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 ↗

Ready to attempt under real exam pressure?

Test your speed and accuracy in the authentic 200-minute timed NTA simulator.

Launch NEET 2021 CBT Test →

← Previous (Q.62) All Chemistry Questions Grid Next (Q.64) →

Spotted a mistake in this solution? Report an error ([email protected])

⚖️ Legal & Fair Dealing Notice: Sourced from publicly released NEET 2021 question papers and NEET answer keys for private study, research, and non-commercial educational practice under Fair Dealing (Section 52(1)(a) of the Indian Copyright Act, 1957). Not affiliated with or endorsed by NTA or NMC.

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?

🔗 Related Questions — Chemical Bonding and Molecular Structure

Covalency & Valence Orbital Availability in Second Period Elements

Highest Occupied Molecular Orbital (HOMO) in Second-Row Diatomics

Bond Order, Molecular Stability and Bond Length Relationships

Evaluation of Molecular Periodic Orders (Dipole Moment and Bond Enthalpy)

Solve 4 NEET Questions a Day — One per Subject