The divalent ion of which of the following el | vedclass

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(D) The magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = 5.92 \ B

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$Mn$

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(D) The magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = 5.92 \ BM$,we have $\sqrt{n(n+2)} = 5.92$,which implies $n \approx 5$.For a divalent ion $(M^{2+})$ to have $5$ unpaired electrons,the electronic configuration must be $3d^5$.Among the given elements,$Mn$ $(Z=25)$ has the ground state configuration $[Ar] 3d^5 4s^2$.Upon forming the $Mn^{2+}$ ion,it loses two $4s$ electrons,resulting in the configuration $[Ar] 3d^5$.This configuration contains $5$ unpaired electrons,consistent with the magnetic moment of $5.92 \ BM$.

The divalent ion of which of the following elements in an aqueous solution has a magnetic moment of $5.92 \ BM$?

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