Cr^2+ is reducing but Mn^3+ is oxidizing in nature although both have…

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

Answer & explanation

Correct answer: option 1

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

Whether an ion acts as a reducing or an oxidising agent depends on which change of configuration gives it more stability. \( \ce{Cr^{2+}} \) and \( \ce{Mn^{3+}} \) are both d⁴ ions, but they move in opposite directions:

Cr²⁺ (Cr: [Ar] 3d⁵4s¹; \( \ce{Cr^{2+}} \): [Ar] 3d⁴) loses one electron to become Cr³⁺, [Ar] 3d³. In an octahedral field the three electrons half-fill the t_2g level (t_2g³), a stable arrangement. Because \( \ce{Cr^{2+}} \) gives up an electron, it is a reducing agent: \( \ce{Cr^2+ -> Cr^3+ + e-} \).
Mn³⁺ (Mn: [Ar] 3d⁵4s²; \( \ce{Mn^{3+}} \): [Ar] 3d⁴) gains one electron to become Mn²⁺, [Ar] 3d⁵, the half-filled d subshell. Because \( \ce{Mn^{3+}} \) takes up an electron, it is an oxidising agent: \( \ce{Mn^3+ + e- -> Mn^2+} \).

So the change for chromium is d⁴ to d³.

Why the other options are incorrect

Option (2) d⁴ to d⁵ — Incorrect. This is the change for \( \ce{Mn^{3+}} \) → \( \ce{Mn^{2+}} \), which explains why \( \ce{Mn^{3+}} \) is oxidising.
Option (3) d² to d³ — Incorrect. \( \ce{Cr^{2+}} \) is d⁴, not d²; a d² → d³ change would be the reduction of \( \ce{V^{3+}} \) to \( \ce{V^{2+}} \).
Option (4) d⁹ to d¹⁰ — Incorrect. This is the reduction of \( \ce{Cu^{2+}} \) to \( \ce{Cu^{+}} \); it has nothing to do with chromium.

NCERT (2026-27 edition):
Trends in the M²⁺/M Standard Electrode Potentials – “… Cr²⁺ is reducing as its configuration changes from d⁴ to d³, the latter having a half-filled t_2g level (see Unit 5). On the other hand, the change from Mn³⁺ to Mn²⁺ results in the half-filled (d⁵) configuration which has extra stability.”