Coordination number, voids (tetrahedral, octahedral): Worked examples

The scenario matrix

Every void/coordination question is one of these cells. The worked examples below each carry a tag like (Cell C2) so you can see the whole board getting covered.

#	Cell class	What makes it tricky	Example
C1	Count atoms + voids in a given cell	corner/face/edge sharing fractions	Ex 1
C2	All voids of one type filled → formula	ratio = filled sites : packed atoms	Ex 2, Ex 3
C3	Fraction of voids filled → formula	multiply by the fraction, then reduce	Ex 4 (½), Ex 5 (⅓)
C4	Both void types partly filled	add two cation contributions	Ex 6
C5	Radius-ratio classify (normal case)	read the range table	Ex 7
C6	Radius-ratio boundary / degenerate	exactly on a limit ⇒ which side?	Ex 8
C7	Derive a limiting ratio from geometry	the square/cube touching condition	Ex 9
C8	Real-world: density from voids filled	connect $Z$ , mass, volume	Ex 10
C9	Exam twist: missing/removed atoms	subtract shared contributions	Ex 11
C0	Degenerate: zero voids filled / $r / R \to 0$	limiting behaviour, empty holes	Ex 12

We link the machinery to Close packing in solids (hcp, ccp, bcc), Radius ratio rule and ionic structures, NaCl, ZnS, CaF2 crystal structures, Density of a unit cell and Packing efficiency and unit cell dimensions as each cell needs it.

Cell C1 — Count atoms and voids in a unit cell

Worked example

(Cell C1) In an fcc (ccp) unit cell, count the atoms, the octahedral voids, and the tetrahedral voids. Forecast: guess the three numbers now. (Hint: one of them is not equal to the other two.)

Step 1 — Count packed atoms $N$ . Corners are shared by 8 cells, faces by 2. $N = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4$ Why this step? An atom sitting on a shared boundary only "belongs" fractionally to this cell — you must weight each by how many cells split it.

Step 2 — Octahedral voids. One sits fully inside at the body centre; twelve sit at edge centres, each edge shared by 4 cells. $Oct = 1 + 12 \times \frac{1}{4} = 1 + 3 = 4$ Why this step? Only the body centre and edge centres are equidistant from exactly 6 atoms — the geometric requirement for an octahedral hole.

Step 3 — Tetrahedral voids. Slice the cube into 8 mini-cubes; each mini-cube centre is fully inside and is hugged by 4 atoms. $Tet = 8$ Why this step? Each mini-cube centre lies at equal distance from 4 fcc atoms forming a tetrahedron — see the figure.

In the figure: orange spheres are the 8 corner + 6 face-centre fcc atoms. The green marker at the very middle is the single octahedral void (body centre). The blue triangle marker in the lower-front mini-cube is one of the 8 tetrahedral voids. Notice the green void sits alone in the centre while the blue voids fill each of the 8 sub-cubes — that is exactly why oct = 4 but tet = 8.

Verify: Oct $= N = 4$ ✓, Tet $= 2 N = 8$ ✓. The rule "1 oct + 2 tet per atom" holds.

Cell C2 — All voids of one type filled

Worked example

(Cell C2 · Ex 3 — the classic CaF₂ case) In fluorite, the cations Ca²⁺ form a ccp array and the anions F⁻ occupy all tetrahedral voids. Find the formula. Forecast: here the cation packs (unusual!). Guess whether F beats Ca 2:1 or 1:2.

Step 1 — Count the packing species. Ca²⁺ forms the ccp skeleton, so Ca $= N$ . Why this step? Whatever forms the close-packed array sets $N$ — it need not be the anion.

Step 2 — Count the anions in tetrahedral voids. Tetrahedral voids $= 2 N$ ; all filled, so F $= 2 N$ . Why this step? "All tetrahedral" = 100 % of the $2 N$ tet holes occupied.

Step 3 — Reduce the ratio. $Ca : F = N : 2 N = 1 : 2 \Rightarrow CaF_{2}$ Why this step? Formula is the whole-number ratio of occupied sites; divide both by $N$ .

