2

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

Bartholomew696969
  • 163
  • 5
  • 1
    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 $D{3h}$). – orthocresol Mar 22 '23 at 10:00
  • 1
    Past me wrote about this here: https://chemistry.stackexchange.com/a/58633/16683 – orthocresol Mar 22 '23 at 10:02
  • You write "the HOMO in C3v is bonding (2A1), whereas the HOMO in C3v is non-bonding (1A2'')," but that does not work… – Martin - マーチン Mar 22 '23 at 22:07

0 Answers0