a book for std. xii/12th science chemistry · pdf filetarget publications std. xii sci.:...

A book for Std. XII/12th Science Chemistry Numericals/Problems

Std. XII Sci.

Chemistry Numericals

Prof. Santosh B. Yadav (M. Sc., SET, NET)

Department of Chemistry R. Jhunjunwala College, Ghatkopar

Salient Features:

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. Self evaluative in nature.

Target PUBLICATIONS PVT. LTD. Mumbai, Maharashtra Tel: 022 – 6551 6551

Website: www.targetpublications.in www.targetpublications.org email : [email protected]

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 © 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

TARGET Publications Std. XII Sci.: Chemistry Numericals

Solid State 1

01 Solid state Formulae 1. Density of unit cell:

d = 3A

z.Ma .N

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.

i. Radius Ratio rr

+

−

⎛ ⎞⎜ ⎟⎝ ⎠

:

The critical radius ratio of the void (cation) and sphere (anion), is calculated by solid geometry.

∴ Radius ratio = rr

+

− = Cation radiusAnion radius

ii. Coordination Number (CN) : The number of spheres (atoms, molecules or ions) directly surrounding a single sphere in a

crystal, is called coordination number. 3. Crystal structures of some elements and their coordination number’s (CN):

Crystal structure Example Coordination No. bcc Li, Na, K, Rb, Cs, Ba 8

fcc or ccp Ni, Cu, Ag, Au, Pt 12 hcp (Hexagonal closed packed) Zn, Mo, Cd, V, Be, Mg 21

4. Relation between radius ratio, coordination number and geometry :

Radius ratio rr

+

−

⎛ ⎞⎜ ⎟⎝ ⎠

Coordination

number Geometry Examples

0.155 to 0.225 3 Planar triangular B2O3

0.225 to 0.414 4 Tetrahedral ZnS 0.414 to 0.732 6 Octahedral NaCl

0.732 to 1.0 8 Cubic CsCl

TARGET PublicationsStd. XII Sci.: Chemistry Numericals

Solid State2

5. Characteristics of some typical crystal structure : Crystal Type of unit

cell Examples Radius

ratio CN

Cation Anion CsCl bcc CsCl, CsBr, TiCl 0.93 8 8 NaCl fcc AgCl, MgO 0.52 6 6 ZnS fcc ZnS 0.40 4 4 CaF2 fcc CaF2, SrF2, CdF2 0.73 8 4

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) = ?

Formula: r = 3a4

Calculation: For BCC From formula,

a = r 43× = 224 4

1.7320× = 517.3 pm

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 To find: Edge length of unit cell (a) = ?

Formula: r = a2 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

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 ratio = Radiusof thecationRadiusof theanion

Calculation: From formula,

Radius ratio = rr

+

−

0.22 = r1.4

+

r + = 0.22 × 1.4 ∴ r + = 0.308 Å

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:



Solid State 3


Radius ratio = rr

+

− = 2Be

2S

r

r+

−

= 59170

= 0.347

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 ratio = rr

+

− = 96618

= 0.1553

Since the radius ratio lies in 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 = ?

Formula:



Radius ratio = rr

+

− = 94146

= 0.6438

Since the radius ratio lies in 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 To find: Radius (r) =?

Formula: Radius (r) = 3a4

Calculation: For BCC From formula,

Radius (r) = 3a4

= 1.7320 4.294× = 1.86 Å

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 ( Ar

+ ) = 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–) = ?


Solid State4

Formula:


Calculation: i. For,

Limiting value for rr

+

− for tetrahedral hole

is 0.225 – 0.414 From formula, Radius of the tetrahedral hole ( Ar

+ ) = Radius ratio × r–

= 0.225 × 195 = 43.875 pm ii. For cation A+ with radius = 82 pm From formula,

Radius ratio = rr

+

− = 82

195 = 0.4205

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−.

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 thecationRadius of theanion

Calculation: ZnS has tetrahedral arrangement.

The range of rr

+

− for stable four fold

coordination is 0.225 to 0.414 Hence the radius of anion can be calculated by

taking rr

+

− = 0.225

∴ r– = r0.225

+

= 50

0.225

= 222.22 pm

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 ratio = rr

+

− =

2Mg

2S

r

r+

−

= 65184

= 0.3533

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: Given: Density (d) = 2.7 g/cm3 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

ii. Density (d) = A

z MN V××


Solid State 5

Calculation: From formula (i), V = (4.04 × 10−8 cm)3 = 6.594 × 10−23 cm3 From formula (ii),

M = AN V dz× ×

= 23 236.023 10 6.594 10 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: Given: Radius (r) = 248 pm = 2.48 × 10−8cm z = 2 (BCC) Atomic mass of Pd = 106 To find: Density (d) = ?

