Tetrahedral and Octahedral Voids in Crystals Explained

When identical spheres pack as tightly as possible (in FCC or HCP arrangements), small empty spaces remain between them — these are the voids (interstitial sites). The two important kinds are tetrahedral voids, the gap between four touching spheres, and octahedral voids, the gap between six touching spheres. Smaller atoms or ions can fit into these voids, which is how countless compounds get their structures.

Voids matter because they decide where the smaller ion sits in an ionic crystal: in NaCl the Cl⁻ ions form an FCC array with Na⁺ in all octahedral voids; in ZnS the S²⁻ ions form an FCC array with Zn²⁺ in half the tetrahedral voids. The whole family of ionic structures is just "packing + which voids are filled".

Key Concepts

• Tetrahedral void: surrounded by 4 spheres at the corners of a tetrahedron. Coordination number 4. A sphere of radius up to 0.225R (R = packing-sphere radius) fits without distortion.
• Octahedral void: surrounded by 6 spheres at the corners of an octahedron. Coordination number 6. Fits a sphere of radius up to 0.414R.
• Void counts in close packing: if there are N spheres in a close-packed array, there are 2N tetrahedral voids and N octahedral voids. (Per FCC unit cell: 8 tetrahedral, 4 octahedral.)
• Radius ratio rule: the ratio r⁺/r⁻ of cation to anion radius predicts the void occupied — below 0.225 fits tetrahedral, 0.225–0.414 fits octahedral, above 0.414 needs larger sites (e.g. cubic, as in CsCl).

Formulas

• Tetrahedral void radius: r_void = 0.225 R. Derivation: the void centre to a sphere centre is R + r_void, and geometry of a tetrahedron gives (R + r_void)/R = 1.225.
• Octahedral void radius: r_void = 0.414 R. From the geometry of an octahedron: (R + r_void)/R = √2 ≈ 1.414.
• Void counts: tetrahedral voids = 2N, octahedral voids = N, where N = number of close-packed spheres. Assumes ideal close packing of equal spheres.

Worked Example

In an FCC crystal of oxide ions O²⁻ (radius 140 pm), will Mg²⁺ (radius 72 pm) fit comfortably in an octahedral void?

1. Radius ratio = r⁺/r⁻ = 72/140 = 0.514.
2. Octahedral voids accept ratios from 0.414 upward, so 0.514 fits — the Mg²⁺ ion sits in the octahedral void without pushing the oxide ions apart.
3. Check tetrahedral: the limit is 0.225, so Mg²⁺ is far too big for a tetrahedral void (72 pm > 0.225 × 140 = 31.5 pm).

Common Misconceptions

• "Voids are defects." No — tetrahedral and octahedral voids are a natural geometric feature of every close-packed arrangement, present even in a perfect crystal. Filling them is how ionic compounds form, not a flaw.
• "There are equal numbers of both void types." Wrong — there are twice as many tetrahedral voids (2N) as octahedral voids (N). Forgetting the factor of 2 is the most common exam error.
• "The bigger ion always goes into the void." Opposite — the larger ions form the close-packed array, and the smaller ions occupy the voids between them.

Real-World Applications

• Ionic solids: NaCl (all octahedral voids filled), ZnS (half the tetrahedral voids filled), fluorite CaF₂ (all tetrahedral voids filled) — all explained by void occupancy.
• Alloys: carbon atoms sit in the octahedral voids of FCC iron (austenite), which is the basis of steel.
• Battery materials: lithium ions move through octahedral and tetrahedral voids in cathode materials like LiCoO₂; void pathways decide charging speed.

In short, close packing leaves tetrahedral voids (2N, fit 0.225R) and octahedral voids (N, fit 0.414R).