What is the hybridisation of each ofthe carbon atoms present in the compoundHC  C – CH = CH2

To determine the hybridization of each carbon atom in the compound HC≡C-CH=CH2, we need to follow these steps:

Step 1: Count the number of sigma bonds (single bonds) and pi bonds (double and triple bonds) around each carbon atom.

In the given compound, we have:

• The first carbon atom (C1) is bonded to one hydrogen atom (H) and one carbon atom (C2) through a triple bond (≡).
• The second carbon atom (C2) is bonded to one carbon atom (C1) through a triple bond (≡) and one hydrogen atom (H).
• The third carbon atom (C3) is bonded to one carbon atom (C2) through a double bond (=) and one hydrogen atom (H).
• The fourth carbon atom (C4) is bonded to one carbon atom (C3) through a single bond (-) and two hydrogen atoms (H).

Step 2: Determine the hybridization of each carbon atom based on the number of sigma and pi bonds.

• For C1: It has one sigma bond and two pi bonds, so it is sp hybridized.
• For C2: It has one sigma bond and two pi bonds, so it is sp hybridized.
• For C3: It has one sigma bond and one pi bond, so it is sp2 hybridized.
• For C4: It has one sigma bond, so it is sp3 hybridized.

Therefore, the hybridization of each carbon atom in the compound HC≡C-CH=CH2 is as follows:

• C1: sp hybridized
• C2: sp hybridized
• C3: sp2 hybridized
• C4: sp3 hybridized