Q.63 Chemistry ⭐ 360° Option Elimination Active

NEET 2021 Chemistry Question 63 Solution

Topic: Chemical Bonding and Molecular Structure · Official NTA Booklet Code M1

Question

Match List - I with List - II. List - I List - II (a) PCl5 (i) Square pyramidal (b) (ii) Trigonal planar (c) BrF5 (iii) Octahedral (d) (iv) Trigonal bipyramidal Choose the correct answer from the options given below.

(1)(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
(2)(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
(3)(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
(4)(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

✓

Verified Against NTA Final Answer Key

Option (1) : (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)

📚 Chemical Bonding and Molecular Structure Easy

📋 Given Parameters & Problem Setup

List I Molecules (a) PCl5, (b) SF6, (c) BrF5, (d) BF3

List II Geometries (i) Square pyramidal, (ii) Trigonal planar, (iii) Octahedral, (iv) Trigonal bipyramidal

⚡

The 30-Second Exam Hall Shortcut: ⚡ VSEPR Electron Pair Counting Match

⏱️ Save ~25 seconds

$PCl_5$ : 5 bp + 0 lp $\to$ Trigonal bipyramidal (iv); $SF_6$ : 6 bp + 0 lp $\to$ Octahedral (iii); $BrF_5$ : 5 bp + 1 lp $\to$ Square pyramidal (i); $BF_3$ : 3 bp + 0 lp $\to$ Trigonal planar (ii)! Instant 10-second matching to Option (1)!

PCl_5

: Valence electrons

= 5 + 5(7) = 40

. Steric number

= 5

(5 bp, 0 lp)

\implies sp^3d \implies

Trigonal bipyramidal -> (a)-(iv).

SF_6

: Steric number

= 6

(6 bp, 0 lp)

\implies sp^3d^2 \implies

Octahedral -> (b)-(iii).

BrF_5

: Steric number

= 6

(5 bp, 1 lp)

\implies sp^3d^2 \implies

Square pyramidal -> (c)-(i).

BF_3

: Steric number

= 3

(3 bp, 0 lp)

\implies sp^2 \implies

Trigonal planar -> (d)-(ii).

5 Match combination: (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii) -> Option (1).

🛡️

Why You Can 100% Trust This Shortcut: Canonical Gillespie-Nyholm VSEPR theory and hybridization rules.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Analyzing PCl5 (a)

Phosphorus has 5 valence electrons, all bonded to chlorine atoms forming 5 $\sigma$ -bonds with zero lone pairs ( $sp^3d$ hybridization). The ideal geometry is trigonal bipyramidal:

PCl_5 \implies 5\text{ bp} + 0\text{ lp} \implies \mathbf{\text{Trigonal bipyramidal (iv)}}

Step 2

Analyzing SF6 (b)

Sulfur has 6 valence electrons, all forming 6 $\sigma$ -bonds with fluorine atoms with zero lone pairs ( $sp^3d^2$ hybridization). The regular symmetrical geometry is octahedral:

SF_6 \implies 6\text{ bp} + 0\text{ lp} \implies \mathbf{\text{Octahedral (iii)}}

Step 3

Analyzing BrF5 (c)

Bromine has 7 valence electrons: 5 form $\sigma$ -bonds with fluorine and 2 form 1 lone pair ( $5\text{ bp} + 1\text{ lp} = 6$ electron pairs, $sp^3d^2$ hybridization). According to VSEPR, the lone pair occupies one axial site, giving a square pyramidal shape:

BrF_5 \implies 5\text{ bp} + 1\text{ lp} \implies \mathbf{\text{Square pyramidal (i)}}

Step 4

Analyzing BF3 (d)

Boron has 3 valence electrons, forming 3 $\sigma$ -bonds with fluorine atoms with zero lone pairs ( $sp^2$ hybridization). The molecular shape is trigonal planar with bond angles of $120^\circ$ :

BF_3 \implies 3\text{ bp} + 0\text{ lp} \implies \mathbf{\text{Trigonal planar (ii)}}

⭐

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

✓ 100% Correct

Verdict & Scientific Reason:

Flawless 1-to-1 correspondence with all VSEPR geometries.

💡

Why this is airtight: None; textbook VSEPR matching.

2 2

✗ Incorrect

Verdict & Scientific Reason:

Mismatches $PCl_5$ with trigonal planar and $BrF_5$ with trigonal bipyramidal.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing steric numbers 5 and 3.

3 3

✗ Incorrect

Verdict & Scientific Reason:

Mismatches $PCl_5$ with octahedral.

⚠️

Student Fallacy (Why candidates pick this trap): Assigning 6 coordination to phosphorus.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Swaps geometries of $BrF_5$ and $BF_3$ .

⚠️

Student Fallacy (Why candidates pick this trap): Swapping square pyramidal with trigonal planar.

⚠️

NTA Trap Alert: BrF5 Square Pyramidal vs. Trigonal Bipyramidal Trap

Students frequently count 5 fluorine atoms in $BrF_5$ and immediately think "5 bonds = trigonal bipyramidal", forgetting the essential LONE PAIR on bromine ( $5+1=6$ electron pairs $\implies$ square pyramidal).

🛡️ How to avoid the -1 mark: Count total electron pairs:

BrF_5

has

5\text{ bp} + 1\text{ lp} = 6

pairs

\implies

Square pyramidal.

📖

Core NEET Syllabus Standard

NCERT Chemistry · Chemical Bonding and Molecular Structure Standard Benchmark

Core Scientific Principle Tested:

"Molecular shape is determined by the total number of bond pairs and lone pairs around the central atom."

Syllabus Focus: VSEPR Theory and Shapes of Simple Molecules

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Chemistry Question 63 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ VSEPR Electron Pair Counting Match

Step-by-Step Rigorous Workflow & Mathematical Proof

Analyzing PCl5 (a)

Analyzing SF6 (b)

Analyzing BrF5 (c)

Analyzing BF3 (d)

360° Option Elimination Engine Key Feature

NTA Trap Alert: BrF5 Square Pyramidal vs. Trigonal Bipyramidal Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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