The hybridization of carbon atoms in C-C single bond of CH≡C-CH=CH₂ is: A. sp³-sp³ B. sp²-sp³ C. sp-sp² D. sp³-sp

To determine the hybridization of the carbon atoms in the molecule \( \text{CH}\equiv \text{C}-\text{CH}=\text{CH}_2 \), let's break down the molecule and analyze each carbon atom.

The molecule is propene, which has the following structure:

\[ \text{CH} \equiv \text{C} - \text{CH} = \text{CH}_2 \]

**1. Hybridization of Carbon in \( \text{CH} \equiv \text{C} \)**

- The first carbon (which is part of the \( \text{CH} \) group) forms a triple bond with the second carbon and a single bond with hydrogen. In a triple bond, the carbon uses one \( \text{s}p \) hybrid orbital and two unhybridized \( \text{p} \) orbitals to form two \( \pi \) bonds. This means the first carbon is \( \text{sp} \) hybridized.

**2. Hybridization of Carbon in \( \text{CH} \)**

- The second carbon (the middle one) forms a triple bond with the first carbon and a double bond with the third carbon. To form these bonds, this carbon also needs to use one \( \text{s}p \) hybrid orbital to make the sigma bond with the first carbon and another \( \text{s}p \) hybrid orbital to make the sigma bond with the third carbon. The remaining \( \text{p} \) orbitals form the \( \pi \) bonds in the triple bond with the first carbon and the double bond with the third carbon. Thus, this carbon is \( \text{sp}^2 \) hybridized.

**3. Hybridization of Carbon in \( \text{CH}_2 \)**

- The third carbon (which is part of the \( \text{CH}_2 \) group) forms a double bond with the second carbon and single bonds with two hydrogen atoms. For this, the carbon uses three \( \text{s}p^2 \) hybrid orbitals to form the sigma bonds and the remaining \( \text{p} \) orbital to form one \( \pi \) bond with the second carbon. Therefore, this carbon is \( \text{sp}^2 \) hybridized.

**Summary of Hybridization:**

- The first carbon ( \( \text{CH} \equiv \text{C} \) ) is \( \text{sp} \) hybridized.
- The second carbon ( \( \text{CH} \) ) is \( \text{sp}^2 \) hybridized.
- The third carbon ( \( \text{CH}_2 \) ) is \( \text{sp}^2 \) hybridized.

Based on the given choices for the hybridization of the C-C single bond:

The answer is **C. \( \text{sp} - \text{sp}^2 \)** because it matches the hybridization state of the carbons involved in the single bond between the \( \text{CH} \) and \( \text{CH}_2 \) groups.