Multiple options may be correct (+4 / partial / −1). Nothing on this page is looked up from an answer key — identify P, Q and R on the bench, then measure each claim.
Xenon and fluorine give three different fluorides, and which one you get is decided by how much fluorine you give the xenon. Set the mixture and the conditions, then run the reactor. The paper hands you one specific recipe; find it on the map.
Every step is balanced atom by atom in front of you. Nothing is asserted: if the fluorine does not add up, the step will not go green.
A fluorinating agent hands fluorine to something else and is itself reduced. Offer Q a substrate and watch the xenon oxidation state.
Pick a species and the bench builds it from its valence electrons: bonds first, then whatever is left over as lone pairs. Drag the model to turn it. The geometry name is derived from the counts, not typed in.
Commit to a prediction, then check it against what the bench measured. The verdict column is computed from your own bookkeeping — there is no stored answer key on this page.
Written as you work. Only things you actually demonstrated appear here.
Test all four statements on the bench above and the derivation opens by itself.
Statements tested: 0 of 4 — open XeF₄, XeF₆ and XeO₃ on the bench and run the fluorinating assay
Xenon and fluorine give three fluorides and the ratio picks which one. The paper's 1 : 5 mixture at 873 K and 7 bar is the standard preparation of xenon tetrafluoride.
For contrast: 2 : 1 at 673 K and 1 bar gives XeF2, and 1 : 20 at 573 K under 60–70 bar gives XeF6. P = XeF4.
Dioxygen difluoride is a violent low-temperature fluorinating agent. It donates two fluorine atoms and leaves as O2.
Count the fluorine: 4 on the left in XeF4 plus 2 from O2F2 is 6. Xenon goes from +4 to +6. Q = XeF6.
Partial hydrolysis would stop at XeOF4 or XeO2F2. The word complete takes it all the way to the trioxide.
Xenon stays at +6 — this is a substitution, not a redox step. R = XeO3.
Steric number 6, so the electron geometry is octahedral; the two lone pairs take the axial positions, leaving the four fluorines square planar. (A) is correct.
Seven electron pairs cannot arrange themselves octahedrally. The lone pair caps one triangular face and pushes the three fluorines around it apart, giving a distorted (capped) octahedron with C3v symmetry. A perfect octahedron would need six equivalent bonds and no lone pair. (B) is incorrect.
Xenon sits at its highest common oxidation state, +6, held there by six fluorines. Anything that will accept fluorine takes it, and the xenon is reduced — usually all the way to the free element.
All three xenon fluorides fluorinate; the hexafluoride is the strongest of them. (C) is correct.
Three bond pairs and one lone pair: steric number 4, tetrahedral electron geometry, and the molecule you can see is a pyramid — the same shape as NH3, with the lone pair squeezing the O–Xe–O angle below 109.5° to about 103°. (D) is correct.
Marking: +4 only for choosing all three; +3 for any two of them; +2 … −1 for including (B).
| Simulation ID | ADV-2026-P2-CHE-Q06 |
| Exam / Year / Paper | JEE Advanced · 2026 · Paper 2 |
| Subject | Chemistry (Inorganic) |
| Chapter | The p-Block Elements (Group 18) |
| Topic | Noble gas compounds |
| Subtopics | Preparation of xenon fluorides · VSEPR and steric number · Lone pairs and distorted geometries · Hydrolysis of XeF6 · Xenon fluorides as fluorinating agents · Oxidation states of xenon |
| Question type | Multiple correct MCQ (+4 / partial / −1) |
| Difficulty | Moderate |
| Answer | — run the experiment |