The spin magnetic moment of an octahedral com | vedclass

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Question

(C) For an octahedral complex with weak field ligands,$CFSE = (-0.4n_{t_{2g}} + 0.6n_{e_{g}}) \Delta_{0}$.For $d^{2}$ configuration: $CFSE = (-0.4 	i

Vedclass · Accepted Answer

Both $(A)$ and $(B)$

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(C) For an octahedral complex with weak field ligands,$CFSE = (-0.4n_{t_{2g}} + 0.6n_{e_{g}}) \Delta_{0}$.For $d^{2}$ configuration: $CFSE = (-0.4 	imes 2) \Delta_{0} = -0.8 \Delta_{0}$. The number of unpaired electrons $n = 2$. Magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{2(4)} = \sqrt{8} \ BM$.For $d^{7}$ configuration: $CFSE = (-0.4 	imes 5 + 0.6 	imes 2) \Delta_{0} = (-2.0 + 1.2) \Delta_{0} = -0.8 \Delta_{0}$. The number of unpaired electrons $n = 3$. Magnetic moment $\mu = \sqrt{3(5)} = \sqrt{15} \ BM$.Since both configurations result in $CFSE = -0.8 \Delta_{0}$,the correct answer is $(C)$.

The spin magnetic moment of an octahedral complex having $CFSE = -0.8 \Delta_{0}$ and surrounded by weak field ligands can be:

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