Tetrahedral Voids: Geometry, Location, and Applications

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:

Identify tetrahedral voids in close-packed structures.

State the number per atom and per unit cell.

Use radius ratios to predict tetrahedral void occupancy.

Apply to real crystal structures.

Properties of Tetrahedral Voids

Property	Value
Coordination number	4
Number per atom	2 (twice the number of atoms)
Radius ratio ( $r_{void}/r_{atom}$ )	0.225
Shape	Tetrahedron

Radius Ratio

Tetrahedral voids are smaller than octahedral voids. The maximum sphere that fits has radius $0.225R$ . If the radius ratio $r^+/r^-$ 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 = $8/4 = 2$ 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)

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

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

Fluorite ( $CaF_2$ )

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

4 $Ca^{2+}$ per cell, 8 $F^-$ per cell → ratio 1:2 → formula $CaF_2$ ✓

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. $8 \times \frac{1}{4} = 2$ Y atoms per cell. Formula: $X_2Y$ or equivalently $X_4Y_2$ .

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

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).
Mixing up radius ratios — Tetrahedral: 0.225. Octahedral: 0.414. Tetrahedral voids are smaller.
Forgetting partial occupancy — In real crystals, not all voids are filled. The fraction filled determines the formula. $ZnS$ fills half, $CaF_2$ 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 ( $r = 0.414R$ vs $r = 0.225R$ ). This is why larger ions tend to occupy octahedral voids and smaller ions occupy tetrahedral voids.

Octahedral Voids — The larger holes in close-packed structures.
Cubic Close Packing — The FCC structure containing both void types.
Hexagonal Close Packing — HCP also contains tetrahedral voids.

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