Ionic crystals — NaCl, CsCl, ZnS, fluorite, antifluorite structures: Question bank

Four symbols you must own before the traps

Everything below leans on four shorthand pieces of notation. Each is defined in plain words here so no line uses a symbol you have not met.

Definition The four symbols

Radius ratio $ρ$ (Greek letter "rho"): the cation radius divided by the anion radius, $ρ = \frac{r _{+}}{r _{-}}$ where $r_{+}$ is the radius of the positive ion and $r_{-}$ that of the negative ion. It is a pure number (units cancel): a small $ρ$ means a tiny cation in a hole between big anions; $ρ$ near 1 means the two ions are nearly the same size.
Coordination number, written CN: the count of nearest oppositely-charged neighbours touching a given ion. "CN 6" means 6 anions surround each cation; "CN 6:6" gives the cation number first, anion number second.
Formula units per unit cell, written $Z$ : how many complete formula units (e.g. how many NaCl "pairs") are contained inside one repeating box of the crystal, after sharing corner/edge/face ions with neighbouring boxes. For NaCl $Z = 4$ ; for CsCl $Z = 1$ .
Lattice constant / edge length $a$ : the length of one side of the cubic unit cell — the little repeating box whose corners, edges, faces and centre carry the ions. Every "contact" statement like $a = 2 (r_{+} + r_{-})$ compares this box edge to ion radii, telling us which line inside the box the two ions touch along.

Figure — Ionic crystals — NaCl, CsCl, ZnS, fluorite, antifluorite structures

The first figure shows the trigonal (CN 3) triangle case: a tiny cation sits at the centre of an equilateral triangle of 3 anions. At the limit the anions touch each other along each triangle side ( $= 2 r_{-}$ ) and the cation touches each anion (centre-to-corner $= r_{+} + r_{-}$ ). For an equilateral triangle the centre-to-corner distance is $\frac{1}{3} \times$ side, so $r_{+} + r_{-} = \frac{2 r _{-}}{3}$ , giving $ρ = \frac{2}{3} - 1 = 0.155$ . This is the smallest CN in the hard-sphere spectrum.

The second figure shows the octahedral (CN 6) square slice: a cation sits at the centre of a square of 4 anions (a plane cut through the octahedral hole). At the limiting case the edge is anion–anion contact ( $= 2 r_{-}$ ) and the diagonal is anion–cation–anion contact ( $= 2 r_{-} + 2 r_{+}$ ). Pythagoras on the square gives diagonal $= 2 \times$ edge: $2 r_{-} + 2 r_{+} = 2 (2 r_{-}) \Rightarrow ρ = 2 - 1 = 0.414.$

The third figure shows the cubic (CN 8) contact: the cation sits on the body diagonal of a cube of 8 anions. Anions touch along the cube edge ( $= 2 r_{-}$ ); the cation stretches along the body diagonal ( $= 2 r_{-} + 2 r_{+} = 3 \times$ edge), giving $ρ = 3 - 1 = 0.732$ . The same "just-touching" trick on a tetrahedral hole (4 anions at alternate cube corners) gives $ρ = 3/2 - 1 = 0.225$ for CN 4 — so the full spectrum is $0.155, 0.225, 0.414, 0.732$ for CN $3, 4, 6, 8$ .

The fourth figure sketches the attraction–repulsion trade-off as a cation grows: attraction energy keeps falling as more anions crowd in, but anion–anion repulsion switches on sharply once anions touch. Nature settles on the largest CN whose total energy is still lowest — this is the quantitative reason CN jumps at each cutoff rather than rising continuously.

True or false — justify

CsCl has a body-centred cubic (BCC) lattice.

False — BCC needs the same atom at corners and centre. Here corners are Cl⁻ and the centre is Cs⁺, so it is a simple cubic lattice with a two-ion basis, and one box holds one formula unit (

Z = 1

In NaCl the Cl⁻ ions alone form an FCC arrangement.

True — the anions occupy corners and face centres of the cube (

8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4

), and the Na⁺ then fill all octahedral holes of that FCC framework because

ρ \approx 0.52

falls in the CN 6 window.

Fluorite and antifluorite are different geometries.

