Chemistry Chemical Bonding NUMS 2023
PMDC Verified Question 36 of 102
Which of the following has \( \text{sp}^3 \) hybridization?
A
\( \text{BF}_3 \)
B
\( \text{C}_2\text{H}_4 \)
C
\( \text{BeCl}_2 \)
D
\( \text{CH}_4 \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: \( \text{CH}_4 \)
Concept:

An \( \text{sp}^3 \) hybridized atom requires 4 electron domains (sigma bonds + lone pairs) to achieve a tetrahedral layout.

Formula:

$$ \text{Steric Number} = 4 \implies \text{sp}^3 $$

Solution:

  • In methane (\( \text{CH}_4 \)), the central Carbon atom forms exactly 4 single sigma bonds with Hydrogen atoms.


  • It has 0 lone pairs.


  • 4 domains require the mixing of one s and three p orbitals, resulting in \( \text{sp}^3 \) hybridization and a perfect tetrahedral shape.


Why other options are incorrect:

  • Option A: \( \text{BF}_3 \) has 3 bonds = \( \text{sp}^2 \).
  • Option B: \( \text{C}_2\text{H}_4 \) (Ethene) has a double bond (3 domains per C) = \( \text{sp}^2 \).
  • Option C: \( \text{BeCl}_2 \) has 2 bonds = \( \text{sp} \).

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