Question

In $CHC{l}_3$, carbon is $s{p}^3$ hybridised. What is the number of hybrid orbitals of carbon that have exactly 25% $‘s^{\prime }$ character in molecule?

Answer 1

In $CHC{l}_3$, the hybridisation of $C$ atom is $s{p}^3$. Thus, each hybrid orbital is expected to have 25% s-character and 75% p-character. But the hybrid orbital used for C-Cl bonding will be having s-character less than 25% due to higher electronegativity of Cl compared to that of C. Also the s-character of the hybrid orbitals used for C-H bonding is observed to be higher than 25%.