8.

Classification of Ionic Structures

In any solid of the type AxBy, the ratio of the coordination number of A to that of
B would be y : x.

8.1 Rock Salt Structure

NaCl exhibits this type of structure. In the rock Cl

Na
+
–

salt structure, Cl− ions exists in FCC pattern and

Na+ ions occupy all octahedral voids. There are
4 effective Na+ ions and 4 effective Cl− ions in a
unit cell of NaCl. So, the general formula is
Na4Cl4 or NaCl as per the effective ions in a unit
cell. The co−ordination number of Na+ ion is 6
and co−ordination number of Cl− ion is also 6.
So, the general formula (using co−ordination
number of ions) is Na6Cl6 or NaCl.. The other
substances having this kind of a structure are
halides of all alkali metals except cesium halides
and oxides of all alkaline earth metals except
beryllium oxide.

8.2 Zinc Blende Structure

This structure is shown by ZnS. Sulphide ions

are face centered cubic position and Zinc is
present in alternate tetrahedral voids.
Formula is Zn4S4, i.e, ZnS. Coordination
number of Zn is 4 and that of sulphide is also
4. Other substance that exists in this kind of a
structure is BeO.

8.3 Fluorite Structures

This structure is exhibited by CaF2. Calcium ions are face centered cubic position and
fluoride ions are present in all the tetrahedral voids. There are four calcium ions and
eight fluoride ions per unit cell. Therefore the formula is Ca4F8, (i.e, CaF2). The
coordination number of fluoride ions is four (tetrahedral voids) and thus the
coordination number of calcium ions is eight. Other substances which exist in this
kind of structure are UO2, and ThO2.
 Unit cell representation of CaF 2 structure

8.4 Anti-Fluorite Structure

This structure is shown by Li2O. Oxide ions are face centered cubic
position and lithium ions are present in all the tetrahedral voids.
There are four oxide ions and eight lithium ions per unit cell. As it
can be seen, this unit cell is just the reverse of Fluorite structure, in
the sense that, the positions of cations and anions is interchanged.
Other substances which exist in this kind of a structure are Na2O,
K2O and Rb2O. Anti-fluorite structure

8.5 Cesium Halide Structure

This structure is shown by CsCl. Chloride ions are primitive

cubic while the Cesium ion occupies the center of the unit cell.
There is one chloride ion and one cesium ion per unit cell.
Therefore the formula is CsCl. The coordination number of
cesium is eight and that of chloride ions is also eight. Other
Cesium halide structure
substances which exist in this kind of a structure are all halides
of cesium.

8.6 Spinel and Inverse Spinel Structure

Spinel is a mineral (MgAl2O4). Generally they can be represented as M2+M23+O4,

where M2+ is present in one-eighth of tetrahedral voids in a FCC lattice of oxide ions
and M3+ ions are present in half of the octahedral voids. M2+ is usually Mg, Fe, Co,
Ni, Zn and Mn; M3+ is generally Al, Fe, Mn, Cr and Rh. Examples are ZnAl2O4, Fe3O4,
FeCr2O4 etc. Many substances of the type M M O also have this structure. In an
4+ 2+
2 4

inverse spinel the ccp is of oxide ions, M2+ is in one-eight of the tetrahedral voids
while M3+ would be in one-eight of the tetrahedral voids and one-fourth of the
octahedral voids.
Example 1

Iron crystallizes in FCC lattice. The figures given below shows the iron atoms
in four crystallographic planes.

Draw the unit cell for the corresponding structure and identify these planes in
the diagram. Also report the distance between two such crystallographic
planes in each case in terms of the edge length (a) of the unit cell.
Solution:

Distance between two Distance between two Distance between two

such planes = a such planes = a such planes = a

A C

3a a
Distance between two such planes = =
3 3
Note: Second diagram is FCC; Third diagram is Diagonal plane
Example 2:

The figures given show the location of atoms in three crystallographic planes in an
f.c.c. lattice. Draw the unit cell for the corresponding structure and identify these
planes in your diagram.

Answer
Hint: Crystallographic planes in the simple definition are a set of parallel and equally
spaced planes that pass through the centers of atoms in a simple cube. In the fcc
lattice there are atoms present at every corner of the cube along with one atom at
the center of each face.

Complete answer: In the question above, we are given the location of atoms in a
plane and we have to draw the unit cells that might give us these arrangements and
identify those planes. We might have to imagine where these atoms may be present
in a simple cube to draw the planes correctly. Let us try to draw by looking at the
image one by one:
The first diagram looks like it is the face of one FCC lattice because it has four atoms
at the corners and one at the center. Like in this figure:

This plane is a face plane.

For the second diagram, you can say that it is not in a single face. The plane should
pass the center of each face to get the diagram correctly. Let us try to draw a plane
for this set of atoms. And the plane comes out to be like this:
This is a triangular plane
Now for the third diagram it seems that the plane is a diagonal plane. The planes will
pass through the center of the top and the bottom face. And the unit cell for this
diagram would be:

This plane is a diagonal face plane.

