Download Std. XII / 12th Chemistry Numericals

A book for Std. XII/12th Science Chemistry Numericals/Problems
Written according to the New Text book (2012-2013) published by the Maharashtra State Board
of Secondary and Higher Secondary Education, Pune.
Std. XII Sci.
Chemistry Numericals
Prof. Santosh B. Yadav
(M. Sc., SET, NET)
Department of Chemistry
R. Jhunjunwala College, Ghatkopar
Salient Features:
9
9
9
9
9
Completely exam oriented solved problems.
Formulae bank for every topic.
Practice problems with hints for every subtopic.
Problems from various competitive exams.
236 Solved problems, 637 Problems for practice
and 104 Multiple Choice Questions.
9 Self evaluative in nature.
Target PUBLICATIONS PVT. LTD.
Mumbai, Maharashtra
Tel: 022 – 6551 6551
Website: www.targetpublications.in
www.targetpublications.org
email : [email protected]
Std. XII Sci.
Chemistry Numericals
©
Target Publications Pvt Ltd.
Sixth Edition: November 2012
Price: ` 120/-
Printed at:
India Printing Works
42, G.D. Ambekar Marg,
Wadala,
Mumbai – 400 031
Published by
Target PUBLICATIONS PVT. LTD.
Shiv Mandir Sabhagriha,
Mhatre Nagar, Near LIC Colony,
Mithagar Road,
Mulund (E),
Mumbai - 400 081
Off.Tel: 022 – 6551 6551
email: [email protected]
PREFACE
The desire to learn Chemistry remains diminished unless and until the student masters Physical chemistry. Physical chemistry
is a field of science which mainly consists of problems and hence it calls for a deep knowledge of formulas and ability to
solve numerical problems quickly and efficiently.
Hence to ease this task we bring to you “Std. XII Sci. Chemistry Numericals” a book containing adequate solved problems
for every chapter classified into subtopics that provides an indepth knowledge of the procedure to tackle the problems. At the
end of each topic and sub-topic practice problems are provided to test the student’s preparation and increase his confidence.
Additional and multiple choice questions are also provided to increase the knowledge and ability of the student. Board
problems and various competitive exams problems of the last many years have been included to provide the importance of
questions.
To end on a candid note, I would like to make a humble request to each and every student: Preserve this book as a Holy
Grail as it helps you in the complete and thorough preparation from the examination point of view. There is always a room
for improvement, hence I welcome all suggestions and regret any errors that may have occurred in the making of this book.
Best of luck to all the aspirants!
Your’s faithfully
Publisher
Contents
No.
Topic Name
Page No.
1.
Solid State
1
2.
Solutions and Colligative Properties
20
3.
Chemical Thermodynamics and Energetics
62
4.
Electrochemistry
116
5.
Chemical Kinetics
163
6.
IUPAC Name and Nomenclature
234
Std. XII Sci.: Chemistry Numericals
TARGET Publications
Solid state
01
Formulae
1.
Density of unit cell:
z.M
d= 3
a .NA
where, a is edge of unit cell
NA = Avogadro number (6.023 × 1023)
M = Molar mass
z = number of atoms per unit cell
For fcc, z = 4
for bcc, z = 2
for simple cubic, z = 1
2.
Radius rule and coordination number for ionic crystals:
In simple ionic crystals, the cations commonly occupy the voids or holes. The voids are empty
spaces left between anionic spheres.
⎛ r+ ⎞
i.
Radius Ratio ⎜ − ⎟ :
⎝r ⎠
The critical radius ratio of the void (cation) and sphere (anion), is calculated by solid
geometry.
r+
Cation radius
Radius ratio = − =
r
Anion radius
∴
ii.
3.
Coordination Number (CN) :
The number of spheres (atoms, molecules or ions) directly surrounding a single sphere in a
crystal, is called coordination number.
Crystal structures of some elements and their coordination number’s (CN):
Crystal structure
bcc
fcc or ccp
hcp (Hexagonal closed packed)
4.
Example
Li, Na, K, Rb, Cs, Ba
Ni, Cu, Ag, Au, Pt
Zn, Mo, Cd, V, Be, Mg
Coordination No.
8
12
21
Relation between radius ratio, coordination number and geometry :
⎛ r+ ⎞
− ⎟
⎝r ⎠
Radius ratio ⎜
0.155 to 0.225
0.225 to 0.414
0.414 to 0.732
0.732 to 1.0
Solid State
Coordination
number
Geometry
Examples
3
4
6
8
Planar triangular
Tetrahedral
Octahedral
Cubic
B2O3
ZnS
NaCl
CsCl
1
Std. XII Sci.: Chemistry Numericals
5.
TARGET Publications
Characteristics of some typical crystal structure :
Crystal
CsCl
NaCl
ZnS
CaF2
Type of unit
cell
bcc
fcc
fcc
fcc
Examples
CsCl, CsBr, TiCl
AgCl, MgO
ZnS
CaF2, SrF2, CdF2
Solved Examples
Type 1: Radius Ratio of ionic compound/
The Formula of compound
Example 1.1
Barium has a radius of 224 pm and crystallizes
in a body-centred cubic structure. What is the
edge length of the unit cell?
Solution:
Given:
Radius (r) = 224 pm
To find:
Edge length of unit cell (a) = ?
3a
Formula: r =
4
Calculation:
For BCC
From formula,
r × 4 224 × 4
=
= 517.3 pm
a=
1.7320
3
Example 1.2
Aluminium crystallizes in cubic close packed
structure. Its metallic radius is 125 pm. What is
the edge length of unit cell?
Solution:
Given:
Radius (r) = 125 pm
Edge length of unit cell (a) = ?
To find:
a
Formula: r =
2× 2
Calculation:
Since Al crystallizes in Face centred cubic
(FCC) structure
From formula,
a = r×2× 2
= 125 × 2 × 1.4142
∴
a = 353.5 pm
2
Radius
ratio
0.93
0.52
0.40
0.73
CN
Cation Anion
8
8
6
6
4
4
8
4
Example 1.3
In silicates the oxygen atom forms a tetrahedral
void. The limiting radius ratio for tetrahedral
void is 0.22. The radius of oxide is 1.4 Å. Find
out the radius of cation.
Solution:
Given:
Radius of oxide (r−) = 1.4 Å
Radius ratio = 0.22
To find:
Radius of cation (r+) = ?
Formula:
Radius of the cation
Radius ratio =
Radius of the anion
Calculation:
From formula,
Radius ratio =
r+
r−
r+
1.4
+
r = 0.22 × 1.4
r + = 0.308 Å
0.22 =
∴
Example 1.4
The radius of Be2+ cation is 59 pm and that of
S2− is 170 pm. Find out the coordination number
and structure of BeS.
