Question

Which of the following organic compounds has the same hybridization as its combustion product,$\text{C}{{\text{O}}_{2}}$?

Answer 1

Identify the number of bonds and type of bonds($\sigma ,\pi$) present. Hybridisation of an organic compound is given by the VBT. Valence Bond theory decides hybridisation based on the number of $\sigma$bonds and lone pair of electrons attached to the atom of consideration.

Complete answer:
According to valence bond theory, electrons in a molecule occupy atomic orbitals and not molecular orbitals. The atomic orbitals overlap on the bond formation and the strength of the bond depends on the extent of overlap.
Postulates of Valence Bond Theory:
- Covalent bonds are formed when two valence orbitals belonging to two different atoms overlap onto each other. Due to overlapping the electron density in that region increases, thereby increasing the stability of the molecule thus formed.
- The presence of many unpaired electrons in the valence shell of an atom enables the atoms to form multiple bonds with each other. However, the paired electrons present in the valent shell do not take part in formation of chemical bonds.
- Covalent bonds are directional and parallel to the region corresponding to atomic orbitals that are going to overlap.
- Sigma bonds and pi bonds differ in the pattern that the atomic orbitals overlap in, i.e. sigma bonds undergo head on overlap however pi bonds undergo sideways overlapping.
For calculation of hybridization we consider only sigma bonds and lone pair of electrons and not the pi bonds.
We will now determine the hybridization of the compounds mentioned in the options as well as $\text{C}{{\text{O}}_{2}}$.
Ethane:
Total number of $\sigma$ bonds per carbon atom = 4
Lone pairs = 0
Hybridisation = $s{{p}^{3}}$
Ethene:
Total number of $\sigma$ bonds per carbon atom = 3
Lone pairs = 0
Hybridisation = $s{{p}^{2}}$
Ethyne:
Total number of $\sigma$ bonds per carbon atom = 2
Lone pairs = 0
Hybridisation = $sp$
Ethanol:
Total number of $\sigma$ bonds per carbon atom = 4
Lone pairs = 0
Hybridisation = $s{{p}^{3}}$
$\text{C}{{\text{O}}_{2}}$:
Total number of $\sigma$ bonds per carbon atom = 2
Lone pairs = 0
Hybridisation = $sp$
Therefore, we can conclude that the organic compound that has the same hybridisation as its combustion product is ethyne.

The correct answer is option (C).

Note:
Valence bond theory was successful in determining the hybridization of atoms in a molecule. However, the theory had some limitations ,like:
- Unable to explain the tetravalency of carbon
- No insight or information on the energies of electrons
- Incorrect assumption that electrons are localized in specific areas only
- No distinction between weak and strong ligands ( hybridization of complex compounds)