Hybrid orbitals

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. Let us consider first the 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 $1s^22s^2$, 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. Unfortunately, this scheme does not work entirely because the two Be-H bonds would be different due to their construction from different combinations of orbitals. A solution to this problem was proposed by Linus Pauling in the 30s in the form of orbital hybridization.

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

$$\chi({\bf r}) = C_1 \psi_{1s}({\bf r}) + C_2 \psi_{2p_x}({\bf r}) + C_3 \psi_{2p_y}({\bf r}) + C_4\psi_{2p_z}({\bf r})$$

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).

Thus, for Be, we 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

$$\chi_1({\bf r}) = {1 \over \sqrt{2}}\left[\psi_{2s}({\bf r}) + \psi_{2p_z}({\bf r})\right]$$

$$\chi_2({\bf r}) = {1 \over \sqrt{2}}\left[\psi_{2s}({\bf r}) - \psi_{2p_z}({\bf r})\right]$$

These orbitals appear as shown in the figure below:

Figure: $sp$ hybrid orbitals

Given that the two sp hybrid orbitals are mirror images of each other, they can overlap with the 1s orbital of H 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 (A = left H, B = Be, C = right H)

$$\psi({\bf r}_1,{\bf r}_2,s_1,s_2) = C\left[\psi_{1s}({\bf r... ...arrow}(s_2)- \psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)\right]$$

and the other bond will be described by

$$\psi({\bf r}_3,{\bf r}_4,s_3,s_4) = C\left[\chi_2({\bf r}_3... ...arrow}(s_4)- \psi_{\uparrow}(s_4)\psi_{\downarrow}(s_3)\right]$$

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 $1s^2 2s 2p_x 2p_y$. Now, we can create hybrids by mixing the 2s, $2p_x$ and $2p_y$ orbitals to create what are called $sp_2$ hybrids. These are given by

$$\chi_1({\bf r}) = {1 \over \sqrt{3}} \left[\psi_{2s}({\bf r}) + \sqrt{2}\psi_{2p_x}({\bf r})\right]$$

$$\chi_2({\bf r}) = {1 \over \sqrt{6}} \left[\sqrt{2}\psi_{2s}({\bf r}) - \psi_{2p_x}({\bf r}) + \sqrt{3}\psi_{2p_y}({\bf r})\right]$$

$$\chi_3({\bf r}) = {1 \over \sqrt{6}} \left[\sqrt{2}\psi_{2s}({\bf r}) - \psi_{2p_x}({\bf r}) - \sqrt{3}\psi_{2p_y}({\bf r})\right]$$

The origin of these strange coefficients lies in the mathematical are known as group theory, which is used to analyze all of the transformations that can be made to a molecule without changing how it appears. The $sp_2$ hybrids allow bonding at 120$^{\circ}$ degrees, and these orbitals appear as shown in the figure below:

Figure: $sp2$ hybrid orbitals

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

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 $1s^2 2s 2p_x 2p_y 2p_z$. We can now hybridize the 2s orbital with each of the 2p orbitals to create four hybrids:

$$\chi_1({\bf r}) = {1 \over 2} \left[\psi_{2s}({\bf r}) + \psi_{2p_x}({\bf r}) + \psi_{2p_y}({\bf r}) + \psi_{2p_z}({\bf r})\right]$$

$$\chi_2({\bf r}) = {1 \over 2} \left[\psi_{2s}({\bf r}) - \psi_{2p_x}({\bf r}) - \psi_{2p_y}({\bf r}) + \psi_{2p_z}({\bf r})\right]$$

$$\chi_3({\bf r}) = {1 \over 2} \left[\psi_{2s}({\bf r}) + \psi_{2p_x}({\bf r}) - \psi_{2p_y}({\bf r}) - \psi_{2p_z}({\bf r})\right]$$

$$\chi_4({\bf r}) = {1 \over 2} \left[\psi_{2s}({\bf r}) - \psi_{2p_x}({\bf r}) + \psi_{2p_y}({\bf r}) - \psi_{2p_z}({\bf r})\right]$$

These orbitals are shown in the figure below:

Figure: $sp3$ hybrid orbitals