Hydrogen atom does not show hybridisation, thus number of hybrid orbitals in :
(a) \(\mathrm{C}_6 \mathrm{H}_6\)
All six C-atoms show \(s p^2\)-hybridisation (i.e. 3-orbitals by each \(\mathrm{C}\)-atom)
\(\therefore\) Total number of hybrid orbitals \(=6 \times 3=18\)
(b) \(\left(\mathrm{C H}_3\right)_4 \mathrm{C A l l}\) five \(\mathrm{C}\)-atoms show \(s p^3\)-hybridisation (i.e. 4-orbitals by each C-atom).
\(\therefore\) Total number of hybrid orbitals \(=5 \times 4=20\).
(c) \(\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{O}\)
Two C-atoms belong to \(\mathrm{CH}_3\) group show \(s p^3\)-hybridisation (i.e. 4-orbital by each C-atom).
One C-atom, bonded with O-atom and \(\left(\mathrm{CH}_3\right)_2\)-groups, show \(s p^2\)-hybridisation (i.e.3-hybrid orbitals).
One O- atom also show \(s p^2\)-hybridisation
Thus, total number of hybrid orbitals
\(=8+3+3=14\)
(d)
- C-1 show \(s p\)-hybridisation (i.e. 2-hybrid orbitals)
- C- 2 and 3 show \(s p^2\)-hybridisaton (i.e. 3-hybrid orbital by each C-atom)
- C-4 show \(s p^3\)-hybridisation (i.e. 4-hybrid orbitals.
- N-atom show \(s p\)-hybridisation (i.e. 2-hybrid orbitals)
Thus, total number of hybrid orbitals
\(=2+6+4+2=14\)
Hence, option (2) is correct.
