Which of the following pairs will have the sa | vedclass

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(B) The effective magnetic moment $(\mu_{eff})$ is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired e

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$Cr^{2+}$ and $Fe^{2+}$

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(B) The effective magnetic moment $(\mu_{eff})$ is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.For $Cr^{2+}$: Electronic configuration is $[Ar] 3d^4$. Number of unpaired electrons $(n)$ = $4$.For $Fe^{2+}$: Electronic configuration is $[Ar] 3d^6$. Number of unpaired electrons $(n)$ = $4$.Since both $Cr^{2+}$ and $Fe^{2+}$ have $4$ unpaired electrons,they will have the same effective magnetic moment.Therefore,the correct option is $B$.

Which of the following pairs will have the same effective magnetic moment?

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