Solution:
Given:
Radius of cation Be2+(r+)= 59 pm
Radius of anion S2− (r−) = 170 pm
To find:
i.
The coordination number of
Be2+ S2− = ?
ii.
Structure of BeS = ?
Formula:
Radius of the cation
Radius ratio =
Radius of the anion
Solid State
TARGET Publications
Formula:
Calculation:
From formula,
Radius ratio =
r + rBe2 +
59
=
=
= 0.347
−
r
170
r 2−
S
Since the radius ratio lies in between
0.225 – 0.414
The coordination number of Be2+ S2− is 4 And
the structure of BeS is tetrahedral.
Example 1.5
If the radius of cation is 96 pm and that of anion
is 618 pm. Determine the coordination number
and structure of the crystal lattice.
Solution:
Given:
Radius of cation (r+) = 96 pm
Radius of anion (r−) = 618 pm
To find:
i.
Coordination number = ?
ii.
Structure of the crystal lattice
=?
Formula:
Radius of the cation
Radius ratio =
Radius of the anion
Calculation:
From formula,
r+
96
=
= 0.1553
−
r
618
radius ratio lies in
Radius ratio =
Since the
between
0.155 – 0.225
The coordination number of crystal is 3
And the structure of crystal lattice is Trigonal
planar.
Example 1.6
The radius of calcium ion is 94 pm and that of
an oxide ion is 146 pm. Find the coordination
number of calcium.
Solution:
Given:
Radius of cation (r+) = 94 pm
Radius of anion (r–) = 146pm
To find:
The coordination number of
calcium = ?
Solid State
Std. XII Sci.: Chemistry Numericals
Radius ratio =
Radius of the cation
Radius of the anion
Calculation:
From formula,
94
r+
=
= 0.6438
−
146
r
radius ratio lies in
Radius ratio =
Since the
between
0.414 – 0.732
The coordination number of calcium is 6.
Example 1.7
Sodium metal crystallizes in body centered
cubic lattice with cell edge = 4.29 Å. What is
the radius of sodium atom?
Solution:
Given:
Edge length of unit cell (a)
= 4.29 Å
Radius (r) =?
To find:
3a
Formula: Radius (r) =
4
Calculation:
For BCC
From formula,
3a 1.7320 × 4.29
Radius (r) =
=
= 1.86 Å
4
4
Example 1.8
Br− ion forms a close packed structure. If the
radius of Br− ions is 195 pm. Calculate the
radius of the cation that just fits into the
tetrahedral hole. Can a cation A+ having a radius
of 82 pm be slipped into the octahedral hole of
the crystal A+Br-?
Solution:
Given:
Radius of anion Br– (r– ) = 195 pm
Radius of cation ( rA+ ) = 82 pm
To find:
i.
The radius of the cation that just fits into
the tetrahedral hole (r+) = ?
ii.
Whether the cation A+ having a radius of
82 pm can be slipped into the octahedral
hole of the crystal (A+ Br–) = ?
3
Std. XII Sci.: Chemistry Numericals
Formula:
Radius of the cation
Radius ratio =
Radius of the anion
Calculation:
i.
For,
r+
Limiting value for − for tetrahedral hole
r
is 0.225 – 0.414
From formula,
Radius of the tetrahedral hole
( rA+ ) = Radius ratio × r–
= 0.225 × 195
= 43.875 pm
For cation A+ with radius = 82 pm
From formula,
82
r+
Radius ratio = − =
= 0.4205
195
r
As it lies in the range 0.414 – 0.732,
hence the cation A+ can be slipped into
the octahedral hole of the crystal A+Br−.
ii.
Example 1.9
A solid AB has ZnS type structure. If the radius
of cation is 50 pm, calculate the maximum
possible value of the radius of anion B−.
Solution:
Given:
Radius of cation (r+) = 50 pm
To find:
Radius of anion (r−) = ?
Formula:
Radius ratio =
Radius of the cation
Radius of the anion
Calculation:
ZnS has tetrahedral arrangement.
r+
The range of
for stable four fold
r−
coordination is 0.225 to 0.414
Hence the radius of anion can be calculated by
r+
taking − = 0.225
r
50
r+
=
∴ r– =
0.225
0.225
= 222.22 pm
4
TARGET Publications
Example 1.10
Determine the structure and coordination
number of MgS on the basis of radius ratio in
which radius of Mg2+ and S2– is 65 pm and 184
pm respectively.
Solution:
Given:
Radius of cation Mg2+ (r+) = 65 pm
Radius of anion S2− (r−) = 184 pm
To find:
i.
The coordination number of
MgS = ?
ii.
Structure of MgS = ?
Formula:
Radius of the cation
Radius ratio =
Radius of the anion
Calculation:
From formula,
Radius ratio =
r 2+
65
r+
Mg
=
=
= 0.3533
−
184
r
r 2−
S
Since the radius ratio lies in between
0.225 – 0.414
The coordination number of MgS is 4.
And the structure of MgS is Tetrahedral.
Type 2: Density of the unit cell
Example 2.1
Al crystallizes in FCC structure. Calculate the
molar mass of Al atoms, if length of the unit cell
is 404 pm and density of Al is 2.7 g/cm3.
Solution:
Density (d) = 2.7 g/cm3
Given:
Length of unit cell (a) = 404 pm
= 4.04 × 10−8 cm
z = 4 (FCC)
To find:
Atomic mass of element (M) =?
Formula: i.
V = a3
z×M
ii.
Density (d) =
NA × V
Solid State
Std. XII Sci.: Chemistry Numericals
TARGET Publications
Calculation:
From formula (i),
V = (4.04 × 10−8 cm)3
= 6.594 × 10−23 cm3
From formula (ii),
N ×V×d
M= A
z
6.023 × 1023 × 6.594 × 10−23 × 2.7
=
4
M = 26.81 amu
Example 2.2
If the radius of palladium is 248 pm and the
lattice type is body centered cubic, what is the
theoretical density of palladium ?
Solution:
Radius (r) = 248 pm
Given:
= 2.48 × 10−8cm
z = 2 (BCC)
Atomic mass of Pd = 106
To find:
Density (d) = ?
Formula:
i.
Atomic Radius (r) =
ii.
V = a3
iii.
Density (d) =
Calculation:
For BCC
From formula (i),
1.732 × a
4
−8
2.48 × 10 cm × 4
a=
1.732
= 5.727 × 10−8 cm
From formula (ii),
V = (5.727 × 10−8 cm)3
= 18.78 × 10−23 cm3
From formula (iii),
2 × 106
d=
6.023 × 1023 × 18.78 × 10−23
= 1.87 g/cm3
2.48 × 10−8 cm =
Solid State
z×M
NA × V
3a
4
Example 2.3
Polonium exist as a simple cube. The edge of its
unit cell is 334.7 pm. Calculate its density.
