Q24 / 180 · KCET 2010 · Chemistry
The spin only magnetic moment of $\mathrm{Mn}^{4+}$ ion is nearly
- A$3 \mathrm{BM}$
- B$6 \mathrm{BM}$
- Correct answerC$4 \mathrm{BM}$
- D$5 \mathrm{BM}$
Solution
The electronic configuration of $\mathrm{Mn}$ is
$$
\begin{gathered}
{ }_{25} \mathrm{Mn}=[\mathrm{Ar}] 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{2} \\
\mathrm{Mn}^{4+}=[\mathrm{Ar}] 3 \mathrm{~d}^{3}
\end{gathered}
$$
Thus, three unpaired electrons are present.
$\therefore$ Spin only magnetic moment, $\mu=\sqrt{n(n+2)}$ $\therefore$
$$
n=3
$$
$$
\begin{aligned}
\mu &=\sqrt{3(3+2)} \\
&=\sqrt{15}=3.87
\end{aligned}
$$
$=4 \mathrm{BM}$
Hence, the correct answer is option: 'C'.