Match Shapes of XeO₃, XeF₂, XeO₂F₂ and XeOF₄ – Chemical Bonding JEE Main 2026 PYQ

Quick Summary

Question Type: VSEPR theory and molecular shape matching
Chapter: Chemical Bonding – Shapes of Molecules
Difficulty: ⭐⭐⭐ Medium
Time to Solve: 3-5 minutes
Key Formula: Shape depends on steric number = bond pairs + lone pairs
Correct Answer: (C) A-II, B-III, C-IV, D-I
Why: XeO₃ is trigonal pyramidal like NH₃, XeF₂ is linear like [I₃]⁻, XeO₂F₂ is see-saw like SF₄, and XeOF₄ is square pyramidal like BrF₅.

The Question

JEE Main 2026 – Chemical Bonding

Match List-I with List-II according to shape.

List-I
A. XeO₃
B. XeF₂
C. XeO₂F₂
D. XeOF₄

List-II
I. BrF₅
II. NH₃
III. [I₃]⁻
IV. SF₄

Choose the correct answer from the options given below:

(A) A-II, B-III, C-I, D-IV
(B) A-II, B-I, C-III, D-IV
(C) A-II, B-III, C-IV, D-I
(D) A-III, B-II, C-IV, D-I

Quick Answer

Correct Option: (C) A-II, B-III, C-IV, D-I

Reasoning: Using VSEPR theory:

• XeO₃ has 3 bond pairs and 1 lone pair, so its shape is trigonal pyramidal like NH₃
• XeF₂ has 2 bond pairs and 3 lone pairs, so its shape is linear like [I₃]⁻
• XeO₂F₂ has 4 bond pairs and 1 lone pair, so its shape is see-saw like SF₄
• XeOF₄ has 5 bond pairs and 1 lone pair, so its shape is square pyramidal like BrF₅

Video Solution

If you want the full explanation in a clear step-by-step teaching format, watch the video solution below:

Watch Full Video Solution on YouTube

Understanding the Concept

Why VSEPR Theory Works Here

The shape of a molecule depends on the number of bond pairs and lone pairs around the central atom. In xenon compounds, the central Xe atom often has an expanded octet, so we must count the total electron domains carefully.

Steric number = Bond pairs + Lone pairs

Where:

• Bond pairs determine how many atoms are attached to the central atom
• Lone pairs distort the geometry and change the final molecular shape
• The observed shape is based on both bond pair-bond pair and lone pair-bond pair repulsions

The Key Principle

To solve shape-matching questions:

1. Count bond pairs and lone pairs on the central atom
2. Find the steric number
3. Use VSEPR theory to identify the final molecular shape

Detailed Step-by-Step Solution

Step 1: Apply VSEPR Theory

• Shape depends on bond pairs and lone pairs
• Steric number = bond pairs + lone pairs
• Same steric number can still give different shapes if lone pairs are different

Step 2: Analyze Each Xenon Compound

Download Handwritten Solution

For XeO₃:

• Xe forms 3 bonds with oxygen
• One lone pair remains on Xe
• Steric number = 4
• Shape = trigonal pyramidal
• Match = NH₃ (II)

For XeF₂:

• Xe forms 2 bonds with fluorine
• Three lone pairs remain on Xe
• Steric number = 5
• Shape = linear
• Match = [I₃]⁻ (III)

For XeO₂F₂:

• Xe forms 4 bonds in total
• One lone pair remains on Xe
• Steric number = 5
• Shape = see-saw
• Match = SF₄ (IV)

For XeOF₄:

• Xe forms 5 bonds in total
• One lone pair remains on Xe
• Steric number = 6
• Shape = square pyramidal
• Match = BrF₅ (I)

Step 3: Write the Correct Matching

Comparing all shapes:

• A → II
• B → III
• C → IV
• D → I

Step 4: Final Answer from Options

The correct matching is:

• A-II, B-III, C-IV, D-I
• So the correct option is (C)

Final Answer

Option (C): A-II, B-III, C-IV, D-I ✓

XeO₃ matches NH₃, XeF₂ matches [I₃]⁻, XeO₂F₂ matches SF₄, and XeOF₄ matches BrF₅ based on VSEPR shapes.

Essential Formulas for This Topic

Primary Rules

1. Steric Number Rule:
   • Steric number = bond pairs + lone pairs
   • It helps identify electron-pair geometry
   • Molecular shape depends on lone pair arrangement too
2. Common VSEPR Shapes:
   • SN = 4 with 1 lone pair → trigonal pyramidal
   • SN = 5 with 3 lone pairs → linear
   • SN = 5 with 1 lone pair → see-saw
   • SN = 6 with 1 lone pair → square pyramidal
3. Matching Logic:
   • XeO₃ ↔ NH₃
   • XeF₂ ↔ [I₃]⁻
   • XeO₂F₂ ↔ SF₄
   • XeOF₄ ↔ BrF₅

Important Constants

• Lone pairs distort ideal geometry more strongly than bond pairs
• Xenon compounds commonly show expanded octet
• Same steric number can give different molecular shapes
• Always count lone pairs on the central atom carefully

Common Mistakes to Avoid

❌ Mistake 1: Ignoring Lone Pairs

Wrong Thinking: “Shape depends only on the number of bonded atoms.”

