Ions which have maximum total spin multiplicity?

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$\begin{aligned} {Fe}^{+ 2} \end{aligned}$

$\begin{aligned} {Cu}^{+ 2} \end{aligned}$

${Cr}^{+ 2}$

${Mn}^{+ 2}$

answer is D.

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Detailed Solution

The formula of spin multiplicity is n +1 where, n is total number of unpaired electron. since ${Mn}^{+ 2}$ has 5 unpaired electron. So the spin multiplicity is 5+1=6.

Thus, the option D is correct.

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