Which ion has the maximum magnetic moment?

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Question

(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.

Vedclass · Accepted Answer

$Fe^{3+}$

Vedclass · Answer

(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.      Ion    Outer Configuration    Unpaired Electrons $(n)$    Magnetic Moment ($B$.$M$.)        $V^{3+}$    $3d^2$    $2$    $\sqrt{2(2+2)} = 2.83$        $Mn^{3+}$    $3d^4$    $4$    $\sqrt{4(4+2)} = 4.90$        $Fe^{3+}$    $3d^5$    $5$    $\sqrt{5(5+2)} = 5.92$        $Cu^{2+}$    $3d^9$    $1$    $\sqrt{1(1+2)} = 1.73$  Comparing the values,$Fe^{3+}$ has the highest number of unpaired electrons $(n=5)$,resulting in the maximum magnetic moment of $5.92 	ext{ B.M.}$ Therefore,the correct option is $(C)$.

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