The 'Spin only' Magnetic moment for $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}$ is $\_\_\_\_$ $\times 10^{-1} \mathrm{BM}$. ( given = Atomic number of Ni: 28)
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Solution
$\mathrm{NH}_3$ act as WFL with $\mathrm{Ni}^{2+}$
$$\mathrm{Ni}^{2+}=3 \mathrm{~d}^8
$$
No. of unpaired electron $=2$
$$
\begin{aligned}
\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)} & =\sqrt{8}=2.82 \mathrm{BM} \\
& =28.2 \times 10^{-1} \mathrm{BM} \\
\mathrm{x} & =28
\end{aligned}
$$
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No. of unpaired electron $=2$
$$
\begin{aligned}
\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)} & =\sqrt{8}=2.82 \mathrm{BM} \\
& =28.2 \times 10^{-1} \mathrm{BM} \\
\mathrm{x} & =28
\end{aligned}
$$