I came across the following passage from Structure and Properties chapter of Morrison-Boyd Organic Chemistry:

What gives the covalent bond its strength? It is the increase in electrostatic attraction. In the isolated atoms, each electron is attracted by-and attracts-one positive nucleus;in the molecule, each electron is attracted by two positive nuclei.

However, I don't think it refers to the force holding each atom together. It rather, merely describes the increase in the electrostatic force of attraction between the electrons and the nuclei. I believe that bond strength is a measure of the difficulty in pulling apart the component atoms, not the electrons from the positive nuclei.

What exactly is the pattern or picture of the forces on the nuclei and the electrons, due to one another, that holds the component atoms together? (I am aware that the decrease in overall energy or increase in stability is definitely not a reason to account for the strength of covalent bond, but rather a consequence of the action of such forces.)

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    $\begingroup$ I think the strength of the covalent bond has more to do with the types of atoms that form them. Covalent bonds tend to form between atoms in the upper right hand corner of the periodic table - the same atoms that have high electron affinity. In other words, the explanation should be the same as the one for the high electron affinity of those atoms, and since I don't know it, I leave this as a comment instead of an answer. $\endgroup$ Dec 3, 2017 at 18:49
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    $\begingroup$ Sorry, I should have said "electronegativity". Looking at ptable.com/#Property/Electronegativity reveals it has the pattern I was thinking of, not electron affinity. $\endgroup$ Dec 3, 2017 at 19:03

4 Answers 4


It probably helps to define what a covalent bond really is. Covalent bonds occur when one or more Atomic Orbitals (AO) of the participating atoms constructively interact and form a (bonding) Molecular Orbital (MO). The figure below schematises the formation of a $\sigma_{ss}$ MO when two hydrogen atoms combine to form a dihydrogen molecule:

Sigma orbital

The $+$ signs indicate the wave functions of the $1s$ orbitals of both atoms have the same sign (phase) (two atoms with opposing phases form an anti-bonding MO, not pictured here). The resulting $\sigma_{ss}$ MO is approximately a linear combination of the individual wave functions. Again, schematically:

$$\psi_{1s,1}+\psi_{1s,2}\to \psi_{\sigma}$$

The MO, in accordance with Pauli, always 'contains' $2$ electrons of opposite spin. So the MO can be said to be 'full'.

To the right is also schematised the electron probability density $\psi^2$ and note that this density is very significant on the nuclear axis, between both nuclei. This causes the intra-nuclear Coulombic repulsion force to greatly reduce and the molecular arrangement to be stable, meaning that pulling it apart would cost energy. This energy is often referred to as the bond strength (in $\mathrm{kJ/mole}$).

$\sigma$ MOs can also arise between different types of AOs, like $\sigma_{sp}$ and $\sigma_{pp}$ MOs. And between $p$ and $p$ AOs, so-called $\pi$ MOs can form. These are needed to form the stronger double bonds and the even stronger triple bonds.

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    $\begingroup$ "is the linear combination": this is an approximation only - it might be worth pointing this out. In general one must solve the full bipartite problem to get the combined system eigenstates. The linear superposition approximation is the basis for deriving, for example, the Fermi golden rule and I believe it is pretty good for calculating bond energies. But it swiftly breaks down as the overlap gets stronger: it's used, for example, in calculating approximate eigenmodes of coupled waveguides and it's lousy for anything quantitative in that application in all but the weakest coupling. Part of... $\endgroup$ Dec 4, 2017 at 0:50
  • $\begingroup$ ... the reason is that with bond energies, often a couple of significant figures is enough, but the waveguide analogy to bond energy is the propagation constant, where even small errors are important because they mean large phase errors over long propagation distances. $\endgroup$ Dec 4, 2017 at 0:51
  • $\begingroup$ BTW: This is an excellent answer which I upvoted: I just think one should be careful to warn of the approximation because its something one needs to know if one is doing any quantitative work. It's fine for discussing general principles of course. $\endgroup$ Dec 4, 2017 at 1:01
  • $\begingroup$ @WetSavannaAnimalakaRodVance: thank you! I think I've been careful not to represent this as the 'Gospel truth'... Thanks also for your additions. $\endgroup$
    – Gert
    Dec 4, 2017 at 1:23
  • $\begingroup$ No problem. I was severely burnt by failing to appreciate how loose the approximation can be about 20 years ago, and my lack of understanding cost me about a month of software development, which I had to junk! $\endgroup$ Dec 4, 2017 at 1:58

Assume that we talk about two hydrogen atoms, say $H^1$ and $H^2$, and a (di)hydrogen molecule, $H_2$, which consists of those two hydrogen atoms, that is, $H^1H^2$. One side note, the word "force" is not explicitly mentioned in the paragraph that you have quoted. Now the paragraph first says:

In the isolated atoms, each electron is attracted by -and attracts- one positive nucleus

With our notation this reads as: there is an electrostatic attraction between nucleus and electron of $H^1$, let's call this $E^1$ and there is an electrostatic attraction between nucleus and electron of $H^2$, let's call this $E^2$. By the way, for an hydrogen atom this energy is almost exactly 13.6 eV (to make chemists happy one can say -13.6 eV).

The paragraph continues:

in the molecule, each electron is attracted by two positive nuclei.

With our notation this reads as: when the molecule is formed the electron of $H^1$ is attracted by two nuclei, let's call this energy $E^{12}$ and the electron of $H^2$ is attracted by two nuclei, let's call this energy $E^{21}$.

