Transition and InnerTransition Elements - Result Question 42
42. The correct statement(s) about $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ is/are [atomic number of $\mathrm{Cr}=24$ and $\mathrm{Mn}=25$ ]
(2015 Adv.)
(a) $\mathrm{Cr}^{2+}$ is a reducing agent
(b) $\mathrm{Mn}^{3+}$ is an oxidising 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
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Answer:
Correct Answer: 42. (a, b, c)
Solution: In aqueous solution $\mathrm{Cr}^{2+}\left(3 d^4\right)$ acts as a reducing agent, oxidising itself to $\mathrm{Cr}^{3+}\left(3 d^3\right)$ that gives a completely half-field $t_{2 g}$ level in octahedral ligand field of $\mathrm{H}_2 \mathrm{O}$.
(b) $\mathrm{Mn}^{3+}\left(3 d^4\right)$ is an oxidising agent as it is reduced to $\mathrm{Mn}^{2+}\left(3 d^5\right)$, a completely half-filled stable configuration.
(c) Both $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ have $d^4$ configuration.
(d) $3 d^4 \mathrm{Cr}^{2+}(a q) \xrightarrow{\text { R.A }} \mathrm{Cr}^{3+}(a q)+e$
Hence $(d)$ is wrong statement.
BETA
