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

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$Fe^{3+}$

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(C) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.$1$. For $Mn^{6+}$ $([Ar]3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = 1.73 	ext{ BM}$.$2$. For $Ni^{2+}$ $([Ar]3d^8)$: $n = 2$,$\mu = \sqrt{2(2+2)} = 2.83 	ext{ BM}$.$3$. For $Fe^{3+}$ $([Ar]3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = 5.91 	ext{ BM}$.$4$. For $Ag^{+}$ $([Kr]4d^{10})$: $n = 0$,$\mu = 0 	ext{ BM}$.Thus,$Fe^{3+}$ has the maximum number of unpaired electrons $(n=5)$ and consequently shows the maximum magnetic moment.

Which of the following species shows the maximum magnetic moment?

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