Question

During the complete combustion of methane, what change in hybridization does the carbon atom make?
A.$s{p^3}$ to $sp$
B.$s{p^3}$ to $s{p^2}$
C.$s{p^2}$ to $sp$
D.$s{p^2}$ to $s{p^3}$

Answer 1

Hybridization: Mixing of atomic orbitals which results in the formation of new hybrid orbitals during the chemical bond formation, are known as hybridization. Hybridization occurs when electrons of s and p orbitals react to each other, then hybridization takes place to balance the energy levels.

Complete step by step solution:
We know that there is a difference between the energies of s and p-orbitals. s-orbitals have less energies in comparison with p-orbitals. To balance this energy difference hybridization takes place. After the hybridization the energies of all the orbits formed are the same. For example: in $s{p^3}$ hybridization one s and three p-orbitals takes place in the hybridization. Hence after hybridization four hybrid orbitals are formed which have the same energies.
Methane is $C{H_4}$ molecule. In this compound one s-orbitals mix with three p-orbitals. so initially its hybridization is $s{p^3}$ and forms four hybrid orbitals (that is why methane has four hydrogen connected to carbon by single bonds).
During complete combustion of methane carbon dioxide and water are formed. The reaction is as follows: $C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O$
Now, in carbon dioxide one s-orbital electron forms a hybrid with one p-orbital electron forming $sp$ hybridization. So during the complete combustion of methane the hybridization of carbon changes from $s{p^3}$ to $sp$.

Hence, option A is correct.

Note:
Number of hybrid orbitals formed after the hybridization is equal to the total number of orbitals taking part in the hybridization. For example if hybridization is $s{p^2}$ hybridized then one s and two p- orbitals take part in the hybridization and total three hybrid orbitals will be formed after the hybridization.