Solution:
Given:
Edge length (a) = 334.7
= 3.347 × 10−8 cm
Atomic mass of Po = (M) = 210
z=1
(Simple cube)
Avogadro’s number = NA
= 6.023 × 1023
To find:
Density (d) = ?
Formula: i.
V = a3
z×M
ii.
Density (d) =
NA × V
Calculation:
From formula (i),
V = (3.347 × 10−8 cm)3
= 3.7494 × 10−23 cm3
From formula (ii),
1 × 210
⎛
⎞
d= ⎜
23
−23 ⎟
⎝ 6.023 × 10 × 3.7494 × 10 ⎠
= 9.30 g/cm3
Example 2.4
Gallium crystallizes in a simple cubic lattice.
The density of gallium is 5.904 g/cm3.
Determine a value for atomic radius of gallium.
Solution:
Given:
Density (d) = 5.904 g/cm3
Atomic mass of Ga (M) = 69.7
z = 1 (Simple cube)
Avogadro’s number (NA)
To find:
= 6.023 × 1023
Atomic radius (r) = ?
Formula:
i.
Density (d) =
ii.
V = a3
a
r=
2
iii.
z×M
NA × V
5
Std. XII Sci.: Chemistry Numericals
Calculation:
From formula (i),
1 × 69.7
5.904 =
6.023 × 1023 × V
V = 1.96 × 10−23
From formula (ii),
a = 3 1.96 × 10−23 = 2.7 × 10−8
For Simple cube structure
From formula (iii),
2.7 × 10−8
r =
2
= 1.35 × 10−8 cm
= 135 pm
Example 2.5
You are given a small bar of an unknown metal.
You find the density of the metal to be 11.5
g/cm3. An X-ray diffraction experiment
measures the edge of the face-centred cubic unit
cell as 4.06 × 10−10 m. Find the gram-atomic
mass of this metal and tentatively identify it.
Solution:
Given:
Density (d) = 11.5 g/cm3
z = 4 (FCC)
Edge Length (a) = 4.06 × 10−10 m
= 4.06 × 10−8 cm
To find:
Atomic mass (M) =?
Formula: i.
V = a3
z×M
ii.
Density (d) =
NA × V
Calculation:
From formula (i),
V = (4.06 × 10-8 cm)3
= 6.69234 × 10−23 cm3
From formula (ii),
d × NA × V
M=
z
11.5 × 6.023 × 1023 × 6.69234 × 10−23
M=
4
= 115.88 amu
This weight is close to that of Indium.
6
TARGET Publications
Example 2.6
The edge length of the unit cell of Ta, is
330.6 pm; the unit cell is body-centred cubic.
Tantalum has a density of16.69 g/cm3
i.
Calculate the mass of a tantalum atom.
ii.
Calculate the atomic mass of tantalum in
g/mol.
Solution:
Given:
Edge length of the unit cell (a)
= 330.6 pm = 330.6 × 10−10 cm
= 3.306 × 10−8 cm
Density (d) = 16.69 g/cm3
z = 2 (FCC)
To Find: i.
Mass of a tantalum atom = ?
ii.
Atomic mass of tantalum in
g/mol = ?
Formula: i.
V = a3
Mass
ii.
Density =
Volume
z×M
ii.
Density (d) =
NA × V
Calculation:
From formula (i),
V = (3.306 × 10−8 cm)3
= 3.6133 × 10−23 cm3
i.
Mass of the 2 tantalum atoms in the bodycentered cubic unit cell
From formula (ii),
Mass = Density × Volume
= 16.69 × 3.6133 × 10–23
= 6.0307 × 10−22 g
The mass of one atom of Ta
6.0307 × 10−22
=
2
= 3.015 × 10−22 g
ii.
Atomic mass of tantalum in g/mol
From formula (iii),
N ×V×d
M= A
z
6.023 × 1023 × 3.6133 × 10−23 × 16.69
=
2
Atomic mass of Ta = 181.6 g/mol
Solid State
TARGET Publications
Example 2.7
Nickel crystallizes in a face-centred cubic
lattice. If the density of the metal is
8.908 g/cm3, what is the unit cell edge length in
pm?
Solution:
Density (d) = 8.908 g/cm3
Given:
z = 4 (FCC Lattice)
Atomic mass of Ni (M) = 58.6934
NA = 6.023 × 1023
To find:
Edge length of unit cell (a) = ?
z×M
Formula: i.
Density (d) =
NA × V
ii.
V = a3
Calculation:
From formula (i),
4 × 58.6934
4 × 58.6934
V =
=
6.023 × 1023 × 8.908
NA × d
= 4.376 × 10−23 cm3
From formula (ii),
a = 4.376 × 10−23
= 3.524 × 10−8 cm
= 352.4 pm
Example 2.8
A metal crystallizes in a face-centred cubic
lattice. The radius of the atom is 0.197 nm. The
density of the element is 1.54 g/cm3. What is
this metal?
Solution:
Radius of atom (r) = 0.197 nm
Given:
= 1.97 × 10−8 cm
Density (d) = 1.54 g/cm3
z = 4 (FCC Lattice)
NA = 6.023 × 1023 atoms
To Find: Name of metal = ?
a
Formula: i.
r=
2 2
ii.
V = a3
z×M
iii. Density (d) =
NA × V
Solid State
Std. XII Sci.: Chemistry Numericals
Calculation:
For FCC Lattice
From formula (i),
a
1.97 × 10−8 =
2 2
a = 5.572 × 10−8 cm
From formula (ii),
V = (5.572 × 10−8 cm)3 = 1.72995 × 10–22 cm3
From formula (iii),
N ×V×d
M= A
z
=
6.023 × 1023 × 1.72995 × 10−22 × 1.54
4
M = 40.11 g/mol
The metal is calcium.
Example 2.9
Metallic iron crystallizes in a type of cubic unit
cell. The unit cell edge length is 287 pm. The
density of iron is 7.87 g/cm3. How many iron
atoms are there within one unit cell?
Solution:
Given:
Edge length of unit cell (a)
= 287 pm = 287 × 10−10 cm
= 2.87 × 10–8 cm
Density of iron (d) = 7.87 g/cm3
NA = 6.023 × 1023 atoms mol−1
Atomic mass of iron (M) = 55.845
To find:
Number of iron atoms (z) = ?
Formula: i.
V = a3
z×M
ii.
Density (d) =
NA × V
Calculation:
From formula (i),
V = (2.87 × 10−8 cm)3
= 2.364 × 10−23 cm3
From formula (ii),
d × NA × V
z=
M
7.87 × 6.023 × 1023 × 2.364 × 10−23
=
55.845
z = 2.006
z = 2 atoms per unit cell.
