xii chemistry numericals

8/11/2019 Xii Chemistry Numericals

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A book for Std. XII/12th

Science Chemistry Numericals/Problems


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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.orgemail :[email protected]

Written according to the New Text book (2012-2013) published by the Maharashtra State Board

of Secondary and Higher Secondary Education, Pune.


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Std. XII Sci.Chemist ry Numericals

Target Publications Pvt Ltd.

Sixth Edition: November 2012

Price: 120/-

Printed at:

India Printing Works42, G.D. Ambekar Marg,Wadala,Mumbai 400 031

Published byTargetPUBLICATIONS PVT. LTD.Shiv Mandir Sabhagriha,Mhatre Nagar, Near LIC Colony,Mithagar Road,Mulund (E),Mumbai - 400 081Off.Tel: 022 6551 6551email: [email protected]


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PREFACE

The desire to learn Chemistry remains diminished unless and until the student masters Physical chemistry. Physical chemist

is a field of science which mainly consists of problems and hence it calls for a deep knowledge of formulas and ability

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 problem

for every chapter classified into subtopics that provides an indepth knowledge of the procedure to tackle the problems. At th

end of each topic and sub-topic practice problems are provided to test the students preparation and increase his confidenc

Additional and multiple choice questions are also provided to increase the knowledge and ability of the student. Boar

problems and various competitive exams problems of the last many years have been included to provide the importance

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 Ho

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!

Yours faithfully

Publisher


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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


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TARGET Publications td. XII c .: hem stry Numer cals

Solid State 1

01 Solid stateFormulae

1. Density of unit cell:d = 3

A

z.M

a .N

where, a is edge of unit cellNA= Avogadro number (6.023 10

23)M = Molar massz = number of atoms per unit cellFor fcc, z = 4for bcc, z = 2for 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 emptyspaces left between anionic spheres.

i. Radius Ratior

r

+

:

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

Radius ratio =r

r

+

=

Cation radius

Anion 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 numbers (CN):

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

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

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

Radius ratiorr

Coordination

numberGeometry Examples

0.155 to 0.225 3 Planar triangular B2O30.225 to 0.414 4 Tetrahedral ZnS0.414 to 0.732 6 Octahedral NaCl0.732 to 1.0 8 Cubic CsCl


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TARGET Publicationstd. XII c .: hem stry Numer cals

Solid State2

5. Characteristics of some typical crystal structure :

Crystal Type of unitcell

Examples Radiusratio

CNCation Anion

CsCl bcc CsCl, CsBr, TiCl 0.93 8 8NaCl fcc AgCl, MgO 0.52 6 6

ZnS fcc ZnS 0.40 4 4CaF2 fcc CaF2, SrF2, CdF2 0.73 8 4

Solved Examples

Type 1: Radius Ratio of ionic compound/The Formula of compound

Example 1.1Barium has a radius of 224 pm and crystallizesin a body-centred cubic structure. What is theedge length of the unit cell?Solution:Given: Radius (r) = 224 pmTo find: Edge length of unit cell (a) = ?

Formula: r =3a

4

Calculation:For BCCFrom formula,

a =r 4

3

=224 4

1.7320

= 517.3 pm

Example 1.2Aluminium crystallizes in cubic close packedstructure. Its metallic radius is 125 pm. What isthe edge length of unit cell?Solution:

Given: Radius (r) = 125 pmTo find: Edge length of unit cell (a) = ?

Formula: r =a

2 2

Calculation:

Since Al crystallizes in Face centred cubic

(FCC) structureFrom formula,

a = r 2 2

= 125 2 1.4142 a = 353.5 pm

Example 1.3

In silicates the oxygen atom forms a tetrahedralvoid. The limiting radius ratio for tetrahedralvoid is 0.22. The radius of oxide is 1.4 . Findout the radius of cation.Solution:

Given: Radius of oxide (r) = 1.4

Radius ratio = 0.22

To find: Radius of cation (r+

) = ?Formula:

Radius ratio =Radiusof thecation

Radiusof theanion

Calculation:

From formula,

Radius ratio =r

r

+

0.22 =r

1.4

+

r

+

= 0.22

1.4 r + = 0.308

Example 1.4

The radius of Be2+ cation is 59 pm and that ofS2is 170 pm. Find out the coordination numberand 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 ofBe2+ S2= ?

ii. Structure of BeS = ?Formula:


Radiusof theanion


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Solid State 3

Calculation:

From formula,

Radius ratio =r

r

+

=

2Be

2S

r

r

+

=59

170= 0.347

Since the radius ratio lies in between0.225 0.414

The coordination number of Be2+ S2 is 4 Andthestructure of BeS is tetrahedral.

