Visual walkthrough — Ionic crystals — NaCl, CsCl, ZnS, fluorite, antifluorite structures

We build one idea: the limiting radius ratio — the exact cation-to-anion size at which the packing is perfect. Below it the cation rattles; above it the anions must part.

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Step 1 — Draw the two kinds of sphere

WHAT. We picture the ions as hard spheres: a big anion (negative, radius $r_{-}$) and a small cation (positive, radius $r_{+}$). "Hard" means they cannot overlap — like billiard balls.

WHY this picture. All of ionic packing is a game of touching without overlapping. If we allow squishy spheres, no clean geometry survives. Treating them as rigid balls is what lets simple triangles and Pythagoras give exact answers. See Coordination number for how "neighbours" is counted once these balls are stacked.

WHAT IT LOOKS LIKE. Look at the figure: the pale-blue ball is the anion, radius $r_{-}$ measured from centre to rim; the pink ball is the cation, radius $r_{+}$. Two touching spheres put their centres a distance (sum of radii) apart — that single fact is the engine of everything below.

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Step 2 — What "the limiting case" means

WHAT. We define the limiting ratio $\rho =r_{+}/r_{-}$ as the exact value where two things happen at once: the cation just touches every surrounding anion, AND those anions just touch each other.

WHY this exact moment. Away from this balance one of two failures occurs, and drawing both is the whole point:

• Cation too small → it does not reach the anions; it sits loose in the cavity ("rattles"). Unstable.
• Cation too big → it shoves the anions apart so they no longer touch. Fine — but now a bigger arrangement (more neighbours) becomes possible, so nature switches to it.

The changeover between one coordination number and the next happens precisely at the limiting ratio.

WHAT IT LOOKS LIKE. Three cavities side by side: rattling (too small, gap shown), perfect (both contacts, the case we compute), and pushed-apart (too big, anions separated). Only the middle one is the limit.

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Step 3 — The octahedral case: slice it into a square

WHAT. For coordination number 6 (octahedral), the cation sits with six anions around it — four in a flat square around its "equator," one above, one below. We take the equatorial square slice: four anion centres at the corners, cation centre in the middle.

WHY a slice, and why the square. A 3-D octahedron is hard to reason about; but the tightest squeeze is in that square plane, so a 2-D square captures the limiting geometry exactly. This is the classic trick the parent note used — we are now drawing it. The connection to real holes lives in Close packing FCC HCP and voids.

WHAT IT LOOKS LIKE. A square. Along each edge, two anions kiss → edge length $=2r_{-}$. Along the diagonal, you cross anion → cation → anion → so diagonal $=r_{-}+(r_{+}+r_{-})+r_{+}=2r_{-}+2r_{+}$. Trace the yellow diagonal in the figure and read those pieces off.

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Step 4 — Pythagoras turns the square into a number

WHAT. For any square, the diagonal is $\sqrt{2}$ times the edge. We plug in our two lengths.

WHY $\sqrt{2}$, why Pythagoras here. We need to relate the diagonal (which carries $r_{+}$) to the edge (which carries only $r_{-}$). The only guaranteed relation between a square's diagonal and edge is Pythagoras: $d^2=s^2+s^2=2s^2$, so $d=\sqrt{2}s$. That single geometric fact is what converts a picture into an equation for $\rho$.

WHAT IT LOOKS LIKE. The right triangle formed by two edges and the diagonal — half the square, shaded in the figure. Its legs are both the edge; its hypotenuse is the diagonal.

$\underset{\text{diagonal}}{\underbrace{2r_{-}+2r_{+}}}=\sqrt{2}\underset{\text{edge}}{\underbrace{(2r_{-})}}$

Divide every term by $2r_{-}$ (harmless — it just measures everything in units of the anion):

$\underset{2r_{-}/2r_{-}}{\underbrace{1}}+\underset{=\rho }{\underbrace{\frac{r_{+}}{r_{-}}}}=\sqrt{2}$

$\boxed{\rho =\sqrt{2}-1=0.414}$

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Step 5 — The cubic case: same idea, one dimension up

WHAT. For coordination number 8 (cubic, e.g. CsCl), the cation sits at the centre of a cube of eight anions. Now the anions touch along a cube edge, and the cation lies on the cube's body diagonal.

WHY step up to a cube / body diagonal. With eight neighbours arranged at cube corners, the cation–anion contact runs from a corner through the centre to the opposite corner — that is the body diagonal. So we need the 3-D analogue of Step 4.

WHAT IT LOOKS LIKE. A cube. Edge $=2r_{-}$ (two anions kiss). Body diagonal $=2r_{-}+2r_{+}$ (anion–cation–anion straight through the middle), shown as the yellow line piercing the cube.

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Step 6 — Why the body diagonal is $\sqrt{3}\times$ edge

WHAT. For a cube of side $s$, the body diagonal is $\sqrt{3}s$. We apply Pythagoras twice.

WHY twice. First across the bottom face: face diagonal $=\sqrt{s^2+s^2}=\sqrt{2}s$. Then stand that face diagonal up against the vertical edge $s$: body diagonal $=\sqrt{(\sqrt{2}s{)}^2+s^2}=\sqrt{2s^2+s^2}=\sqrt{3}s$. Two right triangles chained together — that is the only new idea versus Step 4.

