Coordination number, voids (tetrahedral, octahedral): Visual walkthrough

Before any algebra, three words we will use on every line:

Why is "touching = centres a fixed distance apart" the whole game? Because a snug fit means the guest just kisses every wall sphere. That single condition — no gap, no overlap — turns geometry into one equation each time. That is the tool we reach for, and nothing fancier is needed.

Step 1 — What a "snug fit" even means

WHAT. We fix the meaning of a perfect fit before touching a triangle or a cube.

WHY. The radius ratio $r / R$ is defined at the exact moment the small sphere touches all its neighbours — the limiting case. Bigger than that and the walls are forced apart (not close-packed any more); smaller and the guest rattles. So we always solve for the touch-everywhere moment.

PICTURE. In the figure, one big sphere (radius $R$ ) touches one small sphere (radius $r$ ). The green segment joins their two centres. The single rule we will use everywhere: along any line where a small sphere touches a big one, the distance between their centres is exactly $R + r$ — the two reaches added up, no gap and no overlap.

Figure — Coordination number, voids (tetrahedral, octahedral)

$centre-to-centre d = reach of big sphere R + reach of small sphere r (when they just touch)$

$d$ — the straight distance between the two centres.
$R$ — how far the big sphere's surface reaches from its own centre.
$r$ — how far the small sphere's surface reaches.
If $d > R + r$ there is a gap; if $d < R + r$ they overlap (impossible for hard spheres).

Step 2 — The octahedral void is really a flat square

WHAT. We reduce the scary 3-D octahedron (6 spheres) to a flat 2-D square of 4 spheres.

WHY. An octahedron has a top sphere, a bottom sphere, and four "equator" spheres forming a square. The top and bottom sit directly above and below the centre, so they contribute nothing new to the sideways squeeze — the tightest constraint is the equatorial square. Solving a square is far easier than a solid, and it gives the exact answer. That is why we choose to look at the equatorial cross-section and not the full solid.

PICTURE. Four big spheres sit at the corners of a square; the small guest sits dead centre. Neighbouring big spheres touch each other along each edge. The small sphere touches two opposite big spheres along the diagonal.

Step 3 — Two facts about that square

WHAT. We read two lengths straight off the picture: the edge and the diagonal.

WHY. The square gives us two independent measurements of the same figure. Setting them consistent with each other is what pins down $r / R$ — one equation, one unknown ratio.

PICTURE (same square, two coloured lines). The magenta line is an edge; the orange line is the full diagonal through the centre.

Fact A — the edge. Two big spheres touch along an edge, so by Step 1 (both radii are $R$ ): $edge = R + R = 2 R$

Each $R$ is one big sphere reaching to the touch point in the middle of the edge.

Fact B — the diagonal. Walking the diagonal from one corner sphere's centre to the opposite corner sphere's centre, we pass: big-sphere reach $R$ , then across the small sphere $2 r$ (its full width, since the guest is centred), then big-sphere reach $R$ : $diagonal = R + 2 r + R = 2 R + 2 r$

First $R$ — from a corner centre to where big meets small.
$2 r = r + r$ — straight through the small sphere, in one side and out the other.
Last $R$ — from small to the opposite corner centre.

Step 4 — The diagonal of a square is fixed by Pythagoras

WHAT. We compute the diagonal a second way, using only the edge length.

WHY. We now have the diagonal written two ways: once from the spheres (Step 3B) and once from pure geometry. Pythagoras is the right tool because a square's diagonal, edge, and edge form a right triangle — that is exactly the situation the theorem was built for, and it converts the side $2 R$ into a diagonal with no extra unknowns.

PICTURE. Half the square is a right triangle: two legs of length $2 R$ (the edges) and the hypotenuse is the diagonal.

$diagonal^{2} = (2 R)^{2} + (2 R)^{2} = 8 R^{2} \Rightarrow diagonal = 8 R = 22 R$

$(2 R)^{2}$ appears twice — one for each leg of the right triangle.
$8 = 22$ — the two equal legs $2 R$ get combined by the theorem into the hypotenuse.

Step 5 — Solve for the octahedral ratio

WHAT. Set the two diagonal expressions equal and isolate $r / R$ .

