Question

The correct statement(s) about $C{r}^{2+}$ and $M{n}^{3+}$ is(are):

$[AtomicnumbersofCr=24andMn=25]$

• A. $C{r}^{2+}$ is a reducing agent
• B. $M{n}^{3+}$ is an oxidizing agent
• C. Both $C{r}^{2+}$ and $M{n}^{3+}$ exhibit $d^4$ electronic configuration
• D. When $C{r}^{2+}$ is used as a reducing agent, the chromium ion attains $d^5$ electronic configuration

Answer 1

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

The correct options are
A $C{r}^{2+}$ is a reducing agent
B $M{n}^{3+}$ is an oxidizing agent
C Both $C{r}^{2+}$ and $M{n}^{3+}$ exhibit $d^4$ electronic configuration

(A) $C{r}^{2+}$ is a reducing agent. It itself loses electrons (and hence acts as a reducing agent) and forms higher oxidation states. In fact, +6 oxidation state of Cr is very stable.

Hence, the option (A) is correct.

(B) $M{n}^{3+}$ is an oxidizing agent. It itself gains electrons (and hence acts as an oxidizing agent) to form $M{n}^{2+}$ with $3d^5$ electronic configuration which is stable as it is half-filled.

Hence, the option (B) is correct.

(C) Cr and Mn have outer ionic configurations of $3d^{5}4s^1$ and $3d^{5}4s^2$ respectively.
Hence, both $C{r}^{2+}$ and $M{n}^{3+}$ exhibit $d^4$ electronic configuration.

Hence, the option (C) is correct.

(D) When $C{r}^{2+}$ is used as a reducing agent, the chromium ion attains $d^3$ (as in $C{r}^{3+}$) electronic configuration.

Hence, the option (D) is incorrect.