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(B) The magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.Electronic conf

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$Mn^{2+}$

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(B) The magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.Electronic configurations of the ions are:$Cr^{2+} (3d^4) \implies n = 4$$Mn^{2+} (3d^5) \implies n = 5$$Cu^{2+} (3d^9) \implies n = 1$$Co^{2+} (3d^7) \implies n = 3$Since $Mn^{2+}$ has the maximum number of unpaired electrons $(n=5)$,it will show the highest magnetic moment.

Which of the following species is expected to show the highest magnetic moment? (At. Nos.: $Cr = 24$,$Mn = 25$,$Co = 27$,$Ni = 28$,$Cu = 29$)

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