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Use Hund's rule to derive the electronic configuration of \(\mathrm{Ce}^{3+}\) ion, and calculate its magnetic moment on the basis of 'spin-only' formula.
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\(\mathrm{Ce}(\mathrm{Z}=58)=[\mathrm{Xe}]^{54} 4 f^1 5 d^1 6 s^2\)
\(\begin{array}{lrl}
\therefore & \mathrm{Ce}^{3+} & =[\mathrm{Xe}]^{54} 4 f^1 \\
\therefore & \mu & =\sqrt{n(n+2)}
\end{array}\)
Here \(n=1\) i.e., only one unpaired electron.
\(\therefore \quad \mu=\sqrt{1(1+2)}=\sqrt{3}=1 \cdot 73 \mathrm{BM} .\)
\(\begin{array}{lrl}
\therefore & \mathrm{Ce}^{3+} & =[\mathrm{Xe}]^{54} 4 f^1 \\
\therefore & \mu & =\sqrt{n(n+2)}
\end{array}\)
Here \(n=1\) i.e., only one unpaired electron.
\(\therefore \quad \mu=\sqrt{1(1+2)}=\sqrt{3}=1 \cdot 73 \mathrm{BM} .\)
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