The correct decreasing order of spin only mag | vedclass

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Question

(C) The spin only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$Cu^{+} : [

Vedclass · Accepted Answer

$Cr^{2+} > Cr^{3+} > Cu^{2+} > Cu^{+}$

Vedclass · Answer

(C) The spin only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$Cu^{+} : [Ar] 3d^{10}$,$n = 0$,$\mu = 0 \ BM$.$Cu^{2+} : [Ar] 3d^{9}$,$n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.$Cr^{2+} : [Ar] 3d^{4}$,$n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.$Cr^{3+} : [Ar] 3d^{3}$,$n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.Comparing the values: $4.90 > 3.87 > 1.73 > 0$.Therefore,the decreasing order is $Cr^{2+} > Cr^{3+} > Cu^{2+} > Cu^{+}$.

The correct decreasing order of spin only magnetic moment values $(BM)$ of $Cu^{+}$,$Cu^{2+}$,$Cr^{2+}$ and $Cr^{3+}$ ions is $:$

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