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As I know, for Hydrogen atom a $1s$ Slater Type Orbital (STO) equation is (I can get it from here):

\begin{equation}\label{STO} \mathrm{STO} = \sqrt{\frac{\zeta^3}{\pi}} e^{-\zeta r}. \end{equation}

With remark from Szabo A., Ostlund N.S. Modern Quantum Chemistry: Intro to Advanced Electronic Structure Theory, Dover, 1996:

For example, the standard exponent for the $1s$ basis function of hydrogen is , $\zeta = 1.24$. This is larger than the $\zeta = 1.00$ exponent of the hydrogen atom, since the hydrogen $1s$ orbital in average molecules is known to be "smaller" or "denser" than in the atom.

Gaussian Type Orbital (GTO) equation for $1s$-orbital (where did this coefficient $\left(\frac{2\alpha}{\pi}\right)^{3/4}$ come from?):

\begin{equation} \mathrm{GTO} = \left(\frac{2\alpha}{\pi}\right)^{3/4} e^{-\alpha r^2} \end{equation}

So, I have good analitical form of Hydrogen STO and GTO, and using formula \begin{equation}\label{} \sqrt{\frac{\zeta^3}{\pi}}e^{-\zeta r} \approx C_1 \left(\frac{2\alpha_1}{\pi}\right)^{3/4}e^{-\alpha_1 r^2} + C_2 \left(\frac{2\alpha_2}{\pi}\right)^{3/4}e^{-\alpha_2 r^2} + C_3 \left(\frac{2\alpha_3}{\pi}\right)^{3/4}e^{-\alpha_2 r^2}. \end{equation} and copyng $\alpha_i$'s and $C_i$ from basissetexchange.org

#-----------------------------------------------------
# Basis Set Exchange
# Version v0.8.13
# https://www.basissetexchange.org
#-----------------------------------------------------
#   Basis set: STO-3G
# Description: STO-3G Minimal Basis (3 functions/AO)
#        Role: orbital
#     Version: 1  (Data from Gaussian09)
#-----------------------------------------------------


BASIS "ao basis" PRINT
#BASIS SET: (3s) -> [1s]
H    S
      0.3425250914E+01       0.1543289673E+00
      0.6239137298E+00       0.5353281423E+00
      0.1688554040E+00       0.4446345422E+00
END

I can plot Hydrogen orbital

enter image description here

Now I want to do the same for a Lithium atom using data

BASIS "ao basis" PRINT
#BASIS SET: (6s,3p) -> [2s,1p]
Li    S
      0.1611957475E+02       0.1543289673E+00
      0.2936200663E+01       0.5353281423E+00
      0.7946504870E+00       0.4446345422E+00
Li    SP
      0.6362897469E+00      -0.9996722919E-01       0.1559162750E+00
      0.1478600533E+00       0.3995128261E+00       0.6076837186E+00
      0.4808867840E-01       0.7001154689E+00       0.3919573931E+00
END

But, I do not understand where I can get a analitical formula for GTO's primitives. All of references, I saw: \begin{equation}\label{GTO} \mathrm{GTO}(x,y,z;\alpha,\ell_1,\ell_2,\ell_3) = N x^{\ell_1} y^{\ell_2} z^{\ell_3} e^{-\alpha r^2} \end{equation}
with unknown normalizing factor $N$. Sometimes I saw for $1p$-orbital \begin{equation} \mathrm{GTO}(p_x) = \left(\frac{128\alpha^5}{\pi^3}\right)^{1/4} x \exp(-\alpha r^2) \end{equation} but I don't understand where the factor $\left(\frac{128\alpha^5}{\pi^3}\right) $ comes from. So, my general question: where can I see the analytical view of the GTO primitive (like for STO's), and how can you find $\zeta$ for atom obital? (how to read such data?)

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    $\begingroup$ I believe this website (ccl.net/cca/documents/basis-sets/basis.html ) might be helpful, even if it does not answer all the questions. For the 1s STO-3G orbital, the $\mathrm{(\frac{2\alpha}{\pi})^{3/4}}$ is in fact, the normalization constant. So I believe that for the p orbitals (or other orbitals) you only need the C and $\alpha$, then you can normalize that. The 'factor' that you see is in fact the algebraic form of the normalization constant. $\endgroup$ – Shoubhik R Maiti Feb 8 at 19:00
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Main differenece betweet STO's and GTO's is a $r$-expotetnt. Normalizing factor of GTO one can get from common procedure: \begin{equation} N^2 \int_0^\infty \left(r^{n-1}e^{-\zeta r^2}\right)^2 r^2 dr =1 \end{equation} Of course, computation is not so simple for $n > 1$, but we can use online WolframAlpha (clickable calculation) and we get the formulas from the question post.