Note:
There is more than one plane of one type in a unit cell. Above only one plane is
shown for the three types but there are more. The only thing you should be sure of is
that the planes should pass through the center of each face because this is a fcc
lattice.

Example 3:

Metallic gold crystallises in FCC lattice. The edge length of the cubic unit
cell is 4.07Å.
(a) What is the closest distance between gold atoms?
(b) How many “nearest neighbours” does each gold atom have at
the distance calculated
in (a)?
(c) What is the density of gold?
(d) Prove that the packing fraction of gold is 0.74.

Solution:
(a) In FCC lattice, the face centered atom touches four corner atoms of
that face.
2a a
 2 a= 4r or, 2r = =
2 2
 a 4.07
Closest distance between two gold atoms = 2r = = = 2.87 Å
2 2

(b) If we consider a face centered gold atom, it has four corner and eight
a
adjacent face centre atoms present at distance. Therefore, there
2
are 12 nearest neighbours.
nM 4  197
(c) Density () = = = 19.4 g/cc
N AV  a 3
6.023  10  ( 4.07  10 −8 ) 3
23

4
4  r3
3 
(d) Packing fraction = =
3
 4r  3 2
 
 
 2
Example 4:
If NaCl is doped with 10−3 mol % SrCl2, calculate the concentration
of cation vacancies?
Solution:
Due to addition of SrCl2, each Sr2+ ion replaces two Na+ ions, but
occupies only one Na+ lattice point. This creates one cationic vacancy.
Number of moles of Sr2+ present in 100 mole of NaCl = 10−3
Number of moles of cation vacancy in 100 mole of NaCl = 10−3
10 −3
Number of moles of cation vacancies in 1 mol = = 10 −5 mol
100

Total cationic vacancies = 10−5  NAV = 10−5  6.02  1023 = 6.02  1018

Example 5:
A substance form face−centered cubic crystals. Its density is 1.984 g
cm−3 and edge length of the unit cell is 630 pm. Calculate the molar
mass of the substance.
Solution:
nM   a 3  N AV
Since = , we get M=
N AV  a 3 n

Substituting the values, we have

1.984  (6.30  10 −8 ) 3  6.023  10 23
M= = 74.7 g mol−1
4

Example 6:

An ionic compound AB has ZnS type structure. If the radius A+ is 22.5 pm,
then the ideal radius of B− would be
 (a) 54.35 pm (b) 100 pm
(c) 145.16 pm (d) None of these

Solution:
Since ionic compound, AB has ZnS type structure, therefore it has
tetrahedral voids, for which ideal radius ratio is 0.225.
rA +
 = 0.225
rB −

22.5
 rB − = = 100 pm
0.225

Example 7:
If the edge length of the unit cell of sodium chloride is 600 pm, and the
ionic radius of Cl− ion is 190 pm, then the ionic radius of Na+ ion is
(a) 310 pm (b) 110 pm
(c) 220 pm (d) none of these
Solution:
a
As, r + rCl− =
Na + 2
600
 r
Na +
+ 190 = = 300
2

r = 300 − 190 = 110 pm

Na +

Example 8:

A compound XY crystallizes in BCC lattice with unit cell edge length

of 480 pm. If the radius of Y− is 225 pm, then the radius of X+ is
(a) 190.68 pm (b) 225 pm
(c) 127.5 pm (d) None of these
Solution:
For bcc structure,
2 (rc + ra ) = 3 a

where rc and ra represents radius of cation and anion respectively.

or, 2 (rc + 225) = 3  480

 rc = 190.68 pm
Example 9:

Example 10:

Example 11:
Example 12:

Example 13:
Example 14:

Example 15:

Example 16:
Correct option (d)
Explanation: Figure (d) represents cross section of an octahedral site. The interstitial
void formed by combination of two triangular voids of the first and second layer is
called octahedral layer.

Example 17:
Example 18:

Example 19:
Example 20:

A metal crystallises into a lattice containing a sequence of layers as ABABAB ... .

What percentage of voids are left in the lattice?
ABAB packing is hexagonal closepacking in which all atoms occupy 74% of the total
space. Hence, 26% of voids are left in the lattice.

Example 21:

Copper crystallises in fcc with a unit cell length of 361 pm. What is the radius of
copper atom

Example 22:

Answer: C
Example 23:

Example 24:
Example 25:

The correct statement(s) regarding defects in solids is (are)

(a) Frenkel defect is usually favoured by a very small difference in the sizes of cation
and anion.
(b) Frenkel defect is a dislocation defect
(c) Trapping of an electron in the lattice leads to the formation of F-center
(d) Schottky defects have no effect on the physical properties of solids.
Answer: BC