MolecularOrbitals - Maple Help

Molecular Orbitals of Hydrogen Fluoride

 Overview We learn in general chemistry a qualitative description of bonding known as the linear combination of atomic orbitals (LCAO). In the LCAO approach, molecular orbitals are constructed by selecting sums and differences of atomic orbitals.  For example, for hydrogen fluoride, we start with the H1s, F1s, F2s, F2px, F2py, and Fp2z orbitals.  A linear combination of H1s and F2pz orbitals creates a bonding σ orbital and antibonding σ∗ orbital.  The remaining F1s, F2s, F2px, and F2py remain non-bonding orbitals for a bond-order of 1 (See Figure 1). Figure 1: LCAO MO Diagram for HF (Author: LeeAnn Sager. Used with permission.)   The purpose of this activity is to use Hartree-Fock self-consistent-field approach to calculate the molecular orbitals of HF in the minimal atomic basis (STO-3G) and to compare the molecular orbitals to the qualitative LCAO approach.

We begin by initializing the QuantumChemistry package in Maple:

 > $\mathrm{with}\left(\mathrm{QuantumChemistry}\right):\mathrm{Digits}≔15:$

Hydrogen

Now, we define our hydrogen fluoride molecule.  Instead of using the experimental bond distance for HF, here we choose an initial guess of 1 Angstrom for the bond distance.

 > $\mathrm{molec}≔\left[\left["H",0,0,0\right],\left["F",0,0,1\right]\right];$
 ${\mathrm{molec}}{≔}\left[\left[{"H"}{,}{0}{,}{0}{,}{0}\right]{,}\left[{"F"}{,}{0}{,}{0}{,}{1}\right]\right]$ (2.1)

Now, we can both optimize the geometry to determine the HF bond distance and calculate the MOs all with the GeometryOptimization command:

 > $\mathrm{molec},\mathrm{output}≔\mathrm{GeometryOptimization}\left(\mathrm{molec},\mathrm{HartreeFock}\right);$
 ${\mathrm{molec}}{,}{\mathrm{output}}{≔}\left[\left[{"H"}{,}{0}{,}{0}{,}{0}\right]{,}\left[{"F"}{,}{-1.35214125}{}{{10}}^{{-11}}{,}{-8.45207005}{}{{10}}^{{-13}}{,}{0.95545921}\right]\right]{,}{table}{}\left(\left[{\mathrm{dipole}}{=}\left[\begin{array}{c}{1.77169432}{}{{10}}^{{-11}}\\ {0.}\\ {-1.25199508}\end{array}\right]{,}{\mathrm{mo_occ}}{=}\left[\begin{array}{c}{2.00000000}\\ {2.00000000}\\ {2.00000000}\\ {2.00000000}\\ {2.00000000}\\ {0.}\end{array}\right]{,}{\mathrm{converged}}{=}{1}{,}{\mathrm{aolabels}}{=}\left[\begin{array}{c}{"0 H 1s"}\\ {"1 F 1s"}\\ {"1 F 2s"}\\ {"1 F 2px"}\\ {"1 F 2py"}\\ {"1 F 2pz"}\end{array}\right]{,}{\mathrm{mo_symmetry}}{=}\left[\begin{array}{c}{"A"}\\ {"A"}\\ {"A"}\\ {"A"}\\ {"A"}\\ {"A"}\end{array}\right]{,}{\mathrm{e_tot}}{=}{-98.57284735}{,}{\mathrm{group}}{=}{"C1"}{,}{\mathrm{mo_coeff}}{=}\left[\begin{array}{cccccc}{-0.00534051}& {0.15042852}& {-0.53370653}& {0.}& {0.}& {1.05434254}\\ {0.99474577}& {-0.25067856}& {-0.07826747}& {0.}& {0.}& {0.08056985}\\ {0.02225996}& {0.94670532}& {0.41089991}& {0.}& {0.}& {-0.51585168}\\ {0.}& {0.}& {0.}& {0.93575279}& {-0.35265666}& {-1.15541533}{}{{10}}^{{-11}}\\ {0.}& {0.}& {0.}& {0.35265666}& {0.93575279}& {0.}\\ {-0.00267276}& {-0.07825318}& {0.69805959}& {1.35526553}{}{{10}}^{{-11}}& {0.}& {0.81643460}\end{array}\right]{,}{\mathrm{populations}}{=}\left[\begin{array}{c}{0.80773900}\\ {1.99912967}\\ {1.94822186}\\ {2.00000000}\\ {2.00000000}\\ {1.24490947}\end{array}\right]{,}{\mathrm{charges}}{=}\left[\begin{array}{c}{0.19226100}\\ {-0.19226100}\end{array}\right]{,}{\mathrm{rdm1}}{=}\left[\begin{array}{cccccc}{0.61499985}& {-0.00249959}& {-0.15401474}& {1.08778088}{}{{10}}^{{-11}}& {0.}& {-0.76863240}\\ {-0.00249959}& {2.11696937}& {-0.49467165}& {0.}& {0.}& {-0.07535536}\\ {-0.15401474}& {-0.49467165}& {2.13117040}& {0.}& {0.}& {0.42538084}\\ {1.08778088}{}{{10}}^{{-11}}& {0.}& {0.}& {2.00000000}& {0.}& {1.43388491}{}{{10}}^{{-11}}\\ {0.}& {0.}& {0.}& {0.}& {2.00000000}& {0.}\\ {-0.76863240}& {-0.07535536}& {0.42538084}& {1.43388491}{}{{10}}^{{-11}}& {0.}& {0.98683578}\end{array}\right]{,}{\mathrm{mo_energy}}{=}\left[\begin{array}{c}{-25.90350585}\\ {-1.45985743}\\ {-0.57365324}\\ {-0.46311953}\\ {-0.46311953}\\ {0.58982373}\end{array}\right]\right]\right)$ (2.2)

How does the Hartree-Fock bond distance in the STO-3G basis compare with the experimental value?

