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Next: Frontier molecular orbitals Up: lecture_15 Previous: lecture_15

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

\begin{displaymath}
\psi_g({\bf r}) = C_A \psi^A_{2s(Z=Z_A)}({\bf r}) +
C_B \psi^B_{2s(Z=Z_B)}({\bf r})
\end{displaymath}

Recall the energies $H_{AA}$ and $H_{BB}$ that were defined for LCAO in the homonuclear case:
$\displaystyle H_{AA}$ $\textstyle =$ $\displaystyle \int \psi_{2s(Z=Z_A)}^A({\bf r})
\hat{H}_{\rm elec}\psi_{2s(Z=Z_A)}^A({\bf r})dV$  
       
$\displaystyle H_{BB}$ $\textstyle =$ $\displaystyle \int \psi_{2s(Z=Z_B)}^B({\bf r})
\hat{H}_{\rm elec}\psi_{2s(Z=Z_B)}^B({\bf r})dV$  

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 $H_{BB} < H_{AA}$ (remember $H_{AA}$ and $H_{BB}$ are both negative). Now, if we calculate the guess of the ground-state energy

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

we find the more general result

\begin{displaymath}
E_g = {H_{AA} C_A^2 + H_{BB}C_B^2 + 2H_{AB}C_A C_B \over C_A^2 + C_B^2 + 2SC_A C_B}
\end{displaymath}

and then perform the minimization of $E_g$:

\begin{displaymath}
{dE_g \over dC_A} = 0\;\;\;\;\;\;\;\;\;\;{dE_g \over dC_B} = 0
\end{displaymath}

we obtain the conditions
$\displaystyle {dE_g \over dC_A}$ $\textstyle =$ $\displaystyle {2C_A H_{AA} + 2C_B H_{AB} \over D}
- {H_{AA} C_A^2 + H_{BB} C_B^2 + 2C_A C_B H_{AB} \over D^2}(2C_A + 2SC_B) = 0$  
       
$\displaystyle {dE_g \over dC_B}$ $\textstyle =$ $\displaystyle {2C_B H_{BB} + 2C_A H_{AB} \over D}
- {H_{AA} C_A^2 + H_{BB} C_B^2 + 2C_A C_B H_{AB} \over D^2}(2C_B + 2SC_A) = 0$  

If we make the simplifying approximation that the overlap $S\approx 0$, then these conditions lead to the following relationship between $C_A$ and $C_B$:

\begin{displaymath}
{C_A \over C_B} =
{H_{AA}-H_{BB} \over 2H_{AB}}
\pm \sqrt{{(H_{AA}-H_{BB})^2 \over 4 H_{AB}^2} + 1}
\end{displaymath}

In order to make the analysis a bit easier, let us expand the square root using the fact that

\begin{displaymath}
\sqrt{1+x} \approx 1 + {1 \over 2}x
\end{displaymath}

Then

\begin{displaymath}
{C_A \over C_B} = {H_{AA}-H_{BB} \over 2H_{AB}}
\pm \left[1 + {(H_{AA}-H_{BB})^2 \over 8H_{AB}^2}\right]
\end{displaymath}

Now, we know that

\begin{displaymath}
\left[1 + {(H_{AA}-H_{BB})^2 \over 8H_{AB}^2}\right] >
\left\vert{H_{AA}-H_{BB} \over 2H_{AB}}\right\vert
\end{displaymath}

First, look at the $\psi_+({\bf r})$. If $H_{AA}>H_{BB}$, $B$ is more electronegative because its energy is lower (think of the Bohr formula $E_n = -Z^2/n^2$, which shows that as $Z$ increases, the energy decreases). Generally, a not so easy calculation of $H_{AB}$ shows that $H_{AB}<0$. Thus, if $H_{AA}>H_{BB}$, then

\begin{displaymath}
{H_{AA}-H_{BB} \over 2H_{AB}}<0
\end{displaymath}

and this means that $C_A/C_B < 1$ or $C_A<C_B$, which is what we would expect if $B$ is more electronegative. Similarly, if $H_{AA}<H_{BB}$, $A$ is more electronegative. We would then find

\begin{displaymath}
{H_{AA}-H_{BB} \over 2H_{AB}}>0
\end{displaymath}

and $C_A/C_B>1$, and $C_A>C_B$, which is what we would expect if $A$ is more electronegative.



Another consequence we can derive from the above result is that if $\vert H_{AA}\vert \gg \vert H_{BB}\vert$ or $\vert H_{AA}\vert \ll \vert H_{BB}\vert$, then the coefficients will be very different, e.g. $C_A \gg C_B$ or $C_A \ll C_B$, in which case, the resulting MO is not really an MO at all, but rather more like the AO with the larger coefficient. Thus, it is clear that $H_{AA}$ and $H_{BB}$ cannot be very different if we are to have a reasonable amount of mixing of the two AOs. This supports the idea that only orbitals with similar energies can be combined in the LCAO scheme.



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

\begin{displaymath}
\psi_+(-{\bf r}) \neq \psi_+({\bf r})\;\;\;\;\;\;\;\;\;\psi_-(-{\bf r}) \neq -\psi_-({\bf r})
\end{displaymath}

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

\begin{displaymath}
1s^2 2s^2 2p_x
\end{displaymath}

while oxygen's is

\begin{displaymath}
1s^2 2s^2 2p_x^2 2p_y 2p_z
\end{displaymath}

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.
\includegraphics[scale=1.0]{BO_correlation.eps}
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

\begin{displaymath}
\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
\end{displaymath}

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:

\begin{displaymath}
\left(\sigma_{2s}\right)^2 \left(\sigma^*_{2s}\right)^2
\lef...
...ght)^2
\left(\sigma_{2p_z}\right)^2\left(\pi^*_{2p_x}\right)^1
\end{displaymath}

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

\begin{displaymath}
\psi_+({\bf r}) = C_A \psi_{1s}^A({\bf r}) + C_B \psi_{2p_z}^B({\bf r})
\end{displaymath}

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

\begin{displaymath}
\psi_-({\bf r}) = C_A \psi_{1s}^A({\bf r}) - C_B \psi_{2p_z}^B({\bf r})
\end{displaymath}

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 of F with the 1s of H in the molecule HF.
\includegraphics[scale=0.5]{HF_LCAO_orbitals.eps}
Figure: MOs and correlation diagram for HF.
\includegraphics[scale=0.5]{HF_MOs.eps}
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: Another look at the correlation diagram in HF.
\includegraphics[scale=0.75]{HF_correlation.eps}
The bond order of HF is easily seen to be 1.


next up previous
Next: Frontier molecular orbitals Up: lecture_15 Previous: lecture_15
Mark E. Tuckerman 2011-11-21