A compound of a metal ion M^{x+} (Z = 24) has | vedclass

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

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$3$

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(D) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = \sqrt{15} \ BM$.Equating the two expressions: $\sqrt{15} = \sqrt{n(n+2)}$.Squaring both sides: $15 = n(n+2)$.$n^2 + 2n - 15 = 0$.Solving the quadratic equation: $(n+5)(n-3) = 0$.Since $n$ must be a positive integer,$n = 3$.Therefore,the number of unpaired electrons is $3$.

$A$ compound of a metal ion $M^{x+}$ $(Z = 24)$ has a spin-only magnetic moment of $\sqrt{15} \ BM$. The number of unpaired electrons in the compound is:

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