LCAO for heteronuclear diatomic molecules

In a heteronuclear diatomic molecule, there is an electronegativity difference between the atoms, which leads to an asymmetric distribution of the electronic probability density, weighted more heavily toward the element with the greater electronegativity.

Consider constructing a MO from two 2s orbitals for nuclei with different $Z$ values. Let atom A have an atomic number $Z_A$ and atom B have an atomic number $Z_B$. For a single electron interacting with the two nuclei, an LCAO guess wave function could be

$$\psi_g({\bf r}) = C_A \psi_{2s(Z=Z_A)}({\bf r}-{\bf r}_A) + C_B \psi_{2s(Z=Z_B)}({\bf r}-{\bf r}_B)$$

Recall the energies $H_{AA}$ and $H_{BB}$ that were defined for LCAO in the homonuclear case:

$$H_{AA} = \int \psi_{2s(Z=Z_A)}({\bf r}-{\bf r}_A) \hat{H}_{\rm elec}\psi_{2s(Z=Z_A)}({\bf r}-{\bf r}_A)dV$$

$$H_{BB} = \int \psi_{2s(Z=Z_B)}({\bf r}-{\bf r}_B) \hat{H}_{\rm elec}\psi_{2s(Z=Z_B)}({\bf r}-{\bf r}_B)dV$$

For the heteronuclear case, $H_{AA} \neq H_{BB}$. If atom B is the more electronegative, then its $Z$ will be larger (in the second period), and the energy will be more negative due to the large Coulomb attraction between the electrons and the nuclei. Thus, in this case $H_{BB} < H_{AA}$. Now, if we calculate the guess of the ground-state energy

$$E_g = {\int \psi_g({\bf r})\hat{H}_{\rm elec}\psi_g({\bf r}) dV \over \int \psi_g^2({\bf r}) dV}$$

and then perform the minimization of $E_g$:

$${dE_g \over dC_A} = 0\;\;\;\;\;\;\;\;\;\;{dE_g \over dC_B} = 0$$

we find two cases for the solutions: For the bonding orbital, $\vert C_B\vert > \vert C_A\vert$, while for the antibonding orbital $\vert C_B\vert < \vert C_A\vert$. Thus, the bonding orbital takes the form

$$\psi_+({\bf r}) = C_A \psi_{2s(Z=Z_A)}({\bf r}-{\bf r}_A) + C_B \psi_{2s(Z=Z_B)}({\bf r}-{\bf r}_B)$$

where $C_A < C_B$, and the antibonding orbital is

$$\psi_-({\bf r}) = C_A' \psi_{2s(Z=Z_A)}({\bf r}-{\bf r}_A) - C_B' \psi_{2s(Z=Z_B)}({\bf r}-{\bf r}_B)$$

with $C_A' > C_B'$. See the pictures from the lecture for a sketch of how the orbitals roughly appear.

We can no longer use the ``g'' and ``u'' designators because the orbitals have no particular symmetry when ${\bf r}\rightarrow -{\bf r}$. That is

$$\psi_+(-{\bf r}) \neq \psi_+({\bf r})\;\;\;\;\;\;\;\;\;\psi_-(-{\bf r}) \neq -\psi_-({\bf r})$$

Thus, we denote $\psi_+({\bf r})$ simply as $\sigma_{2s}$ and $\psi_-({\bf r})$ simply as $\sigma^*_{2s}$.

Let us now construct a correlation diagram for the heteronuclear diatomic BO. Boron has an electronic configuration

$$1s^2 2s^2 2p_x$$

while oxygen's is

$$1s^2 2s^2 2p_x^2 2p_y^2 2p_z$$

Since we are interested in the chemical bond that forms between them, we only consider the valence electrons explicitly, and these are the electrons in the $n=2$ shell. In BO, oxygen is the more electronegative, so its orbitals are lower in energy than those of boron. This must be indicated on the correlation diagram. Thus, the correlation diagram appears as in the figure below:

Figure: Correlation diagram for the heteronuclear diatomic molecule BO

Note that the ordering of the MOs follows the pattern we would expect for boron rather than oxygen. Only high-level calculations can predict this, but physically, the simple explanation is that there is only one electron in the $\sigma_{2p_z}$ orbital, and only one of the two atoms has a large nuclear charge (unlike in O$_2$, where they both do). There is, therefore, insufficient Coulomb attraction for this one electron to pull the energy of the $\sigma_{2p_z}$ orbital below the energy of the $\pi$ bonding orbitals.

The electronic configuration of BO is, therefore

$$\left(\sigma_{2s}\right)^2 \left(\sigma^*_{2s}\right)^2 \lef... ...ight)^2 \left(\pi_{2p_y}\right)^2 \left(\sigma_{2p_z}\right)^1$$

and the bond order is (1/2)(7-2)=5/2. The fraction bond order indicates that the molecule is paramagnetic, as with O$_2$.

As another example, consider the molecule NO. NO has 11 valence electrons and has the electronic configuration:

$$\left(\sigma_{2s}\right)^2 \left(\sigma^*_{2s}\right)^2 \lef... ...right)^2 \left(\sigma_{2p_z}\right)^2\left(\pi_{2p_x}\right)^1$$

The bond order is also 5/2, and the molecule is paramagnetic as well.

What about the molecule HF? Here, the 1s and 2s orbitals of F are so low in energy compared to the 1s orbital in H that they cannot be combined to form MOs. At the same time, the $2p_x$ and $2p_y$ orbitals of F have an insignificant spatial overlap with the 1s orbital in H (assuming that the two nuclei lie along the $z$-axis) that they also do not form MOs. Only the $2p_z$ orbital of F has significant overlap with the 1s orbital in H and can mix with it energetically. Thus, the LCAO guess wave function takes the form

$$\psi_+({\bf r}) = C_A \psi_{1s}({\bf r}-{\bf r}_A) + C_B \psi_{2p_z}({\bf r}-{\bf r}_B)$$

where atom A is H and atom B is F. If H lies to the right of F, then this orbital has antibonding character, while the orbital

$$\psi_-({\bf r}) = C_A \psi_{1s}({\bf r}-{\bf r}_A) - C_B \psi_{2p_z}({\bf r}-{\bf r}_B)$$

has bonding character because there is significant amplitude in the region between the nuclei. See the figure below:

Figure: Illustration of the overlap between the 2s and 2p orbitals in HF

The orbital $\psi_-$ is denoted simply as $\sigma$ while the orbital $\psi_+$ is denoted $\sigma^*$ for bonding and antibonding, respectively. The 2s, $2p_x$ and $2p_y$ orbitals of F do not mix with anything and are, therefore, called nonbonding orbitals. They are denoted $\sigma^{\rm nb}$ and $\pi_x^{\rm nb}$ and $\pi_y^{\rm nb}$, respectively. The orbital ordering is $\sigma^{\rm nb}$ (since it is a low-energy 2s orbital), followed by $\sigma$, then the two nonbonding $\pi$ orbitals and finally $\sigma^*$. Therefore, the correlation diagram for HF is as shown in the figure below:

Figure: Correlation diagram for HF