What will be the value of x in Fe^{x+},if the | vedclass

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(D) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = \sqrt{24} \ BM$,

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$2$

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(D) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = \sqrt{24} \ BM$,we have $\sqrt{n(n+2)} = \sqrt{24}$.Squaring both sides,$n(n+2) = 24$,which simplifies to $n^2 + 2n - 24 = 0$.Solving the quadratic equation: $(n+6)(n-4) = 0$.Since the number of electrons cannot be negative,$n = 4$.The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$.For $Fe^{2+}$,the configuration is $[Ar] 3d^6$,which has $4$ unpaired electrons.Thus,$x = 2$.

What will be the value of $x$ in $Fe^{x+}$,if the magnetic moment,$\mu=\sqrt{24} \ BM$?

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