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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?

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1 Answer 1

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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.

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  • $\begingroup$ 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
    Commented May 23, 2022 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
    Commented May 23, 2022 at 20:13

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