Physical Chemistry , 1st ed.

With the substitutions, the expressions for the average energies become

E 1 

H

1

11





S

H

12

^12

E 2 

H

1

22





S

H

12

^12

(12.45)

where E 1 is less than E 2. Curiously, the sum of molecular orbital energies E 1

and E 2 is not the same as the sum of the two original atomic orbital energies.

The total energy depends on the magnitudes ofH 12 and S 12 ; in the case of H 2 ,

the sum of the two orbital energies has increased slightly. This is illustrated in

Figure 12.17. The totalenergy of the system includes not just the energy of the

orbitals but the repulsive energy due to the two nuclei at some distance R. This

total energy must therefore be evaluated to minimize the energy in terms ofR,

H 12 , and S 12. The resonance and overlap integrals can be evaluated analytically

using elliptical polar coordinates (instead of spherical polar coordinates).* The

expressions one obtains for the integrals upon application of elliptical polar

coordinates are

H 12 E 11 S 12  2 E 11 eR/a^0  1 

a

R

0

 (12.46)

S 12 eR/a^0  1 

a

R

0



3

R

a

2

 (^20)  (12.47)
where E 11 is simply the energy of the atomic hydrogen 1sorbital and a 0 is the
first Bohr radius, 0.529 Å. The only remaining parameter is R. By varying
Rand calculating the energy, we can find the internuclear distance at which the
energy is minimized. When this is done, an Rof 1.32 Å and an energy of
2.82 10 ^19 J with respect to H Hare calculated. (The energy of infi-
nitely separated H His arbitrarily set equal to zero. Our result means that
the H 2 system is calculated as being 2.82 10 ^19 J lower in energy than the
two separated atoms, meaning that it is more stable by that amount.) This can
be compared to experimentally determined properties ofR1.06 Å and
E4.76 10 ^19 J (relative to H H). Not bad for a first approximation.
Example 12.13
Comment on the value ofS 12 as the interatomic distance Rgoes from 0 to .
Solution
When Ris zero, the two nuclei essentially represent a single atom of charge
2 . Due to orthonormality considerations, the overlap integral S 12 should be
equal to 1. As the two nuclei separate, they get farther and farther apart and
the ground-state atomic wavefunctions overlap less and less. At R,each
atom is effectively isolated from the other and the overlap integral should es-
sentially be 0. At intermediate distances,S 12 can have any value between 0 and

1. This analysis does illustrate why S 12 is called an overlap integral. It indi-
   cates a relative amount of overlap between the atomic orbitals.

408 CHAPTER 12 Atoms and Molecules

*In fact, H 2 is a system that can be solved analytically if the Born-Oppenheimer ap-

proximation is imposed on the system first. Otherwise, it is not analytically solvable.

Energy

(H+ 2 , 1)

(H+ 2 , 2)

(H1) (H2)

Figure 12.17 When the AOs of two hydrogen
atoms combine to make MOs of H 2 , the anti-
bonding orbital increases in energy slightly more
than the bonding orbital decreases in energy. (The
energy axis is not to scale.) Expressions like equa-
tions 12.42–12.44 can be used to estimate the dif-
ferences between atomic and molecular orbital
energies.