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CHEM-UA 127: Advanced General Chemistry I

I.

VALENCE BOND THEORY

Valence bond theory is an alternative approximation scheme for constructing many-electron wave functions. Let us consider H2 . Recall that the Lewis structure for a single H atom is H· and for H2 , it is H : H. Thus, each hydrogen brings one unpaired electron to the bond. Let the two protons be denoted A and B and the two electrons 1 and 2. Now, consider the potential energy V = Vee + Ven + Vnn =

  e2 1 1 1 1 1 1 − − − − + 4πǫ0 r12 r1A r1B r2A r2B RAB

But as RAB → ∞, the 1/r12 , 1/r1B , 1/r2A , and 1/RAB terms vanish and the potential energy becomes simply that of two noninteracting hydrogen atoms   e2 1 1 V −→ − + 4πǫ0 r1A r2B Since the potential energy becomes a simple sum of separate energies for electrons 1 and 2, the wave function should simply be a product ψ1s (r1 − rA )ψ1s (r2 − rB ). But as we let RAB → Re , where Re is the equilibrium bond length, the electrons mix, and we can no longer tell if electron 1 belongs to atom A or atom B and the same for electron 2. Thus, we need to construct a combination of products that is consistent with the Pauli exclusion principle. If we just consider the coordinates r1 and r2 of the electrons, then the only wave function we can construct from a product of 1s orbitals is  A  B A B (r1 )ψ1s (r2 ) − ψ1s (r2 )ψ1s (r1 ) ψu (r1 , r2 ) = Cu ψ1s where the “u” designator indicates that this is an odd function. The constant Cu is the overall noramlization constant. Unfortunately, like in the LCAO method, such a wave function is antibonding and is not a good representation of the ground state. If, however, we construct the wave function   A A B B (r1 ) (r2 )ψ1s (r1 )ψ1s (r2 ) + ψ1s ψg (r1 , r2 ) = Cg ψ1s

(where “g” designates that this is an even function), we violate the Pauli exclusion principle, even though such a wave function leads to a stable chemical bond.

What is missing here is the fact that we have not considered the spins of the electrons. Since the electrons are identical, if we exchange coordinates and spins, then the wave function should change sign. Thus, we can make both wave functions above consistent with the Pauli exclusion principle by multiplying by an appropriate spin wave function. We obtain  A  B A B ψu (r1 , r2 , s1 , s2 ) = ψu (x1 , x2 ) = Cu ψ1s (r1 )ψ1s (r2 ) − ψ1s (r2 )ψ1s (r1 ) [ψ↑ (s1 )ψ↓ (s2 ) + ψ↑ (s2 )ψ↓ (s1 )]  A  B A B (r1 )ψ1s (r2 ) + ψ1s (r2 )ψ1s (r1 ) [ψ↑ (s1 )ψ↓ (s2 ) − ψ↑ (s2 )ψ↓ (s1 )] ψg (r1 , r2 , s1 , s2 ) = ψg (x1 , x2 ) = Cg ψ1s

where s1 and s2 are the z-components of spin for electrons 1 and 2, respectively. We can now use ψg as an approximate 2-electron wave function that leads to a stable chemical bond in H2 . The fact that ψu is antibonding can be easily determined by looking for a nodal plane between the two atoms, in this case, in the plane that exactly bisects the line joining the two atoms, midway between them. That this is, indeed, a nodal plane can be seen by considering two points r1 and r2 for the two electrons that are taken to lie in this plane. B B A (r2 ) (r1 ) have the same value for r1 in this plane, and the same for ψ1s By symmetry, the functions ψ1s (r1 ) and ψ1s A and ψ1s (r2 ). Let us assign the following values: A B ψ1s (r1 ) = ψ1s (r1 ) = A B A ψ1s (r2 ) = ψ1s (r2 ) = A′

