What is the magnetic dipole moment of \(Mn^{2+}\;(3d^{5})\)?
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Always remember the spin-only formula:
\[
\boxed{\mu=\sqrt{n(n+2)}\;BM}
\]
where \(n\) is the number of unpaired electrons.
For \(d^{5}\) configuration,
\[
\boxed{n=5,\qquad \mu=\sqrt{35}=5.92\,BM.}
\]
Mn$^{2+}$ has the configuration $[Ar]3d^5$ with all five $d$ electrons unpaired. The spin-only magnetic moment is $\mu = \sqrt{n(n+2)}$ BM $= \sqrt{5 \times 7} = \sqrt{35} \approx 5.92$ BM.
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