Hence it is Face centred cubic structure (FCC)
7
Std. XII Sci.: Chemistry Numericals
Example 2.10
A metal crystallizes into two cubic system-face
centred cubic (FCC) and body centred cubic
(BCC) whose unit cell lengths are 3.5 and 3.0Å
respectively. Calculate the ratio of densities of
FCC and BCC.
Solution:
FCC unit cell length = 3.5Å
Given:
BCC unit cell length = 3.0Å
z1 for FCC = 4
z2 for BCC = 2
To Find: Ratio of densities of FCC and BCC
d
= 1 =?
d2
Formula: i.
ii.
3
V=a
Density (d) =
Calculation:
FCC unit cell length = 3.5 Å
BCC unit cell length = 3.0 Å
From formula (i),
V1 = (3.5 × 10−8)3
V2 = a3 = (3.0 × 10−8)3
From formula (ii),
z ×M
Density in FCC (d1) = 1
N A × V1
Density in BCC (d2) =
TARGET Publications
Problems for Practice
Type 1: Radius Ratio of ionic compound/
The Formula of compound
1.
A cubic solid is made of two elements P
and Q. Atoms of Q are at the corners of
the cube and that of P are at the bodycentre. What is the formula of the
compound? What are the coordination
numbers of P and Q?
2.
The two ions A+ and B– have radius 58
and 210 pm respectively in closed packed
crystal of compound AB. Predict the
coordination number of A+.
3.
The ionic radii of Rb+, Br– are 1.47 and
1.95 respectively. Predict the most
probable type of geometry exhibited by
RbBr on the basis of radius ratio rule.
4.
A solid has NaCl structure. If radius of
the cation is 150 pm. Calculate the
maximum possible value of the radius of
the anion.
5.
Why is coordination number of 12 not
found in ionic crystals?
6.
Gold crystallizes in a FCC lattice, the
observed unit cell length is 4.070 Å.
Calculate the radius of a gold atom.
7.
A compound is formed by two elements
M and N. The element N forms CCP and
rd
1
of tetrahedral
atoms of M occupy
3
voids. What is the formula of the
compound?
8.
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.
9.
A compound forms hexagonal closepacked structure. What is the total number
of voids in 0.5 mol of it? How many of
these are tetrahedral voids?
z×M
NA × V
z2 × M
N A × V2
⎛z ⎞ ⎛V ⎞
d1
= ⎜ 1 ⎟× ⎜ 2 ⎟
d2
⎝ z 2 ⎠ ⎝ V1 ⎠
⎛4⎞
= ⎜ ⎟×
⎝2⎠
⎛ (3.0 × 10−8 )3 ⎞
⎜
−8 3 ⎟
⎝ (3.5 × 10 ) ⎠
⎛ 2.7 ×10−23 ⎞
−23 ⎟
⎝ 4.2875 ×10 ⎠
=2× ⎜
= 2 × 0.6297 = 1.259
d1
= 1.259
∴
d2
8
Solid State
Std. XII Sci.: Chemistry Numericals
TARGET Publications
Type 2: Density of the unit cell
10.
11.
12.
13.
Thallium(I) chloride crystallizes in either
a simple cubic lattice or FCC lattice of
Cl− ion. The density of a given sample of
solid is 9.0 g cm−3 and edge of the unit
cell is 3.95 × 10−8 cm. Predict the category
of unit cell.
Tungsten has a BCC lattice and each
lattice point is occupied by one atom.
Calculate the metallic radius of the
tungsten atom if density of tungsten is
19.30 g/cm3 and its atomic mass is 183.9.
Calculate the X ray density of Aluminium
which forms FCC crystal lattice, if edge
length of unit cell is 4.049 Å.
(Atomic mass of Al = 26.98 g/mol.
Avogadro’s number = 6.023 × 1023)
18.
Platinum crystallizes in FCC crystal with
unit length of 3.9231 Å. Calculate the
density and atomic radius of platinum.
(Atomic mass of
Pt = 195.08)
Additional Problems for Practice
1.
Europium crystallizes in a BCC lattice.
The density of europium is 5.26 g/cm3.
Calculate the unit cell edge length.
(Atomic mass = 152)
Metallic uranium crystallizes in a
body‐centered cubic lattice, with one U
atom per lattice point. How many atoms
are there per unit cell? If the edge length
of the unit cell is found to be
343 pm,
what is the metallic radius of U in pm?
2.
Al crystallizes in FCC structure. Its
metallic radius is 125 pm. What is the
edge length of unit cell? How many unit
cells are there in 1 cm3 of Al.
A solid is made up of two elements P and
Q. Atoms Q are in FCC arrangement,
while P occupy all the tetrahedral sites.
What is the formula of the compound ?
3.
In FCC structure of mixed oxide, the
lattice is made up of oxide ions, one
eighth of tetrahedral voids are occupied
by divalent ions (A2+) while one half of
octahedral voids are occupied by trivalent
ions (B+). What is the formula of the
oxide?
4.
Niobium is found to crystallize with BCC
structure and found to have density of
8.55 g/cm3. Determine the edge length of
unit cell.
5.
A metallic crystal has FCC lattice
structure. Its edge length is 360 pm. What
is the distance of closest approach for two
atoms?
6.
Gold (atomic radius = 0.144 nm)
crystallizes in a face-centred unit cell.
What is the length of a side of the cell?
7.
Given that a solid crystallizes in a bodycentred cubic structure that is 3.05 Å on
each side. What is the volume of one unit
cell in Å?
9
14.
Copper crystal has a face centred cubic
structure. Atomic radius of copper atom is
128 pm. What is the density of copper
metal? (Atomic mass of copper is 63.5)
15.
Krypton crystallizes with a face-centered
cubic unit cell of edge 559 pm.
i.
What is the density of solid
krypton?
ii.
What is the atomic radius of
krypton?
iii. What is the volume of one krypton
atom?
iv. What percentage of the unit cell is
empty space if each atom is treated
as a hard sphere?
16.
17.
At a certain temperature and pressure an
element has a simple body-centred cubic
unit cell. The corresponding density is
4.253 g/cm3 and the atomic radius is
9.492 Å. Calculate the atomic mass for
this element.
Solid State
Std. XII Sci.: Chemistry Numericals
8.
Many metals pack in cubic unit cells. The
density of a metal and length of the unit
cell can be used to determine the type for
packing. For example, gold has a density
of 19.32 g/cm3 and a unit cell side length
of 4.08 Å.
(1 Å = 1 × 10–8 cm.)
i.
How many gold atoms are in
exactly 1 cm3?
ii.
How many unit cells are in exactly
1 cm3?
iii. How many gold atoms are there per
unit cell?
iv. The atoms/unit cell suggests that
gold packs as a (a) simple,
(b)
body-centered or (c) face-centered
unit cell.
9.