Verify (charge balance): Ca²⁺ contributes $+ 2$ ; two F⁻ contribute $2 \times (- 1) = - 2$ ; net $0$ — electrically neutral, exactly as a real solid must be. So $1 : 2$ is both geometrically and chemically forced. ✓ (See NaCl, ZnS, CaF2 crystal structures.)

Cell C3 — A fraction of voids filled

Cell C4 — Both void types partly filled

Worked example

(Cell C4 · Ex 6) In a ccp array of anions X, cations A fill all tetrahedral voids and cations B fill all octahedral voids. Find the formula. Forecast: you now have three species. Expect a formula like $A_{p} B_{q} X_{s}$ .

Step 1 — Anions. $X = N$ . Why this step? ccp species sets the count.

Step 2 — A in tetrahedral. $A = 2 N$ (all $2 N$ tet voids). Step 3 — B in octahedral. $B = N$ (all $N$ oct voids). Why these steps? Each void type has its own capacity: $2 N$ tet and $N$ oct, filled independently.

Step 4 — Ratio. $A : B : X = 2 N : N : N = 2 : 1 : 1 \Rightarrow A_{2} BX$ Why this step? Combine all three occupied-site counts into one reduced ratio.

Verify (site arithmetic tied to the ratio): Divide every count by $N$ : $A = 2$ , $B = 1$ , $X = 1$ , giving exactly $2 : 1 : 1$ . To see this is a chemically real pattern, impose neutrality with $X = O^{2 -}$ : the two cation charges must satisfy $2 q_{A} + q_{B} = + 2$ . Choosing $q_{A} = 0$ is impossible for a cation, but $q_{A} = \frac{1}{2}$ is nonphysical too — so a strict $2 : 1 : 1$ of mono-and-mono cations cannot balance a $- 2$ anion. The honest reading: the site ratio $2 : 1 : 1$ is fixed by geometry alone (that is what we verify here), and nature realises it only for charge-compatible ions (e.g. mixed-valence or a $2 -$ anion split into two $1 -$ anions). The geometric result $A_{2} B X$ stands. ✓

Cell C5 — Radius-ratio, a normal case

Cell C6 — The boundary / degenerate case

Worked example

(Cell C6 · Ex 8) A cation gives exactly $r / R = 0.414$ . Which void does it occupy — tetrahedral or octahedral? Forecast: it sits on the fence. Does it fall left or right?

Step 1 — Locate the value. $0.414 = 2 - 1$ is precisely the lower limit of the octahedral range. Why this step? The number $0.414$ is not random — it is the exact snug-fit for an octahedral hole.

Step 2 — Interpret the boundary physically. At exactly the lower limit the cation just touches all 6 octahedral neighbours with zero slack. Below it, the cation would rattle in the octahedron and prefer the smaller tetrahedral hole; above it, it comfortably sits octahedral. Why this step? The rule is "occupy the hole whose lower bound you have just reached." At the limit, octahedral CN 6 becomes possible for the first time.

Step 3 — Conclusion. By convention we assign the higher coordination (octahedral, CN 6) once $r / R$ reaches $0.414$ .

Verify: $2 - 1 = 0.41421 \dots$ , the octahedral limiting ratio. ✓

Cell C7 — Derive a limiting ratio from geometry

Worked example

(Cell C7 · Ex 9) From first principles, find the limiting $r / R$ for a tetrahedral void. Forecast: you should land on $\approx 0.225$ . Where does that come from geometrically?

Step 1 — Set the picture. Put 4 big spheres at alternate corners of a cube (that quartet forms a tetrahedron); the small sphere sits at the cube's centre. Let the cube edge be $a$ . Why this step? A cube is the cleanest frame in which a tetrahedron's touching distances become plain diagonals.

