Q.56 Chemistry ⭐ 360° Option Elimination Active

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NEET 2021 Chemistry Question 56 Solution

Topic: Solid State · Official NTA Booklet Code M1

Question

Right option for the number of tetrahedral and octahedral voids in hexagonal primitive unit cell are:

(1)8, 4
(2)6, 12
(3)2, 1
(4)12, 6

✓

Verified Against NTA Final Answer Key

Option (4) : 12, 6

📚 Solid State Easy

📋 Given Parameters & Problem Setup

Lattice Type Hexagonal primitive unit cell (hcp)

Target Number of tetrahedral and octahedral voids respectively

⚡

The 30-Second Exam Hall Shortcut: ⚡ For N Atoms: Tetrahedral = 2N, Octahedral = N

⏱️ Save ~20 seconds

In any close-packed lattice containing $N$ constituent particles, the number of octahedral voids is $N$ , and tetrahedral voids is $2N$ . For a hexagonal primitive unit cell (hcp), $N = 6$ . Therefore, Tetrahedral voids $= 2 \times 6 = 12$ and Octahedral voids $= 6$ . The answer is 12, 6! Match Option (4)!

1 Effective number of atoms per hcp unit cell:

Z = N = 6

(

12 \times \frac{1}{6} + 2 \times \frac{1}{2} + 3 = 6

2 Number of octahedral voids

= N = 6

3 Number of tetrahedral voids

= 2N = 2 \times 6 = 12

4 Question asks for "tetrahedral and octahedral voids" in order: 12, 6.

5 Match Option (4).

🛡️

Why You Can 100% Trust This Shortcut: Canonical close-packed sphere geometry (

V_{\text{tet}} = 2N, V_{\text{oct}} = N

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Calculating Effective Atoms per Hexagonal Unit Cell (Z)

A hexagonal primitive (hcp) unit cell contains: - 12 corner atoms, each shared among 6 unit cells: $12 \times \frac{1}{6} = 2$ - 2 face-centered atoms, each shared between 2 unit cells: $2 \times \frac{1}{2} = 1$ - 3 completely enclosed internal atoms: $3 \times 1 = 3$ Total effective constituent particles per unit cell:

Z = N = 2 + 1 + 3 = 6

Step 2

Formula for Interstitial Voids in Close-Packing

For any close-packed arrangement of $N$ spheres: - Number of Octahedral Voids ( $O_V$ ) $= N$ - Number of Tetrahedral Voids ( $T_V$ ) $= 2N$

Step 3

Evaluating Void Counts for hcp

Substituting $N = 6$ :

T_V = 2 \times 6 = 12, \quad O_V = 6

⭐

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:

These are the void counts for a cubic close-packed (fcc/ccp) unit cell ( $N = 4 \implies T_V = 8, O_V = 4$ ), not hcp.

⚠️

Student Fallacy (Why candidates pick this trap): Using fcc parameters instead of hexagonal unit cell.

2 2

✗ Incorrect

Verdict & Scientific Reason:

Inverts the requested order (gives octahedral first, then tetrahedral).

⚠️

Student Fallacy (Why candidates pick this trap): Swapping tetrahedral and octahedral void counts.

3 3

✗ Incorrect

Verdict & Scientific Reason:

Per-atom ratio ( $2:1$ ), not the total count per unit cell.

⚠️

Student Fallacy (Why candidates pick this trap): Reporting the ratio per atom rather than per unit cell.

4 4

✓ 100% Correct

Verdict & Scientific Reason:

Exact void counts for hcp unit cell: 12 tetrahedral and 6 octahedral voids.

💡

Why this is airtight: None; rigorous crystallography analysis.

⚠️

NTA Trap Alert: Order of Voids and FCC vs. HCP Trap

NTA traps candidates in two ways: 1. Option (1) gives $8, 4$ , which is for FCC ( $Z=4$ ). HCP has $Z=6$ . 2. Option (2) inverts the order to $6, 12$ . The question explicitly asks for "tetrahedral and octahedral" (12, 6).

🛡️ How to avoid the -1 mark: HCP has

N=6

. Tetrahedral first (

2N = 12

), then octahedral (

N = 6

📖

Core NEET Syllabus Standard

NCERT Chemistry · Solid State Standard Benchmark

Core Scientific Principle Tested:

"In a close-packed arrangement of $N$ spheres, octahedral voids $= N$ and tetrahedral voids $= 2N$ . In an hcp unit cell, $N = 6$ ."

Syllabus Focus: Number of Voids Filled in Close-Packed Structures

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Chemistry Question 56 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ For N Atoms: Tetrahedral = 2N, Octahedral = N

Step-by-Step Rigorous Workflow & Mathematical Proof

Calculating Effective Atoms per Hexagonal Unit Cell (Z)

Formula for Interstitial Voids in Close-Packing

Evaluating Void Counts for hcp

360° Option Elimination Engine Key Feature

NTA Trap Alert: Order of Voids and FCC vs. HCP Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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