Niobium with atomic
crystallizes in body
structure. If density
85.5 g/cm3. Calculate
Niobium
10.
11.
12.
10
mass 92.9 amu
centered cubic
of Niobium is
atomic radius of
If the length of body diagonal for CsCl
which into a cubic structure with Cl– ions
at the corners and Cs+ ions at centre of
unit cell is 7Å and the radius is 1.69 Å
What is the radius of Cl– ?
Many metals pack in cubic unit cells. The
density of a metal and length of the unit
cell can be used to determine the type for
packing. For example, sodium has a
density of 0.968 g/cm3 and a unit cell side
length (a) of 4.29
i.
How many sodium atoms are in
1 cm3?
ii.
How many unit cells are in 1 cm3?
iii. How many sodium atoms are there
per unit cell?
Chromium crystallizes in a body-centred
cubic structure. The unit cell volume is
2.583 × 10−23 cm3. Determine the atomic
radius os Cr in pm.
TARGET Publications
13.
Sodium has a density of 0.971 g/cm3 and
crystallizes with a body-centred cubic unit
cell.
i.
What is the radius of a sodium
atom?
ii.
What is the edge length of the cell?
Give answers in picometers.
14.
Calcium has a cubic closest packed
structure as a solid. Assuming that
calcium has an atomic radius of
197
pm, calculate the density of solid calcium.
15.
Calculate the length of edge of unit cell
for α-iron belonging to BCC structure.
Take the density of α-iron as
7.86 × 103 kg/m3. (Atomic mass of iron =
55.85)
16.
Metallic copper crystallizes in BCC
lattice. If the length of cubic unit cell is
362 pm then calculate the closest distance
between two copper atoms, also calculate
the density of crystalline copper.
17.
Copper has FCC structure and its atomic
radius is 0.1278 nm. Calculate its density.
(Atomic mass of copper = 63.5)
18.
Vanadium has the iron (monoatomic
FCC) structure. If the length of unit cell
edge is 305 pm, calculate the density of
vanadium.
(Atomic mass of V = 50.94 g/mol)
Questions From Various Exams
1.
The ionic radius of an anion is 2.11 Å .
Find the radius of the smallest cation that
can have stable eight fold coordination
with the above anions.
[GATE-1987]
2.
The chloride ion has a radius of
0.181 nm. Calculate the radius of smallest
cation which can be coordinated with
eight neighbouring chloride ions.
[GATE-1989]
3.
A solid has NaCl structure. If the radius
of the cation is 100 pm, what is the radius
[CBSE 1985]
of the anion?
Solid State
TARGET Publications
Std. XII Sci.: Chemistry Numericals
4.
Predict the closed packed structure of an
ionic compound A+B– in which the radius
of cation = 148 pm and radius of anion =
195 pm. What is the coordination number
of cation?
[CBSE-1998]
12.
A unit cell of sodium chloride has four
formula unit. The edge length of the unit
cell is 0.564 nm. What is the density of
sodium chloride?
[IIT May 1997]
5.
Predict the structure of MgO crystal and
coordination number of its cation in
which radii of cation and anion are equal
to 65 pm and 140 pm respectively.
[CBSE 1998]
13.
6.
The two ions A+ and B– have radius 88
and 200 pm respectively in closed packed
crystal of compound AB. Predict the
coordination number of A+ [CBSE 1990]
The unit cell of an element of atomic
mass 96 and density 10.3 g cm−3 is cube
with edge length 314 pm. Find the
structure
of
the
crystal
lattice.
(Simple cubic, FCC , BCC)
(Avogadro constant = 6.023 × 1023 mol−1)
[CBSE 1995]
14.
The unit cell of an element of atomic
mass 108 and density 10.5 g/cm–3 is a
cube with edge length 409 pm. Find the
structure of the crystal lattice (Simple
cubic,
FCC,
BCC)
(Avogadro
constant(NA) = 6.023 × 1023 mol−1)
[CBSE 1995]
An element (Atomic mass = 60) having
FCC unit cell, has density of 6.23 g cm−3.
What is the edge length of the unit cell?
7.
An ionic compound has unit cell
consisting of A ions at the corners of a
cube and B ions on the centres of face
after cube. What would be the empirical
formula of this compound?
[AIEEE 2005]
8.
In a solid AB having the NaCl structure A
atoms occupies the corners of the cubic
unit cell. If all the face centered atoms
along one of the axes are removed then
the resultant stoichiometry of the solid is
[IIT 2001]
A metallic element crystallizes into lattice
containing a sequence of layers of
ABABAB…. Any packing of spheres
leaves out voids in the lattice. Then
calculate the empty space in percentage
by volume in this lattice.
[IIT 1996]
9.
10.
11.
A substance Ax By crystallizes in FCC
lattice in which atoms A occupy each
corner of the cube and atom B occupy the
centers of each face of the cube. Identify
the composition of AxBy
[IIT 2002]
Chromium metal crystallizes with BCC
lattice. The length of the unit cell edge is
found to be 287 pm. Calculate the atomic
radius. What would be the density of
chromium in g/cm3.
(Atomic mass of Cr = 51.99)
[IIT July 1997]
Solid State
15.
16.
The compound CuCl has ZnS structure
and the edge length of the unit cell is 500
pm.
Calculate
the
density.
(Atomic mass of Cu = 63, Cl = 35.5
Avogadro constant = 6.023 × 1023 mol−1)
[CBSE 1997]
17.
An element A (atomic mass 100) having
BCC structure has unit cell edge 400 pm.
Calculate the density of A and the number
of unit cells for 10 g of A. (Avogadro
Number = 6.023 × 1023 ) [CBSE 1990]
18.
An element of atomic mass 98.5 g/mol
occurs in FCC structure. If its unit cell
edge length is 500 pm and its density is
5.22 g/cm3. What is the value of
Avogadro constant?
[CBSC 1997]
19.
A face centred cubic element (atomic
mass = 60) has a unit cell 400 pm. What
is its density?(N = 6.023 × 1023 mol−1)
[CBSE 1992]
11
Std. XII Sci.: Chemistry Numericals
20.
21.
22.
Copper crystal has face centred cubic
lattice
structure.
Its
density
is
3
8.93 g/cm .What is the length of the unit
cell? (NA = 6.023 × 1023 mol−1; Atomic
mass of Cu = 63.5)
[CBSE 1992]
A metal (At mass = 50) has a BCC crystal
structure. The density of the metal is 5.96
g/cm3. Find the volume of unit cell. (NA =
6.023 × 1023 mol−1 )
[CBSE 1993]
The density of chromium metal is
7.2 cm3. If unit cell is cubic with edge
length of 289 pm, determine the type of
unit cell (Simple/BCC/FCC) At mass of
Cr = 52 amu
[CBSE
1997]
23.
