2
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ Can anyone tell me why it has been downvoted? $\endgroup$
    – MathMan
    Mar 31, 2018 at 21:12
  • $\begingroup$ Dunno, but elaborating should do the post some good. $\endgroup$
    – Mithoron
    Mar 31, 2018 at 23:05

1 Answer 1

2
$\begingroup$

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$\delta$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.