Hybridization: sp3 hybridization in Methane

We have discussed the formation of CH₄ in the last post. We have seen that in CH₄  molecule bonds are formed by one s and three p orbitals of C. That means it must have two kinds of bonds, one which is formed by s orbital and the others which are formed by p orbitals of C. But the chemical and physical evidences indicate that it has 4 identical bonds. To solve this mystery scientist proposed the hypothesis (virtual theory) of Hybridization. There are no evidences that it exists but it gives an explanation and it is extensively used in the study of carbon compounds.

Let’s have a look on CH₄ molecule again in the light of this hypothesis. Carbon has 1 spherical s orbital and 3 dumbbell shaped p orbitals but it wants 4 identical orbitals so it uses hybridization.

This hybridization is similar to the hybridization you use in your garden, when you hybridise white rose with red rose you get pink rose. Pink rose is a combination of them but is different from them. C applies similar technique to get identical orbitals. But there is a condition that C can hybridize only those orbitals which have similar energy and 2s has lesser energy than 2p. So C supplied some energy to 2s orbital and lifts it to the same level of 2p orbitals. Now it can mix its one spherical 2s and three dumbbell shaped 2p orbitals and get four identical hybridised orbitals named as sp³. These hybridised orbitals are equal in energy and identical in shape. Their shape is neither spherical nor dumb-belled but it’s somewhat between them.

Each sp³ hybridi​s​ed orbital accommodates unpaired electron of C. These sp³ hybridised orbital arrange themselves in a tetrahedral shape to keeps equal and maximum distance from each other.

H has one unpaired electron in its s orbital. H comes closer to C so that its s orbital and sp³ hybridised orbital of C could get overlapped with each other. After overlapping of orbitals they share their electrons and make bonds. Thus CH₄ molecule is formed by bonding between four sp³ orbitals of C and s of four H atoms.

In hybridization we have learnt that:

• Only the orbitals which have similar energy can be mixed by hybridization.
• Number of hybridised orbitals is equal to the number of atomic orbitals that participate in hybridization. For example, one s and three p orbitals hybridised to form four sp³ hybridised orbitals.
• Hybridised orbitals of central atom decide the shape of the molecule.
• Bonding occurs by overlapping of hybridised orbitals of central atom and the orbital of other atoms.

Let’s take another example of BF₃ molecule. Central atom B needs 3 orbitals but it has one s and three p orbitals. In the next post we will see that how does Boron use hybridization to get 3 equivalent hybridized orbitals?

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