Ionic Crystal Structure- Practice Problems and Solutions

What You Actually Need to Know About Ionic Crystal Structure Problems

Ionic crystal structure problems show up on every chemistry exam. They're not hard once you understand the patterns. Most students fail because they try to memorize everything instead of learning the calculation methods.

This guide cuts through the nonsense. You'll get real practice problems, real solutions, and the exact steps to solve any ionic crystal problem they throw at you.

Common Ionic Crystal Lattice Types

Most problems focus on five basic structures. Know these cold:

• NaCl (Rock Salt) – FCC lattice with octahedral coordination
• CsCl (Cesium Chloride) – Simple cubic with cubic coordination
• ZnS (Zinc Blende) – FCC lattice with tetrahedral coordination
• CaF2 (Fluorite) – FCC with calcium at corners and FCC with fluoride in tetrahedral holes
• TiO2 (Rutile) – Tetragonal structure common in ceramics

The NaCl structure is the most common exam question. CsCl appears less often but trips students up because it looks like simple cubic when it's not.

Key Formulas You Must Memorize

No shortcuts here. These formulas are the entire foundation:

• Number of ions per unit cell = ions touching the cell corners + ions on faces + ions on edges + ions inside
• Edge length (a) relates to ionic radii through geometry
• Density = (number of formula units × molar mass) / (NA × a³)
• Radius ratio = r(cation) / r(anion)

The radius ratio tells you which structure is stable. Get this wrong and you'll pick the wrong coordination number every time.

Radius Ratio Rules

If your ratio falls outside 0.732, the structure becomes unstable or changes completely.

How to Solve Ionic Crystal Structure Problems

Step 1: Identify the Structure Type

Look at the ions given. NaCl structure = face-centered cubic with octahedral holes. CsCl structure = simple cubic arrangement. ZnS structure = tetrahedral coordination.

Don't guess. Use the radius ratio first.

Step 2: Count Ions in the Unit Cell

For NaCl:

• Na⁺ ions: 12 on edges (each shared by 4 cells) + 1 in center = 4
• Cl⁻ ions: 8 on corners (each shared by 8 cells) + 6 on faces (each shared by 2 cells) = 4

Total: 4 NaCl units per unit cell.

Step 3: Relate Edge Length to Ionic Radii

In NaCl, the ions touch along the edge. The edge length a = 2(r⁺ + r⁻). In CsCl, the ions touch along the body diagonal. The body diagonal = 2(r⁺ + r⁻) = a√3.

This is where most students get stuck. They don't know which diagonal to use. Read the problem carefully. It usually tells you which ions touch.

Step 4: Calculate Density

Density problems combine everything. Use:

ρ = (Z × M) / (NA × a³)

Where Z = number of formula units per unit cell, M = molar mass, NA = Avogadro's number, a = edge length.

Always check your units. a must be in cm. Convert from pm or Å if needed.

Practice Problems and Solutions

Problem 1: NaCl Structure Calculation

Sodium chloride crystallizes in a face-centered cubic lattice with edge length 564 pm. Calculate the density. (Atomic masses: Na = 23.0 g/mol, Cl = 35.5 g/mol)

Solution:

Step 1: Identify Z for NaCl = 4 formula units per unit cell.

Step 2: Convert edge length. a = 564 pm = 564 × 10⁻¹⁰ cm = 5.64 × 10⁻⁸ cm

Step 3: Calculate a³ = (5.64 × 10⁻⁸)³ = 1.79 × 10⁻²² cm³

Step 4: Apply formula.

ρ = (4 × 58.5) / (6.022 × 10²³ × 1.79 × 10⁻²²)

ρ = 234 / 1.08

ρ = 2.17 g/cm³

Actual NaCl density is 2.16 g/cm³. Close enough.

Problem 2: Finding Ionic Radius

Potassium chloride has a CsCl-type structure with edge length 398 pm. Calculate the ionic radius of K⁺ if Cl⁻ radius is 181 pm.

Solution:

CsCl structure: ions touch along the body diagonal.

Body diagonal = a√3 = 398 × 1.732 = 689 pm

Body diagonal = 2(rK⁺ + rCl⁻)

689 = 2(rK⁺ + 181)

689 = 2rK⁺ + 362

2rK⁺ = 327

rK⁺ = 163.5 pm

Simple. Just remember which diagonal applies to which structure.

Problem 3: Radius Ratio Analysis

Predict the coordination number and structure type for MgO. Ionic radii: Mg²⁺ = 72 pm, O²⁻ = 140 pm.

Solution:

Radius ratio = 72 / 140 = 0.514

From the table: 0.414 < 0.514 < 0.732

This falls in the octahedral range. Coordination number = 6. MgO adopts the NaCl structure.

MgO is a classic example. High charge cations with moderate size ratios always go octahedral.

Problem 4: CaF2 Fluorite Structure

Calculate the number of Ca²⁺ and F⁻ ions in one unit cell of CaF2.

Solution:

CaF2 fluorite structure:

• Ca²⁺ ions: 8 on corners + 6 on faces = 8 + 3 = 4 Ca²⁺ ions
• F⁻ ions: All tetrahedral holes = 8 F⁻ ions

Ratio: 4 Ca²⁺ : 8 F⁻ = 1:2 ✓ Matches CaF2 formula

Problem 5: Density from Crystal Data

Zinc sulfide crystallizes in the zinc blende structure with edge length 540 pm. Calculate its density. (Zn = 65.4 g/mol, S = 32.1 g/mol, ZnS formula units per cell = 4)

Solution:

a = 540 pm = 5.40 × 10⁻⁸ cm

a³ = (5.40 × 10⁻⁸)³ = 1.57 × 10⁻²² cm³

ρ = (4 × 97.5) / (6.022 × 10²³ × 1.57 × 10⁻²²)

ρ = 390 / 0.945

ρ = 4.13 g/cm³

Common Mistakes Students Make

• Wrong diagonal – Using edge length when body diagonal is required. Check the structure.
• Counting ions wrong – Forgetting which ions are shared between cells. Corners = 1/8, edges = 1/4, faces = 1/2.
• Unit conversion errors – Forgetting to convert pm to cm. This destroys your density calculation.
• Ignoring the radius ratio – Picking a structure without checking if it's geometrically possible.
• Using wrong coordination number – NaCl is 6:6, not 8:8. CsCl is 8:8, not 6:6.

Quick Reference Table

Final Tips

Stop wasting time reading about ionic bonding theory. You need calculation practice. Work through at least 20 problems before your exam.

When solving, always:

1. Draw the unit cell if possible
2. Identify which ions touch where
3. Set up the geometric relationship
4. Solve for the unknown
5. Check if your answer makes physical sense

That's it. No magic. Just geometry and arithmetic applied correctly.