To figure out what hybridization an atom has is to just count the number of atoms bonded to it and the number of lone pairs. Double and triple bonds still count as being only bonded to one atom. (Chem.Libretexts)

Why does this rule hold? Why are double and triple bonds counted only once?

Wouldn't a double bond have the equivalent number of electrons as 2 lone pairs?

And if orbitals are all about electrons, and there are three times as many electrons in a triple bond, shouldn't it count three times?


The hybridised orbitals are almost always involved in $\sigma$-bonding, and in housing the lone pair(s) of electrons. This helps us determine the geometry of the molecule.

Consider acetylene, $\ce{C_2H_2}$.

The carbon is $\mathrm{sp}$-hybridised. So, it has 2 $\sigma$-bonds - one to a hydrogen, one to another carbon. All this is in the third shell. So, a total of 4 orbitals (1 $\mathrm{s}$, 3 $\mathrm{p}$) undergo hybridisation and form 2 $\mathrm{sp}$-orbitals and 2 $\mathrm{p}$-orbitals. The $\pi$-bonding occurs due to these unhybridised $\mathrm{p}$-orbitals.

Hope this clears your doubt!

PS: The only exception to the first statement that I know is https://en.wikipedia.org/wiki/Aryne.

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The rule fails pretty quickly. For example, it almost systematically fails for anything in the third period and higher or for most oxygen species. However, it is indeed valuable for carbon and nitrogen centres.

If you would count double bonds twice and triple bonds three times, you would always, as a rule (carbenium ions not considered) arrive at $\mathrm{sp^3}$ for carbon. However, the idea of teaching students hybridisation is not only to explain how bonding can actually occur in carbon (which would have a $[\ce{Ne}]\,\mathrm{2s^2\,2p^2}$ in ground state) but also how double and triple bonds differ from single bonds.

In general, double and triple bonds are considered to derive from an underlying σ bond to which a number of π bonds (one or two) have been added. These π bonds are fundamentally different from σ bonds; the bonding orbitals of the latter are always directed towards the bond partner while the principal direction of the formers’ bonding orbitals is perpendicular to the bond axis. This perpendicular arrangement is achieved by forming the π bonds using unhybridised p orbitals, while σ bonds are formed with $\mathrm{sp}^n$ hybrid orbitals.

Therefore, you are actually counting the underlying σ bonds (and lone pairs, where applicable) rather than indiscriminately counting bonds. If you do that, you are ensured to always have sufficient p orbitals remaining unhybridised to form the necessary multiple bonds.

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