Chemistry Chemical Bonding PMC 2021
PMDC Verified Question 61 of 102
Hybridization in ethyne (\( \text{C}_2\text{H}_2 \)):
A
\( \text{sp-sp} \)
B
\( \text{sp}^2\text{-sp} \)
C
\( \text{sp}^2\text{-sp}^2 \)
D
\( \text{sp}^3\text{-sp}^3 \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: \( \text{sp-sp} \)
Concept:

The presence of a triple bond dictates that the carbon atoms are \( \text{sp} \) hybridized.

Formula:

$$ \text{H} - \text{C} \equiv \text{C} - \text{H} $$

Solution:

  • In ethyne, each Carbon atom forms exactly 2 sigma bonds (one with Hydrogen, one with the other Carbon) and has no lone pairs.


  • A steric number of 2 corresponds to \( \text{sp} \) hybridization.


  • Since both carbons are identical, the sigma bond between them is formed by the head-on overlap of an \( \text{sp} \) orbital from one carbon with an \( \text{sp} \) orbital from the other.


Why other options are incorrect:

  • Option C: Applies to ethene (double bond).
  • Option D: Applies to ethane (single bonds).

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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