6
$\begingroup$

For hydrogen, the 3-21G basis set (in CFOUR format) is

H:3-21G
3-21G Split-valence basis set

  1
    0
    2
    3

0.5447178000D+01 0.8245472400D+00 0.1831915800D+00 

0.1562849787D+00 0.00000000 
0.9046908767D+00 0.00000000 
0.00000000 1.0000000 

My interpretation is that this defines two orbitals: $$\phi_1 = 0.1562849787g(0.5447178) + 0.9046908767g(0.8245472400)$$

and

$$\phi_2 = 1.0g(0.18319158)$$

But how are these two orbitals combined to form an orbital for the lone electron?

$\endgroup$

1 Answer 1

6
$\begingroup$

They are combined into the final orbitals $\psi_i$ through a linear combination, $\sum_j c_{ij} \phi_j$. Determining the coefficients $c_{ij}$ of that combination is the entire goal of the calculation.

$\endgroup$
2
  • $\begingroup$ Ah, so essentially $\phi_1$ and $\phi_2$ get their own coefficients in the molecular orbitals, rather than sharing the same one if we only had $\phi=\phi_1 + \phi_2$ as our atomic orbital? $\endgroup$
    – theQman
    May 23 at 1:20
  • $\begingroup$ Exactly! - Also, you're correct in calling the result a molecular orbital even though the calculation is on a single atom. It's just a special case, but in general, the combinations of your atom-centered basis functions indeed gives you the delocalized MOs. $\endgroup$
    – Antimon
    May 23 at 20:13

Your Answer

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

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