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

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$Cr^{2+}$

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(D) The magnetic moment (spin only) is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$1$. For $Mn^{2+}$ $(d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.$2$. For $Ti^{2+}$ $(d^2)$: $n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.$3$. For $V^{2+}$ $(d^3)$: $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.$4$. For $Cr^{2+}$ $(d^4)$: $n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.The value $4.90 \ BM$ is of the order of $5 \ BM$. Thus,$Cr^{2+}$ is the correct answer.

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

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