Ionic crystals — NaCl, CsCl, ZnS, fluorite, antifluorite structures: Worked examples

Before we start, one reminder of the four symbols we reuse everywhere, all defined in the parent:

Recall The four symbols (tap to reveal)

$r_{+}$ = radius of the cation (the small positive ion), in picometres (pm). $1 pm = 1 0^{- 12} m = 1 0^{- 10} cm$ .
$r_{-}$ = radius of the anion (the big negative ion).
$ρ = r_{+} / r_{-}$ = the radius ratio — a pure number with no units, because pm ÷ pm cancels.
$Z$ = number of formula units per unit cell (how many "recipe portions" of the compound fit in one box).
$a$ = the edge length of the cubic unit cell (the box), in pm or cm.

The scenario matrix

Every question this topic can throw is one of these cells. Each worked example below is tagged with the cell(s) it covers.

#	Cell class	What makes it tricky	Covered by
A	Small $ρ$ (0.225–0.414) → CN 4	must not jump to CN 6	Ex 1
B	Mid $ρ$ (0.414–0.732) → CN 6	the "default" NaCl case	Ex 2
C	Large $ρ$ (0.732–1.0) → CN 8	must not call it BCC	Ex 3
D	$ρ$ exactly on a boundary	which side wins?	Ex 4
E	$ρ > 1$ (cation bigger than anion)	flip the ratio!	Ex 5
F	Forward density (data → $ρ_{density}$ )	unit conversion pm→cm	Ex 6
G	Backward (density → find $a$ or $M$ )	rearranging the formula	Ex 7
H	MX₂ / M₂X stoichiometry (fluorite family)	$Z$ counts differently	Ex 8
I	Edge–radius geometry, both contact directions	edge vs body diagonal	Ex 9
J	Exam twist / degenerate input	a "gotcha" that breaks a naive rule	Ex 10

The three ratio cutoffs we lean on all page long (derived in the parent):

Cell A — small ratio, CN 4

Worked example 1 — Predict the structure of CuBr (

$r_{Cu^{+}} = 96$ pm, $r_{Br^{-}} = 196$ pm) Forecast: the cation is roughly half the anion. Guess the CN before reading on — 4, 6, or 8?

Step 1. Form the ratio $ρ = r_{+} / r_{-} = 96/196$ . Why this step? CN is decided by relative size, not absolute size — the ratio is the only thing that matters. $ρ = \frac{96}{196} = 0.490$

Wait — $0.490$ lands in the 0.414–0.732 band, so this looks like CN 6! But CuBr is experimentally zinc-blende (CN 4). Let me re-read the data: use $r_{Cu^{+}} = 60$ pm for the tetrahedral (4-coordinate) radius. This is the lesson of cell A: an ion's tabulated radius shrinks when its coordination number drops (fewer neighbours pull it tighter). Using the 4-coordinate value: $ρ = \frac{60}{196} = 0.306$

Step 2. Locate $0.306$ on the walls. Why this step? We compare against the cutoffs to read off CN. $0.225 < 0.306 < 0.414$ → CN 4 → ZnS (zinc blende) structure.

Verify: $0.306$ is comfortably inside the tetrahedral band (not near either wall), so the prediction is robust; CuBr is indeed zinc-blende type. ✓

Cell B — mid ratio, CN 6 (the NaCl default)

Cell C — large ratio, CN 8 (and the "not BCC" trap)

Cell D — sitting exactly on a boundary

Worked example 4 — A hypothetical salt has

$ρ = 0.414$ exactly. What CN? Forecast: does the boundary value belong to CN 4 or CN 6?

Step 1. Recall what $0.414 = 2 - 1$ means physically: it is the ratio where, in an octahedral (CN 6) arrangement, the anions just touch each other and the cation just touches every anion — the tightest possible CN-6 packing. Why this step? A boundary value is a geometric limit, not an arbitrary number. Understanding it tells us which way it falls.

Step 2. At exactly $ρ = 0.414$ , CN 6 is stable (contact is exactly satisfied, no rattling). Below it the cation would rattle in the octahedral hole and prefer CN 4; above it CN 6 is comfortable. So $ρ = 0.414$ → CN 6 (the lower limit of the octahedral band). Why this step? The convention: each band is inclusive at its lower wall — the wall is the minimum ratio that supports that CN.

Verify: at $ρ = 0.414$ , plug into the octahedral contact equation $2 (r_{+} + r_{-}) = 2 (2 r_{-})$ . With $r_{+} = 0.414 r_{-}$ : left $= 2 (1.414 r_{-}) = 2.828 r_{-}$ ; right $= 2 \cdot 2 r_{-} = 2.828 r_{-}$ . Equal → contact exact → CN 6 valid. ✓

Cell E — cation bigger than anion ( $ρ > 1$ )

Worked example 5 — A student writes

$ρ = r_{-} / r_{+}$ for a compound where the metal ion is huge. Is that ok? Given $r_{+} = 220$ pm, $r_{-} = 133$ pm (say a large cation with fluoride). Forecast: what happens to the rule when the cation is the bigger ion?

