I'm now studying the quantum chemistry calculation and Gaussian basis set.

For example, when using the 6-31G basis set, the C atom has 6 basis functions in 1s orbital:

  • 3+1=4 in $2s$,
  • 3+1=4 in $2p_x$,
  • 3+1=4 in $2p_y$, and
  • 3+1=4 in $2p_z$,

whereas the O atom also has 6 basis functions in 1s orbital:

  • 3+1=4 in $2s$,
  • 3+1=4 in $2p_x$,
  • 3+1=4 in $2p_y$,
  • 3+1=4 in $2p_z$.

That is, the number of basis functions is the same in these two atoms, is it correct?

I think that the O atom has more electrons than the C atom, so the number of basis functions should be different and the O atom should have more basis functions than the C atom, but is this my misunderstanding?

  • $\begingroup$ +1. Typically the number of basis functions is the same within a row (so C and O will have the same number of basis functions as Ne, but not as many as Ar). $\endgroup$ Jun 19, 2021 at 2:33

2 Answers 2


Yes you're absolutely correct that the basis sets for O and C will typically have the same number of orbitals (not just for the 6-31G basis set but also for the famous cc-pVXZ family, and other basis sets).

It's true that O has more electrons than C, but in both cases, the electrons are primarily occupied in the $1s$, $2s$, and $2p$ orbitals. If you're considering to trim down the carbon basis set to have only $2p_x$ and $2p_y$ (i.e. to remove $2p_z$) things can get imbalanced.

B, C, N, O, F, and Ne will typically all have the same number of basis functions for a given basis set label (such as cc-pVDZ), whereas the corresponding atoms in the next row: Al, Si, P, S, Cl, and Ar will have a slightly bigger basis set than the previous atoms, but they will all have the same number of basis functions as each other.

  • $\begingroup$ Thank you for the answer! I found that the number of basis functions is the same even if the number of electrons is different as long as the row is the same. $\endgroup$
    – neco
    Jun 19, 2021 at 12:10

I think that it is a feature of how the basis set is defined. 6-31G is a valence double-zeta basis set of the Pople family. Here, 6 means that the core orbitals are represented with 6 Gaussian functions (GTO) while 31 means the valence orbitals are split into two orbitals with the scheme 3 GTO + 1 GTO. So, I looked at the basis set data for 6-31G, and it seems to be clear how the orbitals are assigned.

For example, look at the basis sets of N, P and As (group 15):

S   6
1         0.4173511460E+04       0.1834772160E-02
2         0.6274579110E+03       0.1399462700E-01
3         0.1429020930E+03       0.6858655181E-01
4         0.4023432930E+02       0.2322408730E+00
5         0.1282021290E+02       0.4690699481E+00
6         0.4390437010E+01       0.3604551991E+00
L   3
1         0.1162636186E+02      -0.1149611817E+00       0.6757974388E-01
2         0.2716279807E+01      -0.1691174786E+00       0.3239072959E+00
3         0.7722183966E+00       0.1145851947E+01       0.7408951398E+00
L   1
1         0.2120314975E+00       0.1000000000E+01       0.1000000000E+01

For nitrogen $\ce{1s->6 GTO; 2s,2p->3GTO + 1GTO}$. A feature of the Pople style basis sets is that the s and p orbitals are fused when possible into an SP orbital (also called "L" orbital). This means that the s and p GTO's from the same orbital have the same exponent, but different coefficients. This was done to reduce the cost of calculations.

S   6
1         0.1941330000E+05       0.1851598923E-02
2         0.2909420000E+04       0.1420619174E-01
3         0.6613640000E+03       0.6999945928E-01
4         0.1857590000E+03       0.2400788603E+00
5         0.5919430000E+02       0.4847617180E+00
6         0.2003100000E+02       0.3351998050E+00
L   6
1         0.3394780000E+03      -0.2782170105E-02       0.4564616191E-02
2         0.8101010000E+02      -0.3604990135E-01       0.3369357188E-01
3         0.2587800000E+02      -0.1166310044E+00       0.1397548834E+00
4         0.9452210000E+01       0.9683280364E-01       0.3393617168E+00
5         0.3665660000E+01       0.6144180231E+00       0.4509206237E+00
6         0.1467460000E+01       0.4037980152E+00       0.2385858009E+00
L   3
1         0.2156230000E+01      -0.2529241139E+00      -0.1776531273E-01
2         0.7489970000E+00       0.3285184468E-01       0.2740581964E+00
3         0.2831450000E+00       0.1081254762E+01       0.7854215630E+00
L   1
1         0.9983170000E-01       0.1000000000E+01       0.1000000000E+01

For phosphorus, $\ce{1s->6 GTO; 2s,2p->6 GTO; 3s,3p->3 GTO + 1 GTO}$

S   6
1         0.1005955000E+06       0.1726750000E-02
2         0.1510482000E+05       0.1323462000E-01
3         0.3440884000E+04       0.6535848000E-01
4         0.9703961000E+03       0.2278042000E+00
5         0.3112852000E+03       0.4774525000E+00
6         0.1066284000E+03       0.3567619000E+00
L   6
1         0.2166679000E+04       0.2271761000E-02       0.3832156000E-02
2         0.5165414000E+03       0.3033475000E-01       0.3023558000E-01
3         0.1678674000E+03       0.1259057000E+00       0.1328632000E+00
4         0.6364638000E+02      -0.6687172000E-02       0.3447648000E+00
5         0.2613673000E+02      -0.6065306000E+00       0.4640368000E+00
6         0.1085439000E+02      -0.4823144000E+00       0.2064824000E+00
L   6
1         0.9506989000E+02      -0.5587423000E-02      -0.6816583000E-02
2         0.3318087000E+02       0.5632506000E-01      -0.2970303000E-01
3         0.1406773000E+02       0.2625835000E+00       0.4704335000E-01
4         0.6153288000E+01      -0.1718349000E+00       0.3645042000E+00
5         0.2721712000E+01      -0.7175645000E+00       0.4945157000E+00
6         0.1185334000E+01      -0.3184598000E+00       0.2149830000E+00
L   3
1         0.1615315000E+01       0.2645372000E+00      -0.2574061000E-01
2         0.5513300000E+00      -0.1952737000E+00       0.3072764000E+00
3         0.2227620000E+00      -0.9595400000E+00       0.7537368000E+00
L   1
1         0.8292300000E-01       0.1000000000E+01       0.1000000000E+01
D   3
1         0.5030227000E+02       0.8144711000E-01
2         0.1384166000E+02       0.3792908000E+00
3         0.4393458000E+01       0.7040401000E+00
D   1
1         0.1310755000E+01       1.0000000

For arsenic, I believe the assignment is: $\ce{1s->6 GTO; 2s,2p->6 GTO; 3s,3p->6 GTO; 4s,4p-> 3 GTO +1 GTO; 3d-> 3 GTO + 1 GTO}$

The same number of orbitals and GTOs are used in the same row. For example, if you look at the 6-31G basis set for silicon, you would see the same type of assignment as in phosphorus. This type of scheme seems to be consistent in other basis set families too, as mentioned in @NikeDattani's answer.

[All basis sets are from Basis Set Exchange: https://www.basissetexchange.org/, and in GAMESS(US) format]

  • 1
    $\begingroup$ Thanks for your answer and very detailed examples! It will be very helpful for my study. $\endgroup$
    – neco
    Jun 19, 2021 at 12:16

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