Std. XII Sci. Success Chemistry - I
2.
Formulae
1.
Density of Unit Cell :
Mass of unit cell
=
Volume of unit cell
zM
g cm–3
a3  N0
a = Edge length of unit cell in cm
z = number of atoms per unit cell
M = Molar mass
N0 = Avagadro’s number (6.023  1023)
=
2.
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Silver crystallises in fcc type and has
density 19.4 g cm3 . If the edge length of
unit cell is 407.2 pm, calculate Avogadro
number.
[Atomic mass of Au is 197 gmol1]
Solution :
Type of crystal = fcc
d = 19.4 g cm3
a = 407.2 pm = 4.072  108 cm
z=4
Avogadro number = NA = ?
Number of atoms per unit cell (z)
Simple cubic (z = 1)
Body centred cubic (z = 2)
Face centred cubic (z = 4)
Edge centred (z = 2)
3.
Mass of Au atom =
 Mass of unit cell = 4 ×
d=a
r=
a
2
Body
Centred
3
d=
.a
2
3
r=
.a
4
197
NA
Volume of cubic unit cell = a3
= (4.072  108)3
= 6.75  10 23 cm 3
Relationship between nearest neighbour distance
(d) and radius of atom (r) and edge of unit cell
(a) for unit cell of a cubic crystals
Simple
cubic
197
g
NA
Density of unit cell = d 
Face
Centred
a
d=
2
a
r=
2 2
Mass of unit cell
Volume of unit cell
4 ×197 / N A
6.75 ×10 13
4 ×197
 NA 
= 6.02  1023 mol1
19.4 × 6.75 ×10 23
19.4 
*3.
Solved Problems
*1.
Niobium is found to crystalline with bcc
structure and found to have density of
8.55 g cm3. Determine the atomic radius of
niobium if its atomic mass is 93. [Mar. 15]
Solution :
d = 8.55 g cm-3, z = 2
M = 93
d

z×M
2 × 93

d × N A 8.55 × 6.022 ×10 23
= 36.1  1024 cm3
1
 a  (36.1×1024 ) 3  3.3×108 cm
For bcc,
3
3
×a 
× 3.3×10 8 = 1.429  108 cm
4
4
= 14.29 nm
Solid State
d
z×M
4 × 63.5

3
8 3
a × N A (3.61×10 ) × 6.022 ×10 23
= 8.96 g cm3
z×M
a3 × NA
 a3 
r
Copper crystallises into a fcc structure and
the unit cell has length of edge
3.61  108cm. Calculate the density of
copper. Atomic mass of copper is 63.5 g
mol1.
Solution :
a = 3.61 × 108 cm
z=4
M = 63.5 g mol1
4.
Face centred cubic lattice of copper has
density of 8.966 g.cm–3. Calculate the
volume of the unit cell.
Given molar mass of copper is 63.5 g.mol –1
and Avogadro number NA is 6.022 × 1023
mol–1.
Solution :
d = 8.966 gcm–3
M = 63.5 g mol–1
NA = 6.022 × 1023 mol–1
1.40
HORIZON Publication
Std. XII Sci. Success Chemistry - I


V=?
For fcc, z = 4
zM
Now, d =
NA  V
4  63.5
8.966 
6.022  10 3  V
4  63.5
V
6.022  10 23  8.966
254

 10 23
54
V = 4.7 × 10–23 cm3
Solution :
Type of crystalline structure of CsCl is bcc.
a = 412.1 pm = 4.121  108 cm
Molar mass of CsCl = 133 + 35.5
= 168.5 g mol 1
Mass of one CsCl molecule =
= 28  10  23 g
In the unit cell of CsCl, there is one Cs ion
at the body centre and 8C1 ions are at 8
corners.
 Number Cs+ in unit cell = 1
Number of Cl  ions =
*5.
Silver crystallises in fcc structure with edge
length of unit cell, 4.07  108 cm and if
density of metallic silver is 10.5 g cm3.
Calculate atomic mass of silver.
Solution:
a = 4.07 × 10-8 cm, d = 10.5 g cm-3
M=?
d
d × a3 × NA
z
10.5× (4.07 ×10 8 ) 3 × 6.022×10 23
=
4
10.5× 67.42×10 24 × 6.022×10 23
=
4
 M
6.
In an ionic solid, the radii of cation and
anion are 0.9858Å and 1.86Å. Predict the
type of geometry and coordination
number.
Solution :
Radius of a cation = r+ = 0.9858 Å
Radius of an anion = r– = 1.86 Å
Type of geometry = ?
Coordination number = ?
r  0.9858

