Chemistry Physical - How to Solve: Solid State (Unit Cell, Packing Efficiency, Voids, Radius Ratio, Defects) – NEET UG Guide

How to Solve: Solid State (Unit Cell, Packing Efficiency, Voids, Radius Ratio, Defects) – NEET UG Guide

Introduction

Mastering Solid State unlocks 5-7 direct questions in NEET Chemistry—worth 20+ marks—and helps you predict crystal structures, drug stability, and even semiconductor behavior in physics. If you can solve unit cell problems in under 2 minutes, you’ll save time for tougher organic and biology questions.

WHAT YOU NEED TO KNOW FIRST

1. Basic 3D geometry – Volume of a cube, sphere, and cylinder.
2. Density formula – Density = Mass / Volume.
3. Avogadro’s number – 6.022 × 10²³ atoms/mol.

KEY TERMS & FORMULAS

1. Unit Cell Types & Atoms per Unit Cell

MEMORISE THIS: SC = 1 atom, BCC = 2 atoms, FCC = 4 atoms.

2. Edge Length (a) & Atomic Radius (r) Relationship

3. Packing Efficiency (PE)

Formula: Packing Efficiency (%) = (Volume occupied by atoms / Volume of unit cell) × 100

MEMORISE THIS: - SC = 52.4% - BCC = 68% - FCC/CCP = 74%

4. Density of Unit Cell

Formula: Density (ρ) = (Z × M) / (a³ × Nₐ) - Z = Number of atoms per unit cell - M = Molar mass (g/mol) - a = Edge length (cm) - Nₐ = Avogadro’s number (6.022 × 10²³ mol⁻¹)

MEMORISE THIS FORMULA.

5. Voids (Interstitial Sites)

MEMORISE THIS: - Tetrahedral voids = 2 × atoms in FCC - Octahedral voids = Equal to atoms in FCC

6. Radius Ratio (r₊ / r₋)

Formula: Radius Ratio = r_cation / r_anion

MEMORISE THIS: | Radius Ratio Range | Coordination Number | Shape | |----------------------|------------------------|----------| | 0.155 – 0.225 | 3 | Triangular planar | | 0.225 – 0.414 | 4 | Tetrahedral | | 0.414 – 0.732 | 6 | Octahedral | | 0.732 – 1.0 | 8 | Cubic |

7. Crystal Defects

MEMORISE THIS: - Schottky = Vacancy pair (ionic solids) - Frenkel = Displaced ion (AgCl, ZnS)

STEP-BY-STEP METHOD

Step 1: Identify the Unit Cell Type

• SC (Simple Cubic) → 1 atom, a = 2r
• BCC (Body-Centered Cubic) → 2 atoms, a = (4r)/√3
• FCC (Face-Centered Cubic) → 4 atoms, a = 2√2 r

Step 2: Calculate Edge Length (a) if Atomic Radius (r) is Given

• Use the correct formula from Step 1.

Step 3: Find Number of Atoms per Unit Cell (Z)

• SC = 1, BCC = 2, FCC = 4

Step 4: Calculate Volume of Unit Cell (a³)

• Convert a to cm if needed (1 Å = 10⁻⁸ cm).

Step 5: Calculate Density (ρ)

• Use ρ = (Z × M) / (a³ × Nₐ)
• Units: g/cm³

Step 6: Find Packing Efficiency (if asked)

• Use PE = (Volume of atoms / Volume of unit cell) × 100
• Volume of atoms = Z × (4/3)πr³

Step 7: Determine Voids (if asked)

• FCC has 8 tetrahedral voids & 4 octahedral voids.
• BCC has no regular voids.

Step 8: Check for Defects (if asked)

• Schottky → Missing ion pair → Lower density
• Frenkel → Displaced ion → Same density

WORKED EXAMPLES

Example 1 – Basic (FCC Unit Cell)

Question: Copper crystallizes in an FCC structure. If the atomic radius of Cu is 1.28 Å, find:
1. Edge length (a)
2. Density (ρ) of Cu (M = 63.5 g/mol)

Step-by-Step Solution:
1. Identify unit cell: FCC → a = 2√2 r
2. Calculate a: a = 2√2 × 1.28 Å = 3.62 Å Convert to cm: 3.62 × 10⁻⁸ cm
3. Atoms per unit cell (Z): FCC → Z = 4
4. Volume of unit cell (a³): (3.62 × 10⁻⁸ cm)³ = 4.74 × 10⁻²³ cm³
5. Density (ρ): ρ = (4 × 63.5) / (4.74 × 10⁻²³ × 6.022 × 10²³) ρ = 8.92 g/cm³

What we did and why: - Used a = 2√2 r for FCC. - Converted Å to cm for correct units. - Applied density formula with Z = 4 for FCC.

Example 2 – Medium (Packing Efficiency & Voids)

Question: A metal crystallizes in FCC. If its atomic radius is 1.44 Å, find:
1. Packing efficiency
2. Number of tetrahedral voids per unit cell

Step-by-Step Solution:
1. Identify unit cell: FCC → Z = 4
2. Volume of atoms: Volume = Z × (4/3)πr³ = 4 × (4/3)π(1.44 × 10⁻⁸ cm)³ Volume = 5.08 × 10⁻²³ cm³
3. Edge length (a): a = 2√2 × 1.44 Å = 4.07 Å = 4.07 × 10⁻⁸ cm
4. Volume of unit cell (a³): (4.07 × 10⁻⁸ cm)³ = 6.74 × 10⁻²³ cm³
5. Packing efficiency: PE = (5.08 × 10⁻²³ / 6.74 × 10⁻²³) × 100 = 75.4% (Close to 74% due to rounding)
6. Tetrahedral voids in FCC: Number of voids = 2 × Z = 2 × 4 = 8

What we did and why: - Calculated volume of atoms and unit cell separately. - Used 2 × Z for tetrahedral voids in FCC.

Example 3 – Exam-Style (Radius Ratio & Defects)

Question: An ionic solid has a radius ratio of 0.52. It shows a defect where equal numbers of cations and anions are missing. What is:
1. The coordination number of the cation?
2. The type of defect?

Step-by-Step Solution:
1. Radius ratio = 0.52 - Range 0.414 – 0.732 → Octahedral coordination (CN = 6)
2. Defect description: - Equal missing cations & anions → Schottky defect

What we did and why: - Used radius ratio table to find coordination number. - Recognized Schottky defect from missing ion pairs.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 20-mark cheat sheet for Solid State in NEET.

1. Unit cells:
2. SC = 1 atom, BCC = 2, FCC = 4.

3. Edge length: SC = 2r, BCC = (4r)/√3, FCC = 2√2 r.

4. Density formula: ρ = (Z × M) / (a³ × Nₐ)

5. Convert a to cm (1 Å = 10⁻⁸ cm).

6. Packing efficiency:

7. SC = 52%, BCC = 68%, FCC = 74%.

8. Voids:

9. FCC has 8 tetrahedral & 4 octahedral voids.

10. Radius ratio:

11. 0.225 → tetrahedral, 0.414 → octahedral.

12. Defects:

13. Schottky = missing pair → lower density.
14. Frenkel = displaced ion → same density.

Now go solve 3 past papers—you’ve got this!