Structure of Ionic Compounds

Q: What is the radius ratio rule in ionic solids?

The radius ratio rule says that the ratio of cation radius to anion radius decides how many anions surround a cation. A ratio of 0.225 to 0.414 gives coordination 4, 0.414 to 0.732 gives 6 and 0.732 to 1 gives 8. It is a first guess, since ions are not perfectly hard spheres.

Q: What is the structure of NaCl?

In sodium chloride the chloride ions form a face-centred cubic lattice and sodium ions fill all the octahedral voids. Each ion has six neighbours of the other kind, so the coordination is 6 to 6, and each unit cell holds four NaCl formula units. The edge length equals twice the sum of the ionic radii.

Q: What is the difference between the NaCl and CsCl structures?

NaCl has fcc chloride ions with sodium in octahedral voids, coordination 6 to 6 and four formula units per cell. CsCl has chloride ions at the corners of a simple cube with caesium at the centre, coordination 8 to 8 and one formula unit per cell. CsCl forms because the caesium ion is large, with radius ratio about 0.93.

Q: Why is CsCl not a body-centred cubic lattice?

A body-centred cubic lattice needs identical particles at the corners and at the body centre. In CsCl the corners hold chloride ions and the centre holds a caesium ion, so the lattice is simple cubic with a two-ion basis.

Q: What is the coordination number in fluorite and antifluorite?

In fluorite, CaF2, calcium ions form fcc and fluoride ions fill all tetrahedral voids, so calcium has coordination 8 and fluoride 4. In antifluorite, such as Na2O, the roles are reversed: oxide ions are fcc, sodium ions fill all tetrahedral voids, and the coordination is 4 for sodium and 8 for oxide.

Q: How does pressure affect the coordination number of an ionic crystal?

Pressure pushes ions closer together and raises the coordination number, so NaCl changes to the CsCl type at very high pressure. Heating expands the lattice and lowers the coordination number, so CsCl changes to the NaCl type at about 760 K.

Q: Is the structure of ionic compounds in the NEET or JEE Main syllabus?

No. The whole Solid State chapter, including ionic structures, was removed from NEET and JEE Main from 2024. JEE Advanced 2026 still lists ionic radii, radius ratio and simple ionic compounds, so NaCl, CsCl, ZnS and CaF2 structures remain important there.

Q: What is asked about ionic structures in JEE Advanced?

JEE Advanced questions ask for coordination numbers, the number of formula units per cell, edge length from ionic radii, density, radius ratio predictions, void occupancy in structures like spinel and fluorite, and nearest-neighbour distances in NaCl, CsCl and ZnS.

ChemistrySolid StateFor JEE aspirants

The structure of ionic compounds follows one idea: the larger ions (usually anions) form a lattice and the smaller ions sit in its voids. The cation-to-anion radius ratio $r_{+} / r_{-}$ decides which void is filled and so the coordination number. This page builds the rock salt (NaCl, 6 : 6), zinc blende (ZnS, 4 : 4), fluorite ( $Ca F_{2}$ , 8 : 4), antifluorite ( $N a_{2} O$ ) and caesium chloride (CsCl, 8 : 8) structures, plus spinel, perovskite and diamond. The structure of ionic compounds is tested only in JEE Advanced now.

On this page1Radius ratio rule2Two coordination numbers3NaCl4ZnS5

Ca F_{2}

and

N a_{2} O

6CsCl7Pressure and temperature8Spinel, perovskite, diamond

Key Formulas - Quick Reference

★ Must learn Radius ratio → CN: $0.225$ - $0.414$ → 4 (tetrahedral); $0.414$ - $0.732$ → 6 (octahedral); $0.732$ - $1$ → 8 (cubic).
★ Must learn Edge relations: NaCl $a = 2 (r_{+} + r_{-})$ ; CsCl $\frac{3}{2} a = r_{+} + r_{-}$ ; ZnS and $Ca F_{2}$ $\frac{3}{4} a = r_{+} + r_{-}$ .
★ Must learn Formula units per cell $Z$ : NaCl 4, ZnS 4, $Ca F_{2}$ 4, $N a_{2} O$ 4, CsCl 1, diamond 8 atoms.
★ Must learn Density $d = \frac{Z M}{N _{A} a ^{3}}$ with $M$ = formula mass and $Z$ = formula units per cell.
Coordination: NaCl 6 : 6, ZnS 4 : 4, $Ca F_{2}$ 8 : 4, $N a_{2} O$ 4 : 8, CsCl 8 : 8.
Charge balance of coordination: (CN of cation) × (number of cations) = (CN of anion) × (number of anions).
High pressure raises CN (NaCl → CsCl type); high temperature lowers it (CsCl → NaCl type).

