Chemistry Chemical Bonding MDCAT 2019
PMDC Verified Question 67 of 102
The structure of Xenon trioxide (\( \text{XeO}_3 \)) is shown below.

Xe O O O


With reference to the Valence shell electron pair repulsion theory (VSEPR), the shape of \( \text{XeO}_3 \) is:
A
Tetrahedral
B
Trigonal pyramidal
C
Bent (or angular)
D
Trigonal planar
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: Trigonal pyramidal
Concept:

The molecular shape is dictated by the arrangement of bond pairs and lone pairs around the central atom according to VSEPR theory.

Formula:

$$ \text{Type} = \text{AB}_3\text{E} $$

Solution:

  • Xenon (Xe) in \( \text{XeO}_3 \) forms three double bonds with oxygen atoms (3 bonding domains).


  • Xe is a noble gas with 8 valence electrons. Sharing 6 electrons with Oxygen leaves 2 electrons, which form 1 lone pair.


  • A molecule with 3 bond pairs and 1 lone pair (\( \text{AB}_3\text{E} \)) adopts a trigonal pyramidal geometry.


Why other options are incorrect:

  • Option A: Would require 4 bond pairs and 0 lone pairs.
  • Option C: Would require 2 bond pairs.
  • Option D: Would require 3 bond pairs and 0 lone pairs.

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.

Want to solve full-length papers under timed exam conditions?

Practice with zero-scroll lockdown sprints, dynamic latency zone timers, live peer selection telemetry, and the automated Amber mistake recovery loop.