In which of the following molecules is the central atom unhybridized:
(a) $\ce{S(CH3)2}$
(b) $\ce{SO2}$
(c) $\ce{SiH4}$
(b) $\ce{PCl3}$

The answer given is (a), but I can't figure out why this would happen.

According to me $\ce{S(CH3)2}$ should have a hybridization of $\mathrm{sp^3}$ since it has two sigma bonds and two lone pairs. In fact, I can't identify a reason for any of these molecules in which hybridization is absent since the only condition I know for the absence of hybridization is when the central atom has a large size (Drago Rule).

Can anybody help me to understand why the central atom is unhybridized here?


2 Answers 2


We can immediately discard $\ce{SiH4}$ — a tetrahedral molecule can never be unhybridised.

We can also pretty quickly discard $\ce{SO2}$. The dated description of this molecule included octet expansion which only worked with hybridisation. But even the more modern description employing charge separation ($\ce{\overset{-}{O}-\overset{+}{S}=O}$) requires some hybridisation to allow for all three bonds to happen, and the Wikipedia-stated structure with an angle of $119^\circ$ does not allow non-hybridisation. (Remember that hybridisation is determined by geometry, not the other way around.)

This leaves both $\ce{PCl3}$ and $\ce{Me2S}$ for discussion. And in general, it is possible for both compounds to have an unhybridised central atom.

Contrary to what is often taught at the high-school level, it is typically most favourable for an atom to remain unhybridised when forming bonds. Bonds would then be formed with p orbitals first, which are typically closer in energy to the bonding partner, allowing for a greater stabilisation. p orbitals are also directional (while s orbitals are spherical) meaning that using the p orbital of the bond’s direction a greater overlap can be generated which in turn means a more stable bond. Likewise, a lone pair prefers residing in a more stable orbital and an s orbital is lower in energy than a p orbital. So whenever an atom has the chance, it will form its bonds unhybridised.

However, most ‘simple’ compounds such as water and ammonia feature bond angles far away from the ideal $90^\circ$ of an unhybridised central atom. This is due to a $90^\circ$ bond angle inducing far too much steric strain between the bonding partners (hydrogen atoms in the water and ammonia molecules). The central atom hybridises to such an extent that the destabilising steric strain is minimised while still keeping the bonding orbitals’ p contribution maximum.

Going from water and ammonia to their heavier homologues hydrogen sulfide and phosphane, the bond angles are tightened to almost exactly $90^\circ$. This is due to the greater sizes of the central atoms which have longer bond lengths per se and thus allow the hydrogens to move closer together without additional steric strain. Both hydrogen sulfide and phosphane have one lone pair that resides in a practically pure s orbital.

Moving on from $\ce{H2S}$ and $\ce{PH3}$ to $\ce{Me2S}$ and $\ce{PCl3}$, we are now increasing the size of the substituents. Whether or not the central atom can remain unhybridised depends on how big the substituents are and whether an angle of approximately $90^\circ$ is too small or not. Merely looking at them, one might assume that chlorine is bigger than a methyl group (the methyl group’s hydrogens are pointing away from sulfur at angles of approximately $109^\circ$) and therefore one might be inclined to assume, given the question in its stated version, that $\ce{Me2S}$ feature an unhybridised sulfur while $\ce{PCl3}$’s phosphorus slightly hybridises to reduce steric strain.

A conclusive answer can only be given by looking at published structures, however. For $\ce{PCl3}$, the Wikipedia page reports a bond angle $\angle(\ce{Cl-P-Cl}) \approx 100^\circ$ — this angle is incompatible with non-hybridisation and thus phosphorus is hybridsed.[1] However, performing a literature search for the bond angle of $\ce{Me2S}$ gave me two sources at the top of the list: one claiming a bond angle $\angle(\ce{C-S-C}) = 99.05^\circ$[2] and one claiming a bond angle of $\angle(\ce{C-S-C}) = 98^\circ 52'$[3] — again neither of which being consistent with non-hybridisation.

Given the physical data of two bond angles practically equal (what is $1^\circ$ anyway?), I conclude the question is based on a false premise and can be challenged.


[1]: https://en.wikipedia.org/wiki/Phosphorus_trichloride

[2]: T. Ijima, S. Tsuchiya, M. Kimura, Bull. Chem. Soc. Jpn. 1977, 50, 2564–2567. DOI: 10.1246/bcsj.50.2564.

[3]: L. Pierce, M. Hayashi, J. Chem. Phys. 1961, 35, 479–485. DOI: 10.1063/1.1731956.

  • $\begingroup$ Thank you so much...I had no idea that the main cause of hybridization was steric hindrance and not the stability due to hybridization. Is hybridization unfavourable even for molecules with no lone pairs like $\ce{SiH4}$ (keeping steric hindrance aside)? $\endgroup$
    – oshhh
    Mar 21, 2017 at 17:31
  • 1
    $\begingroup$ @OsheenSachdev Molecules like $\ce{SiH4}$ complicated for a different reason. On paper, we do the hybridisation scheme and thereby create four equal bonds. Nature uses molecular orbital theory to arrive at two different bonding orbital energies — but still four identical bonds because that is how quantum mechanics work. If you then go and localise the molecular orbitals, you do indeed arrive at something that can be labelled $\mathrm{sp^3}$ for each bonding silicon orbital. $\endgroup$
    – Jan
    Mar 21, 2017 at 17:35
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    $\begingroup$ In those highly symmetric molecules, the solution that gives the highest symmetry is typically observed. Any distortion would require reducing the symmetry which typically means destabilising at least one orbital. $\endgroup$
    – Jan
    Mar 21, 2017 at 17:36

Here dimethyl sulpfide is the correct answer as if you will see in sulphur atom both 3s and 3p are completly filled and in order to become sp hybridised it has to excit it's electrons to the vacant d-orbital. This is only possible if energy difference between 3p and 3d is low. Since carbon has low electronegativity it is unable to bring 3p and 3d close so the energy differece is large enough that excitation cannot take place and the sulphur atom remains unhybridized.

for example- Since fluorine, chlorine, etc are highly electronegative so they bring 3p and 3d close and so excitation is possible and sulphur becomes sp3 hyb.

Hope this answer will help.

  • 1
    $\begingroup$ you wrote "here", where is here? Did you plan to attach an image? $\endgroup$ May 9, 2022 at 13:03

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