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KCET · Chemistry · Structure of Atom

The spin only magnetic moment of \(\mathrm{Mn}^{4+}\) ion is nearly

  1. A \(3 \mathrm{BM}\)
  2. B \(6 \mathrm{BM}\)
  3. C \(4 \mathrm{BM}\)
  4. D \(5 \mathrm{BM}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(4 \mathrm{BM}\)

Step-by-step Solution

Detailed explanation

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, \(\mu=\sqrt{n(n+2)}\) \(\therefore\)
\[
n=3
\]
\[
\begin{aligned}
\mu &=\sqrt{3(3+2)} \\
&=\sqrt{15}=3.87
\end{aligned}
\]
\(=4 \mathrm{BM}\)