Example 1.5

If the radius of cation is 96 pm and that of anionis 618 pm. Determine the coordination numberand structure of the crystal lattice.

Solution:

Given: Radius of cation (r+) = 96 pm

Radius of anion (r

) = 618 pmTo find: i. Coordination number = ?

ii. Structure of the crystal lattice= ?

Formula:


Radiusof theanion

Calculation:

From formula,

Radius ratio =r

r

+

=

96

618

= 0.1553

Since the radius ratio lies in between0.155 0.225

The coordination number of crystal is 3And the structure of crystal lattice is Trigonalplanar.

Example 1.6

The radius of calcium ion is 94 pm and that ofan oxide ion is 146 pm. Find the coordinationnumber of calcium.

Solution:

Given: Radius of cation (r+) = 94 pm

Radius of anion (r) = 146pm

To find: The coordination number ofcalcium = ?

Formula:


Radiusof theanion

Calculation:

From formula,

Radius ratio = rr

+

= 94

146 = 0.6438

Since the radius ratio lies in between0.414 0.732The coordination number of calcium is 6.

Example 1.7

Sodium metal crystallizes in body centeredcubic lattice with cell edge = 4.29 . What isthe radius of sodium atom?Solution:

Given: Edge length of unit cell (a)= 4.29

To find: Radius (r) =?

Formula: Radius (r) =3a

4

Calculation:For BCCFrom formula,

Radius (r) =3a

4=

1.7320 4.29

4

= 1.86

Example 1.8Br ion forms a close packed structure. If theradius of Br ions is 195 pm. Calculate theradius of the cation that just fits into thetetrahedral hole. Can a cation A+having a radiusof 82 pm be slipped into the octahedral hole ofthe 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 intothe tetrahedral hole (r+)= ?ii. Whether the cation A+having a radius of

82 pm can be slipped into the octahedralhole of the crystal (A+ Br) = ?


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Solid State4

Formula:


Radiusof theanion

Calculation:

i. For,

Limiting value for rr

+

for tetrahedral hole

is 0.225 0.414From formula,Radius of the tetrahedral hole

( Ar+ ) = Radius ratio r

= 0.225 195= 43.875 pm

ii. For cation A+with radius = 82 pmFrom 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 intothe octahedral hole of the crystal A+Br.

Example 1.9

A solid AB has ZnS type structure. If the radiusof cation is 50 pm, calculate the maximumpossible value of the radius of anion B.Solution:

Given: Radius of cation (r

+

) = 50 pm

To find: Radius of anion (r) = ?

Formula: Radius ratio =Radiusof thecation

Radiusof theanion

Calculation:

ZnS has tetrahedral arrangement.

The range ofr

r

+

for stable four fold

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

takingr

r

+

= 0.225

r=r

0.225

+

=

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:


Radiusof theanion

Calculation:

From formula,

Radius ratio =r

r

+

=

2Mg

2S

r

r

+

=65

184= 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 108cm

z = 4 (FCC)

To find: Atomic mass of element (M) =?Formula: i. V = a3

ii. Density (d) =A

z M

N V


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Solid State 5

Calculation:

From formula (i),

V = (4.04 108cm)3= 6.594 1023cm3

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 thelattice type is body centered cubic, what is thetheoretical density of palladium ?Solution:

Given: Radius (r) = 248 pm= 2.48 108cm

z = 2 (BCC)Atomic mass of Pd = 106

To find: Density (d) = ?

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

4

ii. V = a3

iii. Density (d) =A

z M

N V

Calculation:For BCCFrom formula (i),

2.48 108 cm =1.732 a

4

a =82.48 10 cm 4

1.732

= 5.727 108cmFrom formula (ii),

V = (5.727 108cm)3

= 18.78 1023cm3

From formula (iii),

d =23 23

2 106

6.023 10 18.78 10

= 1.87 g/cm3

Example 2.3

Polonium exist as a simple cube. The edge of itsunit cell is 334.7 pm. Calculate its density.Solution:

Given: Edge length (a) = 334.7

= 3.347 108

cmAtomic mass of Po = (M) = 210z = 1 (Simple cube)Avogadros number = NA

= 6.023 1023To find: Density (d) = ?Formula: i. V = a3

ii. Density (d) =A

z M

N V

Calculation:

From formula (i),

V = (3.347 108cm)3

= 3.7494 1023cm3From formula (ii),

d =23 23

1 210

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:


Atomic mass of Ga (M) = 69.7

z = 1 (Simple cube)

Avogadros number (NA)

= 6.023 1023

To find: Atomic radius (r) = ?

