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Crystal Structure

Tetrahedral Voids: Geometry, Location, and Applications

Understand tetrahedral voids in close-packed crystal structures. Learn their geometry, count per atom, radius ratio, and role in structures like zinc blende and fluorite.

V
Vectora Team
STEM Education
7 min read
2025-10-11

What Are Tetrahedral Voids?

A tetrahedral void is an empty space in a close-packed structure surrounded by 4 atoms arranged at the corners of a tetrahedron — typically 3 from one layer and 1 from the adjacent layer.

In any close-packed structure, there are exactly 2 tetrahedral voids per atom — making them twice as numerous as octahedral voids.

Learning Goals: By the end of this guide, you should be able to:

  1. Identify tetrahedral voids in close-packed structures.
  2. State the number per atom and per unit cell.
  3. Use radius ratios to predict tetrahedral void occupancy.
  4. Apply to real crystal structures.

Properties of Tetrahedral Voids

PropertyValue
Coordination number4
Number per atom2 (twice the number of atoms)
Radius ratio (rvoid/ratomr_{void}/r_{atom})0.225
ShapeTetrahedron

Radius Ratio

Tetrahedral voids are smaller than octahedral voids. The maximum sphere that fits has radius 0.225R0.225R. If the radius ratio r+/r−r^+/r^- falls between 0.225 and 0.414, the smaller ion typically occupies tetrahedral voids.

Location in the FCC Unit Cell

In an FCC unit cell, tetrahedral voids are located at positions one-quarter of the way along the body diagonals:

  • 8 tetrahedral voids per unit cell (all entirely inside the cell)
  • FCC has 4 atoms per cell → ratio = 8/4=28/4 = 2 tetrahedral voids per atom ✓

Tetrahedral Void Explorer

See tetrahedral voids in 3D within CCP structures. Visualise the 4-atom coordination and compare with octahedral voids side by side.
Explore Tetrahedral Voids

Crystal Structures Using Tetrahedral Voids

Zinc Blende (ZnS)

S2−S^{2-} ions in CCP. Zn2+Zn^{2+} occupies ½ of the tetrahedral voids.

  • 4 S2−S^{2-} per cell, 4 Zn2+Zn^{2+} per cell (half of 8) → ratio 1:1 → formula ZnSZnS ✓
  • Each Zn2+Zn^{2+} has CN = 4, each S2−S^{2-} has CN = 4.

Fluorite (CaF2CaF_2)

Ca2+Ca^{2+} ions in CCP. F−F^- ions occupy all 8 tetrahedral voids.

  • 4 Ca2+Ca^{2+} per cell, 8 F−F^- per cell → ratio 1:2 → formula CaF2CaF_2 ✓

Diamond

Carbon atoms in CCP, with additional carbons filling half the tetrahedral voids. This creates the tetrahedral bonding network responsible for diamond's hardness.

Worked Examples

Example 1: Formula Prediction

Question: In CCP of X, atoms of Y fill ¼ of the tetrahedral voids. What is the formula?

Solution: 4 X atoms per cell. 8×14=28 \times \frac{1}{4} = 2 Y atoms per cell. Formula: X2YX_2Y or equivalently X4Y2X_4Y_2.

Example 2: Void Comparison

Question: A CCP structure has how many total voids per unit cell?

  • Octahedral: 4
  • Tetrahedral: 8
  • Total: 12 voids per unit cell (with 4 atoms)

Common Mistakes

  1. Confusing the count — 2 per atom, not 1 — There are twice as many tetrahedral voids as atoms. This is different from octahedral voids (1:1).

  2. Mixing up radius ratios — Tetrahedral: 0.225. Octahedral: 0.414. Tetrahedral voids are smaller.

  3. Forgetting partial occupancy — In real crystals, not all voids are filled. The fraction filled determines the formula. ZnSZnS fills half, CaF2CaF_2 fills all.

Frequently Asked Questions

Why are there twice as many tetrahedral voids as octahedral voids?

Between two close-packed layers, each atom in the top layer sits over a depression formed by 3 atoms in the bottom layer, creating a tetrahedral void pointing "up" AND another pointing "down". This geometry naturally produces 2 tetrahedral voids for every 1 atom.

Which is larger, a tetrahedral or octahedral void?

Octahedral voids are larger (r=0.414Rr = 0.414R vs r=0.225Rr = 0.225R). This is why larger ions tend to occupy octahedral voids and smaller ions occupy tetrahedral voids.