WHAT IT LOOKS LIKE. The figure shows the two stacked right triangles: the shaded face triangle giving $\sqrt{2}s$, then the vertical triangle giving $\sqrt{3}s$.

$\underset{\text{body diagonal}}{\underbrace{2r_{-}+2r_{+}}}=\sqrt{3}\underset{\text{edge}}{\underbrace{(2r_{-})}}$

Divide by $2r_{-}$:

$1+\rho =\sqrt{3}\Rightarrow \boxed{\rho =\sqrt{3}-1=0.732}$

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Step 7 — Edge case: the tetrahedral limit (and why it's smallest)

WHAT. Coordination number 4 (tetrahedral, e.g. ZnS): the cation sits in a hole surrounded by four anions at the corners of a tetrahedron. The same both-contacts-at-once argument gives $\rho =\sqrt{3/2}-1=0.225$.

WHY it's the smallest cutoff, and why we treat it separately. Fewer neighbours means a smaller, tighter cavity, so an even smaller cation fits — hence the lowest threshold. Its geometry hides inside a cube: put four anions on alternate corners of a cube; the cavity centre is the cube centre. Anion–anion touch is along a face diagonal ($=2r_{-}$), cation–anion is half the body diagonal ($=r_{+}+r_{-}$).

WHAT IT LOOKS LIKE. A cube with four highlighted alternate corners forming the tetrahedron; the face diagonal (anion–anion) and the half-body-diagonal to the centre (cation–anion) are drawn.

Setting body-diagonal $=\sqrt{3}\times$edge and face-diagonal $=\sqrt{2}\times$edge, with face diagonal $=2r_{-}$ and half body diagonal $=r_{+}+r_{-}$:

$r_{+}+r_{-}=\frac{\sqrt{3}}{2}s,2r_{-}=\sqrt{2}s\Rightarrow s=\frac{2r_{-}}{\sqrt{2}}=\sqrt{2}{r}_{-}$

$r_{+}+r_{-}=\frac{\sqrt{3}}{2}\sqrt{2}{r}_{-}=\sqrt{\frac{3}{2}}{r}_{-}\Rightarrow \boxed{\rho =\sqrt{\frac{3}{2}}-1=0.225}$

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Step 8 — Degenerate limit: $\rho \to 1$ and $\rho \to 0$

WHAT. Two extreme sanity checks with no algebra:

• $\rho \to 1$: cation and anion are equal spheres. There is no "hole" to sit in — it is just close packing of one size. This is the ceiling of the CN 8 band; beyond it the ionic-hole language breaks down.
• $\rho \to 0$: an infinitely tiny cation. It fits any gap but touches nothing — physically it would be a bare charge with no defined coordination. Below $0.155$ even three-fold contact fails.

WHY show these. They tell you the rule is a band, not a formula that runs forever. Every real crystal lands inside one band, and Defects in ionic solids appears when ions are pushed near a band edge.

WHAT IT LOOKS LIKE. Two cartoons: equal balls (no hole) on the left; a dust-speck cation lost in a huge cavity on the right.

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The one-picture summary

WHAT. One figure lines up all three cases: the tetrahedral hole ($0.225$), the octahedral square ($0.414$), the body-centred cube ($0.732$) — each with its contact diagonal drawn and its $\sqrt{k}-1$ value. The number line underneath marks the bands and pins ZnS, NaCl, CsCl where they belong.

WHY. Everything above collapses to a single mantra: pick the diagonal, set it to $\sqrt{k}\times (2r_{-})$, subtract 1.

Recall Feynman retelling — the whole walkthrough in plain words

Picture big soft-blue oranges packed together; you want to hide small pink marbles in the gaps. A marble is "happy" only when it touches every orange around it and those oranges still touch each other — that snug moment is the "limiting ratio." I drew that moment three ways. In a square of four oranges (a flat slice of a six-orange cage), the straight-across distance is $\sqrt{2}$ edges, and grinding through the algebra the marble must be $0.414$ oranges wide. In a cube of eight oranges the straight-through distance is $\sqrt{3}$ edges, giving $0.732$. In the tightest four-orange tetrahedron it's $0.225$. Same recipe every time: find the line the marble sits on, say it's $\sqrt{k}$ times the orange-touch line, subtract one. As the marble grows, $k$ climbs $\frac{3}{2}\to 2\to 3$ and it can boss around more oranges — that's why ZnS(4) → NaCl(6) → CsCl(8). At the very end, if the marble is orange-sized there's no gap left, and if it's a speck it touches nothing — so the rule is a set of bands, and every real salt lands in one.

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Connections

• Parent topic — the full structure zoo this derivation feeds.
• Close packing FCC HCP and voids — where these tetrahedral/octahedral holes physically live.
• Coordination number — the "how many neighbours" that $\rho$ predicts.
• Unit cell and Z calculation — turning these cages into counts per cell.
• Defects in ionic solids — what happens near the band edges.