Chemistry Chemical Bonding ETEA 2019
PMDC Verified Question 83 of 102
The bond angle in \( \text{H}_2\text{S} \) is less than \( \text{H}_2\text{O} \). It is due to:
A
Small size of oxygen atom
B
Greater E.N of oxygen atom
C
Oxygen contain two lone pairs of electrons
D
All of the above
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: Greater E.N of oxygen atom
Concept:

When the central atom is highly electronegative, bond pairs are pulled closer to the central nucleus. This causes strong bond pair-bond pair (BP-BP) repulsion, which forces the bond angle to widen.

Formula:

$$ \text{Bond Angle} \propto \text{Electronegativity of Central Atom} $$

Solution:

  • Oxygen (EN = 3.5) is much more electronegative than Sulfur (EN = 2.5).


  • In \( \text{H}_2\text{O} \), the bonding electrons are pulled very close to the Oxygen nucleus. Because they are confined in a tight space, they repel each other strongly, keeping the angle relatively wide (104.5°).


  • In \( \text{H}_2\text{S} \), the bonding electrons are further from the Sulfur nucleus. The BP-BP repulsion is weaker, allowing the lone pairs to crush the bond angle down to about 92°.


Why other options are incorrect:

  • Option A: Size plays a role, but electronegativity is the direct electronic driver of the bond pair positioning.
  • Option C: Both Oxygen and Sulfur contain two lone pairs, so this is not the distinguishing factor.

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