1. The Radius Ratio Rule

In a binary ionic solid there are two kinds of ions of different size. The bigger ions (generally the anions) pack into a lattice; the smaller ions (generally the cations) go into its voids. Which void a cation takes depends on how big it is compared with the anion, measured by the radius ratio $r_{+} / r_{-}$ .

Rule 1: each cation should be surrounded by as many anions as possible (and vice versa), to maximise attraction.
Rule 2: anion-anion and cation-cation contact should be avoided, to minimise repulsion. So the cation must be large enough to touch all its anions and hold them apart.

Figure 1: Why the radius ratio matters. An arrangement is stable while the cation still touches all its anions. Below the limiting ratio the cation rattles and like-charged anions press together, so a lower coordination number is adopted instead.

The critical size is exactly the limiting ratio of the void (see Interstitial Voids). Above it the arrangement is stable; below it the cation rattles, the anions touch and repel, and a lower coordination is preferred.

Figure 2: The radius-ratio rule on one scale. The boundaries are the limiting void ratios; the larger the cation relative to the anion, the more anions fit around it.

Radius ratio $r_{+} / r_{-}$	Void occupied	CN	Example
$< 0.155$	linear	2	$Be F_{2}$ (gas), $H F_{2}^{-}$
$0.155$ to $0.225$	triangular	3	$B_{2} O_{3}$
$0.225$ to $0.414$	tetrahedral	4	ZnS ( $0.40$ ), CuCl
$0.414$ to $0.732$	octahedral	6	NaCl ( $0.52$ ), MgO ( $0.46$ )
$0.732$ to $1$	cubic	8	CsCl ( $0.93$ ), $Ca F_{2}$ ( $0.73$ )

Limits of the rule. The rule treats ions as hard spheres and ignores covalent character, so it is only a first guess. RbCl (

r_{+} / r_{-} = 0.82

) and KCl (

0.73

) should adopt the CsCl structure but crystallise as rock salt; LiI (

0.28

) should be tetrahedral but is rock salt too (its huge iodide ions touch). Polarisation also pulls compounds such as ZnS and AgI towards lower coordination than their ratio suggests. In JEE Advanced, apply the rule unless the question states the real structure.

Key idea

r_{+} / r_{-}

decides the void: larger cation → larger void → higher CN.

2. Two Coordination Numbers

An ionic solid has two coordination numbers: the CN of the cation (number of anions touching it) and the CN of the anion (number of cations touching it). They are linked by the formula, because every cation-anion contact is counted once from each side:

CN_{+} \times (number of cations) = CN_{-} \times (number of anions)

For AB compounds the two are equal (6 : 6, 4 : 4, 8 : 8). For $A B_{2}$ the cation's CN is twice the anion's (8 : 4 in $Ca F_{2}$ ); for $A_{2} B$ it is the reverse (4 : 8 in $N a_{2} O$ ). For $Ba F_{2}$ , $CN (B a^{2 +}) = 8$ , so $CN (F^{-}) = \frac{8 \times 1}{2} = 4$ .

Exam Trick The formula ratio flips the CN ratio.

Ca F_{2}

is 1 : 2 in ions and 8 : 4 in coordination. If you know one CN and the formula, you know the other.

3. Rock Salt Structure (NaCl, 6 : 6)

Figure 3: Rock salt.

C l^{-}

ions form an fcc lattice and

N a^{+}

ions fill every octahedral void (edge centres and body centre). Each ion has 6 of the other kind around it, so the structure is 6 : 6.

Lattice: $C l^{-}$ ions form an fcc lattice; $N a^{+}$ ions fill all the octahedral voids (edge centres and body centre). Equivalently, $N a^{+}$ fcc with $C l^{-}$ in the octahedral voids: two interpenetrating fcc lattices.
Ions per cell: $C l^{-}$ $= 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$ ; $N a^{+}$ $= 12 \times \frac{1}{4} + 1 = 4$ . Cell content $N a_{4} C l_{4}$ , so $Z = 4$ formula units.
Radius ratio: $\frac{r _{N a^{+}}}{r _{C l^{-}}} = \frac{95}{181} = 0.52$ , inside the octahedral range.
Edge relation: ions touch along the edge: $a = 2 (r_{N a^{+}} + r_{C l^{-}})$ . The $C l^{-}$ ions do not touch: $2 a \neq = 4 r_{C l^{-}}$ .
Coordination: each $N a^{+}$ has 6 $C l^{-}$ and each $C l^{-}$ has 6 $N a^{+}$ (octahedral): 6 : 6.
Examples: halides of Li, Na, K and Rb; AgCl, AgBr; MgO, CaO, FeO, NiO.

