$ $=[\mathrm{Ar}]\,3d^{10}$ (i) $\underset{\text{Manganese dioxide}}{2\mathrm{MnO}_2} + \underset{\text{ Potassium hydroxide}}{4\mathrm{KOH}}$ $+ \mathrm{O}_2 \xrightarrow{\Delta}$ $\underset{\text{Potassium manganate}}{2\mathrm{K}_2\mathrm{MnO}_4} + 2\mathrm{H}_2\mathrm{O}$ (ii) $\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6\mathrm{I}^- \rightarrow 2\mathrm{Cr}^{3+}$ $+ 3\mathrm{I}_2 + 7\mathrm{H}_2\mathrm{O}$ OR (i) Manganese exhibits greatest number of oxidation states as it is able to acquire half filled to completely filled configurations. Thus Mn can loose or share electrons from both the orbitals. (ii) $\mathrm{Cr}$ has the highest melting point. As the number unpaired electrons increases upto $d^5$ configuration, it results in the increase in the strength of metallic bonds. To break the metallic bond, significant energy is required thus $\mathrm{Cr}$ with highest number of unpaired electrons (6) has the highest melting point. (iii) Sc shows only +3 oxidation state as if has too few electrons to lose thus, looses one electron from $3d$-electron to attain +3 oxidation state. (iv) $\mathrm{Co}$ is the strongest oxidising agent as the $E$ value for the redox couple $\mathrm{M}^{3+} / \mathrm{M}^{2+}$ is highest.