In the figure: four orange spheres sit on alternate cube corners; the blue sphere is the small guest at the cube centre. The solid green line is a face diagonal where two big spheres touch ( $= 2 R = a 2$ ). The dashed red line is the body diagonal that runs big–small–big ( $= 2 R + 2 r = a 3$ ). The whole derivation is just comparing those two coloured lines.

Step 2 — Big spheres touch across a face diagonal. Two big spheres at opposite corners of a face are in contact, so the face diagonal equals $2 R$ : $a 2 = 2 R \Rightarrow R = \frac{a 2}{2}$ Why this step? "Touching" = centre-to-centre distance is exactly $2 R$ ; on a face that distance is the face diagonal $a 2$ .

Step 3 — Small + big touch along the body diagonal. The body diagonal runs corner-through-centre-to-corner, carrying $R + 2 r + R$ : $a 3 = 2 R + 2 r$ Why this step? The centre sphere lies exactly on the body diagonal, so that diagonal is the sum $R + 2 r + R$ .

Step 4 — Divide to kill $a$ . Dividing Step 3 by $2 R = a 2$ : $\frac{a 3}{a 2} = 1 + \frac{r}{R} \Rightarrow \frac{3}{2} = 1 + \frac{r}{R}$ $\frac{r}{R} = \frac{3}{2} - 1 \approx 0.225$ Why this step? We only want the ratio, so eliminating the arbitrary edge length $a$ leaves a pure number.

Verify: $1.5 - 1 = 0.22474 \dots \approx 0.225$ ✓ — matches the topic note's tetrahedral limit.

Cell C8 — Real-world density problem

Worked example

(Cell C8 · Ex 10) An oxide crystallises with O²⁻ in ccp (edge $a = 4.20 \times 1 0^{- 8}$ cm) and metal M in all octahedral voids. Molar masses: M $= 24.3$ g mol⁻¹, O $= 16.0$ g mol⁻¹. Find the density. Forecast: MgO-like solid; guess whether $ρ$ is nearer 1 or nearer 4 g cm⁻³.

Step 1 — Find $Z$ , the formula units per cell. O²⁻ in ccp gives 4 anions per cell; octahedral voids $= 4$ , all filled ⇒ 4 M cations. So the cell contains 4 formula units of MO ⇒ $Z = 4$ . Why this step? Density counts mass inside one cell, so we need how many MO units live there. (Formula MO comes from Cell C2 logic: 1:1.)

Step 2 — Mass in one cell. $M_{formula} = 24.3 + 16.0 = 40.3 g mol^{- 1}$ $m = \frac{Z \cdot M _{formula}}{N _{A}} = \frac{4 \times 40.3}{6.022 \times 1 0 ^{23}} g$ Why this step? Avogadro's number $N_{A}$ converts "per mole" mass into "per formula unit" mass — see Density of a unit cell.

Step 3 — Volume of the cell. $V = a^{3} = (4.20 \times 1 0^{- 8})^{3} = 7.409 \times 1 0^{- 23} cm^{3}$ Why this step? A cubic cell's volume is edge cubed; the density is mass over this volume.

Step 4 — Density. $ρ = \frac{m}{V} = \frac{4 \times 40.3}{6.022 \times 1 0 ^{23} \times 7.409 \times 1 0 ^{- 23}} \approx 3.61 g cm^{- 3}$ Why this step? $ρ = \frac{Z M}{N _{A} a ^{3}}$ — the master unit-cell density formula.

Verify (units): $\frac{g mol ^{- 1}}{( mol ^{- 1} ) ( cm ^{3} )} = g cm^{- 3}$ ✓. Numeric $\approx 3.61$ g cm⁻³ (real MgO $\approx 3.58$ ). ✓

Cell C9 — Exam twist: an atom removed

Worked example

(Cell C9 · Ex 11) Start from an fcc oxide $M$ (M at corners + face centres, O in all octahedral voids, formula MO). Now remove the two M atoms on one pair of opposite faces. What is the new formula? Forecast: the formula will drift away from 1:1 — toward more or less M?

Step 1 — Original M count. $M = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$ . Why this step? Standard fcc corner + face sharing.

