1
$\begingroup$

I optimized a geometry of $\ce{Ni(H2O)6}$ using quantum chemistry and got a molecule which is symmetrical for the $\ce{Ni-O}$ distances as shown here:

ball and stick model of Ni(H2O)6

In the literature it says that this complex has an $\ce{O_h}$ point group, but this complex looks like it is $\ce{T_h}$. Do I need to revise my group theory or is what I am looking at actually a $\ce{T_h}$ point group?

| improve this question | |
$\endgroup$
  • 4
    $\begingroup$ Welcome to Chemistry StackExchange. It look like $T_h$ as drawn (ignoring bonds to H atoms, and considering all atoms are fixed), but without any H it is $O_h$. Additionally the OH group is going to rotate more or less freely so its average will be the same as if H atoms are ignored. $\endgroup$ – porphyrin Apr 4 '17 at 17:27
  • $\begingroup$ In addition to the Hydrogen atoms swirling around, wasn't the tetrahedral symmetry $T_h$ a subgroup of the octahedral $O_h$, too? $\endgroup$ – Buttonwood Apr 4 '17 at 21:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.