Question

Find the ratio of lone pairs present on $Cl$ atoms to $s{p}^2$-hybrid orbital in $C_3{N}_{3}C{l}_3$ molecule.

Answer 1

There are $9$ lone pairs of electrons present on $Cl$ atom and $3$ lone pairs present in $s{p}^2$ hybrid orbitals of $C_3{N}_3(C{l}_3{)}_3$.
Hence, the ratio of the total number of lone pairs of electrons present on $Cl$ atom to lone pairs present in $s{p}^2$ hybrid orbitals of $C_3{N}_3(C{l}_3{)}_3$ is $\frac{9}{3}=3$.