The spin only magnetic moments (in BM) for fr | vedclass

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(A) The spin only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $Ti^{3+

Vedclass · Accepted Answer

$1.73, 3.87, 0$

Vedclass · Answer

(A) The spin only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $Ti^{3+}$ $(Z=22)$: Electronic configuration is $[Ar] 3d^{1}$. Here,$n = 1$. Thus,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.For $V^{2+}$ $(Z=23)$: Electronic configuration is $[Ar] 3d^{3}$. Here,$n = 3$. Thus,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.For $Sc^{3+}$ $(Z=21)$: Electronic configuration is $[Ar] 3d^{0}$. Here,$n = 0$. Thus,$\mu = 0 \ BM$.Therefore,the magnetic moments are $1.73, 3.87, 0$ respectively.

The spin only magnetic moments (in $BM$) for free $Ti^{3+}, V^{2+}$ and $Sc^{3+}$ ions respectively are (At.No. $Sc: 21; Ti: 22; V: 23$)

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