The magnetic moment of Fe^{2+} is B.M. | vedclass

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

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$4.9$

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(C) The magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n + 2)} 	ext{ B.M. }$,where $n$ is the number of unpaired electrons.The atomic number of $Fe$ is $26$,and its electronic configuration is $[Ar] 3d^6 4s^2$.For $Fe^{2+}$,the configuration is $[Ar] 3d^6$.In the $3d$ subshell,there are $5$ orbitals. According to Hund's rule,the $6$ electrons occupy these orbitals as follows: one orbital has $2$ electrons (paired),and the remaining $4$ orbitals have $1$ electron each (unpaired).Thus,the number of unpaired electrons $n = 4$.Substituting $n = 4$ into the formula: $\mu = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.9 	ext{ B.M. }$

The magnetic moment of $Fe^{2+}$ is $B.M$.

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