The spin-only magnetic moment of [ZCl_{4}]^{2 | vedclass

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(C) The spin-only magnetic moment of a complex is calculated using the formula: $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electr

Vedclass · Accepted Answer

$Co$

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(C) The spin-only magnetic moment of a complex is calculated using the formula: $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.The oxidation state of $Z$ in $[ZCl_{4}]^{2-}$ is $+2$.The spin-only magnetic moments for the given metals in the $+2$ oxidation state are:$(a)$ $Mn^{2+}$: Electronic configuration is $3d^{5}$,$n=5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, BM$.$(b)$ $Ni^{2+}$: Electronic configuration is $3d^{8}$,$n=2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \, BM$.$(c)$ $Co^{2+}$: Electronic configuration is $3d^{7}$,$n=3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, BM$.$(d)$ $Cu^{2+}$: Electronic configuration is $3d^{9}$,$n=1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \, BM$.Since the given magnetic moment is $3.87 \, BM$,$Z$ must be $Co$.

The spin-only magnetic moment of $[ZCl_{4}]^{2-}$ is $3.87 \, BM,$ where $Z$ is $.....$

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