For a Lithium atom we have contraction (6s,3p) -> [2s,1p], which means the first 3s of 6s GTO's cotract to one s "1s"-STO $$\mathrm{STO}(1s) = \sqrt{\frac{\zeta^3}{\pi}} e^{-\zeta_1 r},$$ second 3s of 6s GTO's contract to another s "2s"-STO
$$\mathrm{STO}(2s) = \sqrt{\frac{\zeta^5}{3\pi}} r e^{-\zeta_2 r},$$ 3p GTO's contract to one p - "2p"-STO $$\mathrm{STO}(2p) = \sqrt{\frac{\zeta_2^5}{\pi}} x e^{-\zeta_2 r}$$ (notation in orbitals in quotation marks "1s", "2s" means the same as for the hydrogen atom). So, as we can see, every atomic STO contract by only 3 GTO's (which justifies their name).


For a Litium atom we have for "1s" atomic orbital \begin{equation} \sqrt{\frac{\zeta_1^3}{\pi}} e^{-\zeta_1 r} \approx C_1 \left(\frac{2\alpha_1}{\pi}\right)^{3/4}e^{-\alpha_1 r^2} + C_2 \left(\frac{2\alpha_2}{\pi}\right)^{3/4}e^{-\alpha_2 r^2} + C_3 \left(\frac{2\alpha_3}{\pi}\right)^{3/4}e^{-\alpha_3 r^2} \end{equation} where \begin{align} \alpha_1 = 0.1611957475E+02; C_1 = 0.1543289673E+00;\\ \alpha_2 = 0.2936200663E+01; C_2 = 0.5353281423E+00;\\ \alpha_3 = 0.7946504870E+00; C_3 = 0.4446345422E+00; \end{align} enter image description here

For "2s" atomic orbital \begin{equation} \sqrt{\frac{\zeta_2^5}{3\pi}} r e^{-\zeta_2 r} \approx C_4 \left(\frac{2\alpha_1}{\pi}\right)^{3/4}e^{-\alpha_4 r^2} + C_5 \left(\frac{2\alpha_2}{\pi}\right)^{3/4}e^{-\alpha_5 r^2} + C_6 \left(\frac{2\alpha_3}{\pi}\right)^{3/4}e^{-\alpha_6 r^2} \end{equation} where \begin{align} \alpha_4 = 0.6362897469E+00; C_4 = -0.9996722919E-01;\\ \alpha_5 = 0.1478600533E+00; C_5 = 0.3995128261E+00; \\ \alpha_6 = 0.4808867840E-01; C_6 = 0.7001154689E+00; \end{align} enter image description here For "2p_x" atomic orbital \begin{equation} \sqrt{\frac{\zeta_2^5}{\pi}} x e^{-\zeta_2 r} \approx D_1 \left(\frac{128\alpha_4^5}{\pi^3}\right)^{1/4} xe^{-\alpha_4 r^2} + D_2 \left(\frac{128\alpha_5^5}{\pi^3}\right)^{1/4} xe^{-\alpha_5 r^2} + D_3 \left(\frac{128\alpha_6^5}{\pi^3}\right)^{1/4} xe^{-\alpha_6 r^2} \end{equation} where \begin{align} \alpha_4 = 0.6362897469E+00; D_1 = 0.1559162750E+00;;\\ \alpha_5 = 0.1478600533E+00; D_2 = 0.6076837186E+00; \\ \alpha_6 = 0.4808867840E-01; D_3 = 0 0.3919573931E+00; \end{align} enter image description here (For $p_y$ and $p_z$ --- do the same)

Also, for STO's best fit $\zeta_1 = 2.69$ and $\zeta_2 = 0.81$ are consistent with those given in the article Hehre, W. J., Ditchfield, R., Stewart, R. F., & Pople, J. A. (1970). Self‐Consistent Molecular Orbital Methods. IV. Use of Gaussian Expansions of Slater‐Type Orbitals. Extension to Second‐Row Molecules. The Journal of Chemical Physics, 52(5), 2769–2773. doi:10.1063/1.1673374

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