The spin-only magnetic moment value of the co | vedclass

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Question

(D) The oxidation states of $Mn$ in $MnF_{4}$,$MnF_{3}$,and $MnF_{2}$ are $+4$,$+3$,and $+2$ respectively.The electronic configurations are:$MnF_{4} (

Vedclass · Accepted Answer

$5$

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(D) The oxidation states of $Mn$ in $MnF_{4}$,$MnF_{3}$,and $MnF_{2}$ are $+4$,$+3$,and $+2$ respectively.The electronic configurations are:$MnF_{4} (Mn^{4+}) = [Ar] 3d^{3}$$MnF_{3} (Mn^{3+}) = [Ar] 3d^{4}$$MnF_{2} (Mn^{2+}) = [Ar] 3d^{5}$Among these,$Mn^{3+}$ is the strongest oxidizing agent because it can easily gain an electron to form the stable $d^{5}$ configuration $(Mn^{2+})$.For $Mn^{3+}$ $(3d^{4})$,the number of unpaired electrons $(n)$ is $4$.The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ B.M.$$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.89 \ B.M.$The nearest integer value is $5$.

The spin-only magnetic moment value of the compound with the strongest oxidizing ability among $MnF_{4}$,$MnF_{3}$,and $MnF_{2}$ is $............$ $B.M.$ [nearest integer]

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