The Crystal Field Stabilization Energy (CFSE) | vedclass

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(D) The magnetic moment $\mu = 3.87 \, BM$ corresponds to $n = 3$ unpaired electrons,as $\mu = \sqrt{n(n+2)} \, BM$. For an octahedral complex,the $CF

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$Co^{2+}$

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(D) The magnetic moment $\mu = 3.87 \, BM$ corresponds to $n = 3$ unpaired electrons,as $\mu = \sqrt{n(n+2)} \, BM$. For an octahedral complex,the $CFSE$ is given by the formula: $CFSE = (-0.4 	imes n_{t_{2g}} + 0.6 	imes n_{e_g}) \Delta_0$. For a $d^7$ ion in a weak field (aqua complex),the configuration is $t_{2g}^5 e_g^2$. $CFSE = (-0.4 	imes 5 + 0.6 	imes 2) \Delta_0 = (-2.0 + 1.2) \Delta_0 = -0.8 \Delta_0$. This matches the given $CFSE$ and the number of unpaired electrons $(3)$. Thus,the metal ion is $Co^{2+}$ ($d^7$ configuration).

The Crystal Field Stabilization Energy $(CFSE)$ and magnetic moment (spin-only) of an octahedral aqua complex of a metal ion $(M^{2+})$ are $-0.8\, \Delta_{0}$ and $3.87\, BM$,respectively. Identify $(M^{2+})$:

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