Figure 4: In NaCl the ions touch along the cube edge, so

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

. The chloride ions on a face diagonal do not touch, because the ratio 0.52 is above the limiting 0.414.

Neighbours of $C l^{-}$	Ion	Distance	Number
Nearest	$N a^{+}$	$\frac{a}{2}$	6
2nd	$C l^{-}$	$\frac{a}{2}$	12
3rd	$N a^{+}$ (body centre of a neighbouring cell)	$\frac{3}{2} a$	8
4th	$C l^{-}$	$a$	6

Key idea

NaCl: fcc anions + all octahedral voids,

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

Z = 4

, 6 : 6.

4. Zinc Blende (ZnS, 4 : 4) and Wurtzite

Figure 5: Zinc blende.

S^{2 -}

ions form fcc and

Z n^{2 +}

ions occupy alternate tetrahedral voids (4 of 8). Each ion is tetrahedrally surrounded by 4 of the other: 4 : 4.

Lattice: $S^{2 -}$ ions form fcc; $Z n^{2 +}$ ions occupy alternate tetrahedral voids (4 of the 8, no two in adjacent minicubes).
Ions per cell: 4 $S^{2 -}$ and 4 $Z n^{2 +}$ : $Z = 4$ .
Radius ratio: $\frac{74}{184} = 0.40$ , tetrahedral range, so the anions do not touch.
Edge relation: a $Z n^{2 +}$ is a quarter of the body diagonal from a corner $S^{2 -}$ : $r_{+} + r_{-} = \frac{3}{4} a$ .
Coordination: 4 : 4, both tetrahedral.
Examples: ZnS, CuCl, CuBr, CuI, AgI (low temperature), BeO (wurtzite).

Neighbours of $S^{2 -}$	Ion	Distance	Number
Nearest	$Z n^{2 +}$	$\frac{3}{4} a$	4
2nd	$S^{2 -}$	$\frac{a}{2}$	12
3rd	$Z n^{2 +}$	$\frac{11}{4} a$	12
4th	$S^{2 -}$	$a$	6
5th	$Z n^{2 +}$	$\frac{19}{4} a$	12

ZnS has a second form, wurtzite: the $S^{2 -}$ ions are hexagonal close-packed instead of cubic, and $Z n^{2 +}$ again fills half the tetrahedral voids. Coordination stays 4 : 4. Zinc blende and wurtzite are two polymorphs of ZnS.

Zinc blende

$S^{2 -}$ in ccp (fcc). $Z n^{2 +}$ in half the tetrahedral voids. Cubic. 4 : 4, $Z = 4$ .

Wurtzite

$S^{2 -}$ in hcp. $Z n^{2 +}$ in half the tetrahedral voids. Hexagonal. 4 : 4.

Quick Recall: tap to check

In NaCl, which voids are filled and what fraction?

All the octahedral voids of the fcc

C l^{-}

lattice (4 of 4).

In zinc blende, what fraction of tetrahedral voids hold

Z n^{2 +}

One half (4 of 8).

Ba F_{2}

: CN of

B a^{2 +}

is 8. CN of

F^{-}

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

5. Fluorite ( $Ca F_{2}$ , 8 : 4) and Antifluorite ( $N a_{2} O$ , 4 : 8)

Figure 6: Fluorite.

C a^{2 +}

ions form fcc and

F^{-}

ions fill all 8 tetrahedral voids. Each

F^{-}

has 4

C a^{2 +}

around it and each

C a^{2 +}

has 8

F^{-}

: 8 : 4. Swap the roles of the ions and you get antifluorite,

N a_{2} O

(4 : 8).

