edited bv:
exam que~tionexchange MO and VB Wave Functions for He2 Waf-Kee LI me Chinese University of Hong Kong Shatin. N.T.. Hong - Kong ~
Even though the molecular orhital (MO) andvalence bond (VB) methods become equivalent if worked to their logical extreme, these two treatments do begin from different bases and with different wave functions. Thus, in textbooks covering elementary bonding theory, considerable effort is spent in comoarine the wave functions of these twomethods, often using HZ as &I example. I t is the purpose of this question to point out that, for some systems, the MO and VB methods actually have the same wave function. The simplest, though trivial, example of these systems is the one-electron case of Hz+. A lesser known, but considerably more interesting, example is the Hez system discussed below. Since no complicated mathematics is involved in this question, i t may be instructive to present these results to undergraduate students studying the MO and VB theories; this should help them to gain a deeper understanding of the underlying principles of the two methods. This question is suitable for the undergraduate or first-year graduate courses in physical and inorganic chemistry. Ouedlon
In the MO treatment of Hz, the wave function for the bonding orbital ol, i s g = a b, while that for the antibonding orbital al,* is u = a - b, where a and b represent 1s orbitals at nuclei a and b, respectively. Thus, the MO wave , ~ is (including spin) function for the ~ 1configuration
+
VMo(H,)= [a(l)+ b(l)][a(2)+ b(2)lla(1)8(2)- a(2)8(1)1 where 1 and 2 denote the coordinates of electrons 1and 2, respectively. On the other hand, in the VB scheme, the wave function for the same state of Hz is qve(H,) = [a(l)b(2)+ b(l)o(2)l[a(l)P(2)- a(W(1)I (1) Using the aforementioned notation, write down the determinantal MO and VB wave functions for the a1,2a~,'z
JOHN J. ALEXANDER
Univel~ity01 Cincinnati Cincinnati. Ohio 45221
configuration of He2. Furthermore, show that these two wave functions are identical, aside from a numerical factor. (Hint: a general theorem on determinants states that, if the rows. or columns. of a determinant are related to the rows or columns of another determinant by a linear transformation, the two determinants are identical aside from a numerical factor.) (2) Why do the MO and VB treatments for He2 lead to identical results, though for Hz they do not?
On substitutingthe expressions for g and u, it is seen that each row of ~ M isO a linear combination of two rows from the YVBdeterminant. For instance, row 1of Y Mis~the sums of rows I and 3 of 'Yve. Hence, these two functions are identical, aside from a numerical factor, which can be removed by normalization. (2) If we compare the two wave functionsfor Hz, it is immediately obvious that the ionic terms 0(1)a(2) and b(l)b(2) are included in the MO function, while they are absent in the VB version. That is, there is no correction for electron correlation in the MO function; the two electrons are free to move independently of each other, and both are allowed to be on the same hydrogen nucleus, resulting in an ionic state. However, for Hez, the formation of such ionic states is disallowed by the excluaion principle. Since He in its ground state already has two electrons in its 1s orbital, there is no chance to form an ion by electron build-up, unless 2s or other orbitals are considered. These are disregarded in this first approximation. In other words, for He2,the exclusion principle automatically takes care of the major effects of correlation, which is the main difference between the MO and VB functions for Hz. Hence, the MO and VB wave functions for He2 are identical. Indeed, this situation arises whenever we are considering the interaction of two atoms or ions having only electrons in closed shells.
Volume 67
Number 2
February 1990
131