Formula: i. Density (d) =A

z M

N V

ii. V = a3

iii. r =a

2


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Solid State6

Calculation:

From formula (i),

5.904 =23

1 69.7

6.023 10 V

V = 1.96 1023

From formula (ii),

a = 3 231.96 10 = 2.7 108

For Simple cube structureFrom formula (iii),

r =82.7 10

2

= 1.35 108 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 11.5g/cm3. An X-ray diffraction experimentmeasures the edge of the face-centred cubic unitcell as 4.06 1010 m. Find the gram-atomicmass of this metal and tentatively identify it.Solution:


z = 4 (FCC)

Edge Length (a) = 4.06 1010m

= 4.06 108cm

To find: Atomic mass (M) =?Formula: i. V = a3

ii. Density (d) =A

z M

N V

Calculation:

From formula (i),

V= (4.06 10-8cm)3


M = Ad N V

z

M =23 2311.5 6.023 10 6.69234 10

4

= 115.88 amuThis weight is close to that of Indium.

Example 2.6

The edge length of the unit cell of Ta, is330.6 pm; the unit cell is body-centred cubic.Tantalum has a density of16.69 g/cm3i. Calculate the mass of a tantalum atom.

ii. Calculate the atomic mass of tantalum ing/mol.Solution:

Given: Edge length of the unit cell (a)

= 330.6 pm = 330.6 1010cm

= 3.306 108cmDensity (d) = 16.69 g/cm3

z = 2 (FCC)To Find: i. Mass of a tantalum atom = ?

ii. Atomic mass of tantalum ing/mol = ?

Formula: i. V = a3

ii. Density =Mass

Volume

ii. Density (d) =A

z M

N V

Calculation:

From formula (i),V = (3.306 108cm)3

= 3.6133 1023cm3

i. Mass of the 2 tantalum atoms in the body-centered cubic unit cell

From formula (ii),Mass = Density Volume

= 16.69 3.6133 1023

= 6.0307 1022gThe mass of one atom of Ta

=226.0307 10

2

= 3.015 1022gii. Atomic mass of tantalum in g/mol

From formula (iii),

M =AN V d

z

=23 236.023 10 3.6133 10 16.69

2

Atomic mass of Ta = 181.6 g/mol


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Solid State 7

Example 2.7

Nickel crystallizes in a face-centred cubiclattice. If the density of the metal is8.908 g/cm3, what is the unit cell edge length inpm?Solution:

Given: Density (d) = 8.908 g/cm3z = 4 (FCC Lattice)Atomic mass of Ni (M) = 58.6934NA = 6.023 10

23To find: Edge length of unit cell (a) = ?


z M

N V

ii. V = a3

Calculation:

From formula (i),V =

A

4 58.6934

N d

=

23

4 58.6934

6.023 10 8.908


a = 234.376 10

= 3.524 108cm= 352.4 pm

Example 2.8

A metal crystallizes in a face-centred cubiclattice. The radius of the atom is 0.197 nm. Thedensity of the element is 1.54 g/cm3. What isthis metal?Solution:

Given: Radius of atom (r) = 0.197 nm= 1.97 108cmDensity (d) = 1.54 g/cm3

z = 4 (FCC Lattice)NA= 6.023 10

23atomsTo Find: Name of metal = ?

Formula: i. r =

a

2 2 ii. V = a3

iii. Density (d) =A

z M

N V

Calculation:For FCC LatticeFrom formula (i),

1.97 108 =a

2 2

a = 5.572 108

cmFrom formula (ii),

V = (5.572 108cm)3 = 1.72995 1022cm3From formula (iii),

M = AN V d

z

=23 226.023 10 1.72995 10 1.54

4

M = 40.11 g/molThe metal is calcium.

