Crystal field stabilization energy (CFSE) for | vedclass

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Question

(A) $1$. The central metal ion is $Co^{3+}$. The atomic number of $Co$ is $27$,so the electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.$2$. $CN^-$

Vedclass · Accepted Answer

$-2.4 \, \Delta_0 + 3P$

Vedclass · Answer

(A) $1$. The central metal ion is $Co^{3+}$. The atomic number of $Co$ is $27$,so the electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.$2$. $CN^-$ is a strong field ligand,which causes pairing of electrons in the $d$-orbitals.$3$. For a $d^6$ configuration in an octahedral field with a strong ligand,all $6$ electrons occupy the $t_{2g}$ orbitals: $(t_{2g})^6 (e_g)^0$.$4$. The $CFSE$ formula is $CFSE = (-0.4 	imes n_{t_{2g}} + 0.6 	imes n_{e_g}) \Delta_0 + mP$,where $m$ is the number of electron pairs formed due to the ligand field.$5$. Here,$n_{t_{2g}} = 6$ and $n_{e_g} = 0$. Thus,$CFSE = (-0.4 	imes 6 + 0.6 	imes 0) \Delta_0 = -2.4 \Delta_0$.$6$. In the free $Co^{3+}$ ion $(d^6)$,there is $1$ pair. In the complex $[Co(CN)_6]^{3-}$,there are $3$ pairs in the $t_{2g}$ set. The number of extra pairs formed is $3 - 1 = 2$. However,standard convention for $CFSE$ often includes the total pairing energy $P$ associated with the configuration. Given the options,the correct value is $-2.4 \Delta_0 + 3P$.

Crystal field stabilization energy $(CFSE)$ for the complex $[Co(CN)_6]^{3-}$ will be:

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