Question

(i) What is meant by linear combination of atomic
orbitals?
(ii) Illustrate bonding and antibonding molecular
orbitals based on homonuclear dihydrogen
molecule.​

Answer 1

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Contributed by Mark E. Tuckerman

Professor (Chemistry and Mathematics) at New York University

Linear combination of atomic orbitals (LCAO) is a simple method of quantum chemistry that yields a qualitative picture of the molecular orbitals (MOs) in a molecule. Let us consider H+2 again. The approximation embodied in the LCAO approach is based on the notion that when the two protons are very far apart, the electron in its ground state will be a 1s orbital of one of the protons. Of course, we do not know which one, so we end up with a Schrödinger cat-like state in which it has some probability to be on one or the other.

As with the HF method, we propose a guess of the true wave function for the electron

ψg(r)=CAψA1s(r)+CBψB1s(r)(14.1)

where ψA1s(r)=ψ1s(r−RA) is a 1s hydrogen orbital centered on proton A and ψB1s(r)=ψ1s(r−RB) is a 1s hydrogen orbital centered on proton B. Recall ψ1s(r)=ψ100(r,ϕ,θ) . The positions RA and RB are given simply by the vectors

RA=(0,0,R/2)RB=(0,0,−R/2)(14.2)

The explicit forms of ψA1s(r) and ψB1s(r) are

ψA1s(r)ψB1s(r)=1(πa30)1/2e−|r−RA|/a0=1(πa30)1/2e−|r−RB|/a0

Now, unlike the HF approach, in which we try to optimize the shape of the orbitals themselves, in the LCAO approach, the shape of the ψ1s orbital is already given. What we try to optimize here are the coefficients CA and CB that determine the amplitude for the electron to be found on proton A or proton B.

The guess wave function ψg(r) is not normalized as we have written down. Thus, our guess of the ground-state energy is given by

Eg=∫ψgH^elecψgdV∫ψ2gdV(14.3)

where dV is the electron's volume element, and H^elec is the electronic Hamiltonian (minus the nuclear-nuclear repulsion ke2/R :

H^elec=K^e−ke2(1|r−RA|+1|r−RB|)(14.4)

(we will account for the nuclear-nuclear repulsion later when we consider the energies). Consider the denominator first:

∫ψ2gdV=C2A∫ψ21s(r−RA)dV+C2B∫ψ21s(r−rB)dV+2CACB∫ψ1s(r−RA)ψ1s(r−RB)dV(14.5)

Now, the 1s wave function of hydrogen is normalized so

∫ψ21s(r−RA)=∫ψ21s(r−RB)=1(14.6)

In the cross term, however, the integral

∫ψ1s(r−RA)ψ1s(r−RB)dV(14.7)

is not 1 because the orbitals are centered on different protons (it is only one if the two protons sit right on top of each other, which is not possible). It is also not 0 unless the protons are very far apart. We can calculate the integral analytically, however, it is not that important to do so since there is no dependence on CA or CB . Let us just denote this integral as S . We know that 0≤S≤1 and this is good enough for now. Thus, the denominator is simply

∫ψ2gdV=C2A+C2B+2SCACB(14.8)

As for the numerator

∫ψgH^elecψgdV=C2AHAA+C2BHBB+2CACBHAB(14.9)

where we have defined a bunch of integrals I'm too lazy to do as

HAAHBBHAB=∫ψA1s(r)H^elecψA1s(r)dV=∫ψB1s(r)H^elecψB1s(r)dV=∫ψA1s(r)H^elecψB1s(r)dV

Again, these are integrals we can do, but it is not that important, so we will just keep the shorthand notation. Note that since the two nuclei are the same (they are both protons), we expect HAA=HBB . Since these are equal, we will just call them both HAA . Hence, the guess energy becomes

Eg=HAA(C2A+C2B)+2HABCACBC2A+C2B+2SCACB(14.10)

which is just a ratio of two simple polynomials. Since we know that Eg>E0 , where E0 is the true ground-state energy, we can minimize Eg with respect to the two coefficients CA and CB . Thus, we need to take two derivatives and set them both to 0:

dEgdCA=0dEgdCB=0(14.11)

Defining the denominator simply as D , where D=C2A+C2B+2SCACB , the two derivatives are

dEgdCAdEgdCB=2CAHAA+2CBHABD−HAA(C2A+C2B)+2CACBHABD2(2CA+2SCB)=0=2CBHAA+2CAHABD−HAA(C2A+C2B)+2CACBHABD2(2CB+2SCA)=0

Thus, we have two algebraic equations in two unknowns CA and CB . In fact, these will not determine CA and CB uniquely because they are redundant. However, we also have the normalization of ψg as a third condition, so we have enough information to determine the coefficients absolutely.

The equations can be solved as follows: First we write them as

2CAHAA+2CBHABD2CBHAA+2CAHABD=HAA(C2A+C2B)+