JEE Advanced 2015 · Paper 1 · Chemistry Q34

ChemistryMultiple correct+4 / -2Question 34 of 60

The correct statement(s) about $\ce{Cr^{2+}}$ and $\ce{Mn^{3+}}$ is(are)

[Atomic numbers of Cr = 24 and Mn = 25]

\ce{Cr^{2+}}

is a reducing agent

\ce{Mn^{3+}}

is an oxidizing agent

Both

\ce{Cr^{2+}}

and

\ce{Mn^{3+}}

exhibit

d^4

electronic configuration

When

\ce{Cr^{2+}}

is used as a reducing agent, the chromium ion attains

d^5

electronic configuration

Answer and worked solution

Official answer: A, B, C

(A) Cr²⁺ ( $d^4$ ) is oxidised to the stable Cr³⁺ ( $t_{2g}^3$ ). It is a reducing agent. True.
(B) Mn³⁺ ( $d^4$ ) is reduced to the stable Mn²⁺ ( $d^5$ ). It is an oxidising agent. True.
(C) Both are $d^4$ . True.
(D) Cr²⁺ becomes Cr³⁺, which is $d^3$ , not $d^5$ . False.

(A), (B), (C)

Topics:Transition element properties

JEE Advanced 2015 · Paper 1, Chemistry question 34