False — they are the same geometry with cation and anion roles swapped. In both, one ion forms FCC and the other fills all tetrahedral holes; because there are twice as many tetrahedral holes as FCC sites, the hole-filler is always twice as numerous, so fluorite is MX₂ (cation FCC) and antifluorite is M₂X (anion FCC).

A larger radius ratio always allows a larger coordination number.

Only under the hard-sphere assumption — geometrically a bigger cation (larger

ρ

) can pack more anions before they touch, so

CN

rises with

ρ

. But real ions polarise and covalent character can override packing, so this is a strong trend, not an absolute law (some compounds defy their predicted CN).

In ZnS all tetrahedral holes are filled by Zn²⁺.

False — only half (alternate) tetrahedral holes are filled. There are 8 tetrahedral holes but only 4 Zn²⁺, keeping the 1:1 stoichiometry and CN 4:4; filling all 8 would force a 1:2 ratio like fluorite.

The radius ratio

ρ = 0.414

is the boundary between CN 4 and CN 6.

True — below it a cation "rattles" in the octahedral hole (contact lost) so tetrahedral (CN 4) is preferred; at and above it octahedral (CN 6) becomes stable. It is exactly

2 - 1

from the square-slice Pythagoras (figure s02).

Both cation and anion in NaCl have coordination number 6.

True — CN 6:6 means each Na⁺ touches 6 Cl⁻ and each Cl⁻ touches 6 Na⁺; the coordination is mutual because the stoichiometry is 1:1, so both ions must share neighbours equally.

Coordination number 5 has its own radius-ratio cutoff between CN 4 and CN 6.

False — a symmetric fivefold arrangement of equal spheres with equal cation–anion contact is geometrically impossible, so the hard-sphere table skips straight from CN 4 to CN 6 with no CN 5 cutoff.

Spot the error

"Fluorite CaF₂ has coordination number 6:6 because it is built on an FCC lattice."

The FCC start does not fix CN — what matters is which holes are filled. In fluorite all 8 tetrahedral holes hold F⁻, so each Ca²⁺ touches 8 F⁻ and each F⁻ touches 4 Ca²⁺: the correct answer is 8:4. (A tetrahedral hole is where 4 anions at alternate cube corners meet, the same geometry that gives the

3/2 - 1

cutoff of figure s03.)

"CsCl has

Z = 2

because there is a corner ion and a body-centre ion."

The 8 corner Cl⁻ are each shared among 8 boxes, contributing

8 \times \frac{1}{8} = 1

; the body-centre Cs⁺ is wholly inside, contributing 1. Together that is one formula unit:

Z = 1

, not 2.

"Larger radius ratio means the cation sits in a smaller hole, so CN decreases."

This confuses ratio with hole size. A larger

ρ

means a bigger cation needing a bigger hole surrounded by more anions, so CN increases — trace the growth across figures s02→s03.

"In NaCl the Na⁺ and Cl⁻ touch along the body diagonal."

They touch along the cell edge (recall

a

= edge length of the cubic box), because octahedral holes sit at edge midpoints and the body centre:

a = 2 (r_{+} + r_{-})

. Body-diagonal contact belongs to CsCl, where

3 a = 2 (r_{+} + r_{-})

(compare the cube of figure s03).

"Antifluorite is CaF₂ turned upside down, so its formula is still MX₂."

Swapping cation and anion roles swaps the stoichiometry too: cations now occupy the (twice-as-many) tetrahedral holes, giving M₂X (e.g. Na₂O), not MX₂.

"Since ZnS starts from FCC of S²⁻ with holes filled, it must be denser-packed than NaCl."

Filling half the tetrahedral holes gives CN 4:4, an open structure; NaCl fills all octahedral holes (CN 6:6). Occupancy of the hole set differs, so you cannot rank density from the FCC start alone.

"The radius ratio predicts the exact structure with certainty for every compound."

It is a guideline, not a law. Ions are not perfectly hard spheres and covalent/polarisation effects can override it (some compounds adopt structures the naive ratio would forbid).

Why questions

Why do oppositely charged ions touch while like ions stay apart?

Cation–anion contact maximises electrostatic attraction (lowers energy), while keeping like ions apart avoids the repulsion that would raise energy — the structure balances both, as the energy sketch in figure s04 shows.

Why is the octahedral limiting ratio derived from a square slice, not the full octahedron?