An element crystallizes in a structure
having FCC unit cell of an edge of
200 pm. Calculate its density if 200 g of
this element contains 24 × 1023 atoms
[CBSE 1991]
24.
A metal has FCC crystal structure. The
length of its unit cell is 404 pm. What is
the molar mass of metal atoms if the
density of the metal is 2.72 g/cm3
(NA = 6.023 × 1023)
[CBSE 1993]
25.
26.
12
The density of CsBr which has CsCl
(BCC) structure is 4.4 g/cm3. The unit cell
edge length is 400 pm. Calculate the
interionic distance in crystal of CsBr.
(NA = 6.023 × 1023. At mass of Cs = 133,
[CBSE 1993]
Br = 80)
Potassium fluoride has the NaCl type
structure. The density of KF is
2.48 g/cm3 at 20 °C.
i.
Calculate the unit cell length
ii.
Calculate the nearest neighbour
distance in KF
[CBSE 1999]
TARGET Publications
Multiple Choice Questions
1.
The space occupied by b.c.c. arrangement
is approximately
(A) 50%
(B) 68%
(C) 74%
(D) 56%
2.
The maximum percentage of available
volume that can be filled in a face
centered cubic system by an atom is
(A) 74%
(B) 68%
(C) 34%
(D) 26%
3.
In NaCl lattice, the radius ratio is
r
Na +
r
=
Cl−
(A)
(C)
0.225
0.5414
(B)
(D)
0.115
0.471
4.
Xenon crystallizes in face centre cubic
lattice and the edge of the unit cell is 620
pm, then the radius of Xenon atom is
(A) 219.20 pm
(B) 438.5 pm
(C) 265.5 pm
(D) 536.94 pm
5.
A metallic element crystallizes in simple
cubic lattice. Each edge length of the unit
cell is 3 Å. The density of the element is 8
g / cc. Number of unit cells in 108 g of
the metal is
(A)
(C)
6.
1.33 × 1020
5 × 1023
(B)
(D)
2.7 × 1022
2 × 1024
The density of KBr is 2.75 gm cm−3.
Length of the unit cell is 654 pm.
K
= 39, Br = 80. Then what is TRUE about
the predicted nature of the solid.
(A) Solid has face centered cubic
system with z = 4.
(B) Solid has simple cubic system with
z = 4.
(C) Solid has face centered cubic
system with z = 1
(D) Solid has body centered cubic
system with z = 2
Solid State
TARGET Publications
7.
8.
9.
10.
11.
12.
13.
A compound CuCl has face centered
cubic structure. Its density is 3.4 g cm–3.
The length of unit cell is. (At mass of Cu
= 63.54 and Cl = 35.5)
(A) 5.783 Å
(B) 6.783 Å
(C) 7.783 Å
(D) 8.783 Å
At room temperature, sodium crystallizes
in a body centered cubic lattice with a =
4.24 Å. The theoretical density of sodium
(At. mass of Na = 23) is
(A) 1.002 g cm–3
(B) 2.002 g cm–3
(C) 3.002 g cm–3
(D) 4.002 g cm−3
Std. XII Sci.: Chemistry Numericals
14.
15.
The edge length of the unit cell of NaCl
crystal lattice is 552 pm. If ionic radius of
sodium ion is 95 pm, what is the ionic
radius of chloride ion?
(A) 190 pm
(B) 368 pm
(C) 181 pm
(D) 276 pm
The radius of the Na+ is 95 pm and that of
Cl– ion is 181 pm. Predict the
coordination number of Na+.
(A) 4
(B) 6
(C) 8
(D) Unpredictable
A solid AB has rock salt structure. If the
edge length is 520 pm and radius of A+ is
80 pm, the radius of anion B– would be
(A) 440 pm
(B) 220 pm
(C) 360 pm
(D) 180 pm
NH4Cl crystallizes in bcc lattice with edge
length of unit cell equal to
387 pm. If
–
radius of Cl is 181 pm, the radius of
NH +4 will be
(A) 174 pm
(B) 154 pm
(C) 116 pm
(D) 206 pm
What is the simplest formula of a solid
whose cubic unit cell has the atom A at
each corner, the atom B at each face
centre and C atom at the body centre
(B) A2BC
(A) AB2 C
(D) ABC3
(C) AB3C
Solid State
The packing efficiency of the two−
dimensional square unit cell shown below
is
(A) 39.27 %
(B)
68.02 %
(C)
74.05 %
(D)
78.54 %
L
If ‘a’ stands for the edge length of the
cubic systems: simple cubic, body
centered cubic and face centered cubic,
then the ratio of radii of the spheres in
these systems will be respectively.
(A)
1
3
a:
a:
2
2
(B)
1a :
(C)
(D)
3a:
3
a
2
2a
1
3
1
a:
a:
a
2
4
2 2
1
1
a: 3a:
a
2
2
16.
CsBr crystal has bcc structure. It has an
edge length of 4.3 Å. The shortest
interionic distance between Cs+ and Br−
ions is
(A) 1.86 Å
(B) 3.72 Å
(C) 4.3 Å
(D) 7.44 Å
17.
The number of atoms in 100 g of an fcc
crystal with density d = 10 g / cm3 and
cell edge equal to 100 pm, is equal to
(A) 4 × 1025
(B) 3 × 1025
(C) 2 × 1025
(D) 1 × 1025
18.
An element (atomic mass 100 g / mol )
having bcc structure has unit cell edge
400 pm. Then density of the element is
(A) 10.376 g / cm3
(B) 5.188 g / cm3
(C) 7.289 g / cm3
(D) 2.144 g / cm3
13
Std. XII Sci.: Chemistry Numericals
19.
Copper crystallizes in fcc with a unit cell
length of 361 pm. What is the radius of
copper atom ?
(A) 108 pm
(B) 127 pm
(C) 157 pm
(D) 181 pm
20.
AB crystallizes in a body centered cubic
lattice with edge length ‘a’ equal to
387 pm. The distance between two
oppositely charged ions in the lattice is
(A) 335 pm
(B) 250 pm
(C) 200 pm
(D) 300 pm
21.
A solid has a structure in which ‘W’
atoms are located at the corners of a cubic
lattice, ‘O’ atoms at the centre of edges
and ‘Na’ atoms at the centre of the cube.
The formula for the compound is
(B) NaWO3
(A) NaWO2
(D) NaWO4
(C) Na2WO3
Answers to Additional Problems for Practice
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
14
2 atoms, 8.9 pm
P2Q
AB2O
303.5 pm
255 pm
0.407 nm
28.372 Å
i.
5.9058 × 1022 atoms
ii.
1.47238 × 1022 unit cells
iii. 4 atom/unit cell
iv. FCC
1.43 × 102 pm
181 pm
i.
2.54 × 1022 atoms in 1 cm3
ii.