 Answer HF has a bond distance of 0.91 Angstroms, so the 0.9555 Angstrom is slightly larger with a 5% error.

How do the mo_energies compare with what was expected from Figure 1?

 Answer We see that the 4th and 5th MO energies are degenerate, as expected.

The coefficients of our LCAO-MOs are listed in the output table in the mo_coeff matrix. Each column of the mo_coeff matrix corresponds to an MO, and the column elements correspond to H1s, F1s,  F2s, F2px, F2py, and F2pz, respectively.   Consider the first MO:

 > $\mathrm{output}\left[\mathrm{mo_coeff}\right]\left[...,1\right];$
 $\left[\begin{array}{c}-0.00534050903\\ 0.994745769\\ 0.0222599608\\ 0.000000000\\ 0.000000000\\ -0.00267276436\end{array}\right]$ (2.3)

Note that the coefficient corresponding to the F1s orbital is very nearly 1, as expected from Figure 1.  We can plot the MO using the DensityPlot3D function:

 > $\mathrm{DensityPlot3D}\left(\mathrm{molec},\mathrm{output},\mathrm{orbitalindex}=1,\mathrm{densitycutoff}=0.0001\right);$

Now consider the 2nd MO.  By inspection of the 2nd column of the mo_coeff matrix, we note that the coefficient for the F2s orbital is almost one (with only a little mixing of the H1s and F1s orbitals), again as expected from Figure 1.  However, the density plot of the 2nd MO reveals that there is enough contribution from H1s for it to be a bonding σ orbital.

 > $\mathrm{output}\left[\mathrm{mo_coeff}\right]\left[...,2\right];$
 $\left[\begin{array}{c}0.150428516\\ -0.250678564\\ 0.946705318\\ 0.000000000\\ 0.000000000\\ -0.0782531835\end{array}\right]$ (2.4)
 > $\mathrm{DensityPlot3D}\left(\mathrm{molec},\mathrm{output},\mathrm{orbitalindex}=2,\mathrm{densitycutoff}=0.0001\right);$

An inspection of the 3rd column of mo_coeff suggests that the H1s, F2s, and the F2pz orbitals contribute significantly, giving rise to a σ orbital {Note σ = 0.698 F2pz + (-0.534 H1s) + 0.411 F2s}.  The density plot reveals that the σ orbital has a node in the middle of the bond and hence, is an antibonding orbital.  Importantly, this result differs somewhat from the conventional MO picture, shown above, in which this orbital is a bonding orbital.

 > $\mathrm{output}\left[\mathrm{mo_coeff}\right]\left[...,3\right];$
 $\left[\begin{array}{c}-0.533706534\\ -0.0782674660\\ 0.410899912\\ 0.000000000\\ 0.000000000\\ 0.698059587\end{array}\right]$ (2.5)
 > $\mathrm{DensityPlot3D}\left(\mathrm{molec},\mathrm{output},\mathrm{orbitalindex}=3,\mathrm{densitycutoff}=0.0001\right);$

As expected from Figure 1, the 4th and 5th MOs should be pure F2px and F2py orbitals. We see that for the 4th MO, the coefficient is indeed 1 for the F2px orbital and 0 for all other orbitals, as expected! Similarly for the 5th MO.

 > $\mathrm{output}\left[\mathrm{mo_coeff}\right]\left[...,4\right];$
 $\left[\begin{array}{c}0.000000000\\ 0.000000000\\ 0.000000000\\ 0.935752790\\ 0.352656656\\ 1.35526553{}{10}^{-11}\end{array}\right]$ (2.6)
 > $\mathrm{DensityPlot3D}\left(\mathrm{molec},\mathrm{output},\mathrm{orbitalindex}=4,\mathrm{densitycutoff}=0.0001\right);$
 > $\mathrm{output}\left[\mathrm{mo_coeff}\right]\left[...,5\right];$
 $\left[\begin{array}{c}0.000000000\\ 0.000000000\\ 0.000000000\\ -0.352656656\\ 0.935752790\\ 0.000000000\end{array}\right]$ (2.7)
 > $\mathrm{DensityPlot3D}\left(\mathrm{molec},\mathrm{output},\mathrm{orbitalindex}=5,\mathrm{densitycutoff}=0.0001\right);$

Finally, the 6th MO should be an antibonding σ* orbital composed of σ* = 1.054 H1s + 0.816 F2pz - 0.516 F2s: (Note that this antibonding orbital has two nodes!)

 > $\mathrm{output}\left[\mathrm{mo_coeff}\right]\left[...,6\right];$
 $\left[\begin{array}{c}1.05434254\\ 0.0805698453\\ -0.515851681\\ -1.15541533{}{10}^{-11}\\ 0.000000000\\ 0.816434598\end{array}\right]$ (2.8)
 > $\mathrm{DensityPlot3D}\left(\mathrm{molec},\mathrm{output},\mathrm{orbitalindex}=6,\mathrm{densitycutoff}=0.01\right);$

Why do the F2px and F2py orbitals not mix with the H1s?

 Answer With respect to the molecular axis,  H1s and F2pz orbitals both have σ-symmetry and can therefore overlap to form σ and σ* bonds.  The F2px and F2py orbitals have π-symmetry and therefore cannot overlap with the H1s.