(a) $\mathrm{Cr}=25,\,[\mathrm{Ar}]3d^{4}4s^{2}\rightarrow$ Ground state structure $\mathrm{Cr}^{3+}=[\mathrm{Ar}]3d^{4}$ There are four unpaired electrons present in $\mathrm{Cr}^{3+}$. (b) $\mathrm{Cr}_2\mathrm{O}_7^{2-}+3\mathrm{H}_2\mathrm{S}+8\mathrm{H}^{+}$ $\rightarrow \underset{\text{cromiumion}}{2\mathrm{Cr}^{3+}}+\underset{\text{water}}{7\mathrm{H}_2\mathrm{O}}+\underset{\text{sulphur}}{3\mathrm{S}}$ (c) (i) Electronic configuration- $\mathrm{Mn}=25,\,[\mathrm{Ar}]3d^{5}4s^{2}\rightarrow\mathrm{Mn}^{2+}$ $=[\mathrm{Ar}]3d^{5}\rightarrow\text{ donating }$ $2e^{-}$ $\mathrm{Fe}=26,\,[\mathrm{Ar}]3d^{6}4s^{2}\rightarrow\mathrm{Fe}^{3+}$ $=[\mathrm{Ar}]3d^{5}\rightarrow\text{ donating }$ $3e^{-}$ Mn = after loosing 2 electron $3d$ orbital is half filled and $\mathrm{Mn}^{2+}$ is stable. $\mathrm{Fe}=\text{After}$ loosing three electrons $3d$ orbitals is half filled and more stable. That's why, $\mathrm{Fe}^{3+}$ is stable at +3 oxidation state and $\mathrm{Mn}^{2+}$ is stable in +2 oxidation state. (ii) $\mathrm{E}^{\circ}$ value of copper is +ve i.e, $0.34\,\mathrm{V}$. This is due to presence of high enthalpy of atomization and low enthalpy of hydrogen which make it exceptionally positive. (iii) Electronic configuration of $\mathrm{Eu}^{2+}[\mathrm{Xe}]4f^{7}6s^{2}$ It has the tendency to loose two electron and attain a stable half filled configuration. So, it oxidized by loosing 2 electrons and reduce to other species.