INTEC Chemistry Blog

Hybridization involving other atoms.

Posted on: March 13, 2014

Lets just look a little bit more about C before we discuss the hybridization of other atoms.

Remember, hybridization is important because the hybridization of the atom is related to the molecular geometry(i.e. shape) around that atom, and molecular shape has a very important influence on chemical reactivity and its physical properties.

The hybridization can be deduced by a simple formula:



Because the number of pi bonds are complementary to the number of sigma bonds, you can of course also express that formula in terms of pi bonds.



Take propene,  H2C=CHCH3 as an example. According to the pi based formula,  first carbon (the blue coloured one) in propene, is sp(3 -1) = sp2


This pi bond based formula can also be used to determine hybridization in other atoms such as N and O etc.

Consider ammonia, NH3.


The number of pi bonds is zero, so, by the pi based formula, N’s hybridization is is sp(3 -0) = sp3



In the molecule pyridine, as shown below, the N has 2 single bonds and one pi bond.


so the N in pyridine is sp(3 -1) = sp2 hybridized.


As for oxygen, it is a similar story. In water, O has two single bonds and zero pi bonds, so the hybridization of O in H2O is sp(3 -0) = sp3 hybridized.

In the group of molecules called esters, there are two oxygen atoms (primarily)responsible for making the molecule an ester. Applying the pi based hybridization formula for each oxygen in the ester….


we get sp(3 -1) = sp2 hybridization for the oxygen coloured in green and sp(3 -0) = sp3 hybridization for the oxygen coloured in red.


You may have realised that the N based compounds and the O based compounds above have LONE PAIRS on them. Remember a lone pair are two valence electrons in a single orbital that are NOT used in bonding). Those lone pairs have the same hybridized as we calculated for the atom.


Finally, you may wonder why we didn’t use the original sigma (σ) bond formula for the N and O species. This is because the number of bonds those atoms form by sharing one electron with a different atom, varies. N has 3 single unpaired electrons and O has two, i.e., the valence of N is three and the valency of O is two. Hence the number of σ bonds formed for N and O compared to C is difference so the σ based formula needs to be adjusted each time an atoms with a different valency is used. But the pi (π) based formula doesn’t need to be adjusted, hence it can can be applied more widely 🙂



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