The correct order of spin-only magnetic momen | vedclass

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Question

(C) The spin-only magnetic moment $(\mu)$ depends on the number of unpaired electrons $(n)$ and is given by the formula: $\mu = \sqrt{n(n+2)} \ BM$.Th

Vedclass · Accepted Answer

$Mn^{2+} > Cr^{2+} > Ti^{2+}$

Vedclass · Answer

(C) The spin-only magnetic moment $(\mu)$ depends on the number of unpaired electrons $(n)$ and is given by the formula: $\mu = \sqrt{n(n+2)} \ BM$.The electronic configurations and number of unpaired electrons for the ions are:$1$. For $Mn^{2+}$ $(Z=25)$: $[Ar] 3d^5$. Number of unpaired electrons $(n)$ = $5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \ BM$.$2$. For $Cr^{2+}$ $(Z=24)$: $[Ar] 3d^4$. Number of unpaired electrons $(n)$ = $4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \ BM$.$3$. For $Ti^{2+}$ $(Z=22)$: $[Ar] 3d^2$. Number of unpaired electrons $(n)$ = $2$. $\mu = \sqrt{2(2+2)} = \sqrt{8} \ BM$.Comparing the values,the correct order of spin-only magnetic moment is $Mn^{2+} > Cr^{2+} > Ti^{2+}$.

The correct order of spin-only magnetic moment (in $BM$) for $Mn^{2+}$,$Cr^{2+}$,and $Ti^{2+}$ ions is:

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