Page 284 - Essentials of physical chemistry

246 Essentials of Physical Chemistry

TABLE 11.3
STO-3G One-Electron Energies for Pi-Molecular
Orbitals of Benzene (Hartrees)

E(24) ¼ 0.4940
E(23) ¼ 0.2664
E(22) ¼ 0.2664
E(21) ¼ 0.2699
E(20) ¼ 0.2699
E(17) ¼ 0.4354

Note: STO-3G ) Slater-type-orbitals fitted with a linear
combination of three Gaussian orbitals. See Chapter 17.

l p!p* ffi 2:12607 10 7 m ¼ 2126:07 10 10 m ffi 2126 Å but the experimental transition for
benzene is at 262 nm ¼ 2620 Å. Let us consider what value of the radius would produce the
experimental wavelength.
s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
r ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
8p ma c 3lh 3(2620 10 10 m)(6:62606896 10 34 J s)
,
l p!p* ¼ ) a ¼ 2 ¼ 2 31 8
3h 8p mc 8p (9:10938215 10 kg)(2:99792458 10 m=s)
2
2
8p ma c 10
) a ffi 1:554 10 m ¼ 1:554 Å. Noting that the C–H bonds are about 1 Å
3h
l p!p* ¼
the width across the benzene molecule from H to H would be about 4.8 Å for a radius of 2.4 Å. Thus
we have had to adjust the effective radius to be slightly larger than directly over the C atoms but not
by much and one might imagine that the pi electrons are somewhat attracted to the H atoms.
Actually, we have only adjusted the radius to fit the experimental wavelength. However, it is
pleasing that such a simple model comes so close to the experimental value! Figure 11.5 shows the
coefficients ( 1000) of the pi-orbitals of benzene from an all-electron self-consistent field calcula-
tion where it is evident that the lowest energy orbital is totally symmetric with all the coefficients the
same, while the next orbitals have nodal patterns in a way that corresponds to the POR wave
iu
solutions if we use e ¼ cos(u)þisin(u).
Can this model be extended further? With one additional, fairly severe, assumption, the POR
model can be extended to other aromatic hydrocarbons [4–6]. The main assumption is that
the interior bonds of a poly-aromatic hydrocarbon such as naphthalene are less important than the
delocalized pathway on the outer circumference of the ring structure. A crude demonstration can
be made using a wire coat hanger that has a specific shape but that can be easily bent into a circle
(with a handle). If we break the middle bond in naphthalene and bend the outer bonds into a circular
polygon, we can apply the POR model to the ring. There are ten bonds in the outer edge of
naphthalene and we can use the approximation that their length is about 1.4 Å which is roughly
the bond length in benzene. Thus we can set the circumference of our model ring equal to (10)
(1.4 Å) ¼ 14 Å ¼ C ¼ 2pa. That gives us a radius for our POR ring.
14 A ˚
¼ 2:228 Å. Then we note there are 10 pi-electrons in naphthalene so we expect the
2p
a ¼
POR model to include 5 pairs of electrons. That means the HOMO ! LUMO transition will be
from n ¼ 2to n ¼ 3. Then we can calculate the p ! p* wavelength as
2
2
2
2
hc (3 2 ) 2 3h 2 8p ma c
h
:
DE ¼ E 3 E 2 ¼ hn ¼ ¼ 2 ¼ 2 2 ) l p!p* ¼
l 2ma 2ma (2p) 5h