mu = sqrt{15} is true for the pair :- | vedclass

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Question

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

Vedclass · Accepted Answer

$Co^{+2}, Cr^{+3}$

Vedclass · Answer

(A) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.For $\mu = \sqrt{15}$,we have $n(n+2) = 15$,which implies $n^2 + 2n - 15 = 0$. Solving this quadratic equation,we get $(n+5)(n-3) = 0$,so $n = 3$.Thus,the pair must have $3$ unpaired electrons each.For $Co^{+2}$ $(Z=27)$: $[Ar] 3d^7$,which has $3$ unpaired electrons.For $Cr^{+3}$ $(Z=24)$: $[Ar] 3d^3$,which has $3$ unpaired electrons.Therefore,the pair $Co^{+2}, Cr^{+3}$ satisfies the condition.

$\mu = \sqrt{15}$ is true for the pair :-

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