Example 2.9Metallic iron crystallizes in a type of cubic unitcell. The unit cell edge length is 287 pm. Thedensity of iron is 7.87 g/cm3. How many ironatoms are there within one unit cell?Solution:

Given: Edge length of unit cell (a)= 287 pm = 287 1010 cm= 2.87 108cmDensity of iron (d) = 7.87 g/cm3

NA= 6.023 1023atoms mol1

Atomic mass of iron (M) = 55.845

To find: Number of iron atoms (z) = ?Formula: i. V = a3

ii. Density (d) =A

z M

N V

Calculation:From formula (i),

V = (2.87 108cm)3= 2.364 1023cm3

From formula (ii),

z = Ad N V

M

=

23 23

7.87 6.023 10 2.364 1055.845

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


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Solid State8

Example 2.10

A metal crystallizes into two cubic system-facecentred cubic (FCC) and body centred cubic(BCC) whose unit cell lengths are 3.5 and 3.0respectively. Calculate the ratio of densities of

FCC and BCC.Solution:

Given: FCC unit cell length = 3.5

BCC unit cell length = 3.0

z1for FCC = 4

z2for BCC = 2

To Find: Ratio of densities of FCC and BCC

= 1

2

d

d= ?

Formula: i. V = a3

ii. Density (d) =A

z MN V

Calculation:

FCC unit cell length = 3.5

BCC unit cell length = 3.0

From formula (i),

V1= (3.5 108)3

V2= a3= (3.0 108)3

From formula (ii),

Density in FCC (d1) =

1

A 1

z M

N V

Density in BCC (d2)=2

A 2

z M

N V

1

2

d

d= 1

2

z

z

2

1

V

V

=4

2

8 3

8 3

(3.0 10 )

(3.5 10 )

= 2 23

23

2.7 10

4.2875 10

= 2 0.6297 = 1.259

12

d

d= 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 ofthe cube and that of P are at the body-centre. What is the formula of thecompound? What are the coordinationnumbers of P and Q?

2. The two ions A+ and B have radius 58and 210 pm respectively in closed packedcrystal of compound AB. Predict thecoordination number of A+.

3. The ionic radii of Rb+, Br are 1.47 and1.95 respectively. Predict the most

probable type of geometry exhibited byRbBr on the basis of radius ratio rule.

4. A solid has NaCl structure. If radius ofthe cation is 150 pm. Calculate themaximum possible value of the radius ofthe anion.

5. Why is coordination number of 12 notfound in ionic crystals?

6. Gold crystallizes in a FCC lattice, theobserved unit cell length is 4.070 .Calculate the radius of a gold atom.

7. A compound is formed by two elementsM and N. The element N forms CCP and

atoms of M occupyrd1

3of tetrahedral

voids. What is the formula of thecompound?

8. Ferric oxide crystallizes in a hexagonalclose-packed array of oxide ions with twoout of every three octahedral holesoccupied by ferric ions. Derive theformula of the ferric oxide.

9. A compound forms hexagonal close-packed structure. What is the total numberof voids in 0.5 mol of it? How many ofthese are tetrahedral voids?


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Solid State 9

Type 2: Density of the unit cell

10. Thallium(I) chloride crystallizes in eithera simple cubic lattice or FCC lattice ofClion. The density of a given sample of

solid is 9.0 g cm

3 and edge of the unitcell is 3.95 108 cm. Predict the categoryof unit cell.

11. Tungsten has a BCC lattice and eachlattice point is occupied by one atom.Calculate the metallic radius of thetungsten atom if density of tungsten is19.30 g/cm3and 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. Itsmetallic radius is 125 pm. What is theedge length of unit cell? How many unitcells are there in 1 cm3of Al.

14. Copper crystal has a face centred cubicstructure. Atomic radius of copper atom is128 pm. What is the density of coppermetal? (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 treatedas a hard sphere?

16. At a certain temperature and pressure an

element has a simple body-centred cubicunit cell. The corresponding density is4.253 g/cm3 and the atomic radius is9.492 . Calculate the atomic mass forthis element.

17. Calculate the X ray density of Aluminiumwhich forms FCC crystal lattice, if edgelength of unit cell is 4.049 .(Atomic mass of Al = 26.98 g/mol.Avogadros number = 6.023 1023)

18. Platinum crystallizes in FCC crystal withunit length of 3.9231 . Calculate thedensity and atomic radius of platinum.(Atomic mass of Pt = 195.08)

Additional Problems for Practice

1. Metallic uranium crystallizes in abodycentered cubic lattice, with one Uatom per lattice point. How many atomsare there per unit cell? If the edge lengthof 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 andQ. 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, thelattice is made up of oxide ions, oneeighth of tetrahedral voids are occupiedby divalent ions (A2+) while one half ofoctahedral voids are occupied by trivalentions (B+). What is the formula of theoxide?

4. Niobium is found to crystallize with BCCstructure and found to have density of8.55 g/cm3. Determine the edge length ofunit cell.

