One of the notable features of the transition element is the great variety of oxidation states it may show in its compounds. The elements which give the greatest number of oxidation states occur in or near the middle of the series. Manganese, for example exhibits all the oxidation states from +2 to +7. The lesser number of oxidation states at the extreme ends is either due to the loss or sharing of a few electrons (Sc, Ti) or too many d electrons of higher valence. Early in the first series scandium (II) is virtually unknown and titanium (IV) which is more stable than Ti (III) or Ti (II). On the other end, the only oxidation state of zinc is +2.

\(Cr^{2+}\) is reducing but \(Mn^{3+}\) is oxidizing in nature although both have \(d^4\) configuration. It is due to the reason that \(Cr\) configuration changes from

\(d^4 \text{ to }d^3\)

The correct answer is (1) \(d^4 \text{ to }d^3\).

The electron configuration of \( \text{Cr}^{2+} \) is \( 3d^4 \). It can lose an electron to form \( \text{Cr}^{3+} \), which has the stable \( 3d^3 \) configuration (as it has a half-filled \( t_{2g} \) level, explained in the unit). Hence, it acts as a reducing agent. On the other hand, \( \text{Mn}^{3+} \) also has the \( 3d^4 \) configuration, but it can gain an electron to form \( \text{Mn}^{2+} \), which has the stable \( 3d^5 \) configuration (as it is exactly half-filled). Hence, it acts as an oxidizing agent.

Alternatively, the standard reduction potential (\( E^\circ \)) value for \( \text{Cr}^{3+}/\text{Cr}^{2+} \) is negative (\(-0.41 \, \text{V}\)), whereas the \( E^\circ \) value for \( \text{Mn}^{3+}/\text{Mn}^{2+} \) is positive (\(+1.57 \, \text{V}\)). Hence, \( \text{Cr}^{2+} \) ion can easily undergo oxidation to give \( \text{Cr}^{3+} \) ion and, therefore, acts as a strong reducing agent, whereas \( \text{Mn}^{3+} \) can easily undergo reduction to give \( \text{Mn}^{2+} \) and hence acts as an oxidizing agent.