${\mathrm{Cr}}^{2+}$ and ${\mathrm{Mn}}^{3+}$ both have${\mathrm{d}}^4$ configuration. So,

${\mathrm{Cr}}^{2+}$ and ${\mathrm{Mn}}^{3+}$ both have${\mathrm{d}}^4$ configuration. So, ${\mathrm{Cr}}^{2+}$ is reducing, ${\mathrm{Mn}}^{3+}$ is oxidizing.

When ${\mathrm{Cr}}^{2+}$ is oxidized to ${\mathrm{Cr}}^{3+}$, its conformation changes from ${\mathrm{d}}^4$to ${\mathrm{d}}^3$, making it a reducing agent. The ${\mathrm{d}}^3$ arrangement has a fairly steady half-filled $t_{2g}$ level.