Formula: i. Atomic Radius (r) = 3a4

ii. V = a3

iii. Density (d) = A

z MN V××

Calculation: For BCC From formula (i),

2.48 × 10−8 cm = 1.732 a4×

a = 82.48 10 cm 4

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),

d = 23 23

2 1066.023 10 18.78 10−

×× × ×

= 1.87 g/cm3

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

ii. Density (d) = A

z MN V××


d = 23 23

1 2106.023 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)

= 6.023 × 1023 To find: Atomic radius (r) = ?

Formula: i. Density (d) = A

z MN V××

ii. V = a3

iii. r = a2


Solid State6

Calculation: From formula (i),

5.904 = 23

1 69.76.023 10 V

×× ×

V = 1.96 × 10−23

From formula (ii),

a = 3 231.96 10−× = 2.7 × 10−8 For Simple cube structure From formula (iii),

r = 82.7 10

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

ii. Density (d) = A

z MN V××

Calculation: From formula (i), V = (4.06 × 10-8 cm)3

= 6.69234 × 10−23 cm3 From formula (ii),

M = Ad N Vz

× ×

M = 23 2311.5 6.023 10 6.69234 104

−× × × ×

= 115.88 amu This weight is close to that of Indium.

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

ii. Density = MassVolume

ii. Density (d) = A

z MN 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 body-centered 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

= 226.0307 10

2

−×

= 3.015 × 10−22 g ii. Atomic mass of tantalum in g/mol From formula (iii),

M = AN V dz× ×

= 23 236.023 10 3.6133 10 16.69

2

−× × × ×

Atomic mass of Ta = 181.6 g/mol


Solid State 7

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: Given: Density (d) = 8.908 g/cm3 z = 4 (FCC Lattice)

Atomic mass of Ni (M) = 58.6934 NA

= 6.023 × 1023 To find: Edge length of unit cell (a) = ?


z MN V××

ii. V = a3


V = A

4 58.6934N d×

×= 23

4 58.69346.023 10 8.908

×× ×

= 4.376 × 10−23 cm3 From formula (ii),

a = 234.376 10−× = 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: Given: Radius of atom (r) = 0.197 nm = 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 = ?

Formula: i. r = a2 2

ii. V = a3


z MN V××

Calculation: For FCC Lattice From formula (i),

1.97 × 10−8 = a2 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),

M = AN V dz× ×

= 23 226.023 10 1.72995 10 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

ii. Density (d) = A

z MN V××


V = (2.87 × 10−8 cm)3 = 2.364 × 10−23 cm3 From formula (ii),

z = Ad N VM

× ×

= 23 237.87 6.023 10 2.364 10

55.845

−× × × ×

z = 2.006 z = 2 atoms per unit cell. Hence it is Face centred cubic structure (FCC)


Solid State8

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: Given: FCC unit cell length = 3.5Å BCC unit cell length = 3.0Å z1 for FCC = 4 z2 for BCC = 2 To Find: Ratio of densities of FCC and BCC

= 1

2

dd

= ?

Formula: i. V = a3

ii. Density (d) = A

z MN V××

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),

Density in FCC (d1) = 1

A 1

z MN V

××

Density in BCC (d2) = 2

A 2

z MN V

××

1

2

dd

= 1

2

zz

⎛ ⎞⎜ ⎟⎝ ⎠

× 2

1

VV

⎛ ⎞⎜ ⎟⎝ ⎠

= 4

2⎛ ⎞⎜ ⎟⎝ ⎠

× 8 3

8 3

(3.0 10 )(3.5 10 )

−

−

⎛ ⎞×⎜ ⎟×⎝ ⎠

= 2 ×

23

23

2.7 104.2875 10

−

−

⎛ ⎞×⎜ ⎟×⎝ ⎠

= 2 × 0.6297 = 1.259

∴ 1

2

dd

= 1.259

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 body-centre. 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

atoms of M occupy rd1

3of tetrahedral

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 close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?


Solid State 9

Type 2: Density of the unit cell 10. 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.

11. 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.

12. Europium crystallizes in a BCC lattice. The density of europium is 5.26 g/cm3. Calculate the unit cell edge length. (Atomic mass = 152)

13. 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.

