Chemistry - How can antibonding orbitals be more antibonding than bonding orbitals are bonding?
Solution 1:
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.
Solution 2:
The sum of the energies over all orbitals is constant. The lower energy in a bonded state arises from the fact that not all MOs are occupied by electrons. Because the energy of the electrons is lowered, the total system energy is lowered by bonding. See http://www.dlt.ncssm.edu/tiger/diagrams/bonding/BondingEnergy.gif for an illustration.
Edit: This phenomenon is due to the nuclear-nuclear repulsion of the two (or more) atoms that are binding. See Ball, D.W. Physical Chemistry p. 408
Solution 3:
There is no such thing as conservation of orbital energies at different fixed values of $R$, the internuclear distance. Obviously, at $R \rightarrow \infty$ the energies are the same regardless of the linear combinations you use (there is no interaction and the "molecular" energy is exactly the same as the sum of the "atomic" energies); obviously too, the energies raise a lot if you make $R$ very small: you have the fusion barrier that doesn't allow you to convert your initial atomic orbitals of atoms A and B into the atomic orbitals of atom AB. Energy is obviously conserved in a dynamical process where two atoms approach and you have kinetic energy in the collisional coordinate and so on and so forth (you can immediately see that two atoms cannot form a diatomic molecule is something else does not take away energy from the system).
Solution 4:
Usually, when atoms form bounds they release energy (heating, radiation etc). So, the energy conservation law holds, but this kind of sum rule does not work. Vice versa, one has to spend energy to get dissociation of the molecule.