Destabilization of antibonding MO vs stabilization of bonding MO

My textbook writes that

The net energy stabilization due to the occupation of a bonding MO is equal to the net energy destabilization due to the occupation of the corresponding antibonding MO.

Essentially, it claims that these two quantities are equal.

(Image source: SparkNotes)

Shouldn't the antibonding orbital be destabilized more than the bonding orbital is stabilized, though?

• I agree with your notion. If there was no energy lost by creating bonding and antibonding orbitals and populating them both, helium would easily form compounds (and all our thinking towards antibonding electrons being bad would be null and void). – Jan Oct 21 '15 at 20:28
• – Mithoron Oct 21 '15 at 21:15
• Searching for the truth where there's no one, trying to make scientific sense of what isn't even scientific... All these models are way oversimplified, so there is very little sense in being nitpicking as well as asking about their truth: the general chemistry MOT description is generally miles away from reality. To learn the real things one better studies quantum theory rather than asks all these non-sense like "Isn't the antibonding orbital more antibonding than the bonding orbital is bonding?" – Wildcat Oct 22 '15 at 14:52
• By the way, does the author say a word about $\ce{He2}$ instability? I mean, yeah, at this particular page he says that the energy differences are equal, but maybe he elaborates on that matter a bit later and corrects the initial oversimplified view. – Wildcat Oct 22 '15 at 14:56

TL;DR

"The destabilisation of an antibonding MO is equal to the stabilisation of a bonding MO, relative to the constituent AOs."

Several key assumptions are inherent in this statement.

1. Born–Oppenheimer approximation (not discussed further)
2. "Orbital approximation": wavefunction of $\ce{H2, He2, ...}$ can be expressed as an antisymmetrised product of one-electron wavefunctions (called MOs)
3. Said MOs take the form of the stationary states of $\ce{H2+}$, without modification
4. LCAO approximation: stationary states of $\ce{H2+}$ can be expressed as a linear combination of 1s orbitals on both hydrogens
5. The overlap integral $\langle \mathrm{1s_A} | \mathrm{1s_B} \rangle$ is neglected

If we make all five assumptions, then the statement above is correct.

If we only make assumptions 1 through 4, then MOs will still have the same form as above (i.e. everything else about the MO diagram is correct), but the statement above will be incorrect: the antibonding orbital will be more destabilised than the bonding orbital is stabilised.

The question linked in the comments does give a qualitative explanation, so depending on what you are looking for, it may be worth checking out.

Assumptions

The MOs that you see are one-electron wavefunctions that are obtained by looking at the single-electron species $\ce{H2+}$. Finding the exact stationary states of this system (by analytical solution of the Schrödinger equation) is a bit complicated, so the strategy is to find a set of approximate stationary states for the electron in $\ce{H2+}$, which we can label as $\{|i\rangle\} (i = 1, 2, 3 \cdots)$. These states are called molecular orbitals.

For multi-electron molecules, like $\ce{H2}$, a further approximation is usually invoked, in expressing the electronic wavefunction of $\ce{H2}$ as a properly antisymmetrised product of these states $\{|i\rangle\}$. Therefore, the MOs, which were only "rigorously" derived for $\ce{H2+}$, may be extended to molecules such as $\ce{H2}$. Very loosely speaking, we may assign one stationary state $|i\rangle$ to one electron, which corresponds to placing an electron in an orbital. Because of the Pauli exclusion principle, each stationary state may only be assigned to a maximum of two electrons.

These are obviously crude approximations and it is a surprise that it even works decently well. In particular the issue of electron-electron repulsions in multi-electron systems is glossed over; in this very simple model, the MOs, being derived from a one-electron system, are entirely unaffected by them, which doesn't really make sense. As the comments on the question have alluded to, there are much more sophisticated ways of obtaining MOs nowadays, but since the claim in the question is derived using this model, we have to use it.

MOs of hydrogen molecule ion

Once we have assumed that the stationary states (i.e. MOs) are of the form

$$\psi = c_1\phi_1 + c_2\phi_2$$

where $\psi$ is an MO and $\phi_1, \phi_2$ 1s atomic orbitals on the two hydrogens, then one can use the variational principle to optimise the two coefficients $c_1, c_2$. The idea is that one wants to minimise the energy of the wavefunction $\psi$:

$$\mathcal{E} = \frac{\langle \psi | H | \psi \rangle}{\langle \psi | \psi \rangle}$$

with respect to both $c_1$ and $c_2$. To this end, both the following partial derivatives must be equal to zero:

$$\left(\frac{\partial \mathcal{E}}{\partial c_1}\right)_{\!c_2} = \left(\frac{\partial \mathcal{E}}{\partial c_2}\right)_{\!c_1} = 0$$

The maths will be omitted and if one is interested, one may find it in a variety of textbooks. For example it is covered in Atkins & Friedman's Molecular Quantum Mechanics 5th ed. (search for Rayleigh–Ritz method), or in Coulson's Valence 3rd ed. It is also described thoroughly in this handout from QMUL. The key result is the secular equations, which are a set of generalised eigenvector equations:

$$\mathbf{Hc} - \mathcal{E}\mathbf{Sc} = \mathbf{0}$$

where the $2 \times 1$ vector $\mathbf{c}$ simply contains the coefficients $c_1$ and $c_2$, the elements of the Hamiltonian matrix are $\mathbf{H}_{ab} = \langle a | H | b \rangle$ and the elements of the overlap matrix are $\mathbf{S}_{ab} = \langle a | b \rangle$. The basis set used is $\{|\phi_1\rangle, |\phi_2\rangle\}$. For a non-trivial solution (i.e. $\mathbf{c} \neq \mathbf{0}$) we require the secular determinant to vanish:

$$|\mathbf{H} - \mathcal{E}\mathbf{S}| = 0$$

Now we can use symbols to fill in these matrix elements:

• $H_{11} = H_{22} = \alpha$ (this must be so, as both hydrogen atoms are equivalent by symmetry)
• $H_{12} = H_{21} = \beta$ (assume that $\beta$ is real, so by the hermiticity of $H$, we must have $H_{12} = H_{21}$)
• $S_{11} = S_{22} = 1$ (the atomic orbitals are normalised)
• $S_{12} = S_{21}$ (again, assume that this is real. Assumption 5 is that this goes to zero, but for now we will not neglect it; later we can set it to zero to see what happens.)