5. A metallic crystal has FCC latticestructure. Its edge length is 360 pm. Whatis the distance of closest approach for twoatoms?

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 oneach side. What is the volume of one unitcell in ?


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Solid State10

8. Many metals pack in cubic unit cells. Thedensity of a metal and length of the unitcell can be used to determine the type forpacking. For example, gold has a densityof 19.32 g/cm3and a unit cell side lengthof 4.08 .

(1 = 1 108cm.)

i. How many gold atoms are inexactly 1 cm3?

ii. How many unit cells are in exactly1 cm3?

iii. How many gold atoms are there perunit cell?

iv. The atoms/unit cell suggests thatgold packs as a (a) simple, (b)body-centered or (c) face-centeredunit cell.

9. Niobium with atomic mass 92.9 amucrystallizes in body centered cubicstructure. If density of Niobium is85.5 g/cm3. Calculate atomic radius ofNiobium

10. If the length of body diagonal for CsClwhich into a cubic structure with Cl ionsat the corners and Cs+ ions at centre ofunit cell is 7 and the radius is 1.69 What is the radius of Cl ?

11. Many metals pack in cubic unit cells. Thedensity of a metal and length of the unitcell can be used to determine the type forpacking. For example, sodium has adensity of 0.968 g/cm3and a unit cell sidelength (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 thereper unit cell?

12. Chromium crystallizes in a body-centredcubic structure. The unit cell volume is2.583 1023 cm3. Determine the atomicradius os Cr in pm.

13. Sodium has a density of 0.971 g/cm3andcrystallizes with a body-centred cubic unitcell.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 packedstructure as a solid. Assuming thatcalcium has an atomic radius of 197pm, calculate the density of solid calcium.

15. Calculate the length of edge of unit cellfor -iron belonging to BCC structure.Take the density of -iron as7.86 103kg/m3. (Atomic mass of iron =55.85)

16. Metallic copper crystallizes in BCClattice. If the length of cubic unit cell is362 pm then calculate the closest distancebetween two copper atoms, also calculatethe density of crystalline copper.

17. Copper has FCC structure and its atomicradius is 0.1278 nm. Calculate its density.(Atomic mass of copper = 63.5)

18. Vanadium has the iron (monoatomicFCC) structure. If the length of unit celledge is 305 pm, calculate the density ofvanadium.

(Atomic mass of V = 50.94 g/mol)

Questions From Various Exams

1. The ionic radius of an anion is 2.11 .Find the radius of the smallest cation thatcan have stable eight fold coordinationwith the above anions.

[GATE-1987]

2. The chloride ion has a radius of0.181 nm. Calculate the radius of smallestcation which can be coordinated witheight neighbouring chloride ions.

[GATE-1989]

3. A solid has NaCl structure. If the radiusof the cation is 100 pm, what is the radiusof the anion? [CBSE 1985]


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Solid State 11

4. Predict the closed packed structure of anionic compound A+Bin which the radiusof cation = 148 pm and radius of anion =195 pm. What is the coordination numberof cation? [CBSE-1998]

5. Predict the structure of MgO crystal andcoordination number of its cation inwhich radii of cation and anion are equalto 65 pm and 140 pm respectively.

[CBSE 1998]

6. The two ions A+ and B have radius 88and 200 pm respectively in closed packedcrystal of compound AB. Predict thecoordination number of A+ [CBSE 1990]

7. An ionic compound has unit cellconsisting of A ions at the corners of acube and B ions on the centres of faceafter cube. What would be the empiricalformula of this compound?

[AIEEE 2005]

8. In a solid AB having the NaCl structure Aatoms occupies the corners of the cubicunit cell. If all the face centered atomsalong one of the axes are removed thenthe 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 spheresleaves out voids in the lattice. Thencalculate the empty space in percentageby volume in this lattice. [IIT 1996]

10. A substance Ax By crystallizes in FCClattice in which atoms A occupy eachcorner of the cube and atom B occupy thecenters of each face of the cube. Identifythe composition of AxBy [IIT 2002]

11. Chromium metal crystallizes with BCClattice. The length of the unit cell edge is

found to be 287 pm. Calculate the atomicradius. What would be the density ofchromium in g/cm3.(Atomic mass of Cr = 51.99)

[IIT July 1997]

12. A unit cell of sodium chloride has fourformula unit. The edge length of the unitcell is 0.564 nm. What is the density ofsodium chloride?

