Question

Which one among the following possesses an sp hybridised carbon in its structure
A) \[{\rm{C}}{{\rm{H}}_{\rm{2}}} = {\rm{C}}{\rm{.Cl}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}}\]
B) \[{\rm{C}}{\rm{.C}}{{\rm{l}}_{\rm{2}}} = {\rm{C}}{\rm{.C}}{{\rm{l}}_{\rm{2}}}\]
C) \[{\rm{C}}{{\rm{H}}_{\rm{2}}} = {\rm{C}} = {\rm{C}}{{\rm{H}}_{\rm{2}}}\]
D) \[{\rm{C}}{{\rm{H}}_{\rm{2}}} = {\rm{CH}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}}\]

Answer 1

Hint: An hp hybridised carbon atom is one in which the carbon atom is surrounded by two groups. The surrounding atom can be the atoms that are covalently bonded with the atom and the lone pair of the atom.

Complete Step by Step Answer:
Let’s discuss all the options one by one.
Option A is\[\mathop {{\rm{C}}{{\rm{H}}_{\rm{2}}}}\limits_{s{p^2}} = \mathop {\rm{C}}\limits_{s{p^2}} {\rm{.Cl}} - \mathop {\rm{C}}\limits_{s{p^2}} {\rm{H}} = \mathop {\rm{C}}\limits_{s{p^2}} {{\rm{H}}_{\rm{2}}}\]. Here, we see that all the carbon atoms are surrounded by three groups, so, all the carbon atoms are \[s{p^2}\] hybridised. Therefore, no sp hybridised carbon atom is present in the compound.

Option B is \[\mathop {{\rm{C}}{\rm{.}}}\limits_{s{p^2}} {\rm{C}}{{\rm{l}}_{\rm{2}}} = \mathop {\rm{C}}\limits_{s{p^2}} {\rm{.C}}{{\rm{l}}_{\rm{2}}}\]. Here, both the carbon atoms are \[s{p^2}\]hybridised because they are surrounded by three groups. So, no sp hybridised carbon atom is present in the compound.

Option C is \[\mathop {\rm{C}}\limits_{s{p^2}} {{\rm{H}}_{\rm{2}}} = \mathop {\rm{C}}\limits_{sp} = \mathop {\rm{C}}\limits_{s{p^2}} {{\rm{H}}_{\rm{2}}}\]. Here, C1 is \[s{p^2}\]hybridised because it is surrounded by three groups. C2 is sp hybridised as it is surrounded by two atoms and C3 is \[s{p^2}\]hybridised because it is surrounded by three groups. So, one sp hybridised C atom is present in the compound.

Option D is \[{\rm{C}}{{\rm{H}}_{\rm{2}}} = {\rm{CH}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}}\]. Here, C1 is \[s{p^2}\]hybridised because it is surrounded by two groups. C2 is \[s{p^2}\] hybridised as it is surrounded by three atoms and C3 is \[s{p^2}\]hybridised because it is surrounded by three groups and C4 is \[s{p^2}\] hybridised. So, no sp hybridised C atom is present.
Therefore, option C is right.

Note: It is to be noted that, while counting the surrounding groups of an atom to find its hybridization, the double bond or triple-bonded atom is to be counted as one group and not as two groups.