 0.53
1.86
r
Since radius ratio 0.414 < 0.53 < 0.732, the
geometry is octahedral and coordination
number is 6.
*7. Determine the density of cesium
chloride which crystallises in a bcc type
structure with the edge length 412.1 pm.
The atomic mass of Cs and Cl are 133 and
35.5 respectively.
Solid State
1
×8 1
8
 The unit cell contains one CsCl molecule.
Hence z = 1 and M = 168.5 g mol1
Mass of unit cell = mass of 1 CsCl molecule
= 28  10  23 g
Volume of cubic unit cell = a3
= (4.121  108)3
= 69.98  10 24 cm 3
Mass of unit cell
Density of unit cell =
Volume of unit cell
z×M
a3 × NA
= 106.6 g mol 1
168.5
6.022 ×10 23
=
28×1023
69.98×1024
= 4.0 g cm3
*8.
Unit cell of iron crystal has edge length of
288 pm and density of 7.86 g cm3.
Determine the type of crystal lattice
[Fe = 56].
Solution :
Mass of one Fe atom =
56
6.022 ×10 23
= 9.3  1023 g
Mass of unit cell = z  9.3  1023 g
Volume of unit cell = a3 = (2.88  108)3
= 23.88  1024 cm3
Mass of unit cell
Density of unit cell = d =
Volume of unit cell

7.86 

z
z × 9.3×1023
23.88×1024
7.86 × 23.88×10 24
 2.01  2
9.3×10 23
Since the number of atoms in the unit cell is 2.
 Type of crystal is bcc.
1.41
Std. XII Sci. Success Chemistry - I
*9.
An atom crystallises in fcc crystal lattice and
has a density of 10 g cm-3 with unit cell edge
length of 100 pm. Calculate number of
atoms present in 1 g of crystal.
Solution :
d = 10 g cm-3
a = 100 pm = 100 × 10-10 cm
z=4
it is fcc crystal lattice.
z.M
d 3
a .N A
d.a 3 N A
z
10  (100  10 10 ) 3  6.023  10 23

4
 1.506 g
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11.
A cubic unit cell contains atoms A at the
corners, atoms B at face centres and atom
C at the body centre. What is the
formula of the crystalline compound ?
Solution :
Atoms A are at 8 corners, atoms B at the 6
face centres and one atom C at body centre.
1
8
Total number of atoms of A  × 8  1
1
2
Total number of atoms of B  × 6  3 .
One atom of C at the body centre.
Therefore the unit cell contains one atom of
A, three atoms of B and one atom of C.
Hence the formula of the compound is AB3C.
M

According to definition of Avogadro number
1.506 g of element contains 6.023 × 1023 no. of
atoms.
 1 g of element will contain
1  6.023  10
1.506
23
*12. Atoms C and D form fcc crystalline
structure. Atom C is present at the corners
of the cube and D is at the faces of the cube.
What is the formula of the compound ?
Solution : Atoms C are at 8 corners while atoms
D are at 6 face centres of the cubic unit cell.
At the corner,
 3.99  10 23 atoms
while at each face centre, half of each D atom
is present.
 4  10 23 atoms
1
×8 1
8
1
Number of D atoms = × 6  3 .
2
Number of atoms in the crystal is 4 × 1023
atoms.
*10. An element A and B constitute bcc type
crystalling structure. Element A occupies
body centre position and B is at the corners
of cube. What is the formula of the
compound? What are the coordination
numbers of A and B ?
Solution :
Crystalline structure is bcc type.
Atoms A are at 8 corners and atom B is at
body centre.
 Number of atoms of A in a unit cell
=
1
×81
8
Number of atom B in a unit cell = 1.
Since unit cell contains one atom each of A
and B,
 The formula of the compound is AB.
The coordination number of an atom A at
corner is 8.
The coordination number of an atom B at body
centre is 8.
Solid State
1
th of each C atom is present
8
Number of C atoms =
Thus unit cell contains one C atom and three
D atoms.
Hence the formula of the compound is CD3.
Higher Order
13.
A compound is formed by two elements X
and Y. Atoms of the element Y (as anions)
make ccp and those of the element X (as
cations) occupy all the octahedral voids.
What is the formula of the compound ?
Solution :
The ccp lattice is formed by the element Y.
The number of octahedral voids generated
would be equal to the number of atoms of Y
present in it. Since all the octahedral voids
are occupied by the atoms of X, their number
would also be equal to that of the element Y.
Thus, the atoms of elements X and Y are
present in equal numbers or 1 : 1 ratio.
Therefore the formula of the compound is
XY.
1.42
HORIZON Publication
Std. XII Sci. Success Chemistry - I
14.
Silver crystallizes in fcc lattice. If edge
length of the cell is 4.07 × 10–8 cm and
density is 10.5 g cm–3, calculate the atomic
mass of silver.
Solution :
a = 4.07 × 10–8 cm, d = 10.5 gcm–3
z=4
zM
d 3
a NA