Step 2 — Remove two face atoms. Each face atom contributes $\frac{1}{2}$ . Removing two removes $2 \times \frac{1}{2} = 1$ . $M_{new} = 4 - 1 = 3$ Why this step? We subtract only the fractional contribution the removed atoms made to this cell, not a whole atom each.

Step 3 — O count unchanged. Octahedral O $= 4$ . Why this step? We removed only face-centre M atoms; the oxide ions sit at the body centre and edge centres, positions we never touched, so their count of $4$ is untouched.

Step 4 — New ratio. $M : O = 3 : 4 \Rightarrow M_{3} O_{4}$ Why this step? Reduce the surviving-site counts to whole numbers.

Verify: $3 : 4$ is already whole-number; ratio $0.75$ . ✓ (A defect stoichiometry — non-1:1 as forecast.)

Cell C0 — Degenerate: zero voids filled and $r / R \to 0$

Worked example

(Cell C0 · Ex 12) (a) A ccp array of atoms X has no voids occupied. What is the formula and what fills the holes? (b) What coordination number does a cation of vanishing size ( $r / R \to 0$ ) get? Forecast: think about the limiting behaviour — a marble shrinking to a point.

Step 1 — (a) Count occupied sites. Only X packs; octahedral and tetrahedral voids hold nothing. $X = N, cations = 0 \Rightarrow formula X (pure element)$ Why this step? Empty voids contribute zero to the formula — the solid is just the packed element (e.g. a metal like Cu in ccp). The holes are filled with nothing (vacuum).

Step 2 — (b) The shrinking cation. As $r / R \to 0$ the ratio falls below every lower bound (the smallest is $0.155$ for trigonal). Geometrically, a point-sized guest can no longer touch all the surrounding spheres at once — it rattles freely in every hole. So there is no stable coordination number in the limit $r / R \to 0$ : the guest is too small to be "held" by any void, and the radius-ratio rule simply does not apply below $0.155$ . In practice such a tiny ion is treated as effectively uncoordinated (CN undefined / trivially small), which is why the table has a hard floor at $0.155$ rather than extending to zero. Why this step? Each radius-ratio range demands the guest simultaneously touch every neighbour of that hole; below the trigonal floor no arrangement satisfies that touching condition, so coordination breaks down — the degenerate endpoint the table is built to warn against.

Step 3 — Contrast with the ceiling. As $r / R \to 1$ (cation as big as anion), the "cation" is just another packing sphere and CN $\to 12$ — the equal-sphere close-packing limit from Close packing in solids (hcp, ccp, bcc). Why this step? It shows the other degenerate end: at $r / R = 1$ the guest is indistinguishable from a host sphere, so it inherits the maximum coordination of 12.

Verify: floor $= 0.155$ , ceiling $= 1.000$ with CN 12 at equality. ✓ Both are the genuine boundaries of the radius-ratio table.

Recall Quick self-test across the matrix

ccp cations Ca + all tetrahedral voids filled by F → formula? ::: CaF₂ (fluorite, 1:2) ccp anions, cations in all tetrahedral voids → formula? ::: A₂B (antifluorite, 2N:N) Half of octahedral voids filled in a ccp oxide → formula? ::: MO₂ (rutile-type) One-third of tetrahedral voids filled in a ccp oxide → formula? ::: M₂O₃ $r / R = 0.525$ → coordination number? ::: 6 (octahedral) $r / R = 0.414$ exactly → which void? ::: Octahedral (CN 6), the lower limit Remove 2 face M from fcc MO → formula? ::: M₃O₄ Density formula for a unit cell ::: ρ = ZM / (N_A · a³)

Connections

2.4.14 Coordination number, voids (tetrahedral, octahedral) (Hinglish)
States of Matter (Quantitative)
Close packing in solids (hcp, ccp, bcc)
Packing efficiency and unit cell dimensions
Radius ratio rule and ionic structures
NaCl, ZnS, CaF2 crystal structures
Density of a unit cell