Correct Approach: Lone pairs strongly affect molecular shape. Always count them before deciding the final geometry.

❌ Mistake 2: Confusing Electron Geometry with Molecular Shape

Wrong Approach: Taking trigonal bipyramidal or octahedral directly as the final shape

Correct Approach: First identify electron-pair geometry, then remove lone pair positions to get the actual molecular shape.

❌ Mistake 3: Treating XeF₂ as Bent

Common Error:

• Seeing 2 bonded atoms and assuming bent shape
• Ignoring the 3 equatorial lone pairs in trigonal bipyramidal arrangement

Correct Approach: XeF₂ is linear because the three lone pairs occupy equatorial positions, leaving the two F atoms opposite each other.

❌ Mistake 4: Mixing Up See-Saw and Square Pyramidal

Wrong Thinking: “XeO₂F₂ and XeOF₄ both have one lone pair, so they must have the same shape.”

Correct Understanding:

• XeO₂F₂ has steric number 5, so it is see-saw
• XeOF₄ has steric number 6, so it is square pyramidal
• Different steric numbers lead to different shapes
• Count bond pairs carefully before deciding shape

Key Concept Summary

What You Must Remember

1. Shape depends on steric number: Count bond pairs and lone pairs
2. Lone pairs change molecular shape: They cannot be ignored
3. XeO₃ is trigonal pyramidal: Like NH₃
4. XeF₂ is linear: Like [I₃]⁻
5. XeO₂F₂ is see-saw and XeOF₄ is square pyramidal: Match them carefully

The Golden Rule for Shape-Matching Questions

“First count bond pairs and lone pairs on the central atom, then apply VSEPR theory to get the final molecular shape.”

Frequently Asked Questions

Q1: Why is XeO₃ trigonal pyramidal?

A: XeO₃ has 3 bond pairs and 1 lone pair on Xe, giving steric number 4 and a trigonal pyramidal shape like NH₃.

Q2: Why is XeF₂ linear even though xenon has many electron pairs?

A: XeF₂ has steric number 5 with 3 lone pairs occupying equatorial positions, leaving the two fluorine atoms opposite each other in a linear arrangement.

Q3: Why is XeO₂F₂ see-saw?

A: XeO₂F₂ has 4 bond pairs and 1 lone pair, so steric number is 5 and the shape becomes see-saw, like SF₄.

Q4: Why is XeOF₄ square pyramidal?

A: XeOF₄ has 5 bond pairs and 1 lone pair, so steric number is 6 and the molecular shape is square pyramidal, like BrF₅.

Q5: What is the final correct matching in this question?

A: A-II, B-III, C-IV, D-I.

Prerequisites to Solve This Question

Before attempting this problem, you should understand:

1. VSEPR theory: Bond pair and lone pair repulsion
2. Steric number: How to count total electron domains
3. Common molecular shapes: Linear, trigonal pyramidal, see-saw, square pyramidal
4. Xenon compounds: Expanded octet behavior of xenon
5. Shape comparison: Matching different molecules with similar geometry

After Solving This, You Can:

✅ Predict shapes of xenon compounds using VSEPR theory
✅ Match molecules with similar geometry confidently
✅ Distinguish steric number 5 and steric number 6 cases
✅ Solve JEE Main Chemical Bonding shape questions faster
✅ Avoid common confusion in lone-pair based molecular geometry
✅ Apply VSEPR logic to other p-block compounds

Study Tips for This Topic

For JEE Main:

1. Memorize common shapes: NH₃, SF₄, BrF₅ and [I₃]⁻ are standard reference molecules
2. Count lone pairs carefully: They decide the final shape
3. Practice xenon compounds: These are common in Chemical Bonding
4. Do not rush matching questions: One wrong pair changes the whole option

Common JEE Variants:

• Match molecules according to shape
• Find incorrect shape assignment
• Identify steric number and geometry
• Compare molecular shape with electron-pair geometry

Difficulty Rating & Exam Frequency

Difficulty Level: ⭐⭐⭐ (3/5) – Medium
JEE Main Frequency: High – Molecular shape questions are common in Chemical Bonding
JEE Advanced Frequency: Medium – Often combined with hybridisation and bonding arguments
Topic Importance: Very High – VSEPR theory is a core concept in Chemical Bonding

Written by Nishant Kumar
Chemistry Educator with 10+ Years of Experience Teaching JEE Aspirants
Founder – PadhoLikhoJEE

Last Updated: March 2026
Question Source: JEE Main 2026 PYQ
Topic: Chemical Bonding – Shapes of Molecules