PrayogX › Chemistry › The p-Block Elements (Group 18)
Xenon fluorides: identifying P, Q and R and testing their geometries
JEE Advanced 2026
Paper 2 · Question 6
Chemistry
Noble gas compounds
Moderate
Set the Xe:F2 ratio, temperature and pressure to make XeF4, push it to XeF6 with O2F2 and hydrolyse that to XeO3, then count lone pairs on a rotatable 3D model to test each printed statement.
Open the interactive simulationSub-topics covered
- Preparation of xenon fluorides and the role of the Xe:F2 ratio
- VSEPR and steric number from valence-electron counting
- Lone pairs and distorted geometries
- Complete versus partial hydrolysis of XeF6
- Xenon fluorides as fluorinating agents
- Oxidation states of xenon
What it teaches
- Lone pairs come from valence electrons left over after bonding
- Steric number 7 cannot give a perfect octahedron
- The fluorine-to-xenon ratio selects which fluoride forms
- Hydrolysis substitutes without changing the oxidation state
- A high oxidation state held by fluorine makes a fluorinating agent
Simulation ID
- ADV-2026-P2-CHE-Q06 · revision 1
See this simulation in the library → · open the simulation file on its own