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I've always been told in chemistry lessons that lone pairs repel more due to a greater charge density than bonding pairs. And that makes sense when steric effects are involved when bond orbitals overlap.

But this conflicts with what I have learnt in physics i.e. Coulomb's law, which states that the only thing affecting the force between two charged bodies is the magnitude of the charge and the distance between them. Therefore it seems weird to me that a lone pair should "repel" more because of greater charge density, despite having the same charge magnitude as a bonding pair.

What am I neglecting or if I'm right, does that mean for large atoms (such as iodine, where electron pairs are further apart) the lone pairs and bonding pairs repel equally as per Coulomb's law?

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  • $\begingroup$ Let me elaborate on my question. Coloumbs law states that two charged particles of charge +q and distance d repel with force (k * q^2 / d^2 ) where k = 1 / (4*pi*8.85*10^-11). The 'volume' of the two charged bodies and thus their charge densities are irrelevant provided the distance between their centres are constant . $\endgroup$
    – MY2K
    Commented Mar 31, 2014 at 14:14
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    $\begingroup$ I guess you meant VSEPR. It is a very useful tool to explain certain geometries. For predictions I would not rely on that theory as one has to use other crude approximations like hybridisation. $\endgroup$
    – Martin - マーチン ♦
    Commented May 5, 2014 at 5:00

3 Answers 3

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In bonded pairs of electrons, the repulsion of the negative charges is somewhat reduced by the positive charge of the bonded atom's nuclei. Since lone pairs don't have to deal with this positive charge, naturally their repulsion is stronger.

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    $\begingroup$ Makes perfect sense and fits with Coloumbs law! Thank you for clearing up this matter of confusion and I can't believe I didn't think of that. $\endgroup$
    – MY2K
    Commented Mar 31, 2014 at 19:56
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The charge magnitude is the same in both cases; however, the sizes of the orbitals are different. The lone pair electrons are in the field of one nucleus, while the two electrons of a bonding pair are in the field of two nuclei. As a result, the orbital with the bonding pair tends to be smaller, and it requires less space than the more diffuse orbital of a lone pair. Therefore, the distance between bonding pair orbitals is larger. This reduces the repulsive force between them, because according to Coulomb's law, the electrostatic force $F$ between two charges $q_1$ and $q_2$ is inversely proportional to the distance $r$ between them.

$$|F|=k_{e}\frac{|q_1q_2|}{r^2}$$

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  • $\begingroup$ Interesting, @Jannis, you state that the bonding pair has a higher density and the OP state the lone pair has the higher density. I think a reference is needed here. $\endgroup$
    – LDC3
    Commented Mar 31, 2014 at 13:39
  • $\begingroup$ This is what in found in my textbook (Riedel, Inorganic chemistry) and what seems logical to me. $\endgroup$
    – Jannis Andreska
    Commented Mar 31, 2014 at 13:46
  • $\begingroup$ Everything you have stated is fundamental to VSPR theory but you haven't explained why lone pairs repel more since as i explained in the OP the repulsion is subject to coulombs Law which suggests charge 'density' to be irrelevant in repulsion. $\endgroup$
    – MY2K
    Commented Mar 31, 2014 at 14:08
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    $\begingroup$ The lone pairs repel more because their electrons are contained in larger orbitals. Larger orbitals tend to be closer to each other, especially when they belong to the same atom. Since the distance between charges is considered in Coulomb's Law, this would explain the higher repulsion. $\endgroup$
    – Jannis Andreska
    Commented Mar 31, 2014 at 14:17
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Lone pair-Lone pair repulsion $>$ Lone pair-Bonding pair repulsion $>$ Bonding pair-Bonding pair repulsion.
enter image description here

Lone pairs are attached to same atom, they are free. As expected, they should repulse to greater extent. Bond pairs are attached to two different atoms, the heavy atoms attached to them might make them not repulse to greater extent. Now, lone pair-bond pair repulsion should be intermediate, as they have one lone pair and one bond pair.

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  • $\begingroup$ For better understanding, look at the images of this page, and remember I have just guessed that this might be reason, as SE aims to have exact answer, I would be happy to know any errors, and will for sure try to correct it. $\endgroup$
    – Sensebe
    Commented Mar 31, 2014 at 13:59
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    $\begingroup$ Your answer states the fundamentals of VSPR theory but doesn't explain why lone pairs repel more. Being 'free' doesn't suggest to me any reason to repel more. Please refer to Coloumb's Law which would suggest that charge 'density' should be irrelevant. $\endgroup$
    – MY2K
    Commented Mar 31, 2014 at 14:06

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