Q.75 Chemistry ⭐ 360° Option Elimination Active

NEET 2026 Chemistry Question 75 Solution

Topic: Chemical Bonding and Molecular Structure & Coordination Compounds · Official NTA Booklet Code 50

Question

Match the species in List I with their geometry in List II. List I: A. PCl₅; B. BrF₅; C. BF₄⁻; D. [Ni(CN)₄]²⁻. List II: I. Tetrahedral; II. Square Planar; III. Trigonal bipyramidal; IV. Square pyramidal. Choose the correct answer from the options given below:

✓

Verified Against NTA Final Answer Key

Option (2) : A-III, B-IV, C-I, D-II

📚 Chemical Bonding and Molecular Structure & Coordination Compounds Easy

📋 Given Parameters & Problem Setup

Species A PCl₅

Species B BrF₅

Species C BF₄⁻

Species D [Ni(CN)₄]²⁻

Geometries I. Tetrahedral, II. Square Planar, III. Trigonal bipyramidal, IV. Square pyramidal

⚡

The 30-Second Exam Hall Shortcut: ⚡ Steric Number & Electron Pair Fingerprint

⏱️ Save ~50 seconds

Match PCl₅ (5 bp = TBP) and [Ni(CN)₄]²⁻ (d⁸ strong field = Square Planar) to finish in 10 seconds!

1 A. PCl₅: 5 bond pairs, 0 lone pairs -> Steric Number 5 -> Trigonal bipyramidal (III) -> A-III.

2 This immediately eliminates Option (1)!

3 D. [Ni(CN)₄]²⁻: Ni²⁺ (3d⁸) with strong field CN⁻ forces pairing -> dsp² hybridization -> Square Planar (II) -> D-II.

4 This eliminates Option (3) and Option (4)!

5 Confirm B and C: BrF₅ (5 bp + 1 lp = Square pyramidal, IV); BF₄⁻ (4 bp + 0 lp = Tetrahedral, I).

6 Option (2) is rock-solid: A-III, B-IV, C-I, D-II!

🛡️

Why You Can 100% Trust This Shortcut: Direct application of VSEPR theory for main group species and Crystal Field / Valence Bond Theory for coordination complexes.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Determine Geometry of PCl₅ (Species A)

Phosphorus has 5 valence electrons, all forming single σ bonds with five chlorine atoms. Steric number = 5 (5 bond pairs + 0 lone pairs). Hybridization = sp³d.

\text{Steric Number} = 5 + 0 = 5 \implies \text{Trigonal bipyramidal (III)}

• Match: A -> III.

Step 2

Determine Geometry of BrF₅ (Species B)

Bromine has 7 valence electrons: 5 form single bonds with fluorine, leaving 1 lone pair. Steric number = 5 bond pairs + 1 lone pair = 6. Hybridization = sp³d² with octahedral electron geometry and square pyramidal molecular shape.

\text{Steric Number} = 5 + 1 = 6 \implies \text{Square pyramidal (IV)}

• Match: B -> IV.

Step 3

Determine Geometry of BF₄⁻ (Species C)

Boron has 3 valence electrons + 1 negative charge = 4 electrons, forming four σ bonds with four fluorine atoms. Steric number = 4 bond pairs + 0 lone pairs = 4. Hybridization = sp³.

\text{Steric Number} = 4 + 0 = 4 \implies \text{Tetrahedral (I)}

• Match: C -> I.

Step 4

Determine Geometry of [Ni(CN)₄]²⁻ (Species D)

Nickel is in the +2 oxidation state (3d⁸). Cyanide (CN⁻) is a strong field ligand that forces the 3d electrons to pair up, leaving one vacant 3d orbital. Hybridization is dsp² with a square planar geometry.

\text{Ni}^{2+} (3d^8) + 4\text{CN}^- \xrightarrow{\text{pairing}} dsp^2 \implies \text{Square Planar (II)}

• Match: D -> II.
• Full Match: A-III, B-IV, C-I, D-II (Option 2).

⭐

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

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Swaps geometries of PCl₅ and BrF₅.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing 5 bond pairs with 5 bond pairs + 1 lone pair.

2 B

✓ 100% Correct

Verdict & Scientific Reason:

Exact match: PCl₅ is TBP (III), BrF₅ is square pyramidal (IV), BF₄⁻ is tetrahedral (I), and [Ni(CN)₄]²⁻ is square planar (II).

💡

Why this is airtight: None; flawless VSEPR and CFT alignment.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Assigns tetrahedral to BrF₅ and square pyramidal to [Ni(CN)₄]²⁻.

⚠️

Student Fallacy (Why candidates pick this trap): Miscounting electron pairs.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Assigns square planar to BrF₅.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing square planar (AX₄E₂) with square pyramidal (AX₅E).

⚠️

NTA Trap Alert: Confusing BrF₅ (Square Pyramidal) with PCl₅ (TBP)

Both PCl₅ and BrF₅ have five peripheral halogen atoms, but PCl₅ has NO lone pairs (Trigonal bipyramidal), while BrF₅ has ONE lone pair on bromine (Square pyramidal).

🛡️ How to avoid the -1 mark: Count total valence electrons: PCl₅ has 5 pairs (TBP); BrF₅ has 6 pairs (Square pyramidal).

📖

Core NEET Syllabus Standard

NEET Chemistry Core Syllabus · Chemical Bonding and Coordination Compounds Standard Benchmark

Core Scientific Principle Tested:

"Shape is dictated by steric number and lone pairs: 5+0 = TBP, 5+1 = square pyramidal, 4+0 = tetrahedral, dsp² = square planar."

Syllabus Focus: VSEPR Theory shapes and dsp² square planar coordination geometries

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

PCl₅ = TBP; BrF₅ = Square Pyramid; BF₄⁻ = Tetrahedral; [Ni(CN)₄]²⁻ = Square Planar!

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NEET 2026 Chemistry Question 75 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Steric Number & Electron Pair Fingerprint

Step-by-Step Rigorous Workflow & Mathematical Proof

Determine Geometry of PCl₅ (Species A)

Determine Geometry of BrF₅ (Species B)

Determine Geometry of BF₄⁻ (Species C)

Determine Geometry of [Ni(CN)₄]²⁻ (Species D)

360° Option Elimination Engine Key Feature

NTA Trap Alert: Confusing BrF₅ (Square Pyramidal) with PCl₅ (TBP)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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