da 3 N A
z
10.5  (4.07  108 )3  6.022  1023

4
= 107 g mol–1 = 107 u
M
15.
A cubic solid is made of two element P and
Q. Atoms of Q are at the corners of the
cube and P at the body-centre. What is the
formula of the compound ? What are the
coordination numbers of P and Q ?
Solution :
1
Number of Q atoms = 8   1
8
Number of P atoms = 1
 The formula of the compound is PQ.
The coordination number of both P and Q
is 8.
16.
Copper crystallizes into a fcc lattice with
edge length 3.61 × 10–8 cm. Show that the
calculated density is in agreement with its
measured value of 8.92 g cm–3.
Solution :
z=4
M = 63.5 g mol–1
a = 3.61 × 10–8 cm
zM
4  63.5
d 3

8 3
a N A (3.61 10 )  6.022 10 23
= 8.96 g cm–3
17.
Ferric oxide crystallizes in a hexagonal
close-packed array of oxide ions with two
out of every three octahedral holes
occupied by ferric ions. Derive the formula
of the ferric oxide.
Solution :
O2– ions form cubic close packed lattice.
1
 Number of O2– = 8   1
8
Solid State
2
2
of octahedral voids =
3
3
3
2
Formula of the compound = Fe 2/3 O
= Fe2O3
Number of Fe3+ =

18.
Aluminium crystallizes in a cubic closepacked structure. Its metallic radius is
125 pm.
i.
What is the length of the side of the
unit cell ?
ii. How many unit cells are there in 1.00
cm3 of aluminium ?
Solution :
i.
For ccp structure, edge length,
a  2 2 r  2 2  125 pm = 354 pm
ii. Volume of unit cell = a3
= (354 × 10–8 m)3
 Number of unit cell in 1.00 cm3
1
= 2.26  1022

