Question

The transition metal ion that has 'spin-only' magnetic moment value of $5.96$ is

• A. $M{n}^{2+}$
• B. $F{e}^{2+}$
• C. $V^{2+}$
• D. $C{u}^{2+}$

Answer 1

Answer: (A)

Magnetic moment $(\mu )=\sqrt{N(N+2)}$
where $N=$ Number of unpaired electron. The electronic configuration of
$M{n}^{2+}$ is $[Ar]3d^5.$ It contains $5$ unpaired electron.
Hence, $\mu =\sqrt{5(5+2)}$
$\mu =\sqrt{35}=5.96B.M.$