Magnetic moment 2.83, BM is given by which of | vedclass

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Question

Magnetic moment is given by $\mu=\sqrt{n(n+2)}$

Here, $n=$ number of unpaired electrons

$\Rightarrow 2.83=\sqrt{n(n+2)}$

$\Rightarrow(2.83)^{

Vedclass · Accepted Answer

$Ni^{2+}$

Vedclass · Answer

Magnetic moment is given by $\mu=\sqrt{n(n+2)}$

Here, $n=$ number of unpaired electrons

$\Rightarrow 2.83=\sqrt{n(n+2)}$

$\Rightarrow(2.83)^{2}=n(n+2)$

$0=8.00+n^{2}+2 n$

$n^{2}+ 2n-8=0$

$(n+4)(n-2)=0$

$n=2$      $n=-\,4$

$-\,4$ not be consider

Hence, $Ni^{2+}$ possesses a magnetic moment of $2.83\,\mathrm{B} . \mathrm{M}$

Magnetic moment $2.83\, BM$ is given by which of the following ions ?

(At. nos. $Ti = 22,\,Cr = 24,$ $ Mn = 25,\,Ni = 28$)

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