The correct statements about \(Cr^{2+}\) and \(Mn^{3+}\) are: (atomic number of Cr = 24 and Mn = 25)
(A) \(Cr^{2+}\) is a reducing agent
(B) \(Mn^{3+}\) is an oxidising agent
(C) Both \(Cr^{2+}\) and \(Mn^{3+}\) exhibit \(d^4\) electronic configuration
(D) When \(Cr^{2+}\) is used as reducing agent chromium ion attains \(d^5\) electronic configuration
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B), (C) only
Cr (Z = 24): [Ar]3d54s1, so \( \ce{Cr^{2+}} \) = [Ar]3d4. Mn (Z = 25): [Ar]3d54s2, so \( \ce{Mn^{3+}} \) = [Ar]3d4.
(A) Cr2+ is a reducing agent — True. \( \ce{Cr^{2+}} \) loses an electron to become \( \ce{Cr^{3+}} \) (d3), which has a half-filled t2g level and is very stable.
(B) Mn3+ is an oxidising agent — True. \( \ce{Mn^{3+}} \) gains an electron to become \( \ce{Mn^{2+}} \) (d5), the extra-stable half-filled configuration.
(C) Both have d4 configuration — True. Both ions are 3d4, as worked out above.
(D) Cr2+ as a reducing agent attains d5 — False. A reducing agent is oxidised: \( \ce{Cr^{2+}} \) → \( \ce{Cr^{3+}} \) + e−, so the ion goes from d4 to d3, not d5.
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 t2g 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.”