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(D) The magnetic moment (spin-only) is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For a magnetic mo

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$Cr^{2+}$

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(D) The magnetic moment (spin-only) is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For a magnetic moment of approximately $5 \ BM$,we need $\sqrt{n(n+2)} \approx 5$,which implies $n(n+2) \approx 25$. This corresponds to $n = 5$.Let us calculate the number of unpaired electrons for each ion:$1. V^{2+} (Z=23): [Ar] 3d^3$,$n=3$,$\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87 \ BM$.$2. Ti^{2+} (Z=22): [Ar] 3d^2$,$n=2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \ BM$.$3. Mn^{2+} (Z=25): [Ar] 3d^5$,$n=5$,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92 \ BM$.$4. Cr^{2+} (Z=24): [Ar] 3d^4$,$n=4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \ BM$.Since $4.90 \ BM$ is closest to $5 \ BM$,the correct answer is $Cr^{2+}$.

Which of the following paramagnetic ions has a magnetic moment (spin-only) of the order of $5 \ BM$? $(Mn = 25, Cr = 24, V = 23, Ti = 22)$

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