The exact form of the Hamiltonian, as well as more explicit expressions for these quantities, are given in Atkins MQM. For now they need not concern us. Solving the determinant we obtain two values of $\mathcal{E}$ and two associated vectors $\mathbf{c}$, which can simply be written as the MOs $\psi = c_1\phi_1 + c_2\phi_2$. The MOs (i.e. approximate stationary states) are now quoted without proof, with their eigenvalues (i.e. their approximate energies):

\begin{align} \psi_1 &= \frac{1}{\sqrt{2 + 2S}}(\phi_1 + \phi_2) & & & \left\langle \psi_1 \middle| H \middle| \psi_1 \right\rangle &= \mathcal{E}_1 = \frac{\alpha + \beta}{1 + S_{12}} \tag{1} \\ \psi_2 &= \frac{1}{\sqrt{2 - 2S}}(\phi_1 - \phi_2) & & & \left\langle \psi_2 \middle| H \middle| \psi_2 \right\rangle &= \mathcal{E}_2 = \frac{\alpha - \beta}{1 - S_{12}} \tag{2} \end{align}

$\mathcal{E}_1$ is the ground state energy (i.e. $\mathcal{E}_1 < \mathcal{E}_2$) since $\alpha$ and $\beta$ are both negative quantities (again MQM has more info).

Stabilisation and destabilisation energies

Now we have actually reached the part where the original question can be answered. What is the stabilisation of the bonding MO? The energy of the constituent AOs is $H_{11} = H_{22} = \alpha$. So, the stabilisation must be:

\begin{align} E_{\mathrm{stab}} &= \alpha - \mathcal{E}_1 \\ &= \alpha - \frac{\alpha + \beta}{1 + S_{12}} \\ &= \frac{\alpha S_{12} - \beta}{1 + S_{12}} \end{align}

Similarly, the destabilisation of the antibonding MO is:

\begin{align} E_{\mathrm{destab}} &= \mathcal{E}_2 - \alpha \\ &= \frac{\alpha S_{12} - \beta}{1 - S_{12}} \end{align}

The quantity $S_{12}$ is positive (again it should be in MQM), so $E_{\mathrm{destab}} > E_{\mathrm{stab}}$.

The "incorrect" piece of information comes from a further simplifying assumption - namely, that $S_{12} \ll 1$. That is not the greatest assumption, since the values of overlap integrals can sometimes be quite non-negligible (for a $\ce{H-H}$ bond length of $74~\mathrm{pm}$, $S_{12} = 0.753$!), but it is a useful way to predict approximate energies and the general forms of the MOs.

If $S_{12} \ll 1$, then $1 \pm S_{12} \approx 1$ and equations $(1)$ and $(2)$ become:

\begin{align} \psi_1 &= \frac{1}{\sqrt{2}}(\phi_1 + \phi_2) & & & \left\langle \psi_1 \middle| H \middle| \psi_1 \right\rangle &= \mathcal{E}_1 = \alpha + \beta \tag{3} \\ \psi_2 &= \frac{1}{\sqrt{2}}(\phi_1 - \phi_2) & & & \left\langle \psi_2 \middle| H \middle| \psi_2 \right\rangle &= \mathcal{E}_2 = \alpha - \beta \tag{4} \end{align}

Now it turns out that, with the additional neglect of overlap approximation, $E_{\mathrm{stab}} = E_{\mathrm{destab}} = -\beta$. That's where the "destabilisation of antibonding MO = stabilisation of bonding MO" idea comes from.

The arguments above can be extended to the case where you have constituent AOs which are different. The main difference is that $H_{11} \neq H_{22}$, so you have to write $H_{11} = \alpha_1$ and $H_{22} = \alpha_2$. The result you obtain will be rather similar: the destabilisation of the antibonding MO is greater than the stabilisation of the bonding MO, but in the limit of $S_{12} \to 0$, they become equal.

• "The true wavefunction for H2HX2 does not need to have an electron in one "orbital" or another. But that is a story for another day (and for someone else to say...)" Want to know more about it. Will the hyperlinked book work? – Mockingbird Jul 20 '17 at 17:29
• @Mockingbird it's been a while since I wrote this answer. Fundamentally what I meant is that we cannot solve the Schrodinger equation for $\ce{H2}$. When we construct a total wavefunction from constituent molecular orbitals, it is merely an approximation to the true wavefunction, which we can't obtain. I'd recommend reading Atkins MQM (I removed the other references). – orthocresol Jul 20 '17 at 17:37
• I cleared the earlier comments as I revised the answer thoroughly; I think it is accurate now, but if there are still mistakes please drop a comment. – orthocresol Jul 20 '17 at 19:20
• Why is there a consensus regarding the Schrodinger equation of $\ce{H2}$ can't be solved? – Mockingbird Jul 21 '17 at 2:27
• @Mockingbird no, I don't want to elaborate further. This answer is long enough and I don't want to spend any more time on it, please go read a QM book if you want more detail, or ask a new question – orthocresol Jul 21 '17 at 3:06