2 Substituting these into ψu (x1 , x2 ), we obtain ψu (x1 , x2 ) = Cu [AA′ − A′ A] [ψ↑ (s1 )ψ↓ (s2 ) + ψ↑ (s2 )ψ↓ (s1 )] = 0 Since the wave function has a node midway between the two atoms, it is clearly antibonding and should have a higher energy than the corresponding bonding wave function ψg . A similar argument can be used for the molecule F2 . Each F has an electronic configuration 1s2 2s2 2p2x 2p2y 2pz and the Lewis structure of F2 is Most of the electrons are in lone pairs, but the 2pz electrons, which are unpaired

FIG. 1. Lewis structure of F2 .

in each F come together to form the bond. Thus, the bonding wave function should be a 2-electron wave function constructed from 2pz orbitals. The bonding wave function takes the “gerade” form as in H2 :   A B A B (r1 ) [ψ↑ (s1 )ψ↓ (s2 ) − ψ↑ (s2 )ψ↓ (s1 )] (r2 )ψ2p (r2 ) + ψ2p (r1 )ψ2p ψg (r1 , r2 , s1 , s2 ) = Cg ψ2p z z z z

For HF, the 2pz orbital on F and 1s orbital on H come together to form the bonding wave function. In order to be consistent with the Pauli principle, we need a wave function of the form   H H F F (1) [ψ↑ (s1 )ψ↓ (s2 ) − ψ↑ (s2 )ψ↓ (s1 )] (1) (2) + ψ1s (2)ψ2p ψ(r1 , r2 , s1 , s2 ) = ψ1s (1)ψ2p z z

Looking at the HF example, it becomes clear how much valence bond theory attempts to appear as a “quantum version” of the Lewis dot structure model. Valence bond theory attempts to construct very approximate wave functions for the bonding electrons in a Lewis structure, leaving the orbitals unused in the construction of the valence bond wave functions for the lone pair electrons. In the case of HF, we use the 2pz orbitals of F, which leaves the 2s, 2px and 2py orbitals unused. Since there are three lone pairs, these three orbitals are sufficient to hold each of the lone pairs as spin-up/spin-down couples. Historically, valence bond theory was used to explain bend angles in small molecules. Of course, it was only qualitatively correct in doing this, as the following example shows. Let us construct the valence bond wave functions for the two bonding pairs in H2 O. The construction is illustrated in the figure below: If we consider that the molecule lies in the xy plane, then let the OH1 bond lie along the x-axis and the OH2 bond lie along the y axis. We would construct the following VB wave functions using the 2px and 2py orbitals of oxygen: h i H1 H1 O O (2) + ψ (1) [ψ↑ (s1 )ψ↓ (s2 ) − ψ↑ (s2 )ψ↓ (s1 )] (1)ψ2p (2)ψ ψ1 (1, 2) = C1 ψ1s 2p 1s x x h i H2 H2 O O (2) + ψ1s (1) [ψ↑ (s1 )ψ↓ (s2 ) − ψ↑ (s2 )ψ↓ (s1 )] (1)ψ2p (2)ψ2p ψ2 (1, 2) = C2 ψ1s y y

This would leave the 2s and 2pz orbitals of oxygen left over for the two lone pairs on the oxygen. As the figure above suggests, these wave functions would lead to a 90◦ angle between the two OH bonds, which is clearly not correct. However, the qualitative trend is right. If we consider that the two hydrogens in H2 O have positive partial charges, then Coulomb repulsion between them can be invoked to justify the fact that the angle between the two bonds is larger than 90◦ . II.

