So I understand molecular orbitals and how to do VSEPR models, but I seem to be struggling with understanding localized bonding theory and how to do hybridizations. After drawing the initial lewis structure then molecular orbital, I do not know where to go in terms of whether something has a $\pi$ or $\sigma$ bond.
Also, how do I know when an atom has a carbon attached as an extra sigma bond or not? I understand how to know if it is $\mathrm{sp}$, $\mathrm{sp}^2$, etc. but I do not know what that means.


2 Answers 2


First of all, 'hybridisation' is a hypothetical concept, i.e. orbitals don't really mix to form new orbitals, but 'hybridisations' are very successful in explaining structures of molecules and their physical properties. So knowing that a compound has $\text{sp}^2$ hybridisation will just help you arrange the constituent atoms around the central atom. Physically it has no other meaning. It's just like vector cross product in Physics (for example the torque on a rotating wheel acts along the axis, i mean what does that even mean physically).

Secondly regarding how to know if an atom has a $\pi$ or a $\sigma$ bond: observing many structures(Lewis structures) and recognizing certain trends is the way out.

Some rules of the thumb that will help you draw structures are as follows:
1. Identify the central atom as the one that can make the most bonds.
2. Hydrogen makes only one bond. If Oxygen is present in the same compound, the Hydrogen is generally attached to the Oxygen forming a $\ce{O-H}$ group.
3. Second period elements cannot expand their octets(i.e have more than 8 electrons in valence shell)
4. Mostly second period elements can make $\pi$ bonds.(That really solves part of your query). Although in some rare cases third period elements may also make $\pi$ bonds.
5. Never neglect the lone pairs(or non-bonding electrons) of the central atom.

Remember that these are only thumb rules and may be violated sometimes.

  • $\begingroup$ I don't see why you regard vector cross products as just useful hypotheticals, the direction of the torque vector is the direction the axis of rotation points in, it gives us the rotation sense (clockwise or anticlockwise) $\endgroup$
    – Amadeus
    Nov 18, 2021 at 9:01
  • $\begingroup$ @Amadeus I realize that myself from 9 years ago might have been overly critical about vector cross product. On the other hand, I think I'm gonna leave the remark as is, till I can think of another parallel. $\endgroup$ Nov 19, 2021 at 12:13

I'm not very fond of VBT, but I'll try my best:

In the 20th Century, Chemist Linus Pauling decided to try and "hybridise" atomic orbitals to explain molecular geometry. We you probably know, the s orbital is spherical and the 3 p orbitals are dumbbell-shaped, meaning that the "unhybridised" orbitals SHOULD form bonds 90 degrees to each other (clearly not true).

To explain molecular geometry, Pauling decided to use a couple of mathematical concepts to explain molecular geometry, and this is what he postulated:

  1. The greater the overlap between atomic orbitals, the stronger the bond

  2. Due to the shape of the molecular orbitals, a pure s orbital has bond strength 1 and a pure p orbital has bond strength sqrt3.

  3. The maximum bond strength is obtained when an orbital has 25% s character and 75% p character, and this is what we call "sp3" hybridisation.

In the case of electron deficient molecules (i.e. BF3), sp2 hybridisation is observed instead; this is because hybridised orbitals have higher energy than non-hybridised s orbitals, and the greater the p character, the higher the energy. Hence, in this case forming another bond (i.e. forming a Lewis adduct) would cause the hybridisation to change to sp3, increasing bond strength.

There are a couple more things that Pauling postulated regarding lone pair s character etc., but I don't think you should concern yourself with those.

Last thing- VBT fails miserably once you study it further. For example, the sp3 orbitals are not orthogonal to one another, and hence cannot be stationary-state solutions to the Schrodinger Equation. At the end of the day, Molecular Orbital Theory (MOT) is much more accurate.

Hope this helped.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.