Question

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

• A. $ Mn^{2+} $
• B. $ Fe^{2+} $
• C. $ V^{2+} $
• D. $ Cu^{2+} $

Answer 1

Answer: (A)

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