Constructing MOs for the water molecule

In this section, we will construct approximate molecular orbitals for a water molecule by considering a simple linear triatomic of the general form HXH, where X is a second row element. The linear form clearly does not apply to H$_2$O, but we can still construct a reasonable argument for the energetic ordering and structure of the MOs. We first note that each H will donate a $1s$ orbital in the LCAO scheme, and X will likely donate at least $2s$ and possible $2p$ orbitals, depending on its chemical identity. Let atom A be an H, let atom B be X and let atom C be H. What orbitals can we construct? Note that the molecule is symmetric about the center (the position of X), hence the orbitals have to have the same symmetry.

As with HF, the only $2p$ orbital of X that will overlap with $1s$ of H is the $2p_z$. Hence, consider the combination

$$\psi_1({\bf r}) = \vert C_A\vert \psi_{1s}^A({\bf r}) + \ve... ...\vert\psi_{2s}^B({\bf r}) + \vert C_A\vert\psi_{1s}^C({\bf r})$$

Note that the two coefficients in front of the $1s$ orbitals are the same by symmetry.

How big should the $2s$ orbital of X be compared to the $1s$ orbital of H? This depends on several things. First, is the nuclear charge on X and the second is the electronegativity difference between H and X. The

first determines how quickly the 2s orbitals, remembering that the exponential part is $\exp(-Zr/a_0)$, and the electronegativity difference determines the relative magnitude of $C_A$ compared to $C_B$. In the case of BeH$_2$, the $2s$ orbital on Be multiplied by the coefficient $C_B$ (which is smaller than that of H due to Be's lower electronegativity) causes the $2s$ orbital to be smaller than the $1s$ on H. For H$_2$O, the $2s$ orbital decays at about the same rate in $r$ as does the $1s$ orbital of H, however, the coefficient $C_B$ will be larger than $C_A$ because O is more electronegative. Hence, for O, the $2s$ orbital should be larger than that of the $1s$ on H.

However, this is a technical issue that is less important than the qualitative features of the MOs that we now seek. The orbital $\psi_1({\bf r})$ constructed above is purely bonding as the figure below shows:

Figure: MOs of the linear triatomic HXH.

This orbital is also even, so we can denote it as a $\sigma_{g2s}$ orbital signifying that it is constructed from a $2s$ orbital of X combined with the $1s$ orbital of H (no designator is needed for this $1s$ orbital since this is the only orbital H can donate and hence, is obvious). The only other MO that can be constructed that has the right symmetry is

$$\psi_2({\bf r}) = \vert C_A\vert \psi_{1s}^A({\bf r}) - \ver... ...\vert\psi_{2s}^B({\bf r}) + \vert C_A\vert\psi_{1s}^C({\bf r})$$

which will be antibonding, as shown in the figure above. The orbital is even, so we can denote it as $\sigma^*_{g2s}$.

Next, if we combine a $2p_z$ orbital of X with the $1s$ of H, there are two possibilities that have the right symmetry. The first is

$$\psi_3({\bf r}) = \vert C_A\vert \psi_{1s}^A({\bf r}) - \ver... ...ert\psi_{2p_z}^B({\bf r}) - \vert C_A\vert\psi_{1s}^C({\bf r})$$

which is a bonding orbital as shown in the figure above. since it is odd, we can denote it as $\sigma_{u2p_z}$. Finally, there is

$$\psi_4({\bf r}) = \vert C_A\vert \psi_{1s}^A({\bf r}) + \ver... ...ert\psi_{2p_z}^B({\bf r}) - \vert C_A\vert\psi_{1s}^C({\bf r})$$

which is purely antibonding (see figure above) and odd. Hence, we denote this as $\sigma^*_{u2p_z}$. The orbitals $2p_x$ and $2p_y$ from X are nonbonding and become $\pi_{2p_x}$ and $\pi_{2p_y}$ nonbonding orbitals.

The figure below illustrates the difference between the actual linear case we just analyzed and the truly bent molecule, e.g. H$_2$O. The geometry changes the ordering somewhat, but the qualitative picture we obtain from the linear case makes it a useful construction.

Figure: MOs of linear vs. bent triatomics.