Calculate the 'spin only' magnetic moment of $M ^{2+}(\operatorname{aq})$ ion $(Z=27)$

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$Z=27$

$\Rightarrow[ Ar ] 3 d^{7} 4 s^{2}$

$\therefore M^{2+}=[A r] 3 d^{7}$

i.e., $3$ unpaired electrons

$\therefore n=3$

$\Rightarrow \sqrt{n(n+2)}=\mu$

$\Rightarrow \sqrt{3(3+2)}=\mu$

$\Rightarrow \sqrt{15}=\mu$

$\mu \approx 4 \,BM$

932-s18

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