How would you account for the following:
$(i)$ Of the $d^{4}$ species,$Cr^{2+}$ is strongly reducing while $Mn^{3+}$ is strongly oxidising.
$(ii)$ $Co^{2+}$ is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.
$(iii)$ The $d^{1}$ configuration is very unstable in ions.

(N/A) $(i)$ $Cr^{2+}$ is strongly reducing in nature. It has a $d^{4}$ configuration. While acting as a reducing agent,it gets oxidized to $Cr^{3+}$ (electronic configuration,$d^{3}$). This $d^{3}$ configuration can be written as $t_{2g}^{3}$ configuration,which is a more stable configuration. In the case of $Mn^{3+} (d^{4})$,it acts as an oxidizing agent and gets reduced to $Mn^{2+} (d^{5})$. This has an exactly half-filled $d$-orbital and is highly stable.
$(ii)$ $Co^{2+}$ is stable in aqueous solutions. However,in the presence of strong field complexing reagents,it is oxidized to $Co^{3+}$. Although the $3^{rd}$ ionization energy for $Co$ is high,the higher amount of crystal field stabilization energy $(CFSE)$ released in the presence of strong field ligands overcomes this ionization energy.
$(iii)$ The ions in $d^{1}$ configuration tend to lose one more electron to get into stable $d^{0}$ configuration. Also,the hydration or lattice energy is more than sufficient to remove the only electron present in the $d$-orbital of these ions. Therefore,they act as reducing agents.