[IIT May 1997]

13. The unit cell of an element of atomicmass 96 and density 10.3 g cm3 is cubewith edge length 314 pm. Find thestructure of the crystal lattice.(Simple cubic, FCC , BCC)

(Avogadro constant = 6.023 1023 mol1)[CBSE 1995]

14. The unit cell of an element of atomicmass 108 and density 10.5 g/cm3 is acube with edge length 409 pm. Find thestructure of the crystal lattice (Simple

cubic, FCC, BCC) (Avogadroconstant(NA) = 6.023 10

23 mol1)[CBSE 1995]

15. An element (Atomic mass = 60) havingFCC unit cell, has density of 6.23 g cm3.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 500pm. Calculate the density.(Atomic mass of Cu = 63, Cl = 35.5Avogadro constant = 6.023 1023 mol1)

[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 numberof unit cells for 10 g of A. (AvogadroNumber = 6.023 1023) [CBSE 1990]

18. An element of atomic mass 98.5 g/moloccurs in FCC structure. If its unit celledge length is 500 pm and its density is5.22 g/cm3. What is the value ofAvogadro constant? [CBSC 1997]

19. A face centred cubic element (atomicmass = 60) has a unit cell 400 pm. Whatis its density?(N = 6.023 1023 mol1)

[CBSE 1992]


17/24


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 mol1; 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 mol1)

[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 [CBSE1997]

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 10

23) [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/cm3at 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. arrangementis approximately(A) 50% (B) 68%

(C) 74% (D) 56%

2. The maximum percentage of availablevolume that can be filled in a facecentered 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 cubiclattice and the edge of the unit cell is 620pm, 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 simplecubic lattice. Each edge length of the unit

cell is 3 . 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 cm3.Length of the unit cell is 654 pm. K= 39, Br = 80. Then what is TRUE aboutthe 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 cubicsystem with z = 1

(D) Solid has body centered cubicsystem with z = 2


18/24


Solid State 13

7. A compound CuCl has face centeredcubic structure. Its density is 3.4 g cm3.The length of unit cell is. (At mass of Cu= 63.54 and Cl = 35.5)(A) 5.783 (B) 6.783 (C) 7.783 (D) 8.783

8. At room temperature, sodium crystallizesin a body centered cubic lattice with a =4.24 . The theoretical density of sodium(At. mass of Na = 23) is

(A) 1.002 g cm3 (B) 2.002 g cm3

(C) 3.002 g cm3 (D) 4.002 g cm3

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 ofCl ion is 181 pm. Predict thecoordination 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

+

is80 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 edgelength of unit cell equal to 387 pm. Ifradius 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 ateach corner, the atom B at each facecentre and C atom at the body centre(A) AB2C (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)1

2a :

3

2a :

3

2a

(B) 1a : 3 a : 2 a

(C)1

2a :

3

4a :

1

2 2a

(D)1

2a : 3 a :

1

2a

16. CsBr crystal has bcc structure. It has anedge length of 4.3 . The shortestinterionic distance between Cs+ and Brions is

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

17. The number of atoms in 100 g of an fcccrystal with density d = 10 g / cm3 andcell 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


19/24


Solid State14

19. Copper crystallizes in fcc with a unit celllength of 361 pm. What is the radius ofcopper atom ?(A) 108 pm (B) 127 pm(C) 157 pm (D) 181 pm

20. AB crystallizes in a body centered cubiclattice with edge length a equal to387 pm. The distance between twooppositely 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 Watoms are located at the corners of a cubiclattice, O atoms at the centre of edgesand 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. P2Q3. AB2O4. 303.5 pm5. 255 pm6. 0.407 nm7. 28.372 8. i. 5.9058 1022atoms

ii. 1.47238 1022unit cellsiii. 4 atom/unit celliv. FCC

9. 1.43 102pm10. 181 pm11. i. 2.54 1022atoms in 1 cm3

ii. 1.27 1022unit cellsiii. 2 atoms per unit cell

12. 128 1010pm13. i. 185.5 pm

ii. 428.4 pm

14. 1.54 g/cm3

15. 0.124 nm16. 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

2. r+= 1.32

3. 241.5 pm

4. Cubic, 8

5. Octahedral, 6

6. 6

7. AB38. A3B4

9. 26 %

10. AB311. 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 1022unit cells

18. 6.03 10+23 mol1

19. 6.226 g/cm3

20. 361.5 pm

21. 2.7857 1023 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)


20/24


Solid State 15

Hints to Problems for Practice

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

centreAtoms of Q are present at thecorners of the cube

To find: Formula of the compound = ?Co-ordination numbers of P andQ = ?

