Mathematical Explanation
When examining the linear combination of atomic orbitals (LCAO) for the $\ce{H2+}$ molecular ion, we get two different energy levels, $E_+$ and $E_-$ depending on the coefficients of the atomic orbitals. The energies of the two different MO's are:
$$\begin{align}
E_+ &= E_\text{1s} + \frac{j_0}{R} - \frac{j' + k'}{1+S} \\
E_- &= E_\text{1s} + \frac{j_0}{R} - \frac{j' - k'}{1-S}
\end{align} $$
Note that $j_0 = \frac{e^2}{4\pi\varepsilon_0}$, $R$ is the internuclear distance, $S=\int \chi_\text{A}^* \chi_\text{B}\,\text{d}V$ the overlap integral, $j'$ is a coulombic contribution to the energy and $k'$ is a contribution to the resonance integral, and it does not have a classical analogue. $j'$ and $k'$ are both positive and $j' > k'$. You'll note that $j'-k' > 0$.
This is why the energy levels of $E_+$ and $E_-$ are not symmetrical with respect to the energy level of $E_\text{1s}$.
Intuitive Explanation
The intuitive explanation goes along the following line: Imagine two hydrogen nuclei that slowly get closer to each other, and at some point start mixing their orbitals. Now, one very important interaction is the coulomb force between those two nuclei, which gets larger the closer the nuclei come together. As a consequence of this, the energies of the molecular orbitals get shifted upwards, which is what creates the asymmetric image that we have for these energy levels.
Basically, you have two positively charged nuclei getting closer to each other. Now you have two options:
- Stick some electrons between them.
- Don't stick some electrons between them.
If you follow through with option 1, you'll diminish the coulomb forces between the two nuclei somewhat in favor of electron-nucleus attraction. If you go with method 2 (remember that the $\sigma^*_\text{1s}$ MO has a node between the two nuclei), the nuclei feel each other's repulsive forces more strongly.
Further Information
I highly recommend the following book, from which most of the information above stems:
- P. Atkins and R. Friedman: Molecular Quantum Mechanics, $5^\text{th}$ ed. Oxford University Press, 2011.
stabilizing energy of each bonding is less than the destabilising energy of antibonding. Now how is that possible if their sum has to equal the energies of the combining atomic orbitals
has it? $\endgroup$ – Alex Jul 29 '13 at 21:31