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. 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.

17. 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. 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. 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 body-centred cubic structure that is 3.05 Å on each side. What is the volume of one unit cell in Å?


Solid State10

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 mass 92.9 amu

crystallizes in body centered cubic structure. If density of Niobium is 85.5 g/cm3. Calculate atomic radius of Niobium

10. 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– ?

11. 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? 12. 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.

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 of the anion? [CBSE 1985]


Solid State 11

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]

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] 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]

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] 9. 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]

10. 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]

11. 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]

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] 13. 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] 15. 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?

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]


Solid State12

20. Copper crystal has face centred cubic lattice structure. Its density is 8.93 g/cm3.What is the length of the unit cell? (NA = 6.023 × 1023 mol−1; Atomic mass of Cu = 63.5) [CBSE 1992]

21. 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] 22. 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. 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, Br = 80) [CBSE 1993]

26. 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]

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 +Na

Cl

r

r −

=

(A) 0.225 (B) 0.115 (C) 0.5414 (D) 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) 1.33 × 1020 (B) 2.7 × 1022 (C) 5 × 1023 (D) 2 × 1024 6. 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 13

7. 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 8. 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

9. 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 10. 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 11. 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 12. 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

4NH+ will be (A) 174 pm (B) 154 pm (C) 116 pm (D) 206 pm 13. 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

(A) AB2 C (B) A2BC (C) AB3C (D) ABC3

14. 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 % 15. 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) 12

a : 32

a : 32

a

(B) 1a : 3 a : 2 a

(C) 12

a : 34

a : 12 2

a

(D) 12

a : 3 a : 12

a

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

L


Solid State14

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

(A) NaWO2 (B) NaWO3 (C) Na2WO3 (D) NaWO4 Answers to Additional Problems for Practice 1. 2 atoms, 8.9 pm

2. P2Q 3. AB2O 4. 303.5 pm 5. 255 pm 6. 0.407 nm 7. 28.372 Å 8. i. 5.9058 × 1022 atoms ii. 1.47238 × 1022 unit cells iii. 4 atom/unit cell iv. FCC 9. 1.43 × 102 pm 10. 181 pm 11. i. 2.54 × 1022 atoms in 1 cm3 ii. 1.27 × 1022 unit cells iii. 2 atoms per unit cell 12. 128 × 10−10 pm 13. i. 185.5 pm ii. 428.4 pm 14. 1.54 g/cm3

15. 0.124 nm 16. 313 pm, 4.45 g/cm3

17. 8.98 kg/m3

18. 5.96 g/cm3

Answers to Questions from Various Exams

1. 1.545 Å 2. r+ = 1.32 Å 3. 241.5 pm 4. Cubic, 8 5. Octahedral, 6 6. 6 7. AB3

8. A3B4 9. 26 % 10. AB3

11. 124.27 pm, 7.30 g/mL 12. 2.16 g/cm3

13. Body centred cubic (BCC) lattice. 14. Face centred cubic (FCC) lattice. 15. 400 pm 16. 5.22 g/cm3 17. 5.188 g/cm3, 3.0 × 1022 unit cells 18. 6.03 × 10+23 mol−1 19. 6.226 g/cm3 20. 361.5 pm 21. 2.7857 × 10−23 cm3 22. Body centred cubic (BCC) lattice. 23. 41.7 g/cm3

24. 27 g/mol 25. 346.4 pm 26. 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 15

Hints to Problems for Practice

Problem 1: Given : Atoms of P are present at the body-

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

of atom Q per unit cell = 18

atoms

Number of atoms contributed by 8

corners of atom Q per unit cell = 18

× 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: Given: Radius of Cation A+ (r+) = 58 pm Radius of anion B– (r–) = 210 pm

To find: The coordination number of AB = ? Formula:



Radius ratio = rr

+

− = A

B

r

r+

−

= 58210

= 0.276

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

Problem 3: Given: Radius of Cation Rb+ (r+) = 1.47Å

Radius of anion Br– (r–) = 1.95 Å

To find: Structure of RbBr = ? Formula:



Radius ratio = rr

+

− = Rb

Br

r

r+

−

= 1.471.95

= 0.7538

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:


Calculation: NaCl has octahedral structural arrangement

The range of rr

+

− for stable six fold

coordination is 0.414 to 0.732 Hence the radius of cation can be calculated by

taking rr

+

− = 0.414

From formula,

0.414 = rr

+

−

r– = r0.414

+ = 150

0.414 = 362.32 pm

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.


