Explanation:

In organic chemistry, the process of homolytic fission involves the breaking of a bond in such a way that each atom retains one electron from the bond. This leads to the formation of two radicals, each with an unpaired electron.

The homolytic fission of a C-C bond in ethane (CH3-CH3) can be represented as follows:

CH3-CH3 → CH3• + •CH3

Here, the bond between the two carbon atoms is broken, and each carbon atom retains one electron from the bond, resulting in the formation of two methyl radicals, CH3•.

Hybridization of the Carbon Atom:

To determine the hybridization of the carbon atom in the intermediate, we need to consider the electronic configuration and bonding of carbon.

Before the bond is broken, each carbon atom in ethane is sp3 hybridized. This means that each carbon atom has four sp3 hybrid orbitals, which are used to form sigma bonds with other atoms.

When the C-C bond is broken, each carbon atom retains one electron, resulting in an unpaired electron in each carbon atom's valence shell. This unpaired electron is present in a p orbital, as the hybridization of the carbon atom changes during the homolytic fission.

After the homolytic fission of the C-C bond:

In the intermediate formed after the homolytic fission of the C-C bond, each carbon atom has an unpaired electron in a p orbital. This is because the carbon atom undergoes sp2 hybridization during the formation of the intermediate.

The sp2 hybridization involves the mixing of one s orbital and two p orbitals of the carbon atom, resulting in three sp2 hybrid orbitals. These three sp2 hybrid orbitals are then used to form sigma bonds with other atoms.

Therefore, the correct answer is option 'B' - the carbon atom in the intermediate is sp2 hybridized.