A plane through the octahedral hole cuts through 4 anions in a square with the cation at its centre, so the critical anion–anion (edge) and anion–cation (diagonal) contacts both lie in that 2-D slice — the geometry is fully captured there (figure s02).

Why does nature pick the largest CN that still allows anion–cation contact?

More surrounding anions means more attractions per cation and lower energy — but only until anions touch and repel, at which point energy shoots back up. The lowest-energy compromise is the largest CN before that repulsion switches on, which is exactly the crossover in figure s04.

Why is CsCl not simply "more packed NaCl"?

They differ in hole type filled: CsCl has each ion at the centre of a cube of 8 (CN 8, body-diagonal contact), NaCl has octahedral coordination (CN 6, edge contact). The larger

ρ

of CsCl drives the switch to a different geometry, not just tighter packing.

Why does fluorite have anions in tetrahedral holes while the big cation is on the FCC lattice?

The Ca²⁺ are large enough to form the FCC framework; the small F⁻ then slot into the small tetrahedral gaps. Because an FCC lattice has twice as many tetrahedral holes as lattice points and all are filled, the stoichiometry comes out 1:2.

Why does each radius-ratio cutoff come from a different Pythagorean relation rather than one fixed formula?

Each hole has its own shape, so a different distance is the "just-touching" line: a triangle gives

\frac{2}{3} - 1 = 0.155

, a tetrahedron

3/2 - 1 = 0.225

, a square/octahedron

2 - 1 = 0.414

, and a cube

3 - 1 = 0.732

(figures s01, s02, s03). The method (anions just touch and cation just touches anions) is shared, but the number is not.

Why does the hard-sphere spectrum jump 4 → 6, skipping CN 5?

No symmetric polyhedron of 5 equal spheres keeps every anion the same distance from the cation and mutually touching, so there is no clean limiting ratio for CN 5 — real fivefold coordination appears only in distorted, covalency-driven cases.

Edge cases

What happens if the cation is smaller than the lower limit of its CN range (say

ρ < 0.414

but forced into octahedral)?

The cation "rattles" inside the too-big hole — anion–cation contact is lost, attraction weakens, so the structure prefers a smaller CN (tetrahedral) where contact is restored.

What is the coordination number when

ρ = 1

(ions identical in size)?

ρ = 1

the cation is as large as the anion, comfortably supporting CN 8 (cubic) — the top of the ionic range; even higher CN like 12 appears in metals but not in these ionic hard-sphere structures.

Is there a lower end to the CN spectrum, and what sits there?

Yes — for

ρ

between

0.155

and

0.225

only CN 3 (trigonal-planar, 3 anions around the cation in a triangle) is stable (figure s01), and below

0.155

even CN 2 (linear) becomes the limit. These are rare for simple ionic solids because such tiny

ρ

is uncommon, but they complete the spectrum.

If a compound's

ρ

lands exactly on a cutoff like 0.732, which structure forms?

Right at the limit both geometries have simultaneous anion–anion and anion–cation contact, so either can appear; real cases resolve by small polarisation or lattice-energy differences rather than the ratio alone.

For a 1:2 compound (like CaF₂), can cation and anion coordination numbers be equal?

No — with twice as many anions as cations, each cation must contact more neighbours than each anion. The CNs are tied to stoichiometry:

CN_{+} : CN_{-} = 8 : 4 = 2 : 1

, matching the anion:cation ratio.

Does the antifluorite structure require the anion to be larger than the cation?

Yes in spirit — the anion forms the FCC framework, so it must be big enough to close-pack while the small cations fit the tetrahedral holes; this is why oxides/sulfides of small alkali metals (Na₂O, Li₂O) adopt it.

Can coordination number 5 ever appear in a real ionic solid despite having no radius-ratio cutoff?

Yes, but only as a distorted polyhedron (square pyramid or trigonal bipyramid) stabilised by covalency or asymmetric bonding — it is never predicted by the pure hard-sphere ratio because equal contacts are geometrically impossible for 5 spheres.

Connections

Coordination number — every trap here turns on counting nearest opposite ions.
Close packing FCC HCP and voids — octahedral vs tetrahedral holes decide CN and stoichiometry.
Unit cell and Z calculation — the $Z$ and edge-length $a$ counting behind the CsCl and NaCl traps.
Density of crystals formula — where correct $Z$ matters numerically (D3/D4).