Chemistry 3730 Fall 2000 Assignment 6 Solutions

1.

I first performed a UHF calculation with the STO-3G^* basis set for NaI. The SCF energy and MP2 correction are

$$% latex2html id marker 65 \begin{array}{rcrl} E_\mathrm{SCF}... ... \therefore E_\mathrm{NaI} & = & -4\,402\,394.90 \end{array}$$

To avoid basis-set superposition errors, we repeated the above calculation for iodine with a sodium ghost atom:

$$% latex2html id marker 67 \begin{array}{rcrl} E_\mathrm{SCF}... ...\\ \therefore E_\mathrm{I} & = & -4\,301\,942.49 \end{array}$$

Performing a similar calculation for the sodium atom, we get

$$% latex2html id marker 69 \begin{array}{rcrl} E_\mathrm{SCF}... ...57\\ \therefore E_\mathrm{Na} & = & -100\,403.95 \end{array}$$

The dissociation energy is therefore

$$\Delta E = E_\mathrm{Na}+E_\mathrm{I}-E_\mathrm{NaI} = 48.46\,\mathrm{kcal/mol}\equiv 3.367\times 10^{-19}\,\mathrm{J}.$$

This is a considerably better value than I obtained without considering correlation effects ( $2.677\times 10^{-19}\,\mathrm{J}$), although it is still quite a bit lower than the value obtained by fitting.

2.

(a)

$$% latex2html id marker 75 \begin{array}{rcrl} E_\mathrm{SCF}... ...41\\ \therefore E_\mathrm{min} & = & -49\,929.63 \end{array}$$

(b)

$$% latex2html id marker 77 \begin{array}{rcrl} E_\mathrm{SCF}... ....38\\ \therefore E_\mathrm{TS} & = & -49\,926.65 \end{array}$$

We can now calculate the activation energy by subtracting the ground-state energy from the transition-state energy:

$$\begin{array}{rcrl} E_a^\mathrm{SCF} & = & 2.95 & \mathrm{kcal/mol}\\ E_a^\mathrm{MP2} & = & 2.98 \end{array}$$

The activation barrier is almost exactly the same, with or without correlation. The reason is that the correlation energy is essentially identical in both conformations. Since the bonds in both conformations are identical, this is fully to be expected.

3.

I started by performing a geometry optimization. I then examined the orbital energies. There is a large gap in energy between the core carbon orbitals and the next orbital up (orbital 7), which has an energy of $-31.49\,\mathrm{eV}$. There is also a relatively large gap between unoccupied orbital 34 and orbital 35: The former has an energy of 14.27eV while the latter is at 22.13eV. Thus, if I allow excitations of up to 50eV in the CI calculations, all possible combinations of excitations from the valence orbitals to orbitals with energies under 18.5eV will be considered. The CI calculation then has a line with nonzero intensity at 137.55nm. Lines corresponding to lower excitation energies all have zero intensity. Because most UV-visible spectrometers can only record spectra down to 180nm or so, benzene is unlikely to have a very strong UV-visible spectrum.