The valence electronic configuration of ${\mathrm{Mn}}^{3+}$ is $3{\mathrm{d}}^4$ and that of ${\mathrm{Mn}}^{2+}$ is stable $3{\mathrm{d}}^5$.The valence electronic configuration of ${\mathrm{Fe}}^{3+}$ is $3{\mathrm{d}}^5$ and that of ${\mathrm{Fe}}^{2+}$ is $3{\mathrm{d}}^6$.

The reduction electrode potential of ${\mathrm{Mn}}^{3+}/{\mathrm{Mn}}^{2+}$ is $+1.57 \mathrm{V}$ while that of ${\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}$ is $+0.77 \mathrm{V}$, hence A is wrong. Higher oxidation state of smaller $\mathrm{d}$-block elements is stabilised (or say form compounds) with smaller anion oxide that can be explained by stearic reason hence $\mathrm{B}$ is correct.

The oxidation electrode potential of ${\mathrm{Cr}}^{2+}/{\mathrm{Cr}}^{3+}$ is $+0.41 \mathrm{V}$ hence, it can reduce ${\mathrm{H}}^{+}$ and so liberate ${\mathrm{H}}_2$.

The unpaired electrons in ${\mathrm{V}}^{2+}$ are $3$ hence, the magnetic moment of ${\mathrm{V}}^{2+}$ will be lesser than $4.4 \mathrm{BM}$.

Hence, only $\mathrm{B}$ and $\mathrm{C}$ are correct.