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

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$Cu^{2+}$

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(B) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.$1$. $Co^{2+}$ $(3d^7)$: $n=3$,$\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87 	ext{ BM}$.$2$. $Cu^{2+}$ $(3d^9)$: $n=1$,$\mu = \sqrt{1(3)} = \sqrt{3} \approx 1.73 	ext{ BM}$.$3$. $Fe^{2+}$ $(3d^6)$: $n=4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 	ext{ BM}$.$4$. $Ni^{2+}$ $(3d^8)$: $n=2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 	ext{ BM}$.Comparing the values,$Cu^{2+}$ has the lowest number of unpaired electrons $(n=1)$,hence it exhibits the lowest spin-only magnetic moment.

Which among the following elements in their respective oxidation states develops the lowest spin-only magnetic moment? (Atomic numbers: $Cu=29, Fe=26, Ni=28, Co=27$)

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