The values of the crystal field stabilization | vedclass

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Question

(C) For a high spin $d^{6}$ ion in an octahedral field,the electronic configuration is $t_{2g}^{4} e_{g}^{2}$.$CFSE = (4 	imes -0.4 \Delta_{o}) + (2

Vedclass · Accepted Answer

$-0.4 \Delta_{o}$ and $-0.6 \Delta_{t}$

Vedclass · Answer

(C) For a high spin $d^{6}$ ion in an octahedral field,the electronic configuration is $t_{2g}^{4} e_{g}^{2}$.$CFSE = (4 	imes -0.4 \Delta_{o}) + (2 	imes 0.6 \Delta_{o}) = -1.6 \Delta_{o} + 1.2 \Delta_{o} = -0.4 \Delta_{o}$.For a high spin $d^{6}$ ion in a tetrahedral field,the electronic configuration is $e^{3} t_{2}^{3}$.$CFSE = (3 	imes -0.6 \Delta_{t}) + (3 	imes 0.4 \Delta_{t}) = -1.8 \Delta_{t} + 1.2 \Delta_{t} = -0.6 \Delta_{t}$.Thus,the values are $-0.4 \Delta_{o}$ and $-0.6 \Delta_{t}$.

The values of the crystal field stabilization energies for a high spin $d^{6}$ metal ion in octahedral and tetrahedral fields,respectively,are :

Similar Questions

The total number of unpaired electrons present in the $d^3$, $d^4$ (low spin), $d^5$ (high spin), $d^6$ (high spin), and $d^7$ (low spin) octahedral complex systems is . . . . . . .

Which of the following is a high spin complex? (Where $\Delta_0$ = splitting energy of octahedral complex,$P.E.$ = Pairing energy)

Complete removal of both the axial ligands (along the $z-$ axis) from an octahedral complex leads to which of the following splitting patterns? (relative orbital energies not on scale)

The degeneracy of $d-$orbitals is lost under:
$I$. Strong field ligand
$II$. Weak field ligand
$III$. Mixed field ligand
$IV$. Chelated ligand field
Choose the correct code:

For an octahedral complex of $Co$ $(Z = 27)$,which of the following has the highest value of $\Delta_0$?

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