The magnetic moment of M^{x+} (atomic number | vedclass

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(B) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = \sqrt{15} \ BM$,

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$3, 4$

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(B) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = \sqrt{15} \ BM$,we have $\sqrt{n(n+2)} = \sqrt{15}$,which implies $n(n+2) = 15$.Solving for $n$,$n^2 + 2n - 15 = 0$,which factors as $(n+5)(n-3) = 0$. Since $n$ must be positive,$n = 3$.The atomic number of $M$ is $25$,which corresponds to Manganese $(Mn)$.The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^{5} 4s^{2}$.For $M^{x+}$ to have $3$ unpaired electrons,it must have a $3d^{3}$ configuration.$Mn^{4+}$ has the configuration $[Ar] 3d^{3} 4s^{0}$,which contains $3$ unpaired electrons.Thus,the number of unpaired electrons is $3$ and the value of $x$ is $4$.

The magnetic moment of $M^{x+}$ (atomic number $= 25$) is $\sqrt{15} \ BM$. The number of unpaired electrons and the value of $x$,respectively,are

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