Calculation:

It is given that the atoms of Q are present at thecorners of the cube. Number of atoms contributed by a corner

of atom Q per unit cell =1

8atoms

Number of atoms contributed by 8

corners of atom Q per unit cell = 18

8

= 1 atomIt is also given that the atoms of P are present atthe body-centre.Therefore, number of atoms of P in one unit cell= 1 atomThis means that the ratio of the number of Patoms to the number of Q atoms, P:Q =1:1Hence, 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:


Radiusof theanion

Calculation:

From formula,

Radius ratio =r

r

+

= A

B

r

r

+

=58

210= 0.276

Since the radius ratio lies in between0.225 0.414The coordination number of AB is 4And the structure of AB is Tetrahedral

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

Radius of anion Br(r) = 1.95

To find: Structure of RbBr = ?Formula:

Radius ratio = Radiusof thecationRadiusof theanion

Calculation:

From formula,

Radius ratio =r

r

+

= Rb

Br

r

r

+

=1.47

1.95= 0.7538

Since the radius ratio lies in between 0.732 1.0The coordination number of RbBr is 8And the structure of RbBr is Cubic.

Problem 4:Given: Radius of cation Na+(r+)=150 pmTo find: Radius of anion Cl (r) =?Formula:


Radiusof theanion

Calculation:

NaCl has octahedral structural arrangement

The range ofr

r

+

for stable six fold

coordination is 0.414 to 0.732

Hence the radius of cation can be calculated by

takingr

r

+

= 0.414

From formula,

0.414 =r

r

+

r =r

0.414

+=

150

0.414= 362.32 pm

Problem 5:

Maximum radius ratio in ionic crystals lies inthe range 0.732 1 which corresponds to acoordination number of 8. Hencecoordination number greater than 8 is notpossible in ionic crystals.


21/24


Solid State16

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

Formula: r =a

2 2

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

r =4.070

2 1.4142

= 1.44

Problem 7:

Given: M occupyrd

1

3of tetrahedral voids

To find : Formula of the compound = ?Calculation:The CCP lattice is formed by the atoms of theelement N.Here, the number of tetrahedral voids generatedis equal to twice the number of atoms of theelement N.

The atoms of element M occupyrd1

3of the

tetrahedral voids.

Therefore, the number of atoms of M = 2 1

3=

2

3of the number of atoms of N.

Therefore, ratio of the number of atoms of M to

that of N is M: N =2

3: 1 = 2:3

Thus, the formula of the compound is M2N3

Problem 8:Given: Ferric oxide has hexagonal close-

packed array. Every three octahedralholes are occupied by ferric ions.

To find: Formula of the ferric oxide = ?Calculation:

Let the number of oxide (O2) ions bex.So, number of octahedral voids =x

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

ferric (Fe3+) ions =2

3x

Therefore, ratio of the number of Fe3+ ions to

the number of O2

ions,Fe3+: O2 =

2

3x:x

=2

3: 1

= 2 : 3Hence, the formula of theferric oxide is Fe2O3.

Problem 9:Given : Compound has hexagonal close-

packed structureAvogadros Number = NA

= 6.023 1023To 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 1023And, number of tetrahedral voids= 2 3.011 1023= 6.022 1023Therefore, total number of voids= (3.011 1023) + (6.023 1023)= 9.034 1023

Problem 10:Given: Density = 9.0 g cm3

Edge length (a) = 3.95 108 cmAtomic mass of Th (M) = 232

To find: Category of unit cell = ?Formula: i. V = a3

ii. Density (d) =

A

z M

N V

Calculation:

From formula (i),

V = (3.95 108cm)3

= 6.163 1023cm3


22/24


Solid State 17

From formula (ii),

z = Ad N V

M

=23 239 6.023 10 6.163 10

232

= 1.4= 1

z =1 atom per unitHence it is Simple cubic structure (SC)

Problem 11:Given: Atomic mass of Tungsten (M)

= 183.9Density (d) = 19.30 g/cm3

z = 2 (For BCC)To find: Metallic radius (a) = ?


z M

N V

ii. V = a3Calculation:

From formula (i),

V =A

z M

N d

=23

2 183.9

6.023 10 19.30

= 3.1640 1023 cm3

From formula (ii),

a = 3 233.1640 10

= 3.1628 108cm

Problem 12:Given: Density of Europium (d) = 5.26 g/cm3

Atomic mass (M) = 152To find: Edge length of unit cell (a) = ?Formula:

i. Mass of 1 atom =Atomicmass

Avogadro'snumber

ii. Volume =Mass

Density

iii. Volume = a3

Calculation:

From formula (i),

Eu =23

152

6.023 10

From formula (ii),

V =23

152

6.023 10 5.26

= 4.7978 1023 cm3

From formula (iii),

a = 3 234.7978 10

a = 3.63 108cm

a = 363 pm

Problem 13:

Given: Radius (r) = 125 pm= 1.25 108 cm

z = 4 (FCC)

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

Number of unit cells in 1 cm3 ofAl = ?

