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

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$Fe^{3+}$

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(B) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$1$. For $Cr^{3+}$ $([Ar] 3d^3)$: $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.$2$. For $Fe^{3+}$ $([Ar] 3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.$3$. For $Ti^{3+}$ $([Ar] 3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.$4$. For $Co^{3+}$ $([Ar] 3d^6)$: In high spin,$n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.Comparing these values,$Fe^{3+}$ has the maximum number of unpaired electrons $(n=5)$,hence it has the maximum theoretical magnetic moment.

Which of the following ions has the maximum theoretical magnetic moment?

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