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(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.$1$. Fo

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$Cu^{2+}$

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(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.$1$. For $Cu^{2+}$ $(Z=29)$: Electronic configuration is $[Ar] 3d^9$. Number of unpaired electrons $(n)$ = $1$.$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 	ext{ BM}$.$2$. For $Cr^{3+}$ $(Z=24)$: Electronic configuration is $[Ar] 3d^3$. Number of unpaired electrons $(n)$ = $3$.$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 	ext{ BM}$.$3$. For $Co^{2+}$ $(Z=27)$: Electronic configuration is $[Ar] 3d^7$. Number of unpaired electrons $(n)$ = $3$.$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 	ext{ BM}$.$4$. For $Fe^{2+}$ $(Z=26)$: Electronic configuration is $[Ar] 3d^6$. Number of unpaired electrons $(n)$ = $4$.$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 	ext{ BM}$.Comparing the values,$Cu^{2+}$ has the least magnetic moment.

Which cation from the following develops the least magnetic moment?

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