Chemistry 3730 Fall 2000 Quiz 5 Solutions

1.

Here are my results:

Excitation energy (eV)	Wavelength (nm)	Intensity
16	262.40	0
	226.40	0
	222.48	0
	189.34	0
	183.91	0.011
	161.56	0
26	309.18	0
	227.69	0
	223.14	0
	189.60	0
	184.23	0.012
	163.74	0
	144.17	0
	141.39	0
62	340.46	0
	232.12	0
	227.84	0
	190.35	0
	185.56	0.017
	176.08	0.001
	149.38	0
	146.73	0
73	340.90	0
	232.19	0
	227.91	0
	190.38	0
	185.59	0.017
	176.87	0.001
	149.44	0
	146.82	0

Of course, your results will differ from mine, at least for smaller excitation energies. If we include only a few of the levels near the HOMO-to-LUMO gap, we don't get very accurate results at all. There are at least two tenable strategies:

(a)

Make sure that we include all of the levels which are relatively close to the HOMO-to-LUMO gap, i.e. choose the excitation energy based on the largest gaps in the orbital energies.

(b)

Start with small excitations and increase the CI excitation energy by (say) 10eV at a time until no further change is observed in the spectral region of interest. In the case of fluorobenzene, an excitation energy of 60eV (or so) seems to be sufficient to give good results above 140nm.

2.

I started by optimizing the geometry of N₂ with the 6-311G^* basis set. I then performed a single-point MP2 calculation:

$$% latex2html id marker 42 \begin{array}{rcrl} E_\mathrm{SCF}... ...26\\ \therefore E_\mathrm{N_2} & = & -68\,609.25 \end{array}$$

I then converted one of the nitrogen atoms to a ghost atom. The ground-state electronic configuration of a nitrogen atom is 1s²2s²2p³ so the multiplicity is 4. The energy of a single nitrogen atom is

$$% latex2html id marker 50 \begin{array}{rcrl} E_\mathrm{SCF}... ...4.52\\ \therefore E_\mathrm{N} & = & -34\,200.03 \end{array}$$

The dissociation reaction is $\mathrm{N}_2\rightarrow 2\mathrm{N}$ so

$$\Delta E = 2E_\mathrm{N} - E_\mathrm{N_2} = 209.19\,\mathrm{kcal/mol}.$$

The experimental dissociation energy is $225\,\mathrm{kcal/mol}$, so this calculation doesn't do at all badly.