Solid State16

Problem 6: Given: Edge length of unit cell (a) = 4.070 Å To find: Radius (r) =?

Formula: r = a

2 2

Calculation: Since Au crystallizes in face centred cubic (FCC) structure From formula,

r = 4.0702 1.4142×

= 1.44 Å Problem 7:

Given: M occupy rd1

3of tetrahedral voids

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.

The atoms of element M occupy rd1

3of the

tetrahedral voids.

Therefore, the number of atoms of M = 2 × 13

=

23

of the number of atoms of N.

Therefore, ratio of the number of atoms of M to

that of N is M: N = 23

: 1 = 2:3

Thus, the formula of the compound is M2N3 Problem 8: Given: Ferric oxide has hexagonal close-

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

It is given that two out of every three octahedral holes are occupied by ferric ions. So, number of

ferric (Fe3+) ions = 23

x

Therefore, ratio of the number of Fe3+ ions to the number of O2− ions,

Fe3+ : O2− = 23

x : x

= 23

: 1

= 2 : 3 Hence, the formula of the ferric oxide is Fe2O3. Problem 9: Given : Compound has hexagonal close-

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 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

ii. Density (d) = A

z MN V××


V = (3.95 × 10−8 cm)3

= 6.163 × 10−23 cm3


Solid State 17

From formula (ii),

z = Ad N VM

× ×

= 23 239 6.023 10 6.163 10232

−× × × ×

= 1.4 = 1 z =1 atom per unit Hence it is Simple cubic structure (SC) 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 MN V××

ii. V = a3 Calculation: From formula (i),

V = A

z MN d××

= 23

2 183.96.023 10 19.30

×× ×

= 3.1640 × 10−23 cm3

From formula (ii),

a = 3 233.1640 10−× = 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:

i. Mass of 1 atom = AtomicmassAvogadro 'snumber

ii. Volume = MassDensity

iii. Volume = a 3


Eu = 23

1526.023 10×

From formula (ii),

V = 23

1526.023 10 5.26× ×

= 4.7978 × 10–23 cm3

From formula (iii),

a = 3 234.7978 10−×

a = 3.63 × 10−8 cm a = 363 pm Problem 13: Given: Radius (r) = 125 pm = 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 = ?

Formula: i. r = a2 2

ii. V = a3


1.25 × 10-8 = a2 1.414×

a = 1.25 × 10−8 × 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

= 23

14.418 10×

= 2.266 × 1024 unit cells


Solid State18

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 MN 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),

Edge length (a) = 512

2

= 362 × 10–10 cm

From formula (ii),

V = (3.62 × 10−8 cm)3 = 47.4 × 10−24 cm3 From formula (iii),

d = 23 24

4 63.5(6.023 10 47.4 10 )−

×× × ×

= 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 = ?

Formula: i. V = a3

ii. Density (d) = A

z MN V××

iii. r = a2 2

iv. V = 43πr3


V = (5.59 × 10−8 cm)3 = 1.7468 × 10−22 cm3 From formula (ii),

d = 23 22

4 83.7986.023 10 1.7468 10−

×× × ×

= 3.19 g/cm3 For FCC From formula (iii),

r = 85.59 10 cm

2 2

−× = 85.59 10 cm

2 1.414

−××

r = 1.98 × 10−8 cm From formula (iv),

V = 43

× 3.142 × (1.98 × 10–8)3

= 43× 3.142 × 7.762 × 10–24

= 239.756 10

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:

23

22

4.568 101.7468 10

−

−

××

= 0.2615

= 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 19

Formula: i. r = 3a4

ii. Volume of the unit cell V = a3

iii. Density (d) = 0

z MN V××

Calculation: For BCC From formula (i),

r = 83 9.492 10

4

−× × = 4.11 × 10−8 cm

From formula (ii), V = (4.11 × 10−8 cm)3 = 6.94 × 10−23 cm3 From formula (iii),

M = N0 × V × dz

= 23 236.023 10 6.94 10 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

ii. Density (d) = A

z MN V××


d = 23 23

4 26.986.023 10 6.6381 10−

×× × ×

= 2.699 g/cm3

Problem 18: Given: Edge length (a) = 3.9231 Å = 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

ii. Density (d) = A

z MN 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),

d = 23 23

4 195.86.023 10 6.038 10−

×× × ×

= 21.53 g/cm3

From formula (iii),

r = 83.9231 10 2

4

−× ×

= 138 .7 pm

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