Computing Overlaps between Nonorthogonal Orbitals Carl W. David University of Connecticut, Storrs, CT 06269 The visualization of hydrogenic orbitals remains a source of unending difficulty to students, as witness the continuing flow of papers concerning bow t o draw such orbitals which have a n ~ e a r e don these oaees.'s2 I t is resonable to attack the proble'n; of understandLgthe geometry of orbitals from a different point of view for two reasons: 1. Unless students themselves actually manipulate the functions that represent orbitals, their pasaive act of staring at images yields little or no understanding. 2. An approach that exercises the calculus in a nonthreatening way is worthwhile for reinforcing some simple ideas about calculus and symmetry as it relates to the evaluation of integral^.^ Once required to understand where in space orbitals have positive and negative values, students' visualizations of spatial properties of orbitals will be enhanced. Orbltal Overlap on a Slngle Center I t is a straightforward and useful task to compute the overlap between hydrogenic orbitals on the same center, since these integrals are always zero. Thus, seeing how this orthogonality occurs in an analytical context helps to cement the idea that these orbitals are not "at right angles to each other" but instead are perpendicular in their own Hilbert space. Some integrals are easy to do and can be done by inspection; others are slightly more difficult and require manipulation. Consider the question of determining the overlap between a 1sorbital and a 2p, orbital, both located on a common atom (the origin). We then have:
the integral form -m to +m must, necessarily, vanish. Thus, we have
which is
Clearly, the x integral vanishes. Orbltal Overlap wlth Orbltalo on Different Centers The situation alters drastically when the orbitals are located on different centers, as is the case in diatomic (and higher) molecules. Thus we ask, what is the overlap between a 1sorbital on nucleus A and a 1sorbital on nucleus B, where the two are separated by the internuclear distance, R? Now, when we transform between Cartesian coordinates and spherical polar coordinates, we will not enjoy the luxury of having nuclei a t the origin! The confocal nature of the problem is apparent when one explicitly writes the distances from the two nuclei to a point inspace, P(x,y,z), where presumably an electronlurks. If one nucleus, called A, is a t (0,O,R/2) and the other (B) is a t (0,0,-R/2), then i t is apparent that the distance from nucleUSAto P(x,y,z) is r~ = \ / x Z
r~ = \/x2
T o proceed, we need the orbitals, written out in Cartesian or semi-Cartesian form. The 1s orbital is: %,o,o = % = e-'
+ y2 + - R/2)2
(8)
(2
while the distance from the B nucleus t o the same point P(x,Y,z)would be:
+ y2+ + R/2)'
(9)
(2
which may be seen by looking a t Figure 1. A I s orbital
(2)
+ +
where p = ,/xz y2 22. Further, we have (since x = p sin 8 cos @, x = p sin 8 sin 4, and z = P cos 8):
so that the desired overlap integral is
This integral can be cast into the form:
in which the integrand clearly is the product of an even function, the exponential, and an odd function (x), so that
' Allendoerfer,R. D. J. Chem. Educ 1990, 67,3739. Douglas, J. E. J. Chem. ~duc.1990, 67,42-44. David, C. W. J. Chem. Educ. 1982, 59.288-289.
Figure 1. The coordinate system
41th
nucleus-elemon
distances
shown
explicitly.
Volume 68 Number 2
February 1991
129
centered on nucleus A would have the form: la*
- eJx2+y2+cz-n~z)z -
(10)
while a 1s orbital centered on nucleus B would have the form: ,&,sB
= e~~~
(11)
leading to the rather messy integral for their mutual overlap:
Although such an integral can actually be evaluated, easily in fact if one changes coordinate systems to elliptic confocal, there are other overlap situations in which the integral can be carried out by inspection, which is infinitely preferable to this situation. This integral does not vanish, as can be seen from the fact that the inteerand is evervwhere nositive. Considernowa lsorbitzon nucleuriand sip, orbital on nucleus H.Here. as we shall see. the internal vanishes identically, due to obvious symmet& We hive for the overlap integral: Figum.2. Explicit indication t a t thereare three 8's that can be used in diatomic COmpUtBtiOnS.
Since the appearance of x in the exponential argument is quadratic, the 1: part of the integration of this function will change sign as the premultiplying x itself changes sign. Thus there will be a contribution to the integral at each point x that is exactly counterbalanced by an equivalent negative termat - x , for each and every, and in fact all, points in y and 2. The integral must vanish by symmetry. A similar integral, based on the overlap of a 2p, orbital on nucleus A and a 2py orbital on nucleus B can be speedily evaluated also. We have:
This integral vanishes, due to the change in sign properties of the product ( x y ) .
'
Adamson. A. A Textbook OlPhysIml Chemistry: Aurdemic: New Yoh. 1979: p 771. Castellan, G. W. Physimi Chemistry; Addlson-Wesley: Reading, MA, 1983: p 541.
130
Journal of Chemical Educatlon
It is clear that using 2p, orbitals results in integrals that do not vanish, leading to a bonding possibilities discussed in every t e ~ t ~ . ~
The difficulty in understanding of the above comes from the fact that orhitals are generally displayed in textbooks in spherical polar form, which is fine for atoms, hut inappropriate for diatomic and higher molecules. Thus, students are not aware of what 8 represents when one is discussing an orhital located at a place other than the origin. This is illustrated in Figure 2. Thus, a p, orbital on nucleus A is written either as: xi?
PA COB BAe
(16)
or in the form implied in eq 15. The angle 8~ is to be measured from the A nucleus, not from the origin.
