# How to calculate crystal field stabilisation energy?

How to calculate it? I know that it depends on the coordination compound and the number of electrons present in t2g and eg orbitals and the final answer is expressed in terms of crystal field splitting parameter. But, how to know how many electrons are there in t2g and eg orbitals?

I think rather than a broad answer and example would help. So, consider the compound $\ce{[Ni(CO)4]}$ and how to calculate CFSE for it?

• forgot to mention wether the election is in high spin or low spin..if low spin d electrons will first fill up the t2g level then the eg level.rest is same . – user4206 Jan 18 '14 at 5:23

Step 1: Look up Nickel Carbonyl and find out what geometry it has.

We need the geometry to know how the $d$ orbitals will split in the ligand field. The geometry can also be predicted: late transition metals with strong field ligands tend to be tetrahedral.

Step 2: Find the appropriate crystal field splitting diagram for this geometry.

Step 3: Figure out how many $d$ electrons there are.

Nickel is in Group 10, with a configuration of $\ce{[Ar]}4s^2 3d^8$ or $\ce{[Ar]}4s^1 3d^9$. In coordination compounds we will consider this configuration to become $\ce{[Ar]}4s^2 3d^8 \implies \ce{[Ar]}3d^{10}$. The $s$ electrons are considered to move to the $d$ sub-shell in bonding - bonding need not represent ground state atomic electron configurations just group state molecular electron configurations. For more information on $d$ electron count, go to this Wikipedia page.

Step 4: Fill in the $d$ electrons.

You can do this part yourself. How do you put 10 $d$ electrons in the orbital diagram?

Step 5: Determine $\Delta E$.

$\Delta$ is the energy different between the $e_g$ and $t_{2g}$ sets of orbitals. This Wikipedia page walks through an octahedral complex. In tetrahedral complexes, the energy of the $e_g$ orbitals is lower and the energy of the $t_{2g}$ orbitals is higher. The energies are:

• $e_g \ (d_{x^2-y^2},\ d_{z^2}) \ E=-\frac{3}{5}\Delta_\textrm{tet}$ (stabilized)
• $t_{2g} \ (d_{xy},\ d_{xz},\ d_{yz}) \ E=\frac{2}{5}\Delta_\textrm{tet}$ (destabilized)

Then multiply the numbers of electrons in the orbitals by the stabilization/destabilization values, and sum.

$$\Delta E = 4\times\left(-\frac{3}{5}\Delta_\textrm{tet}\right)+6\times\left(\frac{2}{5}\Delta_\textrm{tet}\right)=?$$

We have to see the $d$-orbital of the metal atom how many electrons there are, but in excited state not in ground state; then see the spins whether it is low spin or high spin because $d$-orbital is divided into $t_{2g}$ and $e_g$ with respect to their energies difference then apply the formula. The formula is

$$\mathrm{no.\ of\ electrons\ in\ t_{2g}\cdot(-0.4)}+\mathrm{no.\ of\ electrons\ in\ e_g\cdot(0.6)}$$

Example: I have $\mathrm{[Co(F)_6]^{3-}}$ now see cobalt has $3d^7$, $4s^2$ system in ground state but in excited state it loses three electrons in the formation of ions and two elctrons from $4s$ and one from $3d$ orbital so thus Cobalt gets $3d^6$ configuration. Now it is of high spin so $4$ electrons go to $t_{2g}$ orbital and $2$ electrons go to $e_g$ orbital. By applying formula,

$4(-0.4)+2(0.6)=-1.6+1.2=-0.4$

• Thanks for your answer. Could you format it a bit better, it's a bit confusing as it stands. – jonsca Dec 19 '13 at 0:18

The process is quite simple. .. to determine CFSE we need to know that the diffidence of energy levels of two sets have been arbiter taken as 10 dq. . Where t2g set is more stable then eg set octahedral and vice versa in tetrahedral complex. -0.4x + 0.6y will help to calculate it. X indicates electrons in t2g and y in eg. Sum up all and find the diffidence ... That gives cfsc Usually tetrahedral splitting is shown in terms of octahedral as 4/9 OS