ATOM HYBRIDIZATION

For polyatomic molecules, the valence bond theory becomes a very poor approximation because the directionalities of the 2s and 2p orbitals is too restrictive to describe molecules with steric numbers ranging between 2 and 4. The

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FIG. 2. Illustration of the use of VB wave functions for H2 O.

example considered above of H2 O illustrates this rather dramatically! Let us consider an even simpler molecule, BeH2 , which has a steric number of 2 and is linear. Let the atoms lie entirely along the z-axis in the arrangement H-Be-H. Although Be has a ground-state electronic configuration of 1s2 2s2 , but if we “promote” one of the 2s electrons to a state with higher energy and allow its electronic structure to be 1s2 2s2pz , then the unpaired electrons in the 2s and 2pz orbitals can combine with the unpaired electrons in each of the hydrogen atoms to form bonds. The energy needed to excite the electron in Be would be “repaid” by the energy gained in the formation of stable bonds. The two valence-bond wave functions we would construct would be h i H1 H1 Be Be (1)ψ2s (2) + ψ1s (2)ψ2s (1) [ψ↑ (s1 )ψ↓ (s2 ) − ψ↑ (s2 )ψ↓ (s1 )] ψ1 (1, 2) = C1 ψ1s h i H2 H2 Be Be (2) + ψ (1) [ψ↑ (s1 )ψ↓ (s2 ) − ψ↑ (s2 )ψ↓ (s1 )] (1)ψ2p (2)ψ ψ2 (1, 2) = C2 ψ1s 2p 1s z z

Unfortunately, even this simple scheme does not work entirely because the two Be-H bonds would be different due to their construction from different combinations of orbitals. By symmetry, however, we can see that the two BeH bonds should be equivalent. A solution to this problem was proposed by Linus Pauling in the 30s in the form of orbital hybridization, a scheme that we still use today. Pauling used the fact that in the first and second periods, the 2s and 2p orbitals have similar energies. Indeed, for H, the energies are exactly the same. Given that these energies are not that different, we can combine s and p orbitals and still have a valid solution of the Schr¨odinger equation. That is, a general orbital χ(r) = C1 ψ2s (r) + C2 ψ2px (r) + C3 ψ2py (r) + C4 ψ2pz (r) is also a solution of the Schr¨odinger equation with the same energy as a 2s or 2p orbitals individually (this is exactly true for H). In the case of BeH2 , the external potential on the electrons in Be by the two hydrogens changes the energy levels and creates a near degeneracy between the 2s and 2pz orbitals, hence, we are now free to combine the into linear combinations that are more suitable to the construction both of valence bond wave functions and MOs via the LCAO procedure. For Be, we now allow the s and p orbitals to mix and create two hybrid orbitals known as sp orbitals. There are two

4 such orbitals we can create 1 χ1 (r) = √ [−ψ2s (r) + ψ2pz (r)] 2 1 χ2 (r) = √ [−ψ2s (r) − ψ2pz (r)] 2 Note that these orbitals are both normalized and orthogonal: Z Z 2 |χ1 (r)| dV = 1 |χ2 (r)|2 dV = 1 Z

χ∗1 (r)χ2 (r)dV = 0

These orbitals appear as shown in the figure below: Given that the two sp hybrid orbitals are mirror images of each

FIG. 3. sp hybrid orbitals.

other, they can overlap with the 1s orbital of H (shown in the figure) and create two equal bonds, as needed for BeH2 . Using the valence bond formulation, now, one of the BeH bonds will be described by a wave function of the form: h i H1 H1 Be (1)χBe ψ1 (1, 2) = C1 ψ1s 1 (2) + ψ1s (2)χ1 (1) [ψ↑ (s1 )ψ↓ (s2 ) − ψ↑ (s2 )ψ↓ (s1 )] h i H2 H2 Be (1)χBe ψ2 (1, 2) = C2 ψ1s 2 (2) + ψ1s (2)χ2 (1) [ψ↑ (s1 )ψ↓ (s2 ) − ψ↑ (s2 )ψ↓ (s1 )]

