The hybridization in the compound $CH_3 - CH_2 - C \equiv H$ is

To solve this problem, we need to look at the bonding around each carbon atom in the molecule.

Note: There appears to be a slight typo in your question. The structure $CH_3 - CH_2 - C \equiv H$ is chemically impossible because carbon must form four bonds. The correct structure for this series is $CH_3 - CH_2 - C \equiv CH$ (butyne).

1. Analyze the concept: Hybridization

Hybridization is determined by the number of sigma ($\sigma$) bonds and lone pairs around a central atom:

• $sp^3$ hybridization: Occurs when a carbon atom is bonded to 4 other atoms (single bonds only).
• $sp^2$ hybridization: Occurs when a carbon atom is bonded to 3 other atoms (one double bond).
• $sp$ hybridization: Occurs when a carbon atom is bonded to 2 other atoms (one triple bond or two double bonds).

2. Step-by-step breakdown

Let's analyze the structure $CH_3 - CH_2 - C \equiv CH$:

• Carbon 1 ($CH_3$): This carbon is bonded to 3 hydrogens and 1 carbon (all single bonds). Since it has 4 sigma bonds, it is $sp^3$ hybridized.
• Carbon 2 ($CH_2$): This carbon is bonded to 2 hydrogens and 2 carbons (all single bonds). Since it has 4 sigma bonds, it is $sp^3$ hybridized.
• Carbon 3 ($-C \equiv$): This carbon is bonded to one carbon on the left and one carbon on the right via a triple bond. It is bonded to only 2 atoms. Therefore, it is $sp$ hybridized.
• Carbon 4 ($\equiv CH$): This carbon is bonded to one carbon and one hydrogen. It is bonded to only 2 atoms. Therefore, it is $sp$ hybridized.

3. Conclusion

The molecule contains carbon atoms that are either $sp^3$ hybridized or $sp$ hybridized. Looking at the options provided, the combination present in the chain is $sp^3$ and $sp$.

Option A is the best fit among the choices provided.

Exam Tip: Always count the number of atoms attached to the carbon. If it's 4 atoms, it's $sp^3$; if it's 3, it's $sp^2$; if it's 2, it's $sp$. Ignore the pi bonds (the second and third lines in double/triple bonds) when counting for hybridization!