1.27 × 1022 unit cells
iii. 2 atoms per unit cell
128 × 10−10 pm
i.
185.5 pm
ii.
428.4 pm
1.54 g/cm3
0.124 nm
313 pm, 4.45 g/cm3
8.98 kg/m3
5.96 g/cm3
TARGET Publications
Answers to Questions from Various Exams
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
1.545 Å
r+ = 1.32 Å
241.5 pm
Cubic, 8
Octahedral, 6
6
AB3
A3B4
26 %
AB3
124.27 pm, 7.30 g/mL
2.16 g/cm3
Body centred cubic (BCC) lattice.
Face centred cubic (FCC) lattice.
400 pm
5.22 g/cm3
5.188 g/cm3, 3.0 × 1022 unit cells
6.03 × 10+23 mol−1
6.226 g/cm3
361.5 pm
2.7857 × 10−23 cm3
Body centred cubic (BCC) lattice.
41.7 g/cm3
27 g/mol
346.4 pm
i.
537.7 pm
ii.
268.9 pm
Answer Key to Multiple Choice Questions
1.
(B)
2.
(A)
3.
(C)
4.
(A)
5.
(C)
6.
(A)
7.
(A)
8.
(A)
9.
(C)
10. (B)
11. (D)
12. (B)
13. (C)
14. (D)
15. (C)
16. (B)
17. (A)
18. (B)
19. (B)
20. (A)
21. (B)
Solid State
TARGET Publications
Hints to Problems for Practice
Problem 1:
Atoms of P are present at the bodyGiven :
centre
Atoms of Q are present at the
corners of the cube
To find:
Formula of the compound = ?
Co-ordination numbers of P and
Q=?
Calculation:
It is given that the atoms of Q are present at the
corners of the cube.
∴
Number of atoms contributed by a corner
1
of atom Q per unit cell = atoms
8
Number of atoms contributed by 8
1
corners of atom Q per unit cell = × 8
8
= 1 atom
It is also given that the atoms of P are present at
the body-centre.
Therefore, number of atoms of P in one unit cell
= 1 atom
This means that the ratio of the number of P
atoms to the number of Q atoms, P:Q =1:1
Hence, the formula of the compound is PQ.
The coordination number of both P and Q is 8
Problem 2:
Radius of Cation A+ (r+) = 58 pm
Given:
Radius of anion B– (r–) = 210 pm
To find:
The coordination number of AB = ?
Formula:
Radius of thecation
Radius ratio =
Radius of theanion
Calculation:
From formula,
r+
58
r+
Radius ratio = − = A =
= 0.276
r−
r
210
B
Since the radius ratio lies in between
0.225 – 0.414
The coordination number of AB is 4
And the structure of AB is Tetrahedral
Solid State
Std. XII Sci.: Chemistry Numericals
Problem 3:
Radius of Cation Rb+ (r+) = 1.47Å
Given:
Radius of anion Br– (r–) = 1.95 Å
To find:
Structure of RbBr = ?
Formula:
Radius of thecation
Radius ratio =
Radius of theanion
Calculation:
From formula,
Radius ratio =
r + 1.47
r+
= Rb =
= 0.7538
−
r − 1.95
r
Br
Since the radius ratio lies in between 0.732 – 1.0
The coordination number of RbBr is 8
And the structure of RbBr is Cubic.
Problem 4:
Given:
Radius of cation Na+ (r+) =150 pm
To find:
Radius of anion Cl– (r–) =?
Formula:
Radius of thecation
Radius ratio =
Radius of theanion
Calculation:
NaCl has octahedral structural arrangement
r+
The range of
for stable six fold
r−
coordination is 0.414 to 0.732
Hence the radius of cation can be calculated by
r+
taking − = 0.414
r
From formula,
r+
0.414 = −
r
+
150
r
=
= 362.32 pm
r– =
0.414 0.414
Problem 5:
Maximum radius ratio in ionic crystals lies in
the range 0.732 – 1 which corresponds to a
coordination
number
of
8.
Hence
coordination number greater than 8 is not
possible in ionic crystals.
15
Std. XII Sci.: Chemistry Numericals
TARGET Publications
Problem 6:
Given:
Edge length of unit cell (a) = 4.070 Å
To find:
Radius (r) =?
a
Formula: r =
2 2
Calculation:
Since Au crystallizes in face centred cubic
(FCC) structure
From formula,
4.070
r=
2 × 1.4142
= 1.44 Å
It is given that two out of every three octahedral
holes are occupied by ferric ions. So, number of
2
ferric (Fe3+) ions = x
3
Therefore, ratio of the number of Fe3+ ions to
the number of O2− ions,
2
Fe3+ : O2− = x : x
3
2
= :1
3
=2:3
Hence, the formula of the ferric oxide is Fe2O3.
Problem 7:
Problem 9:
Compound has hexagonal closeGiven :
packed structure
Avogadro’s Number = NA
= 6.023 × 1023
To find:
Total number of voids = ?
Number of tetrahedral voids = ?
Calculation:
Number of close-packed particles = 0.5 × NA
= 0.5 × 6.023 × 1023
= 3.011 × 1023
Therefore, number of octahedral voids
= 3.011 × 1023
And, number of tetrahedral voids
= 2 × 3.011 × 1023
= 6.022 ×1023
Therefore, total number of voids
= (3.011 × 1023) + (6.023 × 1023)
= 9.034 × 1023
rd
1
of tetrahedral voids
3
To find : Formula of the compound = ?
Calculation:
The CCP lattice is formed by the atoms of the
element N.
Here, the number of tetrahedral voids generated
is equal to twice the number of atoms of the
element N.
rd
1
The atoms of element M occupy
of the
3
tetrahedral voids.
1
Therefore, the number of atoms of M = 2 × =
3
2
of the number of atoms of N.
3
Therefore, ratio of the number of atoms of M to
2
that of N is M: N = : 1 = 2:3
3
Thus, the formula of the compound is M2N3
Given:
M occupy
Problem 8:
Ferric oxide has hexagonal closeGiven:
packed array. Every three octahedral
holes are occupied by ferric ions.
To find:
Formula of the ferric oxide = ?
Calculation:
Let the number of oxide (O2−) ions be x.
So, number of octahedral voids = x
16
Problem 10:
Given:
Density = 9.0 g cm–3
Edge length (a) = 3.95 × 10−8 cm
Atomic mass of Th (M) = 232
To find:
Category of unit cell = ?
Formula: i.
V = a3
z×M
ii.
