The correct statement(s) about $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ is(are) [Atomic numbers of $\mathrm{Cr}=24$ and $\mathrm{Mn}=25$]

(A) $\mathrm{Cr}^{2+}$ is a reducing agent; (B) $\mathrm{Mn}^{3+}$ is an oxidizing agent; (C) Both $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ exhibit $d^{4}$ electronic configuration

(D) When $\mathrm{Cr}^{2+}$ is used as a reducing agent, the chromium ion attains $d^{5}$ electronic configuration

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · d and f Block Elements

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The correct statement(s) about

Cr^{2 +}

and

Mn^{3 +}

is(are)
[Atomic numbers of

Cr = 24

and

Mn = 25

]

(A) $Cr^{2 +}$ is a reducing agent✓ Correct
(B) $Mn^{3 +}$ is an oxidizing agent✓ Correct
(C)Both $Cr^{2 +}$ and $Mn^{3 +}$ exhibit $d^{4}$ electronic configuration✓ Correct
(D)When $Cr^{2 +}$ is used as a reducing agent, the chromium ion attains $d^{5}$ electronic configuration

Answer: (A), (B), (C)

Solution

Electronic configurations:

$Cr^{2 +}$ : [Ar]3d⁴ (loses 2 electrons from 4s²)
$Mn^{3 +}$ : [Ar]3d⁴ (loses 3 electrons)

(A) Correct:

Cr^{2 +}

can be oxidized to

Cr^{3 +}

(d³, stable half-filled t₂g), so it acts as a reducing agent.

(B) Correct:

Mn^{3 +}

can be reduced to

Mn^{2 +}

(d⁵, stable half-filled d-orbital), so it acts as an oxidizing agent.

(C) Correct: Both have d⁴ configuration.

(D) Incorrect: When

Cr^{2 +}

(d⁴) is oxidized, it becomes

Cr^{3 +}

(d³), not d⁵.

Correct answers: (A), (B) and (C).

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JEE Advanced · 2015 · 24 May 2015 · Paper 1 · d and f Block Elements

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