KCET · Chemistry · Structure of Atom
The spin only magnetic moment of \(\mathrm{Mn}^{4+}\) ion is nearly
- A \(3 \mathrm{BM}\)
- B \(6 \mathrm{BM}\)
- C \(4 \mathrm{BM}\)
- D \(5 \mathrm{BM}\)
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, \(\mu=\sqrt{n(n+2)}\) \(\therefore\)
\[
n=3
\]
\[
\begin{aligned}
\mu &=\sqrt{3(3+2)} \\
&=\sqrt{15}=3.87
\end{aligned}
\]
\(=4 \mathrm{BM}\)
\[
\begin{gathered}
{ }_{25} \mathrm{Mn}=[\mathrm{Ar}] 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{2} \\
\mathrm{Mn}^{4+}=[\mathrm{Ar}] 3 \mathrm{~d}^{3}
\end{gathered}
\]
Thus, three unpaired electrons are present.
\(\therefore\) Spin only magnetic moment, \(\mu=\sqrt{n(n+2)}\) \(\therefore\)
\[
n=3
\]
\[
\begin{aligned}
\mu &=\sqrt{3(3+2)} \\
&=\sqrt{15}=3.87
\end{aligned}
\]
\(=4 \mathrm{BM}\)
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