I am currently doing a solid state physics class and am stuck on a concept regarding molecular orbital (MO) hybridization.

As an example consider an HF molecule bonded along the z-axis. This molecule will form a $\sigma_{1s-2p_z}$ bond between the 1s orbital of the H atom and the $2p_z$ orbital of the F. However, the $2p_x$ and $2p_y$ F orbitals can't form a hybridized orbital because their nodal symmetries are not along the bonding axis.

Every explanation I've found of this phenomena has been something to the effect of

AO’s must have the same nodal symmetry (as defined by the molecular symmetry operations), or their overlap is zero.

or, from Wikipedia:

Molecular orbitals arise from allowed interactions between atomic orbitals, which are allowed if the symmetries (determined from group theory) of the atomic orbitals are compatible with each other.

But I haven't been able to find an explanation of what "compatible symmetries" are, how to determine which symmetries are compatible, or the mathematical derivation for why non-compatible symmetries result in zero overlap.

Any illumination on these points to help my understanding would be greatly appreciated.


1 Answer 1


You're slightly conflating MO formation with hybridisation.

MO's form as an interaction between two Atomic Orbitals (AOs) and mathematically can be seen as the linear combination of two AOs.

MOs can be formed when some conditions are met:

1. Symmetry and sufficient overlap:


(a) formation of a $\sigma_{pp}$ bond.

(b) formation of a $\sigma^*_{pp}$ anti-bond.


(c) formation of a $\pi_{pp}$ bond.

(d) formation of a $\pi^*_{pp}$ anti-bond.

No MOs

Lack of symmetry in (a) and (b) (above) prevents bonding (no overlap). (c) forms a $\sigma_{sp}$ bond.

2. Orbital lobes interacting must have same wave function sign (gerade/ungerade):

The cases (b) and (d) above lead to anti-bonds because the orbital lobes are of opposite sign (as indicated by colour).

3. AOs must be of comparable energy levels.

4. MOs can contain a maximum of 2 electrons (with differing spin quantum number, acc. Pauli).

Hybridisation of AOs:

Hybridisation features heavily in Valence Shell Electron Pair Repulsion theory. Hybridisation minimises the potential energy of many molecules by minimising the electrostatic repulsion between the MOs.

It explains the bond angles for compounds of the $EX_n$ ($X$ is a halogen or hydrogen, $E$ is some element) group of binary compounds. For example, methane's ($CH_4$) tetrahedral structure is explained by it, because that bond angle minimises electrostatic repulsion between the MOs (four $\sigma$ MOs from linear combination of (hydrogen) $1s$ AOs and hybridised $sp^3$ carbon AOs).

The hybridisation process can be schematised as electron configurations as follows:

C hybridisation.

For carbon the $sp^3$ hybridised AOs look like below. For clarity the AOs have been presented separately, with white dots representing the C-nucleus:

sp3 orbitals, carbon

([Image source.])6

The molecular shapes and bond angles for $EX_n$ compounds, due to hybridisation of MOs:

EX2 to 6.

EX7 to 8.

Edit: in response to OP's question in the comments.

$s$ and $p_{x,y,z}$ orbitals can combine to bonding $\sigma$ bonds (MOs), as long as one of the $p$ lobes lies on the inter nuclear axis (the axis connecting the nuclei of the bonding atoms) and provided the energy levels of the $s$ and $p$ are not too far apart:

MO of pz and s

What we call that $p$ orbital ($x$,$y$ or $z$) is a matter of convention, as we can choose the coordinate system to our convenience. Conventionally, the one interacting with an $s$ is set to $z$.

Now have a look at this, bonding and anti-bonding between two $1s$ AOs:

Bonding and anti-bonding in dihydrogen

Remember that both wave functions $+ ψ$ and $– ψ$ satisfy the Schrödinger equation. In the top part of the diagram the wave functions have the same sign (either +,+ or -,-) and are said to be ‘in phase’. In the bottom part the wave functions have the opposite sign (either +,- or -,+) and are said to be ‘out of phase’.

Like actual waves, the wave functions can now show reinforcing interference when they are in phase (top) or negative interference when they out of phase (bottom).

In the top situation a bonding molecular orbital has formed. See how in the top part the value for $ψ^2$ between the nuclei (represented as dots) is positive. This means there is considerable probability of finding the electron(s) in that region and this is where they reduce the electrostatic repulsion between the (both positively charged) nuclei. A $σ_{ss}$ ($σ$ for short) has formed and a $H_{2}$ is born.

But in the bottom part for $ψ^2$ between the nuclei is zero and with no electron presence in that region there is nothing to prevent the electrostatic repulsion between the nuclei from ripping the ensemble apart. This orbital is called an anti-bonding molecular orbital, noted as $σ^*$.

The same principle holds for other types of bonds: the interacting lobes must be in phase (of the same sign - i.e. colour in the figures higher up).

Note that $\pi$ bonds don't offer much inter-nuclear electron density and for that reason never occur on their own but always as part of a double bond (one $\sigma$ plus one $\pi$) or triple bond (one $\sigma$ plus two $\pi$).

  • $\begingroup$ Thank you for the response Gert! You're correct that I conflated the two concepts, thank you for the clarification. However, I'm still a bit confused on why specifically the s and $p_x$ orbitals do not overlap. Just looking at the image of the orbitals as well as the equations of the wavefunctions, it appears that they would overlap slightly as long as you are not on the z-axis. Is there a mathematical way to show the superposition of the wavefunctions give zero? $\endgroup$
    – bicarlsen
    Jul 22, 2016 at 8:47
  • $\begingroup$ @bicarlsen: Hi. I've edited the answer to accommodate these questions. $\endgroup$
    – Gert
    Jul 22, 2016 at 13:02
  • 1
    $\begingroup$ @EmilioPisanty: corrected. You can be my proof reader any time. :-) $\endgroup$
    – Gert
    Jul 22, 2016 at 13:08
  • $\begingroup$ Thank you for the additional explanation. I completely understand the bonding / anti-bonding explanation you provided. However, I am still confused about non-bonding orbitals. You state ... orbitals can combine to bonding σ bonds (MOs), as long as one of the p lobes lies on the inter nuclear axis ... What is the justification for this? For instance, if the bond is on the z-axis, why can the s and $p_x$ orbitals not bond? $\endgroup$
    – bicarlsen
    Jul 22, 2016 at 19:13
  • $\begingroup$ A H $1s$ orbital approaches, via the $z$-axis, an atom with three $p$ orbitals. A $\sigma$ bond can form between the $1s$ and $p_z$ AOs. But that same $1s$ could never bond with the $p_x$ or $p_y$ orbitals because there's no overlap possible. However, another $1s$, approaching via the $x$-axis could bond to the $p_x$ orbital. Same via the $y$-axis. I.o.w. there has to be enough electron probability density between the participating AOs for an actual bond to form. $\endgroup$
    – Gert
    Jul 22, 2016 at 19:25

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