Step 1. The radius ratio is defined as $ρ = r_{+} / r_{-}$ = (smaller-radius-role ÷ larger). The rule assumes the cation is the smaller ion sitting in a hole of the anion lattice. Why this step? The whole derivation put the cation inside the anion cage — that geometry only makes sense if the cation is smaller.

Step 2. Here $r_{+} > r_{-}$ , so literally $ρ = 220/133 = 1.65 > 1$ . A ratio above 1 is outside the table entirely — it signals the roles are swapped: the anion is now the one sitting in holes of a cation lattice. Compute the meaningful ratio as $r_{smaller} / r_{larger} = 133/220 = 0.605$ . Why this step? CN is still governed by the smaller ion in the hole of the larger; we identify who is small.

Step 3. $0.605$ is in $0.414$ – $0.732$ → CN 6 for the small ion. This is exactly the antifluorite / fluorite role-swap idea: whichever ion is smaller fills the holes. Why this step? Connects the numeric fix to the structural concept (see the parent).

Verify: $220/133 = 1.65$ and $133/220 = 0.605$ ; $1/1.65 = 0.606 \approx 0.605$ ✓ — the two ratios are reciprocals, confirming we just flipped roles. ✓

Cell F — forward density (data → density)

Worked example 6 — Density of KBr. Given

$a = 660$ pm, $M = 119$ g/mol, $Z = 4$ (rock-salt, from Ex 2). Forecast: heavier than NaCl (2.16) or lighter? (Bigger cell but heavier atoms…)

Step 1. Convert the edge to cm: $a = 660 pm = 660 \times 1 0^{- 10} cm = 6.60 \times 1 0^{- 8} cm$ . Why this step? Density in g/cm³ needs the volume in cm³; pm must become cm first. $1 pm = 1 0^{- 10} cm$ .

Step 2. Cube it: $a^{3} = (6.60 \times 1 0^{- 8})^{3} = 2.875 \times 1 0^{- 22} cm^{3}$ . Why this step? The unit cell is a cube of side $a$ , so its volume is $a^{3}$ .

Step 3. Apply $ρ = \frac{Z M}{N _{A} a ^{3}}$ (see Density of crystals formula): $ρ = \frac{4 \times 119}{6.022 \times 1 0 ^{23} \times 2.875 \times 1 0 ^{- 22}} = \frac{476}{173.1} \approx 2.75 g/cm^{3}$ Why this step? $Z M$ = mass of ions in one cell (in g/mol), $\div N_{A}$ turns mol into actual grams, $\div a^{3}$ turns it into a density.

Verify: experimental KBr density $= 2.75$ g/cm³ ✓ — matches, so $Z = 4$ is confirmed. Units: $\frac{g/mol}{( mol ^{- 1} ) ( cm ^{3} )} = g/cm^{3}$ ✓.

Cell G — backward (density → find edge)

Worked example 7 — NaCl has density

$ρ = 2.165$ g/cm³, $M = 58.5$ g/mol, $Z = 4$ . Find the edge $a$ . Forecast: should land near 564 pm (the value the parent quoted).

Step 1. Rearrange the density formula for $a^{3}$ : $a^{3} = \frac{Z M}{N _{A} ρ}$ Why this step? We know everything except $a$ ; algebra isolates $a^{3}$ .

Step 2. Plug in: $a^{3} = \frac{4 \times 58.5}{6.022 \times 1 0 ^{23} \times 2.165} = \frac{234}{1.3038 \times 1 0 ^{24}} = 1.795 \times 1 0^{- 22} cm^{3}$ Why this step? Straight substitution; keep track of powers of ten.

Step 3. Cube-root: $a = (1.795 \times 1 0^{- 22})^{1/3}$ . Write it as $(179.5 \times 1 0^{- 24})^{1/3} = (179.5)^{1/3} \times 1 0^{- 8}$ . Since $5.6 4^{3} = 179.4$ , we get $a \approx 5.64 \times 1 0^{- 8} cm = 564 pm$ . Why this step? Cube-root undoes the cube; grouping the exponent as a multiple of 3 makes the root clean.

Verify: $564$ pm matches the parent's quoted edge ✓, and back-substituting gives density $2.165$ g/cm³ ✓.

Cell H — MX₂ / M₂X stoichiometry (fluorite family)

Worked example 8 — Density of CaF₂ (fluorite). Given

$a = 546$ pm, $M (CaF_{2}) = 78$ g/mol. Forecast: what is $Z$ here — is it 4 like NaCl?

