Question

The (C.H) bond length order in hydrocarbons in which the hybridisation of carbon is  $ S{{P}^{3}}\text{ or }S{{P}^{2}}\text{ or }SP $ 

(a)  $ SP-S>S{{P}^{2}}-S>S{{P}^{3}}-S $ 

(b)  $ SP-S=S{{P}^{2}}-S=S{{P}^{3}}-S $ 

(c)  $ S{{P}^{2}}-S>S{{P}^{2}}-S>SP-S $ 

(d)  $ S{{P}^{3}}-S>SP-S>S{{P}^{2}}-S $ 

Answer 1

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

Complete answer:

- When orbitals mix up and form a new set of orbitals of the same energy is called hybridisation and the new set of orbitals is called hybrid orbitals. There are different types of hybridisation and the common ones are  $ s{{p}^{3}},\text{ }s{{p}^{2}}\text{ and }sp $ .

-  $ s{{p}^{3}} $  hybridisation involves saturated organic compounds containing only single covalent bonds.  $ s{{p}^{2}} $  hybridisation 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 and  $ s{{p}^{3}} $  hybrid carbon is electropositive in character.  $ s{{p}^{2}} $  hybrid carbon can be both electropositive and electronegative in character.

The s character of hybrid orbitals is inversely proportional to C-H bond length. So, for sp-s, the bond length will be more. Then comes  $ s{{p}^{2}}-s $  and the shortest bond length is between  $ s{{p}^{3}}-s $ .

Thus, the C-H bond length order in hydrocarbon is  $ SP-S>S{{P}^{2}}-S>S{{P}^{3}}-S $.

So, the correct answer is “Option A”.

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.