Lattice: $C a^{2 +}$ ions form fcc; $F^{-}$ ions fill all 8 tetrahedral voids.
Ions per cell: 4 $C a^{2 +}$ , 8 $F^{-}$ : cell content $C a_{4} F_{8}$ , $Z = 4$ formula units of $Ca F_{2}$ .
Coordination: each $F^{-}$ sits in a tetrahedral void, so CN( $F^{-}$ ) = 4; hence CN( $C a^{2 +}$ ) = 8 (a cube of $F^{-}$ ).
Edge relation: $r_{C a^{2 +}} + r_{F^{-}} = \frac{3}{4} a$ .
Radius ratio: the cation is the ion that is coordinated: $\frac{r _{C a^{2 +}}}{r _{F^{-}}} = \frac{99}{136} = 0.73$ , at the cubic limit, which matches CN 8. (Seen from the anion side, the $F^{-}$ ions form a simple cube with $C a^{2 +}$ in every other cubic void.)
Examples: $Ca F_{2}$ , $Sr F_{2}$ , $Ba F_{2}$ , $Cd F_{2}$ , $SrC l_{2}$ .

Neighbours of $C a^{2 +}$	Ion	Distance	Number
Nearest	$F^{-}$	$\frac{3}{4} a$	8
2nd	$C a^{2 +}$	$\frac{a}{2}$	12
3rd	$F^{-}$	$\frac{11}{4} a$	24
4th	$C a^{2 +}$	$a$	6

Neighbours of $F^{-}$	Ion	Distance	Number
Nearest	$C a^{2 +}$	$\frac{3}{4} a$	4
2nd	$F^{-}$	$\frac{a}{2}$	6
3rd	$F^{-}$	$\frac{a}{2}$	12
4th	$C a^{2 +}$	$\frac{11}{4} a$	12

5.1 Antifluorite

Swap the roles: $O^{2 -}$ ions form fcc and $N a^{+}$ ions fill all the tetrahedral voids. This gives $N a_{2} O$ with 4 : 8 coordination (each $N a^{+}$ has 4 $O^{2 -}$ , each $O^{2 -}$ has 8 $N a^{+}$ ). Examples: $L i_{2} O$ , $N a_{2} O$ , $K_{2} O$ , $N a_{2} S$ .

6. Caesium Chloride Structure (CsCl, 8 : 8)

Figure 7: CsCl.

C l^{-}

ions sit at the corners of a simple cube and

C s^{+}

fills the cubic void at the centre, touching all 8 along the body diagonals:

\frac{3}{2} a = r_{+} + r_{-}

Lattice: $C l^{-}$ ions at the corners of a simple cube; one $C s^{+}$ in the cubic void at the body centre. (Or the reverse; the two views are equivalent.)
Ions per cell: 1 $C l^{-}$ ( $8 \times \frac{1}{8}$ ) and 1 $C s^{+}$ : $Z = 1$ .
Radius ratio: $\frac{169}{181} = 0.93$ , cubic range.
Edge relation: ions touch along the body diagonal: $3 a = 2 (r_{+} + r_{-})$ .
Coordination: 8 : 8.
Examples: CsCl, CsBr, CsI, TlCl, and $N H_{4} Cl$ at room temperature.

CsCl is not body-centred cubic. A bcc lattice needs identical particles at the corners and the centre. Here the centre holds a different ion, so the lattice is simple cubic with a two-ion basis.

NaCl (rock salt)

Anions fcc, cations in octahedral voids. $r_{+} / r_{-} = 0.52$ . 6 : 6. $Z = 4$ . $a = 2 (r_{+} + r_{-})$ .

CsCl

Anions simple cubic, cation in the cubic void. $r_{+} / r_{-} = 0.93$ . 8 : 8. $Z = 1$ . $3 a = 2 (r_{+} + r_{-})$ .

Structure	Lattice ion (arrangement)	Voids filled	CN	$Z$	Edge relation
NaCl	$C l^{-}$ fcc	all octahedral	6 : 6	4	$a = 2 (r_{+} + r_{-})$
ZnS (zinc blende)	$S^{2 -}$ fcc	half tetrahedral	4 : 4	4	$\frac{3}{4} a = r_{+} + r_{-}$
$Ca F_{2}$	$C a^{2 +}$ fcc	all tetrahedral ( $F^{-}$ )	8 : 4	4	$\frac{3}{4} a = r_{+} + r_{-}$
$N a_{2} O$	$O^{2 -}$ fcc	all tetrahedral ( $N a^{+}$ )	4 : 8	4	$\frac{3}{4} a = r_{+} + r_{-}$
CsCl	$C l^{-}$ simple cubic	cubic void	8 : 8	1	$\frac{3}{2} a = r_{+} + r_{-}$

Exam Trick Which diagonal do the ions touch along? Octahedral (NaCl): the edge, so

a

. Tetrahedral (ZnS, fluorite): a quarter of the body diagonal,

\frac{3}{4} a

. Cubic (CsCl): half the body diagonal,

\frac{3}{2} a

Key idea

Five structures, one pattern: lattice ion + which voids + how many fixes the formula, CN and edge relation.

