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Magnetic moment of $\mathrm{Gd}^{3+}$ ion $(\mathrm{Z}=64)$ is
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The correct answer is:
7.9 BM
7.9 BM
$\mathrm{Gd}(64)=[\mathrm{Xe}] 4 f^7 5 d^1 6 s^2$
$\mathrm{Gd}^{3+}=[\mathrm{Xe}] 4 f^7 5 d^0 6 s^0$
i.e. no. of unpaired electrons $=7$
$$
\begin{aligned}
\mu & =\sqrt{n(n+2)}=\sqrt{7(7+2)} \\
& =\sqrt{63}=7.93 \mathrm{BM}
\end{aligned}
$$
$\mathrm{Gd}^{3+}=[\mathrm{Xe}] 4 f^7 5 d^0 6 s^0$
i.e. no. of unpaired electrons $=7$
$$
\begin{aligned}
\mu & =\sqrt{n(n+2)}=\sqrt{7(7+2)} \\
& =\sqrt{63}=7.93 \mathrm{BM}
\end{aligned}
$$
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