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Sigma bonds form due to the axial overlap of two orbitals whereas pi bonds form due to their lateral overlapping. Since the former is more effective, we can say that sigma is stronger than pi. We mean that the orbitals overlap to a further extent in sigma bonds and not as much in pi bonds(...or at least that is what I infer). To me this sounds like a circular proof/explanation, so I thought up another way to understand it.

In a sigma bond, the electron cloud of one atom is attracted to the nucleus of the other atom's nucleus.This happens because of the electrostatic forces of attraction between the positively charged protons in the nucleus of one atom and the negatively charged electron cloud of the other atom. The attractive force causes the electron clouds to the atoms to overlap, despite the fact that the electron clouds should repel each other. However, the electrons start to repel each other after a certain extent of overlapping has occurred. (Because the repulsive forces between the two nuclei and the electron clouds negate the attractive forces)

In a pi bond, there is no positively charged nucleus that would attract the electrons, so there must be another factor favoring the bond formation. Since I haven't studied the quantum mechanics of bonding yet, my best guess is that the pi bond (or lateral overlapping) occurs because of the pairing of electrons. So this means that sigma bonds are stronger because of electrostatic forces, which consequently causes more effective overlapping than pi bonds.

Is my reasoning correct to some (if any) extent?

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    $\begingroup$ when you go through perturbation theory you will result in what is known as the klopman-Salem equation. In its simplified form this equation has two terms a Coulombic interaction and an orbital term. \begin{equation} \Delta E = -\frac{Q^2}{R}+\frac{[c_1c_2\beta]^2}{E_{Homo}-E_{Lumo}}\end{equation} this latter term is what we can look at. The c terms are orbital coefficients and beta is a resonance integral (basically how well the orbitals overlap) the denominator is the difference between the orbitals energies. Therefore a large gap means less contribution. $\endgroup$
    – AngusTheMan
    Mar 18, 2015 at 10:34
  • $\begingroup$ From the orbital term. This is a good way to understand bonding. So we have the coulomb part. An overlap part. And an energy difference part. $\endgroup$
    – AngusTheMan
    Mar 18, 2015 at 10:36
  • $\begingroup$ Look at the directionality of the overlap in the 2 cases. In a sigma bond the orbitals overlap is very effective as it is head-on. Pi overlap of orbitals is less effective because it is lateral. $\endgroup$
    – ron
    Mar 18, 2015 at 13:39
  • $\begingroup$ Thanks so much @AngusTheMan and I hope I understand ur first comment when it's finally taught $\endgroup$
    – tkhanna42
    Mar 18, 2015 at 18:09
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    $\begingroup$ Related: chemistry.stackexchange.com/questions/6244/… $\endgroup$
    – LordStryker
    Mar 19, 2015 at 13:20

2 Answers 2

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This is a very short answer and it might as well have just been a comment. It is probably also tempting in it's simplicity. For starters, let's assume, that nuclei are point charges.

When two atoms form bonds, all the electrons experience the attractive force of the nuclei, of both of them. This is independent of the type of orbital the electrons occupy.

In a σ orbital, the path of maximum electron density is between the nuclei. It is also where the electronic field of the nuclei is strongest. For a π orbital the situation is different.

Let's take a step back and consider an atom. The s orbital is spherical symmetric. The maximum electron density is (formally) in the nucleus. For the p orbital, there is a node at the nucleus. The maximum electron density is further away from the opposite charge, causing attraction to be lower. The energy of the orbital will be higher.

Similarly we can see that for π orbitals. The node of these kind of orbitals is exactly where the field of the nuclei is strongest. Hence the attraction is lowered, the energy raised, because the maximum electronic density is further away from the nuclei.

In very short terms: In σ orbitals the maximum electron density is on the bonding axis, while in π (and δ) orbitals it is off the bonding axis.

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The answer to this question is simple: Since sigma bonds are the first bond that is formed b/w 2 atoms, its formed by head-on overlapping of orbital, this kind of overlapping gives us a strong bond as compared to sideways overlapping(which happens in case of pi bonds). Also, If you observe the two ways orbitals arrange themselves in the two bonds.....in sigma bonds the orbitals are closest to the 2 nuclei as compared to the orbitals of pi bond. Being closer to the nuclei makes them strong as each nucleus pulls the e- of the opposite atoms towards itself and the orbitals overlap. In a more relaxed language: In sigma they come straight at each other whereas in pi they bend themselves towards each other. Also, another imp. point: In sigma the 2 orbitals simply overlap whereas in pi bond the 2 have to bend themselves towards each other and that takes energy. Since pi bonds require certain extra energy for formations this can also be another reason why they are weaker.

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    $\begingroup$ Please don't use unnecessary abbreviations like "e-", "imp.", "b/w". Never heard of this "bending energy" theory for π bonds, too. Do you have a reputable source? $\endgroup$
    – orthocresol
    Jan 1, 2021 at 16:12

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