The metal ion (in gaseous state) with the low | vedclass

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(B) The spin-only magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$, where $n$ is the number of unpaired electrons.To find th

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$Ni^{2+}$

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(B) The spin-only magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$, where $n$ is the number of unpaired electrons.To find the lowest magnetic moment, we need to identify the ion with the minimum number of unpaired electrons $(n)$.$1$. $V^{2+}$ $(Z=23)$: Electronic configuration is $[Ar] 3d^3$. Number of unpaired electrons $(n)$ = $3$.$2$. $Ni^{2+}$ $(Z=28)$: Electronic configuration is $[Ar] 3d^8$. Number of unpaired electrons $(n)$ = $2$.$3$. $Cr^{2+}$ $(Z=24)$: Electronic configuration is $[Ar] 3d^4$. Number of unpaired electrons $(n)$ = $4$.$4$. $Fe^{2+}$ $(Z=26)$: Electronic configuration is $[Ar] 3d^6$. Number of unpaired electrons $(n)$ = $4$.Since $Ni^{2+}$ has the lowest number of unpaired electrons $(n=2)$, it has the lowest spin-only magnetic moment value.

The metal ion (in gaseous state) with the lowest spin-only magnetic moment value is .... .

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