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(A) The magnetic moment $(\mu)$ is related to the number of unpaired electrons $(n)$ by the formula: $\mu = \sqrt{n(n+2)} \ BM$. Greater the number of

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$Fe^{2+}$

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(A) The magnetic moment $(\mu)$ is related to the number of unpaired electrons $(n)$ by the formula: $\mu = \sqrt{n(n+2)} \ BM$. Greater the number of unpaired electrons,higher is the magnetic moment. Let us determine the number of unpaired electrons for each ion: $Cr^{3+}$ $(3d^3)$: $n = 3$ $Fe^{2+}$ $(3d^6)$: $n = 4$ $Co^{2+}$ $(3d^7)$: $n = 3$ $Ni^{2+}$ $(3d^8)$: $n = 2$ Since $Fe^{2+}$ has the maximum number of unpaired electrons $(n=4)$,it exhibits the highest magnetic moment.

Which one of the following shows the highest magnetic moment?

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