Why does Z-in take place in $d^{4}$ low spin configuration whereas in Z-out one can get all the 4 electrons in $d_{xz}, d_{yz}$ which would lead to lower energy

  • $\begingroup$ Can anyone tell me why it has been downvoted? $\endgroup$ – MathMan Mar 31 '18 at 21:12
  • $\begingroup$ Dunno, but elaborating should do the post some good. $\endgroup$ – Mithoron Mar 31 '18 at 23:05

First off, I don't think all 4 electrons will be occupying the $d_{xz}$ and $d_{yz}$ orbitals in z-out distortion. If you do that, you'll be changing spin multiplicity on distortion. That doesn't happen. Rather, all the (originally) $t_{2g}$ levels would fill up, leaving you with a paired up electronic configuration in either $d_{xz}$ or $d_{yz}$ for z-out and a spin paired $d_{xy}$ for z-in.

Now, calculate the relative stabilization such that:

a. z-out: $d_{xz}$ and $d_{yz}$ are stabilized by 1/3$\delta$ and $d_{xy}$ is destabilized by 2/3$\delta$.

b. z-in: $d_{xz}$ and $d_{yz}$ are destabilized by 1/3$\delta$ and $d_{xy}$ is stabilized by 2/3$\delta$.

Your electronic configurations would then read ($e_g)^3$($b_{2g})^1$ for z-out, ($b_{2g})^2$($e_g)^2$ for z-in. Now calculate the relative energies. You will find that z-in is favoured by a factor of 1/3$\delta$.


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