WHY. Both expressions describe the same orange line, so they must be equal. That single equality is the payoff of all the setup.

PICTURE. The two diagonal descriptions laid over each other on one line — sphere version on top, Pythagoras version below — visibly the same length.

$sphere version (Step 3B) 2 R + 2 r = Pythagoras (Step 4) 22 R$

Divide every term by $2$ : $R + r = 2 R$ Subtract $R$ from both sides (isolating the guest): $r = 2 R - R = R (2 - 1)$ Divide by $R$ to get the pure ratio: $\frac{r}{R} = 2 - 1 \approx 0.414$

$2 - 1$ — the leftover once you remove one big-sphere radius from the half-diagonal.
$0.414$ — the guest can be up to $41.4%$ of the big sphere and still fit an octahedral hole.

Step 6 — The tetrahedral void lives inside a cube

WHAT. We model the 4-sphere tetrahedral hole using alternate corners of a cube.

WHY. A tetrahedron drawn free-standing is awkward to measure. But if we pick the right 4 corners of a cube, their centres form a perfect regular tetrahedron — and cube diagonals are trivially handled by Pythagoras. We choose the cube because it hands us clean $2$ (face) and $3$ (body) diagonals. The guest sphere sits at the cube's centre, equidistant from all four.

Which four corners? (be explicit). Give every corner coordinates that are either $0$ or $1$ , so a corner is a triple like $(0, 0, 0)$ or $(1, 0, 1)$ . Add up the three numbers. Pick the four corners whose sum is even: $(0, 0, 0), (1, 1, 0), (1, 0, 1), (0, 1, 1)$ These are the "alternate" corners — no two of them share an edge, and any two are separated by a face diagonal. The other four (odd-sum) corners are the ones we leave empty.

PICTURE. A cube with big spheres on those four even-sum corners; the small guest at the very centre $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$ . Two lines are highlighted: a face diagonal (where two big spheres touch) and the body diagonal (which runs through the guest).

Let the cube edge be $a$ .

Big spheres touch across a face diagonal. Take two chosen corners on the same face, e.g. $(0, 0, 0)$ and $(1, 1, 0)$ — the distance between them is a face diagonal, and both host big spheres that touch: $face diagonal = a 2 = R + R = 2 R \Rightarrow R = \frac{a 2}{2}$

$a 2$ — a square face's diagonal (Pythagoras, legs $a$ and $a$ ).
$2 R$ — the two touching big spheres along that diagonal.

Step 7 — The body diagonal carries the guest

WHAT. Measure the body diagonal (corner to opposite corner, through the centre) two ways.

WHY. The guest sits at the cube centre on a body diagonal. We must be careful that the body diagonal we use actually runs from an occupied corner to another occupied corner — otherwise the sphere-version length $R + 2 r + R$ would be nonsense.

Which body diagonal is occupied at both ends? (be explicit). A cube's body diagonal joins a corner to the corner with all three coordinates flipped, e.g. $(0, 0, 0)$ and $(1, 1, 1)$ . Notice their coordinate sums are $0$ (even) and $3$ (odd): one is chosen, one is empty, so this diagonal is NOT the one we want. Instead take, e.g., $(1, 1, 0)$ and $(0, 0, 1)$ : sums $2$ and $1$ — again mixed. Look carefully: the guest at the centre lies on all four body diagonals, and each body diagonal has one even-sum and one odd-sum end. So no single body diagonal is occupied at both ends.

The correct picture is therefore this: from the centre, the guest touches a chosen corner sphere (say $(0, 0, 0)$ ) along half a body diagonal. By symmetry it touches the diametrically opposite chosen sphere too — but that opposite chosen sphere is $(1, 1, 0)$ -type, reached along a different half-diagonal that is the mirror image. Both half-lengths equal $\frac{a 3}{2}$ , so measuring straight through the guest from one chosen sphere, through the centre, to the opposite chosen sphere gives the full length $a 3$ made of two touching contacts: $centre-line through guest a 3 = first chosen sphere R + across the guest 2 r + opposite chosen sphere R = 2 R + 2 r$

$a 3$ — full centre-to-centre distance between two opposite chosen spheres, which is one body-diagonal length (they sit at cube-diagonal separation).
$R + 2 r + R$ — big reach, straight across the guest, big reach.