Formula: i. r =a

2 2

ii. V = a3

Calculation:

From formula (i),

1.25 10-8 =a

2 1.414

a = 1.25 108 2 1.414

= 3.535 108 cm

= 353.5 pmFrom formula (ii),

V = (3.535 108)3

= 4.418 1023 cm3

Number of unit cells in 1 cm3of Al

= 1 cm3/V

=23

1

4.418 10

= 2.266 1024 unit cells


23/24


Solid State18

Problem 14:Given: Atomic radius of Cu atom

= 128 pm= 128 1010 cmz = 4 (FCC)

Atomic mass of Cu = 63.5To find: Density of Cu (d) = ?Formula: i. Face diagonal

= 2edge lengthii. Volume (V) = a3

iii. Density (d) =A

z M

N V

Calculation:

For FCC LatticeIn face centred cubic arrangement face diagonalis four times the radius of atoms face diagonal

= 4 128= 512 pm= 512 1010cmFrom formula (i),

Edge length (a) =512

2 = 362 1010cm

From formula (ii),

V = (3.62 108cm)3= 47.4 1024cm3

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

= 5.59 108 cmz = 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 M

N V

iii. r =a

2 2

iv. V =4

3r3

Calculation:From formula (i),

V = (5.59 108cm)3= 1.7468 1022cm3From formula (ii),

d =23 22

4 83.798

6.023 10 1.7468 10

= 3.19 g/cm3For FCCFrom formula (iii),

r =85.59 10 cm

2 2

=85.59 10 cm

2 1.414

r = 1.98 108cmFrom formula (iv),

V =4

33.142 (1.98 108)3

=4

33.142 7.762 1024

=239.756 10

3

V = 3.25

10

23

cm

3

Volume of the 4 atoms in the unit cell:3.25 1023cm34 = 1.292 1022cm3Volume of cell not filled with Kr:(1.7468 1022) (1.292 1022)= 4.568 1023cm3% of empty space:

23

22

4.568 10

1.7468 10

= 0.2615

= 26.15 %

Problem 16:Given: Densityd= 4.253 g/cm3

Atomic radius (a) = 9.492= 9.492 108cmz = 2 (BCC)

To Find: Atomic mass of element (M) =?


24/24


Solid State 19

Formula: i. r =3a

4

ii. Volume of the unit cell V = a3

iii. Density (d) =0

z M

N V

Calculation:

For BCCFrom formula (i),

r =83 9.492 10

4

= 4.11 108cm

From formula (ii),

V = (4.11 108cm)3= 6.94 1023cm3From formula (iii),

M = N0V d

z

=23 236.023 10 6.94 10 4.253

2

M = 88.89 amu

Problem 17:Given: Edge length (a) = 4.049

= 4.049 108cmAtomic mass of Al (M)= 26.98 g/molz = 4 (FCC)Avogadros number = NA

= 6.023 1023

To find: Density (d) = ?

Formula: i. V = a3

ii. Density (d) =A

z M

N V

Calculation:

From formula (i),

V = (4.049 108cm)3

= 6.6381 1023cm3

From formula (ii),

d =23 23

4 26.98

6.023 10 6.6381 10

= 2.699 g/cm3

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

= 3.9231 108cmAtomic weight of Pt (M)= 195.08z = 4 (FCC)Avogadros number = NA

= 6.023 1023To find: Density (d) = ?

Atomic radius (r) = ?Formula: i. V = a3

ii. Density (d) =A

z M

N V

iii. Atomic Radius (r) =a 2

4

Calculation:

From formula (i),V = (3.9231 108cm)3

= 6.038 1023cm3

From formula (ii),

d =23 23

4 195.8

6.023 10 6.038 10

= 21.53 g/cm3

From formula (iii),

r =83.9231 10 2

4

= 138 .7 pm

xii chemistry numericals

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