The magnetic moment of a metal ion of first t | vedclass

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Question

(d)$\mu = \sqrt {n(n + 2)} $ ($\mu$ = magnetic moment)

($n =$  no. of unpaired electron).

$2.83 = \sqrt {n(n + 2)} $

$n(n + 2) = 8$

$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(d)$\mu = \sqrt {n(n + 2)} $ ($\mu$ = magnetic moment)

($n =$  no. of unpaired electron).

$2.83 = \sqrt {n(n + 2)} $

$n(n + 2) = 8$

${n^2} + 2n - 8 = 0$

$n = 2$.

The magnetic moment of a metal ion of first transition series is $2.83 $ $BM$. Therefore it will have unpaired electrons

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