Question

How would you account for the following:

A. Of the $d^4$ species, $C{r}^{2+}$ is strongly reducing while manganese $\left(III\right)$ is strongly oxidising.

B. Cobalt $\left(II\right)$ is stable in aqueous solution but in the presence of complexing reagents it is easily oxidized.

C. The $d^1$ configuration is very unstable in ions.

Answer 1

$A.C{r}^{2+}$ is strongly reducing in nature. It has $d^4$ configuration. By losing one electron it gets oxidised to $C{r}^{3+}$ (electronic configuration $d^3$ ) which can be written as $t_2^{3}g$ and it is a more stable configuration. On the other hand, $M{n}^{3+}$ has also $d^4$ configuration and by accepting one electron, it gets reduced to $M{n}^{2+}$ (electronic configuration $d^5$ ) and act as strong oxidising agent. It is due to extra stability of half -filled d-orbital in $M{n}^{2+}$.

${}^{}B.Co\left(II\right)$ is stable in aqueous solutions. However, in the presence of strong field complexing reagents, it is oxidized to $Co\left(III\right)$.
Although the $3^{rd}$ ionization energy for $Co$ is high, but the higher amount of crystal field stabilization energy $\left(CFSE\right)$ in $d^6$ configuration released in the presence of strong field ligands overcomes this ionization energy.

$C.$ The ions in $d^1$ configuration tend to lose one more electron to get into more 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.