After calculating the energy of a $\ce{CH4}$ molecule with a 6-31G(d,p) basis set, I obtained the Gaussian/GAMESS log file and describe a part of the log file as follows.

    1  C  1  1S
    2  C  1  2S
    3  C  1  2X
    4  C  1  2Y
    5  C  1  2Z
    6  C  1  3S
    7  C  1  3X
    8  C  1  3Y
    9  C  1  3Z
   10  C  1  4XX
   11  C  1  4YY
   12  C  1  4ZZ
   13  C  1  4XY
   14  C  1  4XZ
   15  C  1  4YZ
   16  H  2  1S
   17  H  2  2S
   18  H  2  2X
   19  H  2  2Y
   20  H  2  2Z

This shows the orbital types of each atom (i.e., C and H), but in 6-31G(d,p) I believe that C does not have 3S, 3X/Y/Z/etc orbitals and H does not have 2S orbital. In addition, I believe that the number of d orbitals (e.g., 4XX and 4YY) is five, but the above file has six d orbitals.

It may be my lack of study, but does the 6-31G(d,p) calculation actually use more orbitals?


3 Answers 3


As Nike mentioned, most basis sets produce more orbitals than the minimal set that you tend to think of for these elements. We can see how this works out for 6-31G(d,p) by looking at how the basis set is defined:

H     0
S    3   1.00
      0.1873113696D+02       0.3349460434D-01
      0.2825394365D+01       0.2347269535D+00
      0.6401216923D+00       0.8137573261D+00
S    1   1.00
      0.1612777588D+00       1.0000000
P    1   1.00
      0.1100000000D+01       1.0000000
C     0
S    6   1.00
      0.3047524880D+04       0.1834737132D-02
      0.4573695180D+03       0.1403732281D-01
      0.1039486850D+03       0.6884262226D-01
      0.2921015530D+02       0.2321844432D+00
      0.9286662960D+01       0.4679413484D+00
      0.3163926960D+01       0.3623119853D+00
SP   3   1.00
      0.7868272350D+01      -0.1193324198D+00       0.6899906659D-01
      0.1881288540D+01      -0.1608541517D+00       0.3164239610D+00
      0.5442492580D+00       0.1143456438D+01       0.7443082909D+00
SP   1   1.00
      0.1687144782D+00       0.1000000000D+01       0.1000000000D+01
D    1   1.00
      0.8000000000D+00       1.0000000

So for $\ce{H}$, we get 3 different types of orbitals: an S orbital formed from 3 primitive Gaussians, an S orbital formed from 1 primitive, and 3 P orbitals formed from 1 primitive. For C, we have an S orbital formed from 6 primitives (meant to represent the core S orbital), 2 sets of SP orbitals (meant to represent the valence S and P orbitals), and 6 D orbitals formed from 1 primitive.

The P orbital for $\ce{H}$ and D orbital for $\ce{C}$ (these are the (d,p) in 6-31g(d,p)) are referred to as polarization functions. These are higher angular functions added to the basis to allow for a more flexible description of the electron density, especially directional interactions like bonds.

As for why there are 6 D orbitals rather than 5, this is because the Pople basis (the ones written X-YZ...G) use Cartesian functions rather than spherical ones. In the Cartesian representation, it takes 6 functions to represent the D orbitals. This discrepancy between the number of Cartesian and spherical functions continues for higher angular momentum functions as well (e.g. there are 10 Cartesian F functions, rather than the typically seen 7 spherical ones).

However, as Susi noted in the comments, the Pople basis sets actually use spherical functions for F orbitals; this can be confusing to handle. Among many other reasons, this is why the Pople basis sets are mainly just of historical significance and that other, more modern basis sets should be used.

  • $\begingroup$ Thank you for your answer. I now understood why the d-orbits are represented as 6 orbitals. $\endgroup$
    – neco
    Commented Jan 29, 2022 at 6:18
  • 2
    $\begingroup$ Actually, some of the Pople basis sets use cartesian D functions with spherical F functions, and this is not consistently supported in most programs. The field of basis set development has advanced from the early days of Pople and coworkers, with systematic families spanning ranges of accuracy are nowadays available. They should be used instead. $\endgroup$ Commented Feb 1, 2022 at 22:37

Yes, a good basis set will contain virtual (unoccupied) orbitals, in addition to the orbitals that are occupied by electrons.

Almost always: the larger the basis set, the more accurate the calculations (but the more computationally demanding they are).

When I calculated the carbon ionization energy with an aug-cc-pCV8Z basis set, we had 1182 spin orbitals for a single carbon atom, and not only D-type orbitals, but also F-type, G-type, H-type, I-type, K-type, and L-type orbitals.

  • $\begingroup$ Thanks for the answer! I mistook 1S and 2S of H as 1S and 2S orbital but I understood that this is the double-zeta basis set of H. I understand completely now. $\endgroup$
    – neco
    Commented Jan 29, 2022 at 6:16

I should complement Nike and Tyberius' answers by the simple statement that e.g. hydrogen only consists of a single S function for the non-interacting atom. When the atomic symmetry is lifted, you get more contributions.

For instance, even though the H$_2$ ground state consists of a doubly occupied $1\sigma^2$ orbital in Hartree-Fock and density functional theory, the expansion of this molecular orbital in terms of atomic orbitals on the two nuclei consists of not just the $Y_0^0$ $s$-type spherical harmonic of the $1s$ ground state of hydrogen, but also of the $Y_1^0$, $Y_2^0$, $Y_3^0$, $Y_4^0$, $\dots$ spherical harmonics which are provided by $p$, $d$, $f$, $g$, $\dots$ functions.

In addition, you need several radial functions per angular momentum. The additional $s$ functions are known as "breathing" functions, since they enable the $1s$ function in the atom to expand or contract, while the higher-angular-momentum functions are indeed known as polarization functions.

If you go to post-Hartree-Fock level of theory, you also need basis functions that are used to describe excited electronic configurations. These are called correlating functions.

  • $\begingroup$ +1. It's good to have this part here, and I especially liked the part about needing the excited electronic configurations for post-SCF methods like CI. Another interesting topic is the existence of a "continuum" after the discrete excited states, but a countably infinite basis set should still be enough to get the exact solution to the non-relativistic Schroedinger equation with FCI. $\endgroup$ Commented Feb 2, 2022 at 0:03
  • $\begingroup$ Yes, it should, if you use a good basis set. After all, the ground state has to be normalizable i.e. in the L^2 space of functions, so it cannot mix with continuum a.k.a. scattering states that are non-normalizable. $\endgroup$ Commented Feb 2, 2022 at 17:38

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