2
$\begingroup$

The 2 major possible geometries of ammonia are D3h and C3v, but it prefers C3v. My thought process is that this is simply Walsh's rule - the HOMO in C3v is bonding (2A1), whereas the HOMO in C3v is non-bonding (1A2''), so the molecule prefers the trigonal pyramidal geometry based on the HOMO being best-stabilised.

Is this the complete explanation or does the 2nd order Jahn-Teller distortion come into effect here too?

I included a Walsh diagram for PH3 (couldn't find one for ammonia) just for completeness' sake.

enter image description here

$\endgroup$
3
  • 1
    $\begingroup$ By the 'second-order JT distortion', are you referring to the mixing of $2a_1$ and $3a_1$ when symmetry is lowered? It's been a while since I studied this stuff, but I don't think your descriptions are mutually exclusive. It's precisely the mixing that makes $2a_1$ more stable in $C_{3v}$ (if it didn't mix, it'd just be a pure p orbital, same as $a''_2$ in $D_{3h}$). $\endgroup$ Commented Mar 22, 2023 at 10:00
  • 1
    $\begingroup$ Past me wrote about this here: chemistry.stackexchange.com/a/58633/16683 $\endgroup$ Commented Mar 22, 2023 at 10:02
  • $\begingroup$ You write "the HOMO in C3v is bonding (2A1), whereas the HOMO in C3v is non-bonding (1A2'')," but that does not work… $\endgroup$ Commented Mar 22, 2023 at 22:07

0

Your Answer

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

Browse other questions tagged or ask your own question.