In carprolactum if 'a' is number of lone pairs of ${\mathrm{e}}^{-}$ and 'b' is number of $\mathrm{s}{\mathrm{p}}^3$ hybridised atoms. What is the value of a + b?

Solution

number of lone pairs $=\mathrm{a}=3$
number of $\mathrm{s}{\mathrm{p}}^3$ hybridised atoms $=\mathrm{b}=\underset{\mathrm{\text{'}}\mathrm{C}\mathrm{\text{'}}}{5}+\underset{\mathrm{\text{'}}\mathrm{N}\mathrm{\text{'}}}{1}=6$
sum of $\mathrm{a}+\mathrm{b}=3+6=9$

Hence, the correct answer is 9.