15
$\begingroup$

I am a little confused about how to decide how many basis functions that are used in a particular basis set for a given molecule.

STO-3G is a so-called "minimal basis set", meaning that only one basis function is used for each atomic orbital in the atoms of which the molecule is made from.

The water molecule has two H atoms and one O atom, with 1s1 and 1s2 2s2 2p6 electron configurations, respectively. When STO-3G is used, would that then mean that three 1s type basis functions, one 2s type basis function, and 3 p-type basis functions are used, totalling at 7 basis functions that are used, each being made up from a linear combination (LC) of three simple Gaussian functions?

The 6-31+G** is a split-valence basis set, which includes both diffuse and polarisation functions. The two asterisks tell that polarization functions are used for both the H atom and the O atom, p and d polarization functions respectively. I am not sure "how many" polarization functions are used per atom - just 1, or one for each of the p-orbitals? (Some basis sets are of the form 6-31G(3pd); does that mean that 3 p and 1 d polarization functions are used, or 9 p functions?) The single "+" sign means that just one diffuse function is used for the O atom.
So, I think that the 6-31+G** for the water molecule equates to a total of 16 basis functions:

  • 1s in oxygen: one LC of 6 GTOs
  • 2s in oxygen: one LC of 3 GTOs + 1 GTO
  • 2p in oxygen: one LC of 3 GTOs + 1 GTO (for each of px, py, and pz)
  • 1s in hydrogen 1: one LC of 3 GTOs + 1 GTO
  • 1s in hydrogen 2: one LC of 3 GTOs + 1 GTO
  • Diffuse: one diffuse basis function for the oxygen atom
  • Polarization: one polarised basis function for the oxygen atom
  • Polarization: one polarised basis function for the hydrogen atom
  • Total: 16

Is this correct? I do not feel confident carrying out this analysis, especially the point concerning how many polarization functions for the p/d-orbitals that should be used.

$\endgroup$
13
  • $\begingroup$ 6-31+G** is a synonym for 6-31+G(d,p), so one polarization function of d-type is added for O atom and one polarization function of p-type for H atoms. $\endgroup$
    – Wildcat
    Nov 23, 2015 at 19:59
  • 2
    $\begingroup$ You could just run a simple HF calculation and print the basis to check the correctness of you thoughts. $\endgroup$
    – Wildcat
    Nov 23, 2015 at 20:02
  • $\begingroup$ 6-31G(3pd) is essentially 6-31G(3p1d), i.e. it is 6-31G plus 3 p-type functions and 1 f-type function on non-hydrogen atoms. $\endgroup$
    – Wildcat
    Nov 23, 2015 at 20:05
  • $\begingroup$ Do not really have easy access to Gaussian. So after doing some searching, a d-type polarization function consists of six (not five) functions, and a p-type polarizastion function is three functions. Hence the correct total should be 23? $\endgroup$
    – Yoda
    Nov 23, 2015 at 20:05
  • $\begingroup$ Ouch. I was not very careful with words. Say, for 6-31+G(d,p) one set of polarization function of d-type is added for O atom, not just one d-type function, of course. There are always just 3 p-type functions, while how many d- and f-type functions are there depends on do you use spherical or Cartesian harmonics. $\endgroup$
    – Wildcat
    Nov 23, 2015 at 20:08

1 Answer 1

11
$\begingroup$

Notation

  • ** is just a synonym for (d,p), so that 6-31+G** basis is just a different name of 6-31+G(d,p) one.
  • The meaning of (d,p) is that one set of d-type polarisation functions is added for heavy atoms and one set of p-type polarisation functions is added for hydrogens.
  • 6-31G(3pd) is essentially 6-31G(3p1d), i.e. it is 6-31G plus 3 sets of p-type polarisation functions and 1 set of f-type polarisation functions on heavy atoms.

How many functions are there in a set?

  • There are always 3 functions of p-type, but the situation with d- and f-type functions is a bit more involved.
  • There are either 5 or 6 functions of d-type and either 7 or 10 functions of f-type depending on is a calculation restricted to use only spherical harmonics or not.

Cartesian vs. spherical harmonics

Basis functions can be written in two forms: as spherical harmonics $$ N(n, \alpha) Y_m^l r^n e^{-\alpha r^2} $$ or Cartesian ones $$ N(l_x, l_y, l_z, \alpha) x^{l_x} y^{l_y} z^{l_z} e^{-\alpha r^2} \, . $$ And for $l \geq 2$ the number of Cartesian harmonics is greater than spherical harmonics. For instance, a set of d-type function can have five spherical harmonics ($d_{3z^2-r^2}$, $d_{xz}$, $d_{yz}$, $d_{xy}$, $d_{x^2-y^2}$) or six Cartesian harmonics ($d_{zz}$, $d_{xx}$, $d_{yy}$, $d_{xz}$, $d_{yz}$, $d_{xy}$). Note that these six Cartesian harmonics can be transformed into five above mentioned spherical harmonics of d-type plus one function of s-type that can be neglected. Similarly, there are 10 Cartesian f-type functions and 7 spherical ones, and the later could be constructed from the former with 3 p-type functions as a by-product.

The choice which harmonics to use can be controlled through the corresponding option in an input file and default choice varies from program to program. If you're specifically interested in Gaussian, then, if I remember correctly, by default Cartesian d-type functions and spherical f-type functions are used for the Pople 6-31G basis set. Thus, there will be 6 functions in a set of d-type functions and 7 functions in a set of f-type functions. This might be the case that Pople basis sets were even originally defined to use Cartesian d-type functions and spherical f-type ones, but I'm not sure.

$\endgroup$
4
  • $\begingroup$ Yes, the Pople-style sets use Cartesian d-type functions, and "pure" f-type functions: "All of the built-in basis sets use pure f functions. Most also use pure d functions; the exceptions are 3-21G, 6-21G, ... 6-31G, ..." $\endgroup$ Nov 23, 2015 at 20:48
  • $\begingroup$ @GeoffHutchison, true for Gaussian, but not for, say, Molpro. $\endgroup$
    – Wildcat
    Nov 23, 2015 at 20:50
  • $\begingroup$ Indeed, I was simply commenting on your statement "if I remember correctly" - this is true for Gaussian. As you say, some programs convert to only-Cartesian or only-spherical. $\endgroup$ Nov 23, 2015 at 20:55
  • $\begingroup$ Yeah, true. That is why it is usually impossible to reproduce Gaussian results obtained with Pople bases in other programs: you just can't have Cartesian d functions and simultaneously spherical f ones. Harmonics are either all Cartesian, or all spherical in most of the programs. $\endgroup$
    – Wildcat
    Nov 23, 2015 at 20:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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