In the above wave functions, it is clear that H1 is on the right and H2 is on the left, based on the directionalities of χ1 and χ2 . For trigonal planar molecules such as BH3 , we start with the electronic configuration of B, which is 1s2 2s2 2px , and we promote one of the 2s electrons to a 2py orbital, so that we have 1s2 2s2px 2py . Suppose the geometry of BH3 is such that one of the hydrogens lies along the positive x axis. The remaining hydrogens would be in the 3rd and 4th quadrants, respectively, as shown in the figure. If we simply combine the 2s with the 2px and 2py orbitals of boron, the resulting hybrid orbitals will not point in the correct direction. For this reason, we will create rotated versions of the px and py orbitals, which, as we will see are tantamount to taking new combinations of 2px and 2py orbitals to

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y

H H

B

x

H FIG. 4. Geometry of BH3 .

combine with the 2s. Since the rotation occurs in the x-y plane, the coordinate that controls this is the azimuthal angle φ. For the px and py orbitals, the φ dependence is ψ2px ∼ cos φ

ψ2py ∼ sin φ

If we rotate 2py by −30 degrees (see figure, blue is positive and red is negative), the φ dependence becomes

FIG. 5. Rotation of a 2py orbital by −30 degrees. (rot)

ψ2py ∼ sin(φ + 30) Using the fact that sin(α ± β) = sin α cos β ± cos α sin β this rotation gives (rot,1)

ψ2py

∼ sin φ cos 30 + cos φ sin 30 "√

# 3 1 ∼ sin φ + cos φ 2 2 # "√ 3 1 ψ2py + ψ2px ∼ 2 2

6 Similarly, consider rotating −ψ2py by +30 degrees. This gives (rot,2)

−ψ2py

− ∼ sin(φ − 30) # "√ 1 3 sin φ − cos φ ∼ − 2 2 √ 1 3 ∼ − ψ2py + ψ2px 2 2

(see figure). So, we now take the hybrid orbitals to be of the form

FIG. 6. Rotation of a −2py orbital by 30 degrees.

χ1 (r) = −aψ2s (r) + bψ2px (rot,1)

(r)

(rot,2)

(r)

χ2 (r) = −cψ2s (r) + dψ2py

χ3 (r) = −cψ2s (r) − dψ2py

The coefficients a, b, and c are determined by requiring that the orbitals are normalized and mutually orthogonal: Z Z 2 |χ1 (r)| dV = 1 χ∗1 (r)χ2 (r)dV = 0 Z Z

2

|χ2 (r)| dV = 1 2

|χ3 (r)| dV = 1

Z

χ∗1 (r)χ3 (r)dV = 0

Z

χ∗2 (r)χ3 (r)dV = 0

Carrying out the algebra, we obtain the following sp2 hybrid orbitals: i √ 1 h χ1 (r) = √ −ψ2s (r) + 2ψ2px (r) 3 i √ 1 h √ χ2 (r) = √ − 2ψ2s (r) − ψ2px (r) + 3ψ2py (r) 6 i √ 1 h √ χ3 (r) = √ − 2ψ2s (r) − ψ2px (r) − 3ψ2py (r) 6

The sp2 hybrids allow bonding at 120◦ degrees, and these orbitals appear as shown in the figure below: The figure also shows the overlaps of these orbitals with the 1s orbitals of H. Finally, we consider the case of methane CH4 . The electronic configuration of C is 1s2 2s2 2px 2py . We now promote one of the 2s orbitals to the 2pz orbital and write C as 1s2 2s2px2py 2pz . We can now hybridize the 2s orbital with

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FIG. 7. sp2 hybrid orbitals.

each of the 2p orbitals to create four hybrids:  1 −ψ2s (r) + ψ2px (r) + ψ2py (r) + ψ2pz (r) 2  1 χ2 (r) = −ψ2s (r) − ψ2px (r) − ψ2py (r) + ψ2pz (r) 2  1 −ψ2s (r) + ψ2px (r) − ψ2py (r) − ψ2pz (r) χ3 (r) = 2  1 χ4 (r) = −ψ2s (r) − ψ2px (r) + ψ2py (r) − ψ2pz (r) 2 χ1 (r) =

These orbitals are shown in the figure below: and are known as sp3 hybrid orbitals.

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FIG. 8. sp3 hybrid orbitals.