Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The compound having both sp and $s{p^2}$ hybridised carbon atom isA) PropeneB) PropyneC) PropadieneD) None of these

Last updated date: 16th Jul 2024
Total views: 62.1k
Views today: 0.62k
Verified
62.1k+ views
Hint: 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.

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