7. Effect of Pressure and Temperature

Coordination number can change with conditions. Pressure squeezes the ions together and favours structures with more neighbours; heating expands the lattice and favours fewer.

Figure 8: Coordination numbers are not fixed. Pressure pushes a structure to higher coordination; heating pulls it back to lower coordination.

Pressure increases CN: NaCl (6 : 6) changes to the CsCl type (8 : 8) under very high pressure; 4 : 4 structures can go to 6 : 6.
Temperature decreases CN: CsCl (8 : 8) changes to the NaCl type (6 : 6) on heating to about 760 K.

Quick Recall: tap to check

Why is CsCl not called bcc?

The body centre holds

C s^{+}

and the corners hold

C l^{-}

; bcc needs identical points.

Edge relation in fluorite?

r_{C a^{2 +}} + r_{F^{-}} = \frac{3}{4} a

What does heating CsCl to about 760 K do?

It converts the 8 : 8 structure into the 6 : 6 rock-salt type.

8. Spinel, Perovskite and Diamond

8.1 Spinel, $A B_{2} O_{4}$

Oxide ions are cubic close-packed. In a normal spinel such as $MgA l_{2} O_{4}$ , the $M g^{2 +}$ ions occupy one eighth of the tetrahedral voids and the $A l^{3 +}$ ions one half of the octahedral voids. Check with $N$ oxide ions: $M g^{2 +}$ $= \frac{1}{8} (2 N) = \frac{N}{4}$ , $A l^{3 +}$ $= \frac{1}{2} N$ , ratio $1 : 2 : 4$ . Another example: $ZnA l_{2} O_{4}$ .

8.2 Perovskite, $AB O_{3}$

In the cubic cell of perovskite ( $CaTi O_{3}$ , $BaTi O_{3}$ , $SrTi O_{3}$ ) the large $A^{2 +}$ cation sits at the corners, oxide ions at the face centres and the small $T i^{4 +}$ at the body centre. Count: A $= 1$ , O $= 3$ , Ti $= 1$ , so $AB O_{3}$ . $T i^{4 +}$ is octahedral (CN 6) and the corner cation has CN 12.

8.3 Diamond

Figure 9: Two more cubic structures. Perovskite (

AB O_{3}

) puts the small

B

cation in an octahedron of oxides. Diamond is zinc blende with every site filled by carbon: 8 atoms per cell, each bonded tetrahedrally to 4.

Carbon atoms occupy the fcc sites and alternate tetrahedral voids (zinc blende with every site carbon): $Z = 4 + 4 = 8$ .
Each C is bonded to 4 others tetrahedrally; nearest C-C distance $d = \frac{3}{4} a$ , so $a = \frac{4}{3} d$ , and $d = 2 r_{C}$ .
Packing efficiency $= \frac{8 \times \frac{4}{3} π r ^{3}}{a ^{3}}$ with $a = \frac{8 r}{3}$ , which gives $\frac{3 π}{16} = 34%$ .

JEE Advanced In an inverse spinel such as magnetite

F e_{3} O_{4}

(

F e^{3 +} [F e^{2 +} F e^{3 +}] O_{4}

), the tetrahedral sites hold half the

F e^{3 +}

ions, and the octahedral sites hold the

F e^{2 +}

ions and the other half of the

F e^{3 +}

. The octahedral crystal-field preference of

F e^{2 +}

d^{6}

ion) drives this swap, and the resulting uncompensated spins make magnetite ferrimagnetic.

Key idea

Spinel = 1/8 Td + 1/2 Oh in ccp oxide; perovskite = corner A, face O, body B; diamond = ZnS with all sites carbon.

Quick Recall: tap to check

Formula units in a diamond cell?

8 carbon atoms (4 fcc + 4 in alternate tetrahedral voids).

Coordination of Ti in

BaTi O_{3}

6 (octahedron of face-centre oxide ions).

Fraction of voids filled in normal spinel

MgA l_{2} O_{4}

M g^{2 +}

\frac{1}{8}

of tetrahedral,

A l^{3 +}

\frac{1}{2}

of octahedral voids.

