The correct statement(s) about $C{r}^{2+}$ and $M{n}^{3+}$ is(are) [Atomic numbers of $Cr=24$ and $Mn=25$]

The correct option is D $M{n}^{3+}$ is an oxidizing agent
$C{r}^{2+}$ is a reducing agent and $M{n}^{3+}$ is an oxidizing agent and both have electronic configuration $d^4$.

$E_{\frac{C{r}^{3+}}{C{r}^{2+}}}^0=0.41V$

$E_{\frac{M{n}^{3+}}{M{n}^{2+}}}^0=1.51V$

Above $E^{\circ }$ values explains reducing nature of $C{r}^{2+}$ and oxidizing behaviour of $M{n}^{3+}$.