Cr^{2+} has a magnetic moment same as that of | vedclass

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(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$, where $n$ is the number of unpaired electrons.For $Cr^{2+}

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$Fe^{2+}$

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(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$, where $n$ is the number of unpaired electrons.For $Cr^{2+}$ $(Z=24)$: Electronic configuration is $[Ar] 3d^4$. Number of unpaired electrons $(n)$ = $4$.For $Fe^{2+}$ $(Z=26)$: Electronic configuration is $[Ar] 3d^6$. Number of unpaired electrons $(n)$ = $4$.For $Cu^{2+}$ $(Z=29)$: Electronic configuration is $[Ar] 3d^9$. Number of unpaired electrons $(n)$ = $1$.For $Co^{2+}$ $(Z=27)$: Electronic configuration is $[Ar] 3d^7$. Number of unpaired electrons $(n)$ = $3$.For $Cu^{+}$ $(Z=29)$: Electronic configuration is $[Ar] 3d^{10}$. Number of unpaired electrons $(n)$ = $0$.Since $Cr^{2+}$ and $Fe^{2+}$ both have $4$ unpaired electrons, they have the same magnetic moment.

$Cr^{2+}$ has a magnetic moment same as that of

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