Question

The $C-H$ bond distance is the longest in:
(A) ${{C}_{2}}{{H}_{6}}$
(B) ${{C}_{2}}{{H}_{2}}Br$
(C) ${{C}_{2}}{{H}_{4}}$
(D) ${{C}_{2}}{{H}_{2}}$

Answer 1

Hint : Hybridization is the concept of mixing of orbitals leading to formation of a new set of orbitals with the same energy. There are types of hybridization out of which the more common hybridizations are $s{{p}^{3}},s{{p}^{2}},sp$ . The higher the s character of the hybrid orbital lower is the bond length between the orbitals.

Complete Step By Step Answer:
When orbitals mix up and form a new set of orbitals of the same energy is called hybridization and the new set of orbitals is called hybrid orbitals. There are different types of hybridization and the common ones are $s{{p}^{3}},s{{p}^{2}},sp$
$s{{p}^{3}}$ Hybridization involves saturated organic compounds containing only single covalent bond $,s{{p}^{2}}$ hybridization is seen in compounds with carbon atoms linked by double bonds and sp is seen in compounds having carbon atoms involved in triple bonds. We know that the electronegativity of s character is maximum. The electronegativity of the hybrid orbital is directly proportional to the s character of the hybrid orbital. So, the sp orbital is more electronegative.
Higher the s− character means that the Bonded pair electrons are held more closely, ensuring a smaller Bond length and a stronger bond. So, sp has the least bond length and it is the strongest
$s{{p}^{3}}$ has the longest bond length and weakest. Thus ${{C}_{2}}{{H}_{6}}$ is $s{{p}^{3}}$ Hybridization
Therefore, correct answer is option A i.e. ${{C}_{2}}{{H}_{6}}$ .

Note :
As the bond length increases in a $C-H$ bond, the bond strength decreases. Less energy is required to break the bond that is the bond energy will be less. Thus, we can say that as the s-character of hybrid orbital increases, the bond energy also increases.