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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.For $

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$3d^5$

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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.For $3d^1$,$n = 1$,$\mu = \sqrt{1(3)} = 1.73 	ext{ B.M.}$For $3d^8$,$n = 2$,$\mu = \sqrt{2(4)} = 2.83 	ext{ B.M.}$For $3d^5$,$n = 5$,$\mu = \sqrt{5(7)} = 5.92 	ext{ B.M.}$For $3d^7$,$n = 3$,$\mu = \sqrt{3(5)} = 3.87 	ext{ B.M.}$Thus,the $3d^5$ configuration has the highest number of unpaired electrons $(n=5)$,resulting in the highest magnetic moment.

For which of the following transition metal ion configurations is the highest magnetic moment observed?

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