PICTURE. The centre-line through the guest, split into $R + 2 r + R$ , with the two occupied end-spheres marked.

Set the two descriptions of this length equal: $2 R + 2 r = a 3$

Step 8 — Solve for the tetrahedral ratio

WHAT. Eliminate $a$ using Step 6's $2 R = a 2$ , then isolate $r / R$ .

WHY. The cube edge $a$ was only scaffolding — the physics is in the ratio $r / R$ . Dividing the two touch-equations kills $a$ and leaves the ratio alone.

PICTURE. The face-diagonal square and body-diagonal line side by side, the shared edge $a$ crossed out to show it cancels.

Divide the body-diagonal equation by $2 R$ : $\frac{2 R + 2 r}{2 R} = \frac{a 3}{2 R} \Rightarrow 1 + \frac{r}{R} = \frac{a 3}{2 R}$ Now substitute $2 R = a 2$ into the right side (the $a$ cancels): $1 + \frac{r}{R} = \frac{a 3}{a 2} = \frac{3}{2}$ Subtract $1$ : $\frac{r}{R} = \frac{3}{2} - 1 \approx 0.225$

$3/2$ — the body diagonal is $3/2$ times the face diagonal.
$0.225$ — a tetrahedral guest can be at most $22.5%$ of the big sphere.

Step 9 — Edge & degenerate cases (never skip these)

WHAT. Check the boundaries of the story so no scenario surprises the reader.

WHY. Numbers like $0.414$ only make sense against what happens just below and just above.

$r \to 0$ (a point-sized guest). Then $r / R \to 0$ . A vanishing sphere fits any hole with room to spare — it just doesn't touch all walls, so it isn't a "snug" fit. Both boxed formulas correctly give $r = 0$ only when $R \to \infty$ ; a real tiny ion simply rattles.
$r / R$ exactly $0.225$ . The guest touches all 4 tetrahedral walls perfectly — the ideal tetrahedral occupant.
$0.225 < r / R < 0.414$ . Too big to be snug in a tetrahedral hole (it now props the walls apart) yet too small for octahedral. It sits tetrahedrally, pushing the lattice open — still counts as CN 4. This is exactly why the radius ratio table uses ranges.
$r / R$ exactly $0.414$ . Perfect octahedral fit — touches all 6 walls.
$r / R \to 1$ (guest as big as the host). No hole can hold it; the "guest" would itself become a packing sphere. This is why the table stops at $1.000$ .

PICTURE. A number line from $0$ to $1$ with the guest sphere drawn at three sizes: rattling, snug-tet ( $0.225$ ), snug-oct ( $0.414$ ).

The one-picture summary

Everything above compressed: the octahedral square gives $2 - 1$ ; the tetrahedral cube gives $3/2 - 1$ . Same tool (touch $\Rightarrow$ centre distance = $R + r$ ), same trick (measure a diagonal two ways, equate, divide out lengths).

Recall Feynman retelling — the whole walkthrough in plain words

We wanted to know how big a marble can hide in the gaps between packed oranges. The rule of the game is simple: when two balls touch, the distance between their middles equals their two radii added up. For the octahedral gap I looked at just the flat square of four oranges around the hole. Along an edge two oranges touch, giving $2 R$ . Across the diagonal I go orange-marble-orange, giving $2 R + 2 r$ . But a square's diagonal is always $2$ times its edge — so $2 R + 2 r = 22 R$ , and after cleaning up, the marble can be $2 - 1 \approx 0.414$ of an orange. For the tetrahedral gap I hid four oranges at the even-corners of a cube (coordinate sums even) with the marble at the cube's centre. Oranges touch across a face diagonal ( $a 2 = 2 R$ ); the marble sits at the centre and touches two opposite oranges along a centre-line of length $a 3 = 2 R + 2 r$ . Dividing one by the other makes the cube's size vanish and leaves $3/2 - 1 \approx 0.225$ . So octahedral holes are roomier ( $0.414 > 0.225$ ) — bigger guests belong there — and that single fact, drawn from two diagonals, decides which ion goes where in real crystals like NaCl and ZnS.

Connections

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