If the magnetic moment (mu) of an oxide of ma | vedclass

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(B) The formula for magnetic moment is $\mu = \sqrt{n(n+2)}$,where $n$ is the number of unpaired electrons.Given $\mu = 4.9 \ BM$,we have $\sqrt{n(n+2

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$Mn_2O_3$

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(B) The formula for magnetic moment is $\mu = \sqrt{n(n+2)}$,where $n$ is the number of unpaired electrons.Given $\mu = 4.9 \ BM$,we have $\sqrt{n(n+2)} = 4.9$.Squaring both sides,$n(n+2) \approx 24$,which gives $n = 4$.In $Mn_2O_3$,the oxidation state of $Mn$ is $+3$.The electronic configuration of $Mn^{3+}$ is $[Ar] \ 3d^4$.Since there are $4$ unpaired electrons,the magnetic moment is $\sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \ BM$.Therefore,the oxide is $Mn_2O_3$.

If the magnetic moment $(\mu)$ of an oxide of manganese is $4.9 \ BM$,which oxide is it?

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