(354  108 )3
If NaCl is doped with 10–3 mol % of SrCl2,
what is the concentration of cation
vacancies ?
Solution :
Introduction of one Sr2+ introduces one cation
vacancy because Sr2+ replaces two Na+ ions.
 Introduction of 10–3 moles of SrCl2 per 100
moles of NaCl would introduce 10–3 mole
cation vacancies.
 Number of vacancies per mole of NaCl
103
=
= 105 mole = 105  6.022 1023
100
= 6.02 × 1018 vacancies
19.
Problems for practice
1.
2.
3.
An ionic solid of tetrahedral structure has a
cation of radius 1.06 Å. Find the radius of
the anion, if ionic radius ratio is 0.32.
Unit cell of NaCl contains 4 NaCl
molecules. If the density of NaCl crystal is
2.17 gcm–3, find the edge length of the cube.
An element germanium crystallizes in bcc
type crystal structure with edge of unit cell
288 pm and the density of the element is 7.2
g cm-3. Calculate the number of atoms
present in 52 g of the crystalline element.
Also calculate atomic mass of the element.
1.43
Std. XII Sci. Success Chemistry - I
4.
5.
6.
7.
8.
Copper crystallizes in fcc type unit cell. The
edge length of unit cell is 360.8 pm. The
density of metallic copper is 8.92 g cm-3.
Determine atomic mass of copper.
Sodium metal crystallizes in bcc structure
with the edge length of unit cell 4.29 × 10-8
cm. Calculate the radius of sodium atom.
Rubidium having atomic radius 2.48 Å and
atomic mass 85.47 u crystallizes in bcc
structure. If the density of rubidium is 1.51 g
cm–3, determine Avogadro number.
If three elements P, Q and R crystallizes in a
cubic solid lattice with P atoms at the corners,
Q atoms at the cube centre and R atoms at the
centre of edges, then write the formula of
compound.
A solid is made of two elements X and Y.
Atoms X are in fcc arrangement and Y atoms
occupy all the octahedral sites and alternate
tetrahedral sites. What is the formula of the
compound
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4.
5.
6.
7.
Answers
1.
3.
5.
7.
3.312 Å
3.023 × 1023, 51.8
1.86 × 10-8 cm
PQR3
2.
4.
6.
8.
5.618 Å
63.07
6.027 × 1023
XY2
Multiple Choice Questions
1.
2.
3.
The constituent particles of a solid have
(a) translatory motion only
(b) rotatory motion only
(c) vibratory motion only
(d) all the above types of motion
A crystalline solid
(a) changes abruptly from solid to liquid
(b) has no definite melting point
(c) undergoes deformation of its geometry
easily
(d) has
an
irregular
3-dimensional
arrangement
Which of the following exists as covalent
crystals in the solid state ?
(a) Sulphur
(b) Phosphorus
(c) Iodine
(d) Silicon
Solid State
8.
9.
10.
11.
12.
Which of the following are the correct axial
distances and axial angles for rhombohedral
system ?
(a) a = b = c,  =  =   90o
(b) a = b  c,  =  =  = 90o
(c) a  b = c,  =  =  = 90o
(d) a  b  c,       90o
A tetrahedral void in a crystal implies that
(a) shape of the void is tetrahedral
(b) molecules forming the void are
tetrahedral in shape
(c) the void is surrounded tetrahedrally by
four spheres
(d) the void is surrounded by six spheres
Number of tetrahedral voids per atom in a
crystal lattice is
(a) 1
(b) 2
(c) 4
(d) 8
The available space filled in hexagonal close
packing pattern in one layer is
(a) 40%
(b) 52.4%
(c) 60.4%
(d) 70%
In a BCC lattice of an ionic compound the
interionic distance is [Given, Edge length = a]
a
a
(a)
(b)
2
4
3a
3a
(c)
(d)
4
2
+
In a crystal of CsCl, Cs ion occupies the
(a) centre of each face
(b) centre of each edge
(c) corners of the cube
(d) body centre of cube
In a closed packed lattice no. of octahedral
voids and no. of tetrahedral voids are in the
ratio of
(a) 2 : 1
(b) 1 : 1
(c) 2 : 3
(d) 1 : 2
The coordination number of a metal
crystallizing in a hexagonal closed-packed
structure is
(a) 12
(b) 4
(c) 8
(d) 6
Schottky defect defins inperfection in the
lattice structure of a
(a) solid
(b) gas
(c) liquid
(d) plasma
1.44
HORIZON Publication
Std. XII Sci. Success Chemistry - I
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
Na+ ions in NaCl structure are present at
(a) corners of the cube
(b) body centre of the cube
(c) edge centres of the cube
(d) both ‘b’ and ‘c’
The number of Cl– ions required to form ccp
lattice of NaCl structure are
(a) 6
(b) 12
(c) 13
(d) 14
Each unit of cell of NaCl consists of 13 Cl–
ions and
(a) 13 Na+
(b) 14 Na+
+
(c) 6 Na
(d) All are wrong
Out of NaCl, ZnS, CaF2 and CsCl, in which
case the cations form ccp structure ?
(a) NaCl
(b) Zns
(c) CsCl
(d) CaF2
The interionic distance for cesium chloride
crystal will be
a
(a) a
(b)
2
3a
2a
(c)
(d)
2
3
Increase in temperature of the crystalline
compound results in
(a) increase in C.N.
(b) decrease in C.N
(c) decomposition of the compound
(d) None of these
As result of Frenkel defect,
(a) there is no effect on the density
(b) there is no effect on the conductivity
(c) there is no effect on the dielectric
constant
(d) there is no effect on all the three above
Frenkel defect is generally observed in
(a) AgBr
(b) AgI
(c) ZnS
(d) All of these
Ferrimagnetic substances have
(a) zero magnetic moment
(b) small magnetic moment
(c) large magnetic moment
(d) any value of magnetic moment
Which of the following metal oxides is
antiferromagnetic in nature ?
(a) MnO2
(b) TiO2
(c) VO2
(d) CrO2
Which one among the following is an
example of ferroelectric substance ?
(a) Quartz
(b) Lead chromate
(c) Barium titanate (d) Rochelle salt
Solid State
24.
25.
26.
Which of the following is NOT
ferromagnetic?
(a) Cobalt
(b) Iron
(c) Manganese
(d) Nickel
Substance which is weakly repelled by a
magnetic field is
(a) H2O
(b) CrO2
(c) Fe3O4
(d) ZnFe2O4
p-type semi-conductors are made by mixing
silicon with impurities of
[Mar. 2015]
(a) germanium
(b) boron
(c) arsenic
(d) antimony
Answers
1.
6.
11.
16.
21.
26.
c
b
a
d
b
a
2.
7.
12.
17.
22.
c
c
a
c
a
3.
8.
13.
18.
23.
d
d
d
b
c
4.
9.
14.
19.
24.
a
d
d
a
c
5.
10.
15.
20.
25.
c
d
b
d
a
1.45