JEE MAIN - Chemistry (2022 - 29th June Evening Shift - No. 19)
Spin only magnetic moment of [MnBr6]4$$-$$ is _________ B.M. (round off to the closest integer)
Answer
6
Explanation
$\begin{aligned} & {\left[\mathrm{MnBr}_6\right]^{4-}} \\\\ & x-6=-4\end{aligned}$
$$ \begin{aligned} x & =+2 \\\\ \mathrm{Mn} & =[\mathrm{Ar}] 3 d^5 4 s^2 \\\\ \mathrm{Mn}^{2+} & =[\mathrm{Ar}] 3 d^5 4 s^0 \end{aligned} $$
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So, number of unpaired electrons $(n)=5$
$$ \begin{aligned} \mu & =\sqrt{n(n+2)} \\\\ \mu & =\sqrt{5(5+2)} \\\\ & =\sqrt{35} \\\\ & =5.91 \text { B.M. } \\\\ & \approx 6 \text { B.M. } \end{aligned} $$
$$ \begin{aligned} x & =+2 \\\\ \mathrm{Mn} & =[\mathrm{Ar}] 3 d^5 4 s^2 \\\\ \mathrm{Mn}^{2+} & =[\mathrm{Ar}] 3 d^5 4 s^0 \end{aligned} $$
So, number of unpaired electrons $(n)=5$
$$ \begin{aligned} \mu & =\sqrt{n(n+2)} \\\\ \mu & =\sqrt{5(5+2)} \\\\ & =\sqrt{35} \\\\ & =5.91 \text { B.M. } \\\\ & \approx 6 \text { B.M. } \end{aligned} $$
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