16: Hybrid atomic orbitals, geometry, and valence bond theory

Valence bond theory is an alternative approximation scheme for constructing many-electron wave functions. Let us consider \(H_2\). Recall that the Lewis structure for a single \(H\) atom is \(H\cdot\) and for \(H_2\), 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

\[\begin{align*} V &= V_{ee}+V_{en}+V_{nn}\\ &=\dfrac{e^2}{4\pi \epsilon_0}\left [ \dfrac{1}{r_{12}}-\dfrac{1}{r_{1A}}-\dfrac{1}{r_{1B}}-\dfrac{1}{r_{2A}}-\dfrac{1}{r_{2B}}+\dfrac{1}{R_{AB}}\right ] \end{align*}\]

But as \(R_{AB}\rightarrow \infty\), the \(1/r_{12}\), \(1/r_{1B}\), \(1/r_{2A}\), and \(1/R_{AB}\) terms vanish and the potential energy becomes simply that of two noninteracting hydrogen atoms

\[V\rightarrow -\dfrac{e^2}{4\pi \epsilon_0}\left [ \dfrac{1}{r_{1A}}+\dfrac{1}{r_{2B}}\right ]\]

Since the potential energy becomes a simple sum of separate energies for electrons 1 and 2, the wave function should simply be a product \(\psi_{1s}(r_1 -r_A)\psi_{1s}(r_2 -r_B)\). But as we let \(R_{AB}\rightarrow R_e\), where \(R_e\) 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 \(r_1\) and \(r_2\) of the electrons, then the only wave function we can construct from a product of 1s orbitals is

\[\psi_u (r_1 ,r_2)=C_u [\psi_{1s}^{A}(r_1)\psi_{1s}^{B} (r_2)-\psi_{1s}^{A}(r_2)\psi_{1s}^{B}(r_1)]\]

where the \(u\) designator indicates that this is an odd function. The constant \(C_u\) 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

\[\psi_g (r_1 ,r_2)=C_g [\psi_{1s}^{A} (r_1)\psi_{1s}^{B} (r_2)+\psi_{1s}^{A}(r_2) \psi_{1s}^{B} (r_1)]\]

(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

\[\begin{align*}\psi_u (r_1 ,r_2 ,s_1 ,s_2 ) &=\psi_u(x_1 ,x_2)=C_u [\psi_{1s}^{A}(r_1)\psi_{1s}^{B}(r_2)-\psi_{1s}^{A}(r_2)\psi_{1s}^{B}(r_1)][\psi_{\uparrow}(s_1)\psi_{\downarrow}(s_2)+\psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)]\\ \psi_g (r_1 ,r_2 ,s_1 ,s_2) &= \psi_g(x_1 ,x_2)=C_g[\psi_{1s}^{A}(r_1)\psi_{1s}{B}(r_2)+\psi_{1s}^{A}(r_2)\psi_{1s}^{B}(r_1)][\psi_{\uparrow}(s_1)\psi_{\downarrow}(s_2)-\psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)]\end{align*}\]

where \(s_1\) and \(s_2\) are the z-components of spin for electrons 1 and 2, respectively. We can now use \(\psi_g\) as an approximate 2-electron wave function that leads to a stable chemical bond in \(H_2\).

The fact that \(\psi_u\) 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 \(r_1\) and \(r_2\) for the two electrons that are taken to lie in this plane. By symmetry, the functions \(\psi_{1s}^{A}(r_1)\) and \(\psi_{1s}^{B}(r_1)\) have the same value for \(r_1\) in this plane, and the same for \(\psi_{1s}^{B}(r_2)\) and \(\psi_{1s}^{A}(r_2)\). Let us assign the following values:

\[\begin{align*}\psi_{1s}^{A}(r_1) &= \psi_{1s}^{B}(r_1)=A\\ \psi_{1s}^{B}(r_2) &= \psi_{1s}^{A}(r_2)=A' \end{align*}\]

Substituting these into \(\psi_u (x_1 ,x_2)\), we obtain

\[\psi_u (x_1 ,x_2)=C_u [AA' -A' A][\psi_{\uparrow}(s_1)\psi_{\downarrow}(s_2)+\psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)]=0\]

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 \(\psi_g\).

A similar argument can be used for the molecule \(F_2\). Each \(F\) has an electronic configuration

\[1s^2 2s^2 2p_{x}^{2}2p_{y}^{2}2p_{z}^{1}\]

and the Lewis structure of \(F_2\) is

Figure \(\PageIndex{1}\): Lewis structure of \(F_2\).

