Methyl nitrile is hydrolyzed to carboxylic acid in the presence of \(H_3O^+\) as catalyst. An intermediate compound \([A]\) is formed during this reaction.

Which of the following is the correct statement for hybridization of \(C-atom\) and number of \(\sigma \) and \(\pi \) bonds in the carbonyl group?

\(sp^2,\, \ 3\sigma ,\, \ 1\pi \)

The correct answer is option 1. \(sp^2,\, \ 3\sigma ,\, \ 1\pi \).

The correct statement for the hybridization of the \(C\)-atom and the number of \(\sigma\) and \(\pi\) bonds in the carbonyl group is: \(sp^2,\, \ 3\sigma ,\, \ 1\pi \)

In a carbonyl group (C=O), the carbon atom is \(sp^2\)-hybridized. This means that the carbon atom utilizes one \(s\)-orbital and two \(p\)-orbitals to form three equivalent \(sp^2\) hybrid orbitals, which are then used to form sigma (\(\sigma\)) bonds with other atoms.

In the carbonyl group, there is one sigma (\(\sigma\)) bond between the carbon atom and the oxygen atom, and one pi (\(\pi\)) bond formed by the overlap of the unhybridized p-orbital of carbon with the p-orbital of oxygen. So, the correct combination is \(sp^2,\, \ 3\sigma ,\, \ 1\pi\).