Question

Which of the following represents the given mode of hybridisation \(sp^2 - sp^2 - sp - sp\) from left to right?

• A. \(H_2C = CH - C \equiv N\)
• B. \(CH \equiv C - C \equiv CH\)
• C. \(H_2C = C = C = CH_2\)
• D. \(CH_2 = CH_2\)

Hint:

Double bond → \(sp^2\), triple bond → \(sp\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The hybridization of an atom is determined by the number of \( \sigma \) bonds and lone pairs (steric number).
- \( sp^{3} \): 4 \( \sigma \) bonds (or 0 \( \pi \) bonds).
- \( sp^{2} \): 3 \( \sigma \) bonds (or 1 \( \pi \) bond).
- \( sp \): 2 \( \sigma \) bonds (or 2 \( \pi \) bonds).
Step 2: Detailed Explanation:
Let's analyze option (A) \( \text{H}_{2}\text{C}_{1} = \text{C}_{2}\text{H} - \text{C}_{3} \equiv \text{N}_{4} \):
- Carbon 1 (\( \text{C}_{1} \)): Formed by one double bond with \( \text{C}_{2} \). It has 3 \( \sigma \) bonds (2 with H, 1 with C). Hybridization \( = sp^{2} \).
- Carbon 2 (\( \text{C}_{2} \)): Formed by one double bond with \( \text{C}_{1} \) and one single bond with \( \text{C}_{3} \). It has 3 \( \sigma \) bonds (1 with \( \text{C}_{1} \), 1 with H, 1 with \( \text{C}_{3} \)). Hybridization \( = sp^{2} \).
- Carbon 3 (\( \text{C}_{3} \)): Formed by one single bond with \( \text{C}_{2} \) and one triple bond with N. It has 2 \( \sigma \) bonds (1 with \( \text{C}_{2} \), 1 with N). Hybridization \( = sp \).
- Nitrogen 4 (N): Formed by one triple bond with \( \text{C}_{3} \). It has 1 \( \sigma \) bond and 1 lone pair. Hybridization \( = sp \).
The sequence is \( sp^{2} - sp^{2} - sp - sp \).
Step 3: Final Answer:
Option (A) matches the required hybridization sequence.