9. Predicting a Structure

Figure 10: Predicting a structure from the radius ratio. Use it as a first guess: real crystals such as KCl and RbCl break the rule.

9.1 The whole concept at a glance

Figure 11: The ionic structures on one page: lattice ion, void filled, coordination, edge relation and examples.

10. Solved Examples

Solved Example 1

A mineral with formula

A B_{2}

crystallises in the cubic close-packed lattice, with the A atoms at the lattice points. What is the coordination number of the A atoms and of the B atoms? What fraction of the tetrahedral sites is occupied by B atoms?

Solution:

For $N$ A atoms there are $2 N$ tetrahedral sites, and the formula needs $2 N$ B atoms, so B fills all (100%) of the tetrahedral sites.

B in a tetrahedral site has CN 4; then CN(A) $= \frac{4 \times 2}{1} = 8$ . This is the fluorite ( $Ca F_{2}$ ) structure, 8 : 4.

Solved Example 2

CsBr has a CsCl-type (body-centred) structure with edge length 4.3 Å. The shortest inter-ionic distance between

C s^{+}

and

B r^{-}

is
(A) 3.72 Å
(B) 1.86 Å
(C) 7.44 Å
(D) 4.3 Å

Solution:

Answer: (A).

r_{+} + r_{-} = \frac{3 a}{2} = \frac{1.732 \times 4.3}{2} = 3.72 \overset{˚}{A}

Solved Example 3

Calculate the number of formula units in each unit cell: (a) MgO in a rock-salt cell, (b) ZnS in the zinc blende structure, (c) platinum in an fcc unit cell.

Solution:

(a) 4, as in NaCl. (b) 4. (c) 4 atoms ( $8 \times \frac{1}{8}$ corners + $6 \times \frac{1}{2}$ face centres).

Solved Example 4

MgO has the structure of NaCl and TlCl has the structure of CsCl. What are the coordination numbers of the ions in each?

Solution:

MgO (rock salt): CN of $M g^{2 +}$ = CN of $O^{2 -}$ = 6.

TlCl (CsCl type): CN of $T l^{+}$ = CN of $C l^{-}$ = 8.

Solved Example 5

A solid AB has the NaCl structure. If the radius of

A^{+}

is 120 pm, what is the maximum possible radius of the anion

B^{-}

Solution:

For an octahedral site, $r_{+} / r_{-}$ must be at least 0.414. The anion is largest when the ratio is at this minimum:

r_{B^{-}} = \frac{r _{A^{+}}}{0.414} = \frac{120}{0.414} = 290 pm

Solved Example 6

The coordination number of

B a^{2 +}

Ba F_{2}

is 8. What must be the coordination number of

F^{-}

Solution:

Each $B a^{2 +}$ touches 8 $F^{-}$ . To balance charge, each $F^{-}$ must be shared among cations so that $CN (B a^{2 +}) \times 1 = CN (F^{-}) \times 2$ , giving CN( $F^{-}$ ) = 4.

Solved Example 7

The radius of the calcium ion is 94 pm and that of the oxide ion is 146 pm. Predict the crystal structure of calcium oxide.

Solution:

\frac{r _{+}}{r _{-}} = \frac{94}{146} = 0.644

The ratio lies between 0.414 and 0.732: octahedral coordination. Since the ions have equal and opposite charges, the oxide ions are also octahedrally surrounded, so CaO has the rock-salt (NaCl) structure, 6 : 6.

Solved Example 8

CsCl crystallises with

C l^{-}

at each corner and

C s^{+}

at the centre of the cubic cell. If

r (C s^{+}) = 1.69

Å and

r (C l^{-}) = 1.81

Å, what is the edge length?

Solution:

The closest $C s^{+}$ - $C l^{-}$ distance is half the body diagonal: $\frac{3}{2} a = 1.69 + 1.81 = 3.50$ Å.

a = \frac{2 \times 3.50}{3} = 4.04 \overset{˚}{A}

Solved Example 9

Calculate the edge length of the NaCl unit cell given that its density is

2.17 \times 1 0^{3}

kg m

^{- 3}

and its molar mass is

58.5 \times 1 0^{- 3}

kg mol

^{- 1}

Solution:

NaCl is fcc with $Z = 4$ formula units per cell.

a^{3} = \frac{Z M}{d N _{A}} = \frac{4 \times 58.5 \times 1 0 ^{- 3}}{2.17 \times 1 0 ^{3} \times 6.022 \times 1 0 ^{23}} = 1.79 \times 1 0^{- 28} m^{3}

$a = 5.64 \times 1 0^{- 10}$ m $= 564$ pm.

