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Structure of Xenon and Interhalogen Compounds (Shapes and Hybridisation)

Chemistry · Chemical Bonding · NEET

Xenon compounds and interhalogen compounds get their shapes from VSEPR: count bond pairs plus lone pairs on the central atom, then place lone pairs where repulsion is lowest. For example, XeF2 is linear (2 bonds + 3 lone pairs), XeF4 is square planar (4 bonds + 2 lone pairs), and ClF3 is T-shaped (3 bonds + 2 lone pairs). Memory hook: "2 lone pairs sit opposite" gives you square planar (XeF4) and T-shape (ClF3) fast.
Shapes from VSEPR (lone pairs decide)XeF2 (linear)FXeF2 bond + 3 lpsp3dXeF4 (square planar)XeFFFF4 bond + 2 lpsp3d2ClF3 (T-shape)ClFFF3 bond + 2 lpsp3dIF7I7 bond + 0 lpsp3d3lp = lone pair. Same central atom, different lone pairs → different shape.
VSEPR shapes of key xenon fluorides and interhalogens. The number of lone pairs (lp) on the central atom decides the final shape: XeF2 linear (3 lp), XeF4 square planar (2 lp), ClF3 T-shaped (2 lp), IF7 pentagonal bipyramidal (0 lp).

Your doubts, answered

How do I find the shape of any xenon fluoride or interhalogen compound?

Use one simple count. Total electron pairs on the central atom = (valence electrons of central atom + number of single-bonded atoms) / 2. For O double bonds, count the O but do not add electrons for it. Example XeF4: Xe has 8 valence electrons, 4 F atoms → (8+4)/2 = 6 pairs. 4 are bond pairs, so 2 are lone pairs. 6 pairs = sp3d2 = octahedral electron geometry; with 2 lone pairs opposite each other the atoms make a square planar shape. This one method works for every molecule below.

Why is XeF4 square planar and not tetrahedral?

XeF4 has 4 bond pairs AND 2 lone pairs = 6 electron pairs, not 4. So the electron geometry is octahedral (sp3d2), not tetrahedral. The 2 lone pairs go to opposite corners (top and bottom) because that keeps them farthest apart. The 4 F atoms are left in one flat plane, giving a square planar molecule. CH4 is tetrahedral because it has 4 bond pairs and 0 lone pairs. Same number of bonds, different number of lone pairs, so different shape.

What is the hybridisation of XeF2, XeF4 and XeF6?

Count total electron pairs, then match: 2 pairs = sp, 3 = sp2, 4 = sp3, 5 = sp3d, 6 = sp3d2, 7 = sp3d3. XeF2 = 5 pairs (2 bond + 3 lone) = sp3d, linear. XeF4 = 6 pairs (4 bond + 2 lone) = sp3d2, square planar. XeF6 = 7 pairs (6 bond + 1 lone) = sp3d3, distorted octahedral. The extra lone pair in XeF6 pushes the shape out of a perfect octahedron, which is why NEET calls it 'distorted'.

How many lone pairs are on Cl in ClF3, and why is it T-shaped?

Cl has 7 valence electrons. It uses 3 to bond with 3 F atoms, leaving 4 non-bonding electrons = 2 lone pairs. So ClF3 is AB3E2: 3 bond pairs + 2 lone pairs = 5 pairs total = sp3d, trigonal bipyramidal electron geometry. The 2 lone pairs sit in the equatorial plane (more room there), leaving the 3 F atoms forming a T-shape. This is a repeated NEET answer (2018, 2026): ClF3 = T-shaped with two lone pairs.

What are the shapes of the interhalogen types XX', XX'3, XX'5 and XX'7?

Learn them by type. XX' (like ClF, ICl) is linear. XX'3 (like ClF3, BrF3) is T-shaped (AB3E2). XX'5 (like BrF5, IF5) is square pyramidal (AB5E). XX'7 (like IF7) is pentagonal bipyramidal (AB7, sp3d3). NEET 2017 matched exactly these four. Notice the central atom is the bigger/less electronegative halogen and the number of small F atoms decides the shape.

Which xenon or interhalogen molecule has the most lone-pair to lone-pair repulsion?

The one with the most lone pairs on the central atom. XeF2 has 3 lone pairs, which is the highest among common molecules like ClF3 (2), SF4 (1) and IF5 (1). More lone pairs means more lone-pair to lone-pair repulsion. NEET 2022 asked exactly this and the answer was XeF2. Quick rank of lone pairs: XeF2 (3) > XeF4 / ClF3 (2) > XeO3 / IF5 (1).