So, finally what your quoted paragraph claims is that $$\textrm {Energy of covalent bond}=E^{12}+E^{21}-E^{1}-E^{2}$$

One final note: you say

I am aware that the decrease in overall energy or increase in stability is definitely not a reason to account for the strength of covalent bond, but rather a consequence of the action of such forces.

This sentence is not correct. The energy of two hydrogen atoms is $2\times13.6=27.2$ $eV$ and the energy of an (di)hydrogen molecule is roughly $31.2$ $eV$ and difference of which gives the energy of the covalent bond between two hydrogen atoms, that is, roughly $4$ $eV$.

  • $\begingroup$ I like the way you analyse the question. +1 from me. $\endgroup$
    – Gert
    Dec 3, 2017 at 23:05

To a chemist Gert's answer is very good. I'm not sure how much chemistry you know so I'll simply a bit.

The most basic notion is that there are two archetypes of bonds, ionic and covalent.

For an ionic bond each atom either gains or loses electrons to become charged. It is assumed that there is no "sharing" of the electrons. So for NaCl the sodium ion has one positive charge and the chlorine ion has one negative charge. It would be assumed that each ion is a hard sphere, so the ions would be a distance apart equal to the radius of the sodium ion plus the radius of the chloride ion.

The other basic archetype of bonding is the covalent bond. For this type of bond each atom contributes one electron to the molecular orbital, which can hold two electrons, and the atoms share the electrons equally. Water, $H_2O$, would be an example of a molecule with covalent bonds. To extend this a bit more, the hydrogen atom and the oxygen atom each have an orbital with one electron before bonding. When bonding the two atomic orbitals combine so that one becomes a molecular bonding orbital, and the other becomes an molecular anti-bonding orbital. The two electrons go into the lower energy molecular bonding orbital and the higher energy molecular anti-bonding orbital is empty.

For the most part neither conceptual extreme really exists and bonds are generally between the two extremes.

  • $\begingroup$ +1, especially for the last sentence. I always thought it foolish to present this stuff as a dichotomy $\endgroup$ Dec 4, 2017 at 1:06
  • $\begingroup$ Yup. Nice answer. $\endgroup$
    – Gert
    Dec 4, 2017 at 1:32

You may not like this answer, but I think at least part of any analysis of this question should be to challenge thinking that relies too heavily on macroscopic notions of force, particularly statements like:

"I am aware that the decrease in overall energy or increase in stability is definitely not a reason to account for the strength of covalent bond, but rather a consequence of the action of such forces"

In fact, if one must, one can effectively define a quantum mechanical force-oriented description of particle dynamics from statements about stability from lowering energy through the Ehrenafest Theorem (more about this later). Therefore, a force description of problems like this one is quite secondary and guaranteed by the Ehrenfest theorem once one has a description thoroughly worked out in energy terms. This is why I encourage you to take descriptions such as Gert's Answer to be the full answer to a problem like this.

When one gets to quantum mechanics, the only rigorously meaningful things from a physics standpoint are system states, their unitary time evolutions and statistical distributions of measurement outcomes defined by the system states that prevail at the time of measurement. A notion of "Force" is quite unnecessary to this description, for the following reason.

Part of the analysis of evolution is to understand that, when talking about expected (in the statistical sense) or mean behaviors, an isolated quantum mechanical system's total energy is constant and therefore it cannot spontaneously evolve to a higher energy level unless the energy difference is supplied to the system through an interaction with the outside world. Therefore, the lower the energy of a system of two mutually coupled particles compared with the sum of the energies of the two particles when separate, the more unlikely it will be for the two particles to split asunder - they need to be supplied that energy difference. Of course they do split when the temperature of a system of hydrogen molecules is high enough: the bipartite systems are interacting with one another and routinely absorb the energy difference through collisions with other $H_2$s.

Back to the Ehrenfest Theorem, which, in its most general form, reads:

$$\frac{d}{dt}\langle \hat{A}(t)\rangle = \left\langle\frac{\partial \hat{A}(t)}{\partial t}\right\rangle + \frac{1}{i \hbar}\left\langle[\hat{A}(t),\hat{H}]\right\rangle$$

where $\hat{H}$ is the quantum system Hamiltonian aka the Energy Observable and $\hat{A}$ is any other observable. This theorem is trivial to grasp in the Heisenberg picture. It also says that the mean of any observable that commutes with the Hamiltonian is conserved; trivially, energy is therefore conserved. Also, if you plug the one particle-in-a-potential $\frac{\hat{p}^2}{2\,m} + \hat{V}$ into the theorem you get a definition of the force on the particle:

$$\mathrm{d}_t \langle\hat{p}\rangle = -\nabla \langle\hat{V}\rangle\stackrel{def}{=} F$$

Thus the notion of "force" arises as a consequence of the conservation of the mean anything that commutes with the Hamiltonian.

A gentle introduction - leading to a pretty full derivation of these ideas, is to be found in section 7.4 of volume II of the Feynman lectures.

  • $\begingroup$ I don't feel like any answer here, really tackled the problem. :( And yours, while interesting, is rather extended comment. $\endgroup$
    – Mithoron
    Dec 11, 2017 at 21:00
  • $\begingroup$ @Mithoron Is there a problem to be tackled, or are you asking for a quantum system to be explained in terms of classical notions? Newton's laws are induced from 400 year old experimental technique; can we really expect them to explain observations made with more advanced techniques of 100 years ago? $\endgroup$ Dec 11, 2017 at 21:27

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