The spin only magnetic moment of left[ ZCl _{ | vedclass

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Question

(c) Spin magnetic moment of a complex, can be calculated,
$\mu= \sqrt{n(n+2)}$ where, $n$ is the number of unpaired electron.
The oxidation state of

Vedclass · Accepted Answer

$Co$

Vedclass · Answer

(c) Spin magnetic moment of a complex, can be calculated,
$\mu= \sqrt{n(n+2)}$ where, $n$ is the number of unpaired electron.
The oxidation state of $Z$ in $\left[ ZCl _{4}\right]^{2-}$ is $+2$.
The spin magnetic moment of given metals in $+2$ state are as follows:
$(a)$ $Mn ^{2+}$
Electronic configuration $=3 d^{5} 4 s^{0}$
$[Image]$
$\mu=\sqrt{5(5+2)}=5.9 \,BM$
$(b)$ $Ni ^{2+}$
Electronic confiruration $=3 d^{8} 4 s^{0}$
$[Image]$
$\mu=\sqrt{2(2+2)}=2.9 \,BM$
$(c)$ $Co ^{2+}$
Electronic configuration $=3 d^{7} 4 s^{0}$
$[Image]$
$\mu=\sqrt{3(3+2)}=3.87\, BM$
$(d)$ $Cu ^{2+}$
Electronic configuration $=3 d^{9} 4 s^{0}$
$[Image]$
$\mu=\sqrt{1(1+2)}=1.9 \,BM$
There are, $Z$ in $\left[ ZCl _{4}\right]^{2-}$ is $Co$, where the spin only magnetic moment is $3.87 \, BM$.

The spin only magnetic moment of $\left[ ZCl _{4}\right]^{2-}$ is $3.87 \,BM,$ where $Z$ is $.....$

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