Most of the electrons are in lone pairs, but the \(2p_z\) electrons, which are unpaired in each \(F\) come together to form the bond. Thus, the bonding wave function should be a 2-electron wave function constructed from \(2p_z\) orbitals. The bonding wave function takes the "gerade" form as in \(H_2\):

\[\psi_g (r_1 ,r_2 ,s_1 ,s_2)=C_g [\psi_{2p_z}^{A}(r_1)\psi_{2p_z}^{B}(r_2)+\psi_{2p_z}^{A}(r_2)\psi_{2p_z}^{B}(r_1)][\psi_{\uparrow}(s_1)\psi_{\downarrow}(s_2)-\psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)]\]

For \(HF\), the \(2p_z\) 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

\[\psi (r_1 ,r_2 ,s_1 ,s_2)=[\psi_{1s}^{H}(1)\psi_{2p_z}^{F}(2)+\psi_{1s}^{H}(2)\psi_{2p_z}^{F}(1)][\psi_{\uparrow}(s_1)\psi_{\downarrow}(s_2)-\psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)] \tag{1})\]

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 \(2p_z\) orbitals of \(F\), which leaves the \(2s\), \(2p_x\) and \(2p_y\) 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 \(H_2O\). The construction is illustrated in Figure \(\PageIndex{2}\) below:

Figure \(\PageIndex{2}\): Illustration of the use of VB wave functions for \(H_2 O\).

If we consider that the molecule lies in the \(xy\) plane, then let the \(OH_1\) bond lie along the x-axis and the \(OH_2\) bond lie along the y axis. We would construct the following VB wave functions using the \(2p_x\) and \(2p_y\) orbitals of oxygen:

\[\begin{align*}\psi_1 (1,2) &= C_1[\psi_{1s}^{H_1}(1)\psi_{2p_x}^{\bigcirc}(2)+\psi_{1s}^{H_1}(2)\psi_{2p_x}^{\bigcirc}(1)][\psi_{\uparrow}(s_1)\psi_{\downarrow}(s_2)-\psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)]\\ \psi_2 (1,2) &= C_2 [\psi_{1s}^{H_2}(1)\psi_{2p_y}^\bigcirc (2)+\psi_{1s}^{H_2}(2)\psi_{2p_y}^{\bigcirc}(1)][\psi_\uparrow(s_1)\psi_\downarrow (s_2)-\psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)] \end{align*}\]

This would leave the \(2s\) and \(2p_z\) 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 \(90^{\circ}\) 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 \(H_2 O\) 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 \(90^{\circ}\).

Atomic 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 example considered above of \(H_2 O\) illustrates this rather dramatically! Let us consider an even simpler molecule, \(BeH_2\), 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 \(1s^2 2s^2\), but if we "promote" one of the \(2s\) electrons to a state with higher energy and allow its electronic structure to be \(1s^2 2s 2p_z\), then the unpaired electrons in the \(2s\) and \(2p_z\) 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

\[\begin{align*}\psi_1 (1,2) &= C_1 [\psi_{1s}^{H_1}(1)\psi_{2s}^{Be}(2)+\psi_{1s}^{H_1}(2)\psi_{2s}^{Be}(1)][\psi_{\uparrow}(s_1)\psi_{\downarrow}(s_2)-\psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)]\\ \psi_2 (1,2) &= C_2 [\psi_{1s}^{H_2}(1)\psi_{2p_z}^{Be}(2)+\psi_{1s}^{H_2}(2)\psi_{2p_x}^{Be}(1)][\psi_\uparrow (s_1)\psi_\downarrow (s_2)-\psi_\uparrow (s_2)\psi_\downarrow (s_1)]\end{align*}\]

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 Schrödinger equation. That is, a general orbital

\[\chi (r)=C_1 \psi_{2s}(r)+C_2 \psi_{2p_x}(r)+C_3 \psi_{2p_y}(r)+C_4 \psi_{2p_z}(r)\]

is also a solution of the Schrödinger equation with the same energy as a \(2s\) or \(2p\) orbitals individually (this is exactly true for \(H\)). In the case of \(BeH_2\), 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 \(2p_z\) 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.

