Which of the following cations has the lowest | vedclass

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

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$V^{3+}$

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(A) The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$1$. For $V^{3+}$ $([Ar] 3d^2)$: $n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.$2$. For $Cr^{3+}$ $([Ar] 3d^3)$: $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.$3$. For $Mn^{2+}$ $([Ar] 3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.$4$. For $Fe^{2+}$ $([Ar] 3d^6)$: $n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.Comparing the values,$V^{3+}$ has the lowest number of unpaired electrons $(n=2)$,resulting in the lowest spin-only magnetic moment.

List-$I$	List-$II$
$(1) \ \text{Cr}$	$(A) \ [\text{Ar}] \ 3d^8 4s^0$
$(2) \ \text{Fe}^{2+}$	$(B) \ [\text{Ar}] \ 3d^{10} 4s^1$
$(3) \ \text{Ni}^{2+}$	$(C) \ [\text{Ar}] \ 3d^6 4s^0$
$(4) \ \text{Cu}$	$(D) \ [\text{Ar}] \ 3d^5 4s^1$
	$(E) \ [\text{Ar}] \ 3d^6 4s^2$

Which of the following cations has the lowest spin-only magnetic moment?

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