JEE MAIN - Chemistry (2024 - 31st January Morning Shift - No. 25)
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 $$\mathrm{Ni}: 28$$)
Answer
28
Explanation
$$\mathrm{NH}_3$$ act as WFL with $$\mathrm{Ni}^{2+}$$
$$\mathrm{Ni}^{2+}=3 \mathrm{~d}^8$$
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No. of unpaired electron $$=2$$
$$\begin{aligned} \mu=\sqrt{\mathrm{n}(\mathrm{n}+2)} & =\sqrt{8}=2.82 \mathrm{~BM} \\ & =28.2 \times 10^{-1} \mathrm{~BM} \\ \mathrm{x} & =28 \end{aligned}$$
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