The crystal field stabilisation energy of [Co | vedclass

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(C) In $[Co(NH_3)_6]Cl_3$,the oxidation state of $Co$ is $+3$. The electronic configuration of $Co^{3+}$ is $3d^6$. Since $NH_3$ is a strong field lig

Vedclass · Accepted Answer

$-2.4 \, \Delta_0$

Vedclass · Answer

(C) In $[Co(NH_3)_6]Cl_3$,the oxidation state of $Co$ is $+3$. The electronic configuration of $Co^{3+}$ is $3d^6$. Since $NH_3$ is a strong field ligand,it causes pairing of electrons in the $d$-orbitals. Thus,all $6$ electrons occupy the $t_{2g}$ orbitals. The crystal field stabilisation energy $(CFSE)$ is calculated as: $CFSE = (n_{t2g} 	imes -0.4 \, \Delta_0) + (n_{eg} 	imes +0.6 \, \Delta_0) + mP$ $CFSE = (6 	imes -0.4 \, \Delta_0) + (0 	imes +0.6 \, \Delta_0) + 3P$ $CFSE = -2.4 \, \Delta_0 + 3P$ Considering only the crystal field splitting energy term,the value is $-2.4 \, \Delta_0$.

The crystal field stabilisation energy of $[Co(NH_3)_6]Cl_3$ is

Similar Questions

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