A complex [CoL_6]^{n+} where L is a neutral l | vedclass

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(D) The magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = 4.5 \ BM$,we have $\sqrt{n(n

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$Co$ must be in $+3$ oxidation state

Vedclass · Answer

(D) The magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = 4.5 \ BM$,we have $\sqrt{n(n+2)} \approx 4.5$,which implies $n(n+2) \approx 20.25$. Solving for $n$,we get $n \approx 4$.For $Co$ in $+3$ oxidation state $(3d^6)$,if the ligand $L$ is weak,the electrons will be arranged as shown in the diagram,resulting in $4$ unpaired electrons.Thus,$Co$ must be in $+3$ oxidation state.

$A$ complex $[CoL_6]^{n+}$ where $L$ is a neutral ligand has a magnetic moment $\mu = 4.5 \ BM$. Hence,

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