Problem Set 5
Chem 1311
1. Use the molecular orbital energies (eV) shown on the next page for azide ion, N3, and the
energies for the nitrogen s (-26.0 eV) and p (-13.4 eV) orbital energies to construct a molecular
orbital energy level diagram. Show the occupancy of the orbitals.
Obviously one nitrogen is the central atom of the triatomic species and the other two are
terminal atoms. Thus there will be an s and three p orbitals on one side of the diagram and two
s orbital and 6 p orbitals on the other side. The two s orbitals will generate two linear
combinations (can you draw them) and the six p orbitals will generate six linear combinations
(three pairs; can you draw them).
The following molecular orbital energy level diagram is drawn roughly to scale for all but the
highest energy orbitals whose calculated energies are actually much too high, one of the
failings of low-level molecular orbital calculations. Note that the symmetry of each the orbital is
shown. Also note that the four lower energy sigma orbitals all have at least some electron
density between the nuclei and as such range from fully bonding to at least partially bonding
(examine the drawings of the orbitals and their mathematical functions).
c6
c5
_*
_n
c4
_
c3
p
p
c2
s
N1
N3
s
N2
c1
2. a) Draw representations of the sigma molecular orbitals that would be expected for azide
ion if the s orbital on the center nitrogen only interacted with the s orbitals on the terminal
nitrogens and the p orbital on the center nitrogen only interacted with the p orbitals on the
terminal nitrogens. Altogether you should depict 6 molecular orbitals.
3
6
2
5
1
4
b) Compare these s-only and p-only orbitals with the orbitals shown on the attached sheet and
suggest which correspond most closely.
Examination of the molecular orbitals for azide reveals that there are no strictly antibonding
orbitals like those shown above. However, careful examination of the azide mo’s will indicate
that several are mixtures of these combinations.
The mo at -17.5 eV can be viewed as a combination of 3 and 5 with the same relative phases
as shown above.
The mo at -14.6 eV can be viewed as a combination 4 and 2 with the relative phases of 2
reversed from that shown above
The mo at 34.7 eV is a combination 3 and 5 with the phases of 3 reversed from what is shown
above.
The highest energy orbital (39.8 eV) is a combination of 2 and 6 with the phase of 6 reversed
from what is shown above.
With regard to the other sigma orbitals, the one at -33.3 eV is exclusively combination 1. The
orbital at -26.8 eV is not a combination of two of the above, but it can be described as a
combination of 2 with the central atom p orbital.