How exactly should we combine three orbitals when we have three orbitals that are of correct symmetry to interact?

Take $\ce{H2O}$ for example. $\mathrm{2s},$ $\mathrm{2p}_z$ and the $A_1$ group $(\ce{H2})$ orbitals can interact. It would seem trivial that the lowest bonding orbital is $\mathrm{2s}$ in phase with $A_1,$ but what about the anti binding orbital? Most resources suggest $\mathrm{2p}_z-A_1$ out-of-phase interaction, but is $\mathrm{2s}$ with $A_1$ out of phase not correct?

Could this be due to we try to construct the anti bonding orbital with least overlap possible?

  • $\begingroup$ Three AOs produce three MOs. $\endgroup$ Commented Oct 29, 2020 at 9:58
  • $\begingroup$ Yea, so there will be a non-bonding orbital in the middle right, but what about the anti bonding orbital ? $\endgroup$
    – Tsz
    Commented Oct 29, 2020 at 10:22
  • $\begingroup$ It depends on the geometry. You might have a non-bonding orbital in the middle and an antibonding one above, or you may have two roughly equal no-so-antibonding orbitals. $\endgroup$ Commented Oct 29, 2020 at 10:31
  • $\begingroup$ What I am asking is how will they mix? If you have two orbitals their linear combination is trivial, one combination is in phase while the other is not. What if there are three? $\endgroup$
    – Tsz
    Commented Oct 29, 2020 at 11:50
  • $\begingroup$ It will be something like $(+++),\;(+0-),\text{ and }(+-+)$. $\endgroup$ Commented Oct 29, 2020 at 12:55

2 Answers 2


You can estimate what happens by using a Huckel MO approach and, even though it is approximate, it shows the trends. With three similar orbitals on three atoms, such as 3 p orbitals these can be combined in two ways depending on whether the molecule is triangular or linear in shape. The MO's produced always sum to the same energy as that of the three p orbitals.

Without going into details, if we place them in a triangular arrangement each of the p orbitals can interact with each of the other two and the MO orbitals produced have relative energies -2 ,+1, +1 so when three electrons are added there is an overall lowering of energy with two in the lowest energy orbital and one in either of the two degenerate orbitals.

If linear, the three p orbitals will only interact significantly with the one next to it, 1 to 2, 2 to 3 for example. The MO orbitals all have different energy with one lowered a lot and two raised less so, so that the total energy is the same as the initial p orbitals. None are degenerate.

The conclusion is that the energy of the MO orbitals produced depend on the interactions between the p orbitals and this often depends on the geometry with closest p orbitals interacting most.

The Huckel method is described in many places, Wikipedia and most phys. chem. textbooks, if you want to try the calculation for yourself.


Another answer addressed the situation of three p orbitals, each on separate atoms. However, the example of $\ce{H2O}$ given in the question is a bit more complicated, since two of the orbitals are on the same atom (s and p on O). The short answer is that we cannot determine qualitatively what the orbitals look like, specifically the nonbonding one that is intermediate in energy.

If we first consider the s and p orbital independently, we have four possible combinations:

  1. s + A1 (where " + " means same phase) [bonding]
  2. s - A1 [antibonding]
  3. p + A1 [bonding]
  4. p - A1 [antibonding]

But we know that only three orbitals actually result, since three went in. SO already we have a problem that we don't know which three to use. It gets even more complicated when we realize that the s and p orbitals can both contribute to the same MO. (This is often described as "s-p mixing", though it formally is a mixing of the MOs, not the AOs.) If we don't worry about the exact value of the coefficients of each orbital and just focus on sign, we have four distinct results:

  1. s + p + A1 [strongly bonding]
  2. s - p + A1 [essentially nonbonding, since the portions of s and p that are oriented towards A1 are out of phase with each other and at least partially cancel out so there is very little overlap with A1]
  3. s + p - A1 [strongly antibonding] here the portions of s and p oriented towards A1 are additive to make a large lobe that is out of phase with A1
  4. s - p - A1 [essentially nonbonding, since the portions of s and p that are oriented towards A1 are out of phase with each other and partially cancel out as in (2)]

Of these, (1) and (3) are good representations of the known bonding and antibonding orbitals in $\ce{H2O}$ that you will find in pictures of MOs of $\ce{H2O}$, although the contribution of p to (1) is small enough that it closely resembles a simple s + A1 orbital.

The challenge is the third orbital. Determining whether it will more closely resemble (2) or (4) is not something that can be done with simple qualitative analysis. We must instead do a more quantitative analysis, which tells us that (4) is more accurate; the density on O has a small lobe pointed towards A1 that is in phase with A1 but overlaps very little and a large lobe that points away from A1 that is out of phase with A1. This quantitative analysis is covered in advance MO theory texts.


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