Q.64 Chemistry ⭐ 360° Option Elimination Active

NEET 2025 Chemistry Question 64 Solution

Topic: The p-Block Elements (Group 18) · Official NTA Booklet Code 45

Question

Match List - I with List - II. List I: A. XeO₃; B. XeF₂; C. XeOF₄; D. XeF₆. List II: I. sp³d; linear; II. sp³; pyramidal; III. sp³d³; distorted octahedral; IV. sp³d²; square pyramidal. Choose the correct answer from the options given below:

✓

Verified Against NTA Final Answer Key

Option (1) : A-II, B-I, C-IV, D-III

📚 The p-Block Elements (Group 18) Easy

📋 Given Parameters & Problem Setup

List I (Xenon Compounds) A. XeO₃, B. XeF₂, C. XeOF₄, D. XeF₆

List II (Hybridization & Shape) I. sp³d; linear, II. sp³; pyramidal, III. sp³d³; distorted octahedral, IV. sp³d²; square pyramidal

⚡

The 30-Second Exam Hall Shortcut: ⚡ Xenon Valence Pair Counting Shortcut

⏱️ Save ~40 seconds

Xe has 8 valence electrons: XeO₃ has 3σ + 1 lone pair = sp³ pyramidal (A-II). XeF₂ has 2σ + 3 lone pairs = sp³d linear (B-I). XeOF₄ has 5σ + 1 lone pair = sp³d² square pyramidal (C-IV). XeF₆ has 6σ + 1 lone pair = sp³d³ distorted octahedral (D-III)!

1 XeO₃: 3 Xe=O double bonds (3σ) + 1 lone pair = Steric No. 4 ⟹ sp³ pyramidal (A-II).

2 Looking at options: Only Option (1) and (2) have A-II and B-I.

3 XeF₂: 2 Xe-F single bonds (2σ) + 3 equatorial lone pairs = Steric No. 5 ⟹ sp³d linear (B-I).

4 XeOF₄: 1 Xe=O (1σ) + 4 Xe-F (4σ) = 5σ + 1 lone pair = Steric No. 6 ⟹ sp³d² square pyramidal (C-IV).

5 XeF₆: 6 Xe-F (6σ) + 1 lone pair = Steric No. 7 ⟹ sp³d³ distorted octahedral (D-III).

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

7 Select Option (1) in under 15 seconds!

🛡️

Why You Can 100% Trust This Shortcut: VSEPR Theory & Noble Gas Fluoride Structures (NCERT Chemistry · Chemical Bonding and p-Block Elements).

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Structure and Hybridization of XeO₃ and XeF₂

Xenon has 8 valence electrons (5s² 5p⁶):

\text{XeO}_3: \; 3\sigma + 1\text{ lp} = 4 \implies sp^3 \text{ (Pyramidal)}

• A matches II (A-II).
• XeF₂: 2 single bonds (2σ) + 3 lone pairs = 5 electron pairs ⟹ sp³d hybridization. The three lone pairs occupy equatorial positions, giving a Linear geometry (B-I).

Step 2

Structure and Hybridization of XeOF₄

In XeOF₄, xenon forms 1 double bond to oxygen (1σ + 1π) and 4 single bonds to fluorine (4σ):

\text{XeOF}_4: \; 5\sigma + 1\text{ lp} = 6 \implies sp^3d^2 \text{ (Square Pyramidal)}

• The four fluorine atoms form the square base, oxygen occupies one axial position, and the lone pair occupies the opposite axial position.
• C matches IV (C-IV).

Step 3

Structure and Hybridization of XeF₆

In XeF₆, xenon forms 6 single bonds to fluorine (6σ) and retains 1 lone pair:

\text{XeF}_6: \; 6\sigma + 1\text{ lp} = 7 \implies sp^3d^3 \text{ (Distorted Octahedral)}

• The lone pair distorts the regular octahedral symmetry, resulting in a distorted octahedral geometry.
• D matches III (D-III).
• Complete match: A-II, B-I, C-IV, D-III, matching Option (1).

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 A

✓ 100% Correct

Verdict & Scientific Reason:

100% correct. XeO₃ (sp³, pyramidal), XeF₂ (sp³d, linear), XeOF₄ (sp³d², sq. pyramidal), XeF₆ (sp³d³, distorted octahedral).

💡

Why this is airtight: None; flawless VSEPR deduction.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Swaps shapes of XeOF₄ and XeF₆.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing steric number 6 with steric number 7.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Completely scrambled geometries.

⚠️

Student Fallacy (Why candidates pick this trap): Assigning square pyramidal to XeO₃.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Scrambled match.

⚠️

Student Fallacy (Why candidates pick this trap): Wrong steric counting.

⚠️

NTA Trap Alert: Distorted Octahedral vs Regular Octahedral in XeF₆

Because XeF₆ has 6 fluorine atoms, students often think it is regular octahedral (sp³d²). But Xenon has 8 valence electrons: 8 - 6 = 2 electrons (1 lone pair), making steric number = 7 (sp³d³), which is DISTORTED octahedral!

🛡️ How to avoid the -1 mark: Never forget the lone pair on XeF₆: 6 bonds + 1 lone pair = 7 pairs = sp³d³!

📖

Core NEET Syllabus Standard

NEET Chemistry Core Syllabus · The p-Block Elements Standard Benchmark

Core Scientific Principle Tested:

"VSEPR steric number dictates hybridization and molecular geometry."

Syllabus Focus: Group 18 Elements: Xenon Compounds and VSEPR Geometries

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

XeO₃ is Pyramidal (II), XeF₂ is Linear (I), XeOF₄ is Square Pyramidal (IV), XeF₆ is Distorted Octahedral (III)!

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NEET 2025 Chemistry Question 64 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Xenon Valence Pair Counting Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Structure and Hybridization of XeO₃ and XeF₂

Structure and Hybridization of XeOF₄

Structure and Hybridization of XeF₆

360° Option Elimination Engine Key Feature

NTA Trap Alert: Distorted Octahedral vs Regular Octahedral in XeF₆

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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