Step 1. Count ions per cell. Ca²⁺ forms FCC: $8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$ Ca²⁺. F⁻ fills all 8 tetrahedral holes, all internal: $8 \times 1 = 8$ F⁻. Why this step? $Z$ = number of formula units, and one formula unit is one $CaF_{2}$ . We have 4 Ca²⁺ and 8 F⁻ = 4 formula units of CaF₂, so $Z = 4$ (not 4 atoms — 4 CaF₂ units).

Step 2. Convert edge: $a = 546 pm = 5.46 \times 1 0^{- 8}$ cm, $a^{3} = 1.628 \times 1 0^{- 22}$ cm³. Why this step? Same unit fix as Ex 6.

Step 3. Density: $ρ = \frac{Z M}{N _{A} a ^{3}} = \frac{4 \times 78}{6.022 \times 1 0 ^{23} \times 1.628 \times 1 0 ^{- 22}} = \frac{312}{98.04} \approx 3.18 g/cm^{3}$ Why this step? $M = 78$ already counts one Ca + two F, so $Z = 4$ formula units is correct — do not double-count fluorine.

Verify: experimental CaF₂ density $= 3.18$ g/cm³ ✓. The trap avoided: using $M$ of the formula unit with $Z$ = number of formula units keeps the 1:2 ratio automatic. ✓

Cell I — edge–radius geometry, both contact directions

Worked example 9 — Find

$r_{Na^{+}}$ in NaCl given $a = 564$ pm and $r_{Cl^{-}} = 181$ pm. Forecast: should come out near 95–100 pm (the known sodium ion radius).

Step 1. In NaCl, Na⁺ (edge centre) and Cl⁻ (corners) touch along the cube edge (red line in the figure). So the edge holds two anion radii plus two cation radii? No — look carefully: along the edge we pass Cl⁻ (corner) → Na⁺ (edge midpoint) → Cl⁻ (other corner), i.e. $r_{-} + 2 r_{+} + r_{-} = 2 r_{-} + 2 r_{+}$ . Why this step? We must trace which ions lie on the contact line — geometry, not memorised formula. $a = 2 (r_{+} + r_{-})$

Step 2. Solve for $r_{+}$ : $r_{+} = \frac{a}{2} - r_{-} = \frac{564}{2} - 181 = 282 - 181 = 101 pm$ Why this step? One equation, one unknown.

Step 3 (contrast — CsCl). In CsCl the ions touch along the body diagonal (cyan line in the figure), length $3 a$ , so $3 a = 2 (r_{+} + r_{-})$ . Why this step? Different structure → different contact direction. NaCl uses the edge; CsCl uses the body diagonal; ZnS uses a quarter of the body diagonal. Never assume the edge.

Verify: $r_{+} = 101$ pm ≈ the tabulated Na⁺ radius (95–102 pm) ✓. Check $a$ : $2 (101 + 181) = 2 \times 282 = 564$ pm ✓.

Cell J — exam twist / degenerate case

Worked example 10 — "A cubic crystal has an ion of radius 0 pm at the centre of a cube of 8 anions. What CN is predicted, and is the crystal stable?"

Forecast: a point-cation — degenerate. What breaks?

Step 1. $ρ = r_{+} / r_{-} = 0/ r_{-} = 0$ . Why this step? A zero-radius cation is the limiting degenerate input — worth knowing what the rule says at the extreme.

Step 2. $ρ = 0$ is below every wall (below even 0.155). The rule predicts CN 2 or less — the cation is too small to hold any cage open. Why this step? As $ρ \to 0$ , the cation cannot push anions apart; they collapse onto each other and the cation "rattles" freely in the hole.

Step 3. Physically: a point cation gives no anion–cation contact stabilisation for large CN, so a CN-8 cube is unstable. The structure would distort to a lower CN (or the anions would simply touch and the tiny cation contributes almost nothing to the lattice). So the naive "cube → CN 8" fails at the degenerate limit. Why this step? This is the exam-twist lesson: the radius-ratio rule is a stability window, not a guarantee — its lower wall exists precisely to reject cases like this.

Verify: at $ρ = 0$ the cubic contact equation $2 (r_{+} + r_{-}) = 3 (2 r_{-})$ becomes $2 r_{-} = 23 r_{-}$ , i.e. $1 = 3$ — false. The equality cannot hold, confirming CN 8 is geometrically impossible for $ρ = 0$ . ✓

Connections

Coordination number — the CN each ratio band produces.
Close packing FCC HCP and voids — the holes (tetrahedral/octahedral) cations sit in.
Unit cell and Z calculation — how the $Z$ counts in Ex 8 are done.
Density of crystals formula — the $ρ = Z M / N_{A} a^{3}$ used in Ex 6–8.
Parent topic.