Question

In the analysis of an organic compound, it was found that it contains 7 carbon atoms, there are two C = C bonds and one $C\equiv C$ bond. This compound is a hydrocarbon. Hydrocarbons have only carbon and hydrogen elements. On structural analysis it was found that it is covalent in nature and expected structure is given above. The ratio between the pure and hybrid orbitals is?

• A. $7:10$
• B. $8:7$
• C. $5:9$
• D. $1:2$

Answer 1

The correct option is A $7:10$

In $sp,s{p}^2$ and $s{p}^3$ hybridization of $C$, the number of hybrid orbitals formed are $2,3$ and $4$ respectively and the number of pure orbitals are $2,1$ and $0$ respectively.
In this compound, $2$ $sp$ hybridized, $4$ $s{p}^2$ hybridized and $1$ $s{p}^3$ hybridized $C$ atom is present. The number of pure orbitals in these atoms are $4+4+0=8$
Also there are $6H$ atoms which contribute $6$ pure orbitals. Hence,the total number of pure orbitals are $8+6=14$
The number of hybrid orbitals are $4+12+4+0=20$
Hence, the ratio of pure orbitals to hybrid orbitals is $14:20$ or $7:10$