Question

The compound having both sp and $s{p^2}$ hybridised carbon atom is
A) Propene
B) Propyne
C) Propadiene
D) None of these

Answer 1

A short trick to identify the hybridization of an atom is to count the total atoms attached to the atom and the number of lone pairs surrounding the atom. If the counts are 4,3,2, the hybridizations are $s{p^3}$, $s{p^2}$ and sp respectively.

Complete Step by Step Answer:
Here, we have to find out the hydrocarbon that has both sp and $s{p^2}$hybridised atoms of carbon.

Let's check all the options one by one.
Option A is propene. Its structure is $\mathop {{\rm{C}}{{\rm{H}}_{\rm{3}}}}\limits_{s{p^3}} - \mathop {{\rm{CH}}}\limits_{s{p^2}} = \mathop {C{H_2}}\limits_{s{p^2}}$ . C1 is $s{p^3}$ hybridised as it forms a bond with four groups, C2 is $s{p^2}$hybridised as it forms a bond with three groups and C3 is $sp$ hybridised as it forms a bond with two groups. So, no sp hybridised carbon is present in the compound.

Option B is propyne. Its structure is $\mathop {{\rm{C}}{{\rm{H}}_{\rm{3}}}}\limits_{s{p^3}} - \mathop {\rm{C}}\limits_{sp} \equiv \mathop {{\rm{CH}}}\limits_{sp}$. C1 is $s{p^3}$ hybridised as it forms a bond with four groups, C2 is $sp$hybridised as it forms a bond with two groups and C3 is $sp$ hybridised as it forms a bond with two groups. So, no $s{p^2}$hybridised carbon is present in the compound.

Option C is propadiene. Its structure is $\mathop {{\rm{C}}{{\rm{H}}_2}}\limits_{s{p^2}} = \mathop {\rm{C}}\limits_{sp} = \mathop {{\rm{C}}{{\rm{H}}_{\rm{2}}}}\limits_{s{p^2}}$ .C1 is $s{p^2}$ hybridised as it forms bond with three groups, C2 is $sp$hybridised as it forms bond with two groups and C3 is $s{p^2}$ hybridised as it forms bond with three groups. So, both sp and$s{p^2}$hybridised carbon atoms are present in the compound.

Therefore, option C is right.

Note: It is to be noted that, the count of lone pairs surrounding an atom should also be counted while identifying the type of hybridization. In the given questions, no lone pairs are present in any of these compounds, so the count of lone pairs is not needed.