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Noble gases were once thought to form no compounds at all. Xenon fluorides are now standard JEE Advanced material: how they are made, what shape they take and how they react with water. This post covers all three, and first explains where the topic sits, because it has moved.
Is this still in the Class 12 syllabus?
Not for CBSE. The CBSE Chemistry curriculum for 2026-27 has ten Class 12 units, and the p-block is not one of them; Class 11 assesses only general s- and p-block trends, and only formatively. The JEE (Main) 2026 and NEET (UG) 2026 syllabi cover groups 13 to 18 at the level of general trends and the unique behaviour of the first element. The JEE Advanced 2026 syllabus is the one that asks for detail: "Group 18: Chemical properties and uses; compounds of xenon with fluorine and oxygen."
So this post is for JEE Advanced. The current NCERT books no longer carry the chapter. A free official text that does is Lesson 22 of the NIOS Senior Secondary Chemistry course, which the facts below follow. Check the current year's brochure before you plan.
Why does xenon react at all?
Noble gases have closed shells and very high ionisation energies, so they have little tendency to gain or lose electrons. NIOS records that Neil Bartlett made the first noble gas compound in 1962, by reacting xenon with PtF₆. Since then, xenon compounds have been made mainly with the two most electronegative elements, fluorine and oxygen. NIOS also notes that helium, neon and argon form no compounds, while krypton forms KrF₂. Our profile of William Ramsay and the noble gases tells how the group was found.
How are the three fluorides made?
NIOS gives these conditions:
Xe + F₂ → XeF₂, with xenon and fluorine in a 2 : 1 ratio, 873 K, 1 bar.
Xe + 2F₂ → XeF₄, in a 1 : 5 ratio, 873 K, 7 bar.
Xe + 3F₂ → XeF₆, in a 1 : 20 ratio, 573 K, 60 bar.
The pattern is easy to remember: more fluorine and higher pressure give more fluorine atoms per xenon. All three are white solids that sublime readily at 298 K.
What shapes does VSEPR predict?
Use the counting method from our post on VSEPR and molecular shapes: electron pairs around xenon = ½(8 + number of fluorines).
XeF₂: ½(8 + 2) = 5 pairs, 2 bonding and 3 lone. The lone pairs take the equatorial positions, so the shape is linear. NIST's experimental geometry for XeF₂ confirms a linear molecule with Xe–F = 1.974 Å.
XeF₄: ½(8 + 4) = 6 pairs, 4 bonding and 2 lone. The lone pairs sit opposite each other, so the shape is square planar. NIST's geometry for XeF₄ gives F–Xe–F angles of 90° and Xe–F = 1.935 Å.
XeF₆: ½(8 + 6) = 7 pairs, 6 bonding and 1 lone. Six atoms and a lone pair cannot sit in a regular octahedron, so the shape is a distorted octahedron.
Oxidation states follow directly: xenon is +2, +4 and +6. The lone pairs go 3, 2, 1 as fluorines go 2, 4, 6.
Which reactions do you need?
NIOS lists how they behave with water.
XeF₂ hydrolyses slowly: 2XeF₂ + 2H₂O → 2Xe + 4HF + O₂.
XeF₄ reacts violently: 6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂.
XeF₆ reacts violently: XeF₆ + 3H₂O → XeO₃ + 6HF.
Partial hydrolysis of XeF₆ gives oxofluorides: XeF₆ + H₂O → XeOF₄ + 2HF, and XeF₆ + 2H₂O → XeO₂F₂ + 4HF.
They also react with strong Lewis acids: XeF₂ + PF₅ → [XeF]⁺[PF₆]⁻ and XeF₄ + SbF₅ → [XeF₃]⁺[SbF₆]⁻. XeF₆ accepts fluoride from MF (M = Na, K, Rb, Cs) to give M⁺[XeF₇]⁻.
Worked example: is XeF₄ hydrolysis a redox reaction?
Track xenon in 6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂.
Four xenon atoms go from +4 to 0: they gain 4 × 4 = 16 electrons.
Two go from +4 to +6: they lose 2 × 2 = 4 electrons.
Six oxygen atoms go from −2 in water to 0 in O₂: they lose 6 × 2 = 12 electrons.
Electrons lost = 4 + 12 = 16 = electrons gained. The equation balances, and xenon both gains and loses electrons, so this is disproportionation.
Compare XeF₆ + 3H₂O → XeO₃ + 6HF. Xenon is +6 on both sides, and no other element changes, so this one is not redox at all.
The same method gives the shape of the products: XeO₃ has ½(8 + 0) = 4 pairs, three bonds and one lone pair, so it is pyramidal; XeOF₄ has ½(8 + 4) = 6 pairs, five bonds and one lone pair, so it is square pyramidal.
Which mistakes cost marks?
Calling XeF₄ tetrahedral. Two lone pairs make it square planar.
Drawing XeF₂ bent. Three equatorial lone pairs leave it linear.
Writing XeF₆ as a regular octahedron. The lone pair distorts it.
Counting double-bonded oxygen as a monovalent atom in the pair count. Oxygen adds zero.
Unbalanced hydrolysis equations. Check xenon, fluorine, hydrogen and oxygen on both sides.
A practice routine
Write the three preparation equations with ratios, then the pair count and shape for each fluoride.
Balance the five hydrolysis equations from memory and check each one atom by atom.
Find the oxidation state of xenon in every species above.
Draw XeO₃ and XeOF₄ and check your shapes against the NIOS figures.
If VSEPR itself is shaky, the Fermi Concept Library covers every concept in chemistry, each with its own lesson. For the rest of the p-block, see our post on groups 15 to 17.
Today, write the pair counts for XeF₂, XeF₄ and XeF₆ without looking, and check the 3, 2, 1 pattern of lone pairs.