\(sp\) Hybridization

For Be, we now allow the s and p orbitals to mix and create two hybrid orbitals known as \(sp\) orbitals. There are two such orbitals we can create

\[\begin{align*}\chi_1 (r) &= \dfrac{1}{\sqrt{2}}[\psi_{2s}(r)+\psi_{2p_z}(r)]\\ \chi_2 (r) &= \dfrac{1}{\sqrt{2}}[\psi_{2s}(r)-\psi_{2p_z}(r)]\end{align*}\]

Note that these orbitals are both normalized and orthogonal:

\[\int |\chi_1 (r)|^2 dV=1 \ ; \ \int |\chi_2 (r)|^2 dV=1 \ ; \ \int \chi_{1}^{*}(r)\chi_2 (r)dV=0\]

These orbitals appear as shown in Figure \(\PageIndex{3}\)

Figure \(\PageIndex{3}\): The Formation of sp Hybrid Orbitals. Taking the sum and difference of an ns and an np atomic orbital where n = 2 gives two equivalent sp hybrid orbitals oriented at 180° to each other.

Given that the two \(sp\) hybrid orbitals are mirror images of each other, they can overlap with the \(1s\) orbital of \(H\) (shown in the figure) and create two equal bonds, as needed for \(BeH_2\). Using the valence bond formulation, now, one of the \(BeH\) bonds will be described by a wave function of the form:

\[\begin{align*}\psi_1 (1,2) &= C_1 [\psi_{1s}^{H_1}(1)\chi_{1}^{Be}(2)+\psi_{1s}^{H_1}(2)\chi_{1}^{Be}(1)][\psi_{\uparrow}(s_1)\psi_{\downarrow}(s_2)-\psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)]\\ \psi_2 (1,2) &= C_2 [\psi_{1s}^{H_2}(1)\chi_{2}^{Be}(2)+\psi_{1s}^{H_2}(2)\chi_{2}^{Be}(1)][\psi_{\uparrow}(s_1)\psi_\downarrow (s_2)-\psi_{\uparrow}(s_2)\psi_\downarrow (s_1)]\end{align*}\]

In the above wave functions, it is clear that \(H_1\) is on the right and \(H_2\) is on the left, based on the directionalities of \(\chi_1\) and \(\chi_2\).

\(sp^2\) Hybrids

For trigonal planar molecules such as \(BH_3\), we start with the electronic configuration of \(B\), which is \(1s^2 2s^2 2p_x\), and we promote one of the \(2s\) electrons to a \(2p_y\) orbital, so that we have \(1s^2 2s 2p_x 2p_y\). Suppose the geometry of \(BH_3\) 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 Figure \(\PageIndex{4}\).

Figure \(\PageIndex{4}\): Geometry of \(BH_3\).

If we simply combine the \(2s\) with the \(2p_x\) and \(2p_y\) orbitals of boron, the resulting hybrid orbitals will not point in the correct direction. For this reason, we will create rotated versions of the \(p_x\) and \(p_y\) orbitals, which, as we will see are tantamount to taking new combinations of \(2p_x\) and \(2p_y\) orbitals to combine with the \(2s\). Since the rotation occurs in the \(xy\) plane, the coordinate that controls this is the azimuthal angle \(\phi\). For the \(p_x\) and \(p_y\) orbitals, the \(\phi\) dependence is

\[\psi_{2p_x}\sim \cos\phi \ ; \ \psi_{2p_y}\sim \sin\phi\]

Figure \(\PageIndex{5}\): Rotation of a \(2p_y\) orbital by \(-30\) degrees.

If we rotate \(2p_y\) by \(-30\) degrees (Figure \(\PageIndex{5}\); blue is positive and red is negative), the \(\phi\) dependence becomes

\[\psi_{2p_y}^{(rot)}\sim \sin(\phi +30)\]

Using the fact that

\[\sin(\alpha \pm \beta)=\sin\alpha \cos\beta \pm cos\alpha sin\beta\]

this rotation gives

\[\begin{align*}\psi_{2p_y}^{(rot,1)} &\sim sin\phi cos30+cos\phi sin30\\ &\sim \left [ \dfrac{\sqrt{3}}{2}sin\phi +\dfrac{1}{2}cos\phi \right ] \\ &\sim \left [ \dfrac{\sqrt{3}}{2}\psi_{2p_y}+\dfrac{1}{2}\psi_{2p_x}\right ]\end{align*}\]

Similarly, consider rotating \(-\psi_{2p_y}\) by \(+30\) degrees (Figure \(\PageIndex{6}\)). This gives

\[\begin{align*}-\psi_{2p_y}^{(rot,2)} &\sim -sin(\phi -30)\\ &\sim -\left [ \dfrac{\sqrt{3}}{2}sin\phi -\dfrac{1}{2}cos\phi \right ] \\ &\sim -\dfrac{\sqrt{3}}{2}\psi_{2p_y}+\dfrac{1}{2}\psi_{2p_x}\end{align*}\]

Figure \(\PageIndex{6}\): Rotation of a \(-2p_y\) orbital by \(30\) degrees.

