I was reading about the molecular shape of compounds. I learned that the electronegativity of the central atom and the terminal atom in a molecule both play a role in determining bond angle. In $\ce{NH3}$ and $\ce{NF3}$, $\ce{F}$ having higher electronegativity than $\ce{H}$, $\ce{NF3}$ has a smaller bond angle compared to $\ce{NH3}$.

Applying the same logic, it was expected that $\ce{CCl4}$ would have a smaller bond angle than that of $\ce{CH4}$. Surprisingly, I found that both of them have the same bond angle. Why is this the case?

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    $\begingroup$ A molecule is very much a 3-dimensional thing. Unless you have its model before your eyes (maybe only as a mental image, but it has to be a good one), you totally can't move further. $\endgroup$ Dec 23, 2021 at 15:58
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    $\begingroup$ You have 4 atoms of one element, singly bonded to carbon, so they get, on average, as far apart as possible, which means tetrahedrally around the central carbon: chemistry.stackexchange.com/a/118202/79678. $\endgroup$
    – Ed V
    Dec 23, 2021 at 16:03
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    $\begingroup$ A tetrahedron is a tetrahedron no matter how small - with apologies to Dr. Seuss… $\endgroup$
    – Jon Custer
    Dec 23, 2021 at 16:23
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    $\begingroup$ @melanieshebel When applying MathJax to formulas, why not to use mhchem \ce{}? It makes generally cleaner MJ code and keeps symbols upright as expected. $\endgroup$
    – Poutnik
    Dec 23, 2021 at 18:51
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    $\begingroup$ The molecules $\ce{CCl4}$ or $\ce{CH4}$ contain six angles $\ce{Cl-C-Cl}$ or six $\ce{H-C-H}$. Why should one of these angles be different from any other one ? $\endgroup$
    – Maurice
    Dec 24, 2021 at 13:08

2 Answers 2


In $\ce{NH3}$ and $\ce{NF3}$, $\ce{F}$ having higher electronegativity than $\ce{H}$, $\ce{NF3}$ has a smaller bond angle compared to $\ce{NH3}$.

Both of these compounds have a lone pair on the central atom. So the bound electrons and the lone pair (if you are using the simple "electrons pair up" model) compete for space.

Applying the same logic, it was expected that CCl4 would have a smaller bond angle than that of CH4.

All electrons around carbon are involved in bonding, so all four pairs are the same. To apply the electronegativity argument, you should compare the distinct bond angles in $\ce{CH2F2}$ or in $\ce{CH2Cl2}$.


Welcome to Stack exchange chemistry.

Consider for a moment what is known as the isolobal concept, there are a series of atoms and groups which all present the same types of orbitals (or at least close to identical orbitals) and the number of electrons.

Consider for a moment a methane molecule, if we were to break a C-H bond then the carbon atom would only have seven valance electrons. The carbon in an alkane such as methane has rehybridized its orbitals to give us four sp^3 orbitals. These are arranged in a tetrahedron around the carbon.

A covalent bond is formed by sharing the one electron in the sp3 orbital of the carbon with an atomic orbital from another atom that has the right geometry to overlap with the sp3 orbital. The sp3 orbital has two pear-shaped lobes, one is large and one is small. These have opposite signs of the wavefunction.

A hydrogen atom in the ground state has a single electron in an s orbital, this is a sphere-shaped orbital that can interact with the sp3 orbital to form both a bonding and an antibonding orbital. We will only concentrate in this answer on the bonding orbitals.

The sphere-shaped s orbital has the right geometry to interact with the sp3 orbital and it can result in the formation of an occupied (2 electrons in it) bonding orbital between the carbon and the hydrogen. This will be a sigma bond (single bond)

If we change to chlorine, then the outermost orbital (for the valence electrons) of the atom has also rehybridized to give us four sp3 orbitals. Three of these are occupied with two electrons while one in an isolated chlorine atom only has one. The orbital with only one electron can interact with the sp3 orbital on the carbon (bearing only one electron) to form two new molecular orbitals. One is antibonding and one is bonding.

If the bonding orbital between the carbon and the chlorine is occupied with two electrons then we have a bond. The C-Cl and C-H bonds will be different in length. But the angle between them will be dictated by the arrangement of the sp3 orbitals around the carbon atom.

If you still do not understand it then I would suggest that you fall back to VSEPR theory. As it is the festive season go and grab an orange and four cocktail sticks. Stab them into the orange in such a way that they are the greatest angle apart. You should find that the tips of them form a triangle-based pyramid (tetrahedron). It will not matter if you put grapes on the points of the cocktail sticks to represent hydrogen atoms in the methane. Or apples to represent the chlorine atoms in carbon tetrachloride. You will still have the same arrangement of the atoms in your model.

You can then hang it on the tree as a decoration or pull it apart and eat the fruits. When I can not lay my hands on my molecule modeling kit made of plastic balls and straws I tend to grab oranges and then draw atoms with a marker pen on the skin.

  • $\begingroup$ you have tried to explain it with M.O theory.But my ques. wasn't based on that. It was actually based on the e.n of the terminal atoms of the two molecules which play a significant role in bond angle.And you didn't even raise a word on that. $\endgroup$
    – Akash
    Dec 23, 2021 at 17:10
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    $\begingroup$ The electronegativity does not matter: see my first comment and the linked tetrahedral figure. Ammonia is polar because it has three hydrogens attached and a lone pair of electrons. Nitrogen trifluoride is also polar and for the same reason. The electronegativity differences determine the respective dipole moments. But with 4 H atoms around the central carbon, they are, on average, as far apart as possible, hence the tetrahedral geometry, and methane is non-polar. Same principle for carbon tetrachloride. Simple as that. $\endgroup$
    – Ed V
    Dec 23, 2021 at 18:18
  • $\begingroup$ The electronegativity is not relevent to the shape of the molecule, I went through two methods of predicting the shape of the molecule. $\endgroup$ Dec 24, 2021 at 16:29

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