# Help with derivation on why antibonding orbitals are more antibonding

I have read How can antibonding orbitals be more antibonding than bonding orbitals are bonding?. But I am intersted in a specific derivation. During lecture my professor stated that the overlap $S_{ij}$ integral can be between -1 and 1. In other words, $$-1<\int\psi^*_i\psi_j\,dx<1$$ He further mentioned that a negative overlap integral means that the orbitals are antibonding. He also mentioned that a positive overlap integral means that the orbitals are bonding, and that an overlap integral equal to 0 means that the orbitals are nonbonding. In other words,

$$S_{ij}<0\Rightarrow\text{antibonding}$$ $$S_{ij}>0\Rightarrow\text{bonding}$$ $$S_{ij}=0\Rightarrow\text{nonbonding}$$

We then solved the Schrodinger equation using the LCAO method. In this case, we were solving for a hydrogen molecule. We only considered the 1s orbitals of each hydrogen atom. In other words,

$$\Psi_{H_2}=c_1\psi_{1s}+c_2\psi_{1s'}$$ Here, $\psi_{1s}$ and $\psi_{1s'}$ are the atomic orbitals of each hydrogen atom. The derivation that the professor used is similar to the one here: http://www.pci.tu-bs.de/aggericke/PC4e/Kap_II/H2-Ion.htm

In the end he got two wavefunctions with different coefficients $$\Psi_{H_2}=\frac{1}{\sqrt{1+S_{ij}}}(\psi_{1s}+\psi_{1s'})$$ $$\Psi'_{H_2}=\frac{1}{\sqrt{1-S_{ij}}}(\psi_{1s}-\psi_{1s'})$$ He then mentioned that since $S_{ij}$ is always positive, the coefficient of $\Psi'_{H_2}$ is larger than $\Psi_{H_2}$. In other words, $$\frac{1}{\sqrt{1-S_{ij}}}>\frac{1}{\sqrt{1+S_{ij}}}$$ And this is why antibonding orbitals are more antibonding than bonding orbitals are bonding.

But, the problem is this: if $S_{ij}$ can be negative (like he mentioned), then we have no guarantee that the coefficient for the antibonding molecular orbital will be larger than that of the bonding molecular orbital. How do we know that the $S_{ij}$ will always be positive in this case? Also, why does this statement not contradict the statement "$S_{ij}$ is negative for antibonding orbitals." What am I missing?

• Possible duplicate of How can antibonding orbitals be more antibonding than bonding orbitals are bonding? – Todd Minehardt Jan 26 '17 at 13:39
• @Todd Minehardt The question only addresses the energy. It also doesn't consider the overlap being negative. If the overlap integral can be negative, then the antibonding energy can be lower. – CoffeeIsLife Jan 26 '17 at 17:47
• OK - retracting VTC. One thing, though: you state that "He mentioned that since $S_{ij}$ is always positive..." and then go on to say "...if $S_{ij}$ can be negative (like he mentioned)...", which are contradictory statements. Can you please clarify? – Todd Minehardt Jan 26 '17 at 18:20
• I don't think I can clarify further since I didn't understand that part very well. One statement should make the other statement untrue. But it doesn't for some reason? And I am trying to figure out why. If $S_{ij}$ can be negative, the term in front of $\psi_{1s}+ \psi_{1s'}$ will be smaller than the term in front of $\psi_{1s} - \psi_{1s'}$. This will be true for the energy as well. Thus, we end up in a situation where the bonding orbital is more bonding than the antibonding orbital is antibonding. Which should be a problem. I am trying to figure out if one statement is false. – CoffeeIsLife Jan 26 '17 at 20:00

For the 1s orbital in hydrogen, the wavefunction is

$$\psi_\mathrm{1s} = N\mathrm{e}^{-r}$$

where $N$ is some normalisation constant. The normalisation constant is not unique: if $N$ is a satisfactory normalisation constant such that $\left<\psi_\mathrm{1s}\middle|\psi_\mathrm{1s}\right> = 1$, then for any value of $\phi$, the normalisation constant $N\mathrm{e}^{\mathrm i\phi}$ is also valid.

However you can make the assumption, without any loss of generality, that the wavefunctions of both 1s orbitals in dihydrogen ($\ce{H2}$) are of the same phase (i.e. both normalisation constants have the same value of $N$ and $\phi$). As a simple example we can just let $N$ be a positive real number. With this, the wavefunction of the 1s orbitals always have a positive value, and the overlap integral

$$S_\mathrm{AB} = \left<\psi_\mathrm{1s,A} \middle| \psi_\mathrm{1s,B} \right> = \int \psi_\mathrm{1s,A}\psi_\mathrm{1s,B}\,\mathrm{d}\tau$$

always has a positive integrand (no imaginary components, so I dropped the complex conjugate). The integral is therefore positive.

(Note that even though both 1s orbitals have the same exponential form, $\psi_\mathrm{1s,A}$ is not equal to $\psi_\mathrm{1s,B}$ since the orbitals are centred at different points in space.)

In fact, this choice of phase is required so that you can express the bonding and antibonding MOs as the specific linear combinations presented. The "in-phase" linear combination, i.e. the bonding MO, only has the form $N(\psi_\mathrm{1s,A} + \psi_\mathrm{1s,B})$ if you let the two orbitals have the same phase. If you had a different choice of normalisation constant, the bonding MO would be represented by a different linear combination $N(\psi_\mathrm{1s,A} + z\psi_\mathrm{1s,B})$ where $z$ is some complex number determined by the relative phases of the 1s orbitals.

As such, by virtue of expressing the MOs in that form, your overlap integral is necessarily positive.

Note that the above discussion only really applies to 1s + 1s overlap. With other orbitals, which have both positive and negative regions, you can end up with a negative overlap integral depending on the relative phases that you choose.

For example in $\ce{BeH2}$ if you consider the middle Be 2p orbital to overlap with both hydrogen 1s orbitals, one "overlap" will be formed with the positive lobe of the 2p orbital and the other "overlap" will be formed with the negative lobe of the 2p orbital. Assuming that you phase both 1s orbitals to be positive, then the first overlap integral will be positive (positive 1s times positive 2p gives a positive integrand) and the second will be negative (positive 1s times negative 2p gives a negative integrand).

In general your choice of phase is arbitrary; it does not affect the final physical result, and is simply chosen to make the calculations easier. That is to say, the eigenvalues (MO energies) that you get out of it will still be the same. The eigenfunctions may take on a different form, but that is physically unimportant.