Density (d) =
NA × V
Calculation:
From formula (i),
V = (3.95 × 10−8 cm)3
= 6.163 × 10−23 cm3
Solid State
Std. XII Sci.: Chemistry Numericals
TARGET Publications
From formula (ii),
d × NA × V
z=
M
9 × 6.023 × 1023 × 6.163 × 10−23
=
232
= 1.4
=1
z =1 atom per unit
Hence it is Simple cubic structure (SC)
Calculation:
From formula (i),
152
Eu =
6.023 × 1023
From formula (ii),
152
V=
6.023 × 1023 × 5.26
= 4.7978 × 10–23 cm3
From formula (iii),
Problem 11:
Given:
Atomic mass of Tungsten (M)
= 183.9
Density (d) = 19.30 g/cm3
z = 2 (For BCC)
To find:
Metallic radius (a) = ?
z×M
Formula: i.
Density (d) =
NA × V
a=
ii.
V = a3
Calculation:
From formula (i),
z×M
V=
NA × d
2 × 183.9
6.023 × 1023 × 19.30
= 3.1640 × 10−23 cm3
From formula (ii),
=
a = 3 3.1640 × 10−23
= 3.1628 × 10−8 cm
Problem 12:
Given:
Density of Europium (d) = 5.26 g/cm3
Atomic mass (M) = 152
To find:
Edge length of unit cell (a) = ?
Formula:
Atomic mass
i.
Mass of 1 atom =
Avogadro 's number
Mass
Density
ii.
Volume =
iii.
Volume = a 3
Solid State
3
4.7978 × 10−23
a = 3.63 × 10−8 cm
a = 363 pm
Problem 13:
Radius (r) = 125 pm
Given:
= 1.25 × 10−8 cm
z = 4 (FCC)
To find:
Edge length of unit cell (a) = ?
Number of unit cells in 1 cm3 of
Al = ?
a
Formula: i.
r=
2 2
ii.
V = a3
Calculation:
From formula (i),
a
1.25 × 10-8 =
2 × 1.414
−8
a = 1.25 × 10 × 2 × 1.414
= 3.535 × 10−8 cm
= 353.5 pm
From formula (ii),
V = (3.535 × 10−8)3
= 4.418 × 10–23 cm3
Number of unit cells in 1 cm3 of Al
= 1 cm3/V
1
=
4.418 × 1023
= 2.266 × 1024 unit cells
17
Std. XII Sci.: Chemistry Numericals
Problem 14:
Given:
Atomic radius of Cu atom
= 128 pm
= 128 × 10−10 cm
z = 4 (FCC)
Atomic mass of Cu = 63.5
To find:
Density of Cu (d) = ?
Formula: i.
Face diagonal
= 2 × edge length
ii. Volume (V) = a3
z×M
iii. Density (d) =
NA × V
Calculation:
For FCC Lattice
In face centred cubic arrangement face diagonal
is four times the radius of atoms face diagonal
= 4 × 128
= 512 pm
= 512 × 10−10cm
From formula (i),
512
Edge length (a) =
2
= 362 × 10–10 cm
From formula (ii),
V = (3.62 × 10−8 cm)3
= 47.4 × 10−24 cm3
From formula (iii),
4 × 63.5
d=
(6.023 × 1023 × 47.4 × 10−24 )
= 8.897 g/cm3
Problem 15:
Given:
Edge length (a) = 559 pm
= 5.59 × 10−8 cm
z = 4 (FCC)
Atomic mass of Krypton (M)
=
83.798
To find:
i.
Density of solid krypton = ?
ii.
Atomic radius of krypton =?
iii. Volume of one krypton atom
=?
iv. % of the unit cell which is
empty space = ?
18
TARGET Publications
Formula:
3
i.
V=a
ii.
Density (d) =
iii.
r=
iv.
z×M
NA × V
a
2 2
4
V = πr3
3
Calculation:
From formula (i),
V = (5.59 × 10−8 cm)3 = 1.7468 × 10−22 cm3
From formula (ii),
4 × 83.798
d=
6.023 × 1023 × 1.7468 × 10−22
= 3.19 g/cm3
For FCC
From formula (iii),
5.59 × 10−8 cm
5.59 × 10−8 cm
=
r=
2 × 1.414
2 2
−8
r = 1.98 × 10 cm
From formula (iv),
4
V = × 3.142 × (1.98 × 10–8)3
3
4
= × 3.142 × 7.762 × 10–24
3
9.756 × 10−23
=
3
V = 3.25 × 10−23 cm3
Volume of the 4 atoms in the unit cell:
3.25 × 10−23 cm3 × 4 = 1.292 × 10−22 cm3
Volume of cell not filled with Kr:
(1.7468 × 10−22) − (1.292 × 10−22)
= 4.568 × 10−23 cm3
% of empty space:
4.568 × 10−23
= 0.2615
1.7468 × 10−22
= 26.15 %
Problem 16:
Given:
Density d = 4.253 g/cm3
Atomic radius (a) = 9.492Å
= 9.492 × 10−8 cm
z = 2 (BCC)
To Find: Atomic mass of element (M) =?
Solid State
TARGET Publications
Formula:
i.
ii.
iii.
3a
4
Volume of the unit cell V = a3
z×M
Density (d) =
N0 × V
r=
Calculation:
For BCC
From formula (i),
3 × 9.492 ×10−8
= 4.11 × 10−8 cm
r=
4
From formula (ii),
V = (4.11 × 10−8 cm)3 = 6.94 × 10−23 cm3
From formula (iii),
d
M = N0 × V ×
z
6.023 × 1023 × 6.94 × 10−23 × 4.253
=
2
M = 88.89 amu
Problem 17:
Given:
Edge length (a) = 4.049 Å
= 4.049 × 10−8 cm
Atomic mass of Al (M)
= 26.98 g/mol
z = 4 (FCC)
Avogadro’s number = NA
= 6.023 × 1023
To find:
Density (d) = ?
Formula: i.
V = a3
z×M
ii.
Density (d) =
NA × V
Std. XII Sci.: Chemistry Numericals
Problem 18:
Edge length (a) = 3.9231 Å
Given:
= 3.9231 × 10−8 cm
Atomic weight of Pt (M)= 195.08
z = 4 (FCC)
Avogadro’s number = NA
= 6.023 × 1023
To find:
Density (d) = ?
Atomic radius (r) = ?
Formula: i.
V = a3
z×M
ii.
Density (d) =
NA × V
iii.
Atomic Radius (r) =
a 2
4
Calculation:
From formula (i),
V = (3.9231 × 10−8 cm)3
= 6.038 × 10−23 cm3
From formula (ii),
4 × 195.8
d=
6.023 × 1023 × 6.038 × 10−23
= 21.53 g/cm3
From formula (iii),
3.9231 × 10−8 × 2
r=
4
= 138 .7 pm
Calculation:
From formula (i),
V = (4.049 × 10−8 cm)3
= 6.6381 × 10−23 cm3
From formula (ii),
4 × 26.98
d =
6.023 × 1023 × 6.6381 × 10−23
= 2.699 g/cm3
Solid State
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