Solved Example 10

Given

r (M g^{2 +}) = 65

pm and

r (O^{2 -}) = 140

pm, predict the coordination number of

M g^{2 +}

and the structure of MgO.

Solution:

$\frac{r _{+}}{r _{-}} = \frac{65}{140} = 0.46$ , between 0.414 and 0.732: octahedral, CN 6. MgO has the rock-salt structure, as observed.

Solved Example 11

In NaCl the edge length is 5.64 Å. If

r (N a^{+}) = 0.95

Å, the radius of

C l^{-}

is
(A) 1.87 Å
(B) 1.82 Å
(C) 2.82 Å
(D) 3.77 Å

Solution:

Answer: (A). $a = 2 (r_{+} + r_{-})$ gives $r_{+} + r_{-} = 2.82$ Å, so $r_{-} = 2.82 - 0.95 = 1.87$ Å. Option (C) forgets to subtract the cation radius.

Solved Example 12

Zinc blende has edge length 5.41 Å. Calculate its density (ZnS,

M = 97.45

g mol

^{- 1}

Solution:

d = \frac{4 \times 97.45}{( 5.41 \times 1 0 ^{- 8} ) ^{3} \times 6.022 \times 1 0 ^{23}} = 4.09 g cm^{- 3}

This matches the measured density of sphalerite (4.09 g cm $^{- 3}$ ).

Solved Example 13

In zinc blende, the fraction of tetrahedral voids occupied by

Z n^{2 +}

ions is
(A)

\frac{1}{4}

(B)

\frac{1}{2}

\frac{1}{8}

Solution:

Answer: (B). 4 $S^{2 -}$ per cell create 8 tetrahedral voids; 4 $Z n^{2 +}$ fill half of them. In fluorite, by contrast, all tetrahedral voids are filled.

Solved Example 14

Diamond has a cubic cell of edge 3.567 Å. Find the C-C bond length.

Solution:

Each C in a tetrahedral void is a quarter of the body diagonal from its fcc neighbour:

d_{C - C} = \frac{3}{4} a = \frac{1.732 \times 3.567}{4} = 1.545 \overset{˚}{A}

Practice Questions

$r (L i^{+}) = 60$ pm and $r (I^{-}) = 216$ pm. What CN does the radius ratio predict for LiI?Answer: $0.28$ : CN 4 predicted (LiI is actually rock salt; the rule fails).
KBr has the rock-salt structure with $r (K^{+}) = 1.33$ Å and $r (B r^{-}) = 1.95$ Å. Find the edge length.Answer: $a = 2 (1.33 + 1.95) = 6.56$ Å.
CsI has the CsCl structure; $r (C s^{+}) = 1.69$ Å, $r (I^{-}) = 2.16$ Å. Find a.Answer: $a = \frac{2 \times 3.85}{3} = 4.45$ Å.
Give the coordination numbers of $N a^{+}$ and $O^{2 -}$ in $N a_{2} O$ .Answer: $N a^{+}$ 4, $O^{2 -}$ 8.
KCl (rock salt) has a = 6.29 Å and M = 74.55 g mol $^{- 1}$ . Find its density.Answer: $1.99$ g cm $^{- 3}$ .
How many unit cells are there in 1.00 g of NaCl?Answer: $\frac{1.00}{58.5} \times \frac{6.022 \times 1 0 ^{23}}{4} = 2.57 \times 1 0^{21}$ .
Name the voids filled and the fractions in normal spinel.Answer: $\frac{1}{8}$ of tetrahedral voids by $A^{2 +}$ , $\frac{1}{2}$ of octahedral voids by $B^{3 +}$ .

Common Mistakes to Avoid

Watch out

Calling CsCl bcc. Its centre and corners hold different ions, so it is simple cubic with two ions.
Using $a = 2 (r_{+} + r_{-})$ for CsCl or ZnS. That holds only for rock salt; CsCl uses $\frac{3}{2} a$ and ZnS $\frac{3}{4} a$ .
Writing the radius ratio upside down. It is always $r_{cation} / r_{anion}$ for the ion being coordinated.
Saying $Z n^{2 +}$ fills all tetrahedral voids in ZnS. It fills half; all voids are filled only in fluorite and antifluorite.
Taking $Z = 8$ for NaCl because it has 8 ions in total. $Z$ counts formula units: 4.
Mixing fluorite (8 : 4, cation fcc) with antifluorite (4 : 8, anion fcc).
Using the atomic mass of one element in $d = Z M / (N_{A} a^{3})$ . Use the formula mass (58.5 for NaCl).
Assuming higher temperature raises CN. Heating lowers CN (CsCl → NaCl type); pressure raises it.

