A Level Chemistry 3.1.3 Bonding shapes of molecules

How do I work out the number of bonding pairs AND lone pairs around the central atom in the molecule.
I have a certain rule for when its single bonds (e.g. NH3):

Identify the central atom bonded to all other atoms.
Find the number of outer shell electrons of the central atom using its group number.
Add one electron for each bonded atom.
If the species is an ion, add one electron for each negative charge or subtract one electron for each positive charge.
Divide the total number of electrons by two to get the number of electron pairs.
Subtract the number of bonds from the number of electron pairs to determine the number of lone pairs.

This works all fine for single bonds but for >double bonds it doesn't work, e.g. CO2, using this rule apparently carbon has 2 bonding pairs and one lone pair (but obviously it has 2 bonding pairs ONLY). So first of all how do I tell whn there a double bonds and how to work out how many bonding pairs and lone pairs surrounding the atom

Reply 1

nxvastel

Original post

by Jimmy Butler IV

How do I work out the number of bonding pairs AND lone pairs around the central atom in the molecule.
I have a certain rule for when its single bonds (e.g. NH3):
Identify the central atom bonded to all other atoms.
Find the number of outer shell electrons of the central atom using its group number.
Add one electron for each bonded atom.
If the species is an ion, add one electron for each negative charge or subtract one electron for each positive charge.
Divide the total number of electrons by two to get the number of electron pairs.
Subtract the number of bonds from the number of electron pairs to determine the number of lone pairs.
This works all fine for single bonds but for >double bonds it doesn't work, e.g. CO2, using this rule apparently carbon has 2 bonding pairs and one lone pair (but obviously it has 2 bonding pairs ONLY). So first of all how do I tell whn there a double bonds and how to work out how many bonding pairs and lone pairs surrounding the atom

Riiiiiiiight.

Have you tried counting double bonds as two bonds? So, using CO₂ as an example, the central atom carbon has two C=O double bonds. So, each double bond is two electrons, so carbon has used up four electrons in bonding. As carbon is in Group 4, it has used up all its electrons in bonding, and has no lone pairs. CO₂ is a non-polar, linear molecule.

I think you have messed up the idea of "bonds" and "bonding pairs". In CO₂, you can count carbon as having four bonds, two for each double bond when counting individual electrons. However, as each double bond is considered as one bonding pair (as per VSPER theory), use bonding pairs only when considering geometry.

Note: some atoms, such as S, P, Cl, Br, I, Xe (et cetera) can be hypervalent. They can have more than eight electrons in their valence shell - and so can form more bonds "than usual". Think of PH₅, which has a trigonal bipyramidal shape; or SF₆, which is octahedral. You can even have tetrahedral examples - ClO₄^-, the perchlorate anion, is a good example of chlorine having 14 valence electrons and a tetrahedral structure.

Any questions and cool ideas, feel free to spitball them out. I'm more than happy to answer anything.

Reply 2

Jimmy Butler IV

Original post

by nxvastel

Riiiiiiiight.
Have you tried counting double bonds as two bonds? So, using CO₂ as an example, the central atom carbon has two C=O double bonds. So, each double bond is two electrons, so carbon has used up four electrons in bonding. As carbon is in Group 4, it has used up all its electrons in bonding, and has no lone pairs. CO₂ is a non-polar, linear molecule.
I think you have messed up the idea of "bonds" and "bonding pairs". In CO₂, you can count carbon as having four bonds, two for each double bond when counting individual electrons. However, as each double bond is considered as one bonding pair (as per VSPER theory), use bonding pairs only when considering geometry.
Note: some atoms, such as S, P, Cl, Br, I, Xe (et cetera) can be hypervalent. They can have more than eight electrons in their valence shell - and so can form more bonds "than usual". Think of PH₅, which has a trigonal bipyramidal shape; or SF₆, which is octahedral. You can even have tetrahedral examples - ClO₄^-, the perchlorate anion, is a good example of chlorine having 14 valence electrons and a tetrahedral structure.
Any questions and cool ideas, feel free to spitball them out. I'm more than happy to answer anything.

Thanks for answering, can you explain how you worked out CO2's shape again??? I am stil confused

Reply 3

nxvastel

Original post

by Jimmy Butler IV

Thanks for answering, can you explain how you worked out CO2's shape again??? I am stil confused

Once you learn how to count the bonding pairs/lone pairs of the central atom, deducing the shape of the molecule is just a matter of memorisation. As the course progresses, you will definetly get pretty used to it - it will be daunting to memorise so many geometries and names at first.

The following table sums it up pretty well. (Source: PMT Blog)

Screenshot 2026-02-03 064612.png

The key intuition here is electrons repel. The covalent bonds that have electrons in them want to be as far apart from each another as possible.

Remember when counting valence electrons we count double bonds as two electrons (from the central atom), but when deducing VSPER geometry, we count a double bond as one region of electron density. One lone pair also counts as one region of electron density.

Since CO₂ has two double bonds, it has two regions of electron density. Think about it: how can these two regions of electrons spread apart as far as possible? And as you can see from the table, intuition suffices - they are spread as far as apart, with a 180° O=C=O angle.

We can look a couple more examples. Consider BH₃, borane. Stop here if you want to try to work out how many electron pairs and lone pair this molecule has. (Boron is Group 3.)

If you have correctly worked it out, B is the central atom in BH₃. It has three covalent single bonds to each hydrogen, with no lone pairs - meaning it has three regions of electron density. How can these bonds spread out as far as possible?

Simply, if 360° makes up a circle, the furthest the B-H bonds can spread apart is 120°. The molecule adopts a geometry called "trigonal planar" - "trigonal" because the molecule forms a triangle-esque shape, and "planar" because the molecule is flat.

More complex geoemtries beyond three regions of electron density have 3-dimensional shapes, in order to get the electron pairs to spread out the furthest.

Lone pairs are considered to have stronger repulsion, and repel other bonds within the molecule by ca. 2.5° more. You might understand this better if you search up the geoemtry of methane (CH₄) versus ammonia (NH₃) - both have four regions of electron density, but ammonia has a lone pair on nitrogen. It has a smaller H-N-H bond angle than methane's H-C-H. (CH₄ = 109.5°, NH₃ = 107°, which is 2.5° less).

Both CH₄ and NH₄ are considered tetrahedral - although you may sometimes find "tetrahedral" molecules with a lone pair like NH₃ be called "trigonal pyramidal".

There's quite a bit of rules for VSEPR, and I can't have the time to explain them all here. If you have any specific examples in mind for questions you don't quite understand, please ask them here.

Also, may I ask if you are currently in Y12 going through this topic, or is it studying out of self interest?