The magnetic moment of X^{n+} is sqrt{24} B. | vedclass

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(C) The magnetic moment formula is $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.Given $\mu = \sqrt{24} \ B.M.$,we have $

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$4, 2$

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(C) The magnetic moment formula is $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.Given $\mu = \sqrt{24} \ B.M.$,we have $\sqrt{n(n+2)} = \sqrt{24}$,which implies $n(n+2) = 24$.Solving for $n$,we get $n^2 + 2n - 24 = 0$,so $(n+6)(n-4) = 0$. Since $n$ must be positive,the number of unpaired electrons is $4$.The atomic number of $X$ is $26$,which corresponds to $Fe$. The electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$.For $Fe^{2+}$,the configuration is $[Ar] 3d^6$,which has $4$ unpaired electrons.Thus,the number of unpaired electrons is $4$ and the value of '$n$' is $2$.

The magnetic moment of $X^{n+}$ is $\sqrt{24} \ B.M.$ Find the number of unpaired electrons and the value of '$n$' respectively. (Atomic number of $X = 26$)

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