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The spin only magnetic moment of $\mathrm{Ni}^{2+}($ in $\mathrm{BM})$ in aqueous solution will be
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Verified Answer
The correct answer is:
$2.84$
Electronic configuration of $\mathrm{Ni}=[\mathrm{Ar}] 3 d^{8} 4 s^{2}$
$\therefore$ Number of unpaired electrons $=2$
$$
\begin{aligned}
\therefore \text { Magnetic moment }(\mu) &=\sqrt{n(n+2)} \\
&=\sqrt{2(2+2)}=\sqrt{8}=2.84 .
\end{aligned}
$$
$\therefore$ Number of unpaired electrons $=2$
$$
\begin{aligned}
\therefore \text { Magnetic moment }(\mu) &=\sqrt{n(n+2)} \\
&=\sqrt{2(2+2)}=\sqrt{8}=2.84 .
\end{aligned}
$$
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