Q.66 Chemistry · Section A ⭐ 360° Option Elimination Active

NEET 2024 Chemistry Question 66 Solution

Topic: Chemical Bonding and Molecular Structure · Official NTA Booklet Code T1

Question

Match List I (Compound) with List II (Shape/geometry). List I: A. NH₃; B. BrF₅; C. XeF₄; D. SF₆. List II: I. Trigonal Pyramidal; II. Square Planar; III. Octahedral; IV. Square Pyramidal. Choose the correct answer from the options given below:

✓

Verified Against NTA Final Answer Key

Option (3) : A-I, B-IV, C-II, D-III

📚 Chemical Bonding and Molecular Structure Easy

📋 Given Parameters & Problem Setup

Molecule A NH3 (Ammonia)

Molecule B BrF5 (Bromine pentafluoride)

Molecule C XeF4 (Xenon tetrafluoride)

Molecule D SF6 (Sulfur hexafluoride)

⚡

The 30-Second Exam Hall Shortcut: ⚡ Steric Number & Lone Pair Direct Mapping

⏱️ Save ~40 seconds

NH3: 3 bp + 1 lp -> Trigonal Pyramidal (I); BrF5: 5 bp + 1 lp -> Square Pyramidal (IV); XeF4: 4 bp + 2 lp -> Square Planar (II); SF6: 6 bp + 0 lp -> Octahedral (III)!

1 NH3: Steric number = 3 + 1 = 4 (sp³). One lone pair compresses tetrahedral geometry into Trigonal Pyramidal (A-I).

2 BrF5: Br has 7 valence e⁻. 5 bonding pairs + 1 lone pair = steric number 6 (sp³d²). One axial lone pair yields Square Pyramidal (B-IV).

3 XeF4: Xe has 8 valence e⁻. 4 bonding pairs + 2 lone pairs = steric number 6 (sp³d²). Two axial lone pairs yield Square Planar (C-II).

4 SF6: S has 6 valence e⁻. 6 bonding pairs + 0 lone pairs = steric number 6 (sp³d²). Symmetrical Octahedral (D-III).

5 Match: A-I, B-IV, C-II, D-III -> Option (3) in 5 seconds!

🛡️

Why You Can 100% Trust This Shortcut: NCERT Chemistry · Chemical Bonding and Molecular Structure (VSEPR theory and shapes of molecules).

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

VSEPR Calculation for Ammonia (NH3)

Central atom Nitrogen has 5 valence electrons. It forms 3 single bonds with hydrogen atoms and retains 1 non-bonding lone pair: Steric Number = 3 + 1 = 4 (sp³ hybridization). The electron-pair geometry is tetrahedral, but the molecular shape (positions of nuclei) is Trigonal Pyramidal.

\text{SN} = 3\;\text{bp} + 1\;\text{lp} = 4 \implies \text{Trigonal Pyramidal}

Step 2

VSEPR Calculation for Bromine Pentafluoride (BrF5)

Central atom Bromine has 7 valence electrons. Forming 5 Br-F bonds consumes 5 electrons, leaving 2 electrons (1 lone pair): Steric Number = 5 + 1 = 6 (sp³d² hybridization). Octahedral electron geometry with one lone pair gives a Square Pyramidal shape.

\text{SN} = 5\;\text{bp} + 1\;\text{lp} = 6 \implies \text{Square Pyramidal}

Step 3

VSEPR Calculation for Xenon Tetrafluoride (XeF4)

Central atom Xenon has 8 valence electrons. Forming 4 Xe-F bonds leaves 4 non-bonding electrons (2 lone pairs): Steric Number = 4 + 2 = 6 (sp³d² hybridization). To minimize lone pair-lone pair repulsions, the two lone pairs occupy trans axial positions (180° apart), leaving the 4 fluorines in an equatorial plane: Square Planar shape.

\text{SN} = 4\;\text{bp} + 2\;\text{lp} = 6 \implies \text{Square Planar}

Step 4

VSEPR Calculation for Sulfur Hexafluoride (SF6)

Central atom Sulfur has 6 valence electrons, all shared in 6 S-F bonds with 0 lone pairs: Steric Number = 6 + 0 = 6 (sp³d² hybridization). Completely symmetrical regular Octahedral shape.

\text{SN} = 6\;\text{bp} + 0\;\text{lp} = 6 \implies \text{Octahedral}

⭐

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 (1)

✗ Incorrect

Verdict & Scientific Reason:

Assigns octahedral shape to NH3.

⚠️

Student Fallacy (Why candidates pick this trap): Gross mismatch of basic geometries.

2 (2)

✗ Incorrect

Verdict & Scientific Reason:

Mismatches all shapes.

⚠️

Student Fallacy (Why candidates pick this trap): Random pairing.

3 (3)

✓ 100% Correct

Verdict & Scientific Reason:

Perfect alignment with VSEPR steric numbers and shapes.

💡

Why this is airtight: None.

4 (4)

✗ Incorrect

Verdict & Scientific Reason:

Assigns square planar to NH3.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing 4-coordinate with 3-coordinate.

⚠️

NTA Trap Alert: Confusing Electron Geometry with Molecular Shape

XeF4 has an octahedral electron-pair arrangement (steric number 6), but its actual observed molecular geometry is Square Planar because lone pairs are invisible in shape determination.

🛡️ How to avoid the -1 mark: Always distinguish electron domain geometry from molecular geometry (shape of atoms).

📖

Core NEET Syllabus Standard

Chemistry Core Chemical Bonding · Chemical Bonding and Molecular Structure Standard Benchmark

Core Scientific Principle Tested:

"Molecular geometry is determined by the spatial arrangement of bonded atoms that minimizes electron-pair repulsions (lone pair-lone pair > lone pair-bond pair > bond pair-bond pair)."

Syllabus Focus: VSEPR Theory: Molecular Geometries of Polyatomic Species

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Six pairs make Octahedral base: 0 lp is Octahedral, 1 lp is Square Pyramid, 2 lp is Square Flat!

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NEET 2024 Chemistry Question 66 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Steric Number & Lone Pair Direct Mapping

Step-by-Step Rigorous Workflow & Mathematical Proof

VSEPR Calculation for Ammonia (NH3)

VSEPR Calculation for Bromine Pentafluoride (BrF5)

VSEPR Calculation for Xenon Tetrafluoride (XeF4)

VSEPR Calculation for Sulfur Hexafluoride (SF6)

360° Option Elimination Engine Key Feature

NTA Trap Alert: Confusing Electron Geometry with Molecular Shape

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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