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Fluorite Structure (CaF2)

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Fluorite (CaF2)
A cubic crystal structure where each Ca2+ ion is surrounded by eight F- ions in a face-centered cubic arrangement.
Fluorite adopts the CaF₂ structure type with a cubic close-packed (ccp) arrangement of calcium cations. The fluoride anions occupy all tetrahedral holes in this lattice, resulting in 8:4 coordination. The structure belongs to the Fm3̄m space group with a coordination number of 8 for Ca2+ and 4 for F-. The unit cell contains 4 formula units of CaF₂. This structure is commonly found in compounds with large cations and small anions where the radius ratio is greater than 0.732. Many other fluorides and oxides adopt this structure type.

Problem 1

Question: In the fluorite (CaF₂) structure, Ca²⁺ ions form an FCC lattice. How many Ca²⁺ ions are present per unit cell?

[Hint: Count corner and face-centered atoms with their contributions]

Solution

Calculation:

Corner atoms: 8 × (1/8) = 1

Face-centered atoms: 6 × (1/2) = 3

Total Ca²⁺ ions = 1 + 3 = 4 Ca²⁺ ions per unit cell

Problem 2

Question: If the edge length of a CaF₂ unit cell is 5.46 Å, calculate the volume of the unit cell in cm³.

[Given: 1 Å = 10⁻⁸ cm]

Solution

Calculation:

a = 5.46 Å = 5.46 × 10⁻⁸ cm

Volume = a³ = (5.46 × 10⁻⁸)³

V = 162.99 × 10⁻²⁴ cm³

V = 1.63 × 10⁻²² cm³

Problem 3

Question: What is the coordination number of Ca²⁺ and F⁻ ions in the fluorite structure?

Solution

Answer:

Coordination number of Ca²⁺ = 8 (surrounded by 8 F⁻ ions in cubic arrangement)

Coordination number of F⁻ = 4 (surrounded by 4 Ca²⁺ ions in tetrahedral arrangement)

The coordination ratio is 8:4

Challenging Numerical Problem

Question: A fluorite crystal (CaF₂) has a cubic unit cell with edge length a = 5.46 Å. Given that the density of fluorite is 3.18 g/cm³:

(a) Calculate the number of formula units of CaF₂ per unit cell using the density data.

(b) Determine the theoretical packing efficiency of the fluorite structure, assuming that the anions are in contact along the body diagonal and cations are in contact with anions along the face diagonal.

(c) If 12.5% of the tetrahedral holes in the Ca²⁺ FCC lattice remain vacant due to a defect (Frenkel defect), calculate the new density of the defective crystal.

[Given: Atomic mass of Ca = 40 u, F = 19 u, N_A = 6.022 × 10²³ mol⁻¹]

Solution

(a) Number of formula units per unit cell:

Using the density formula: ρ = (Z × M)/(a³ × N_A)

Where: Z = number of formula units, M = molar mass of CaF₂ = 40 + 2(19) = 78 g/mol

a = 5.46 Å = 5.46 × 10⁻⁸ cm, ρ = 3.18 g/cm³

Rearranging: Z = (ρ × a³ × N_A)/M

Z = (3.18 × (5.46 × 10⁻⁸)³ × 6.022 × 10²³)/78

Z = (3.18 × 162.99 × 10⁻²⁴ × 6.022 × 10²³)/78

Z = 311.71/78 = 4 formula units

(b) Packing efficiency:

In fluorite: 4 Ca²⁺ ions (FCC) and 8 F⁻ ions (all tetrahedral holes)

For FCC Ca²⁺: body diagonal = a√3, contains 2r_Ca + 2r_F

Face diagonal relation: 4r_F = a√2 (F⁻ ions touch along face diagonal)

From radius ratio limit: r_Ca/r_F ≈ 0.732

Volume occupied = 4 × (4/3)πr_Ca³ + 8 × (4/3)πr_F³

Substituting r_Ca = 0.732 r_F and solving:

Packing efficiency = (Volume occupied/a³) × 100 = ≈ 73.0%

(c) Density with Frenkel defect:

Frenkel defect: vacancy created but no atoms leave the crystal

12.5% tetrahedral holes vacant = 12.5% of F⁻ ions displaced to interstitial positions

Total mass remains same, volume remains same

Therefore, density remains: ρ = 3.18 g/cm³ (unchanged)

Note: Frenkel defects don't change density as mass and volume are conserved.

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