⚠️ The NEET trap
✗ XeF4 is octahedral / tetrahedral because it has 4 bonds.
✓ XeF4 is square planar. It has 6 electron pairs (4 bond + 2 lone), so it is sp3d2 with an octahedral electron geometry, but the molecular shape (only the atoms) is square planar because the 2 lone pairs sit opposite each other.
🧠 Shape means the arrangement of ATOMS only. Ignore where the electron pairs point; look at where the F atoms end up. 2 lone pairs opposite → flat square → square planar.

Real NEET questions

NEET 2019

Match the Xenon compounds in Column-I with their structure in Column-II: (a) XeF4, (b) XeF6, (c) XeOF4, (d) XeO3; Column-II: (i) pyramidal, (ii) square planar, (iii) distorted octahedral, (iv) square pyramidal.

A · a-i, b-ii, c-iii, d-iv
B · a-ii, b-iii, c-iv, d-i ✓
C · a-ii, b-iii, c-i, d-iv
D · a-iii, b-iv, c-i, d-ii
Solution: Count lone pairs by VSEPR. XeF4 = 4 bond + 2 lone (AB4E2) = square planar (ii). XeF6 = 6 bond + 1 lone (AB6E) = distorted octahedral (iii). XeOF4 = 5 bonded groups + 1 lone (AB5E) = square pyramidal (iv). XeO3 = 3 bond + 1 lone (AB3E) = pyramidal (i). So a-ii, b-iii, c-iv, d-i, which is option B.
NEET 2018

In the structure of ClF3, the number of lone pairs of electrons on the central atom Cl is:

A · Four
B · Two ✓
C · One
D · Three
Solution: Cl has 7 valence electrons. Three electrons form bonds to three F atoms, leaving 4 non-bonding electrons = 2 lone pairs. ClF3 is AB3E2 (T-shaped). Answer: Two.
NEET 2022

Amongst the following, which one will have maximum 'lone pair-lone pair' electron repulsion?

A · ClF3
B · IF5
C · SF4
D · XeF2 ✓
Solution: More lone pairs on the central atom means more lone-pair to lone-pair repulsion. XeF2 is AB2E3 with 3 lone pairs, the highest here (ClF3 has 2, SF4 has 1, IF5 has 1). So XeF2 has the maximum lone pair-lone pair repulsion.
NEET 2017

Match the interhalogen compounds of Column I with the geometry in Column II: (a) XX', (b) XX'3, (c) XX'5, (d) XX'7; Column II: (i) T-shape, (ii) Pentagonal bipyramidal, (iii) Linear, (iv) Square pyramidal, (v) Tetrahedral.

A · (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
B · (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii) ✓
C · (a)-(v), (b)-(iv), (c)-(iii), (d)-(ii)
D · (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Solution: XX' (e.g. ClF) is linear (iii). XX'3 (e.g. ClF3, AB3E2) is T-shaped (i). XX'5 (e.g. BrF5, AB5E) is square pyramidal (iv). XX'7 (e.g. IF7, AB7) is pentagonal bipyramidal (ii). So (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii), which is option B.

Solved Chemical Bonding NEET PYQs

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Frequently asked

Is XeF6 octahedral or distorted octahedral?

Distorted octahedral. XeF6 has 6 bond pairs plus 1 lone pair = 7 electron pairs (sp3d3). The extra lone pair pushes the F atoms out of a perfect octahedron, so NEET marks it as distorted octahedral, not regular octahedral.

What is the shape and hybridisation of XeO3?

XeO3 is sp3 and pyramidal (trigonal pyramidal). Xe forms 3 bonds to O and keeps 1 lone pair (AB3E), so 4 electron pairs total. This is confirmed by NEET 2025.

Why does XeF2 have a linear shape even with 3 lone pairs?

XeF2 has 5 electron pairs (2 bond + 3 lone), sp3d, trigonal bipyramidal electron geometry. All 3 lone pairs go to the equatorial positions where they have the most room. The 2 F atoms are left at the axial (top and bottom) positions, giving a straight line = linear shape.

Which interhalogen is pentagonal bipyramidal?

IF7 (iodine heptafluoride). It has 7 bond pairs and 0 lone pairs (AB7, sp3d3), so all 7 F atoms spread into a pentagonal bipyramid. It is the only common AB7 interhalogen you need for NEET.

How is XeF2 isostructural with IBr2-?

Both have 22 valence electrons (isoelectronic) and are AB2E3 (2 bond pairs + 3 lone pairs), so both are linear (isostructural). NEET 2017 tested this exact pair.

Related concepts:
VSEPR Theory: How to Predict Molecular ShapeExpanded Hybridisation: sp3d, sp3d2, sp3d3 (PCl5, SF6, XeF6)How to Find Hybridisation of Any Central Atom (Quick Trick)What Are Isoelectronic and Isostructural Species?Hydrogen Bonding: Intramolecular vs Intermolecular
Practise Chemistry for NEET:
Chemistry MCQs (free, chapter-wise)Chemical Bonding PYQs