So, we now take the hybrid orbitals to be of the form

\[\begin{align*}\chi_1 (r) &= a\psi_{2s}(r)-b\psi_{2p_x}\\ \chi_2 (r) &= c\psi_{2s}(r)+d\psi_{2p_y}^{(rot,1)}(r)\\ \chi_3 (r) &=c\psi_{2s}(r)-d\psi_{2p_y}^{(rot,2)}(r)\end{align*}\]

The coefficients \(a\), \(b\), and \(c\) are determined by requiring that the orbitals are normalized and mutually orthogonal:

\[\begin{align*}\int |\chi_1 (r)|^2 dV=1 \ &; \ \int \chi_{1}^{*}(r)\chi_2 (r)dV=0\\ \int |\chi_2 (r)|^2 dV=1 \ &; \ \int \chi_{1}^{*}(r)\chi_3 (r)dV=0\\ \int |\chi_3 (r)|^2 dV=1 \ &; \ \int \chi_{2}^{*}(r)\chi_3 (r)dV=0\end{align*}\]

Carrying out the algebra, we obtain the following \(sp^2\) hybrid orbitals:

\[\begin{align*}\chi_1 (r) &= \dfrac{1}{\sqrt{3}}[\psi_{2s}(r)-\sqrt{2}\psi_{2p_x}(r)]\\ \chi_2 (r) &= \dfrac{1}{\sqrt{6}}[\sqrt{2}\psi_{2s}(r)+\psi_{2p_x}(r)+\sqrt{3}\psi_{2p_y} (r)]\\ \chi_3 (r) &= \dfrac{1}{\sqrt{6}}[\sqrt{2}\psi_{2s}(r)+\psi_{2p_x}(r)-\sqrt{3}\psi_{2p_y}(r)]\end{align*}\]

The \(sp^2\) hybrids allow bonding at \(120^\circ\) degrees, and these orbitals appear as shown in Figure \(\PageIndex{7}\):

Figure \(\PageIndex{7}\): Formation of sp² Hybrid Orbitals. Combining one ns and two np atomic orbitals gives three equivalent sp² hybrid orbitals in a trigonal planar arrangement; that is, oriented at 120° to one another.

The figure also shows the overlaps of these orbitals with the \(1s\) orbitals of \(H\).

\(sp^3\) Hybrids

Finally, we consider the case of methane \(CH_4\). The electronic configuration of \(C\) is \(1s^2 2s^2 2p_x 2p_y\). We now promote one of the \(2s\) orbitals to the \(2p_z\) orbital and write \(C\) as \(1s^2 2s2p_x 2p_y 2p_z\). We can now hybridize the \(2s\) orbital with each of the \(2p\) orbitals to create four hybrids:

\[\begin{align*}\chi_1 (r) &= \dfrac{1}{2}[\psi_{2s}(r)+\psi_{2p_x}(r)+\psi_{2p_y}(r)+\psi_{2p_z}(r)]\\ \chi_2 (r) &= \dfrac{1}{2}[\psi_{2s}(r)-\psi_{2p_x}(r)-\psi_{2p_y}(r)+\psi_{2p_z}(r)]\\ \chi_3 (r) &= \dfrac{1}{2}[\psi_{2s}(r)+\psi_{2p_x}(r)-\psi_{2p_y}(r)-\psi_{2p_z}(r)]\\ \chi_4 (r) &= \dfrac{1}{2}[\psi_{2s}(r)-\psi_{2p_x}(r)+\psi_{2p_y}(r)-\psi_{2p_z}(r)]\end{align*}\]

The large lobes of the hybridized orbitals are oriented toward the vertices of a tetrahedron, with 109.5° angles between them (Figure \(\PageIndex{8}\)). Like all the hybridized orbitals discussed earlier, the sp³ hybrid atomic orbitals are predicted to be equal in energy.

Figure \(\PageIndex{8}\): Formation of sp³ Hybrid Orbitals. Combining one ns and three np atomic orbitals results in four sp³ hybrid orbitals oriented at 109.5° to one another in a tetrahedral arrangement.

In addition to explaining why some elements form more bonds than would be expected based on their valence electron configurations, and why the bonds formed are equal in energy, valence bond theory explains why these compounds are so stable: the amount of energy released increases with the number of bonds formed. In the case of carbon, for example, much more energy is released in the formation of four bonds than two, so compounds of carbon with four bonds tend to be more stable than those with only two. Carbon does form compounds with only two covalent bonds (such as CH₂ or CF₂), but these species are highly reactive, unstable intermediates that form in only certain chemical reactions.

Contributors

• Mark Tuckerman (New York University)