Frequently Asked Questions

What is the radius ratio rule in ionic solids?

The radius ratio rule says that the ratio of cation radius to anion radius decides how many anions surround a cation. A ratio of 0.225 to 0.414 gives coordination 4, 0.414 to 0.732 gives 6 and 0.732 to 1 gives 8. It is a first guess, since ions are not perfectly hard spheres.

What is the structure of NaCl?

In sodium chloride the chloride ions form a face-centred cubic lattice and sodium ions fill all the octahedral voids. Each ion has six neighbours of the other kind, so the coordination is 6 to 6, and each unit cell holds four NaCl formula units. The edge length equals twice the sum of the ionic radii.

What is the difference between the NaCl and CsCl structures?

NaCl has fcc chloride ions with sodium in octahedral voids, coordination 6 to 6 and four formula units per cell. CsCl has chloride ions at the corners of a simple cube with caesium at the centre, coordination 8 to 8 and one formula unit per cell. CsCl forms because the caesium ion is large, with radius ratio about 0.93.

Why is CsCl not a body-centred cubic lattice?

A body-centred cubic lattice needs identical particles at the corners and at the body centre. In CsCl the corners hold chloride ions and the centre holds a caesium ion, so the lattice is simple cubic with a two-ion basis.

What is the coordination number in fluorite and antifluorite?

In fluorite, CaF2, calcium ions form fcc and fluoride ions fill all tetrahedral voids, so calcium has coordination 8 and fluoride 4. In antifluorite, such as Na2O, the roles are reversed: oxide ions are fcc, sodium ions fill all tetrahedral voids, and the coordination is 4 for sodium and 8 for oxide.

How does pressure affect the coordination number of an ionic crystal?

Pressure pushes ions closer together and raises the coordination number, so NaCl changes to the CsCl type at very high pressure. Heating expands the lattice and lowers the coordination number, so CsCl changes to the NaCl type at about 760 K.

Is the structure of ionic compounds in the NEET or JEE Main syllabus?

No. The whole Solid State chapter, including ionic structures, was removed from NEET and JEE Main from 2024. JEE Advanced 2026 still lists ionic radii, radius ratio and simple ionic compounds, so NaCl, CsCl, ZnS and CaF2 structures remain important there.

What is asked about ionic structures in JEE Advanced?

JEE Advanced questions ask for coordination numbers, the number of formula units per cell, edge length from ionic radii, density, radius ratio predictions, void occupancy in structures like spinel and fluorite, and nearest-neighbour distances in NaCl, CsCl and ZnS.

Ready to master Solid State?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test

1. The Radius Ratio Rule

2. Two Coordination Numbers

3. Rock Salt Structure (NaCl, 6 : 6)

4. Zinc Blende (ZnS, 4 : 4) and Wurtzite

5. Fluorite (CaF2​, 8 : 4) and Antifluorite (Na2​O, 4 : 8)

5.1 Antifluorite

6. Caesium Chloride Structure (CsCl, 8 : 8)

7. Effect of Pressure and Temperature

8. Spinel, Perovskite and Diamond

8.1 Spinel, AB2​O4​

8.2 Perovskite, ABO3​

8.3 Diamond

9. Predicting a Structure

9.1 The whole concept at a glance

10. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What is the radius ratio rule in ionic solids?

What is the structure of NaCl?

What is the difference between the NaCl and CsCl structures?

Why is CsCl not a body-centred cubic lattice?

What is the coordination number in fluorite and antifluorite?

How does pressure affect the coordination number of an ionic crystal?

Is the structure of ionic compounds in the NEET or JEE Main syllabus?

What is asked about ionic structures in JEE Advanced?

Ready to master Solid State?

5. Fluorite ( $Ca F_{2}$ , 8 : 4) and Antifluorite ( $N a_{2} O$ , 4 : 8)

8.1 Spinel, $A B_{2} O_{4}$

8.2 Perovskite, $AB O_{3}$