Covalent Solid Censtituants are atoms → Bonding is covalent ...

Q. No. 1 Which is not a property of solids

Option 1 Solids are always crystalline in nature

Option 2 Solids have high density and low compressibility

Option 3 The diffusion of solids is very slow

Option 4 Solids have definite volume

Correct Answer 1

Explanation Solids can be exptalline/amorphous

Q. No. 2 Which of the following is a pseudo solid

Option 1 CaF2

Option 2 Glass

Option 3 NaCl

Option 4 All of these

Correct Answer 2

Explanation Amorphous are pseudo solid is glass.

Q. No. 3 Which of the following is non-crystalline solid

Option 1 CsCl

Option 2 NaCl

Option 3 CaF2

Option 4 Glass

Correct Answer 4

Explanation Glass is amorphous solid

Q. No. 4 Among solids the highest melting point is established by

Option 1 Covalent solids

Option 2 Ionic solids

Option 3 Pseudo solids

Option 4 Molecular solids

Correct Answer 1

Explanation Covalent Solid Censtituants are atoms

Bonding is covalent bonding

having very highm.p.

Q. No. 5 Particles of quartz are packed by

Option 1 Electrical attraction forces

Option 2 Vander Waal’s forces

Option 3 Covalent bond forces

Option 4 Strong electrostatic attraction forces

Correct Answer 3

Explanation Covalent bond forces.

Q. No. 6 Which of the following is an example of covalent crystal solid

Option 1 Si

Option 2 NaF

Option 3 Al

Option 4 Ar

Correct Answer 1

Explanation NaF ionic

Ar Molecular Al Molecular Si Covalent

Q. No. 7 Which of the following is an example of metallic crystal solid

Option 1 C

Option 2 Si

Option 3 W

Option 4 Agcl

Correct Answer 3

Explanation C covalent Si covalent AgCl ionic

W metallic

Q. No. 8 Which of the following is an example of ionic crystal solid

Option 1 Diamond

Option 2 LiF

Option 3 Li

Option 4 Silicon

Correct Answer 2

Explanation Diamond Covalent LiF ionic Li metallic Silicon Covalent

Q. No. 9 NaCl is an example of

Option 1 Covalent solid

Option 2 Ionic solid

Option 3 Molecular solid

Option 4 Metallic solid

Correct Answer 2

Explanation Na Cl ionic solid+ -

Q. No. 10 Diamond is an example of

Option 1 Solid with hydrogen bonding

Option 2 Electrovalent solid

Option 3 Covalent solid

Option 4 Glass

Correct Answer 3

Explanation Diamond covalent solid

Q. No. 11 Graphite is

Option 1 A good conductor ☒

Option 2 Sp3 hybridized ☐

Option 3 An amorphous solid ☐

Option 4 Soft and slippery ☒

Explanation Graphite is conductor and is soft having sp2 hybredisation

Q. No. 12 Diamond is

Option 1 A covalent solid ☒

Option 2 Non-conductor ☒

Option 3 Soft and slippery ☐

Option 4 Sp2 hybridized ☐

Explanation Diamond is covalent solid is non-conductor having sp3 hybridisation is hard.

Q. No. 13 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) A : Crystalline solids are anisotropic. R : The constituent particles are very closely packed.

Option 1 If both assertion and reason are correct and reason is the correct explanation of assertion.

Option 2 If both assertion and reason are true but reason is not the correct explanation of assertion

Option 3 If assertion is true but reason is false

Option 4 If reason is true but assertion is false

Correct Answer 2

Explanation Crystalline solid are anisotropic In amorphous substances, as in liquids, the arrangement of particles is random and disordered. all directions are equivalent and properties are independent of direction on the other hand, the particles in a arrangement of particles may be different in different directions. anisotropy in crystal is due to different arrangements of particles in different directions

Q. No. 14 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) A : Amorphous solids are isotropic. R : Amorphous solids do not possess rigidity.





Correct Answer 3

Explanation Assertion is true but reason is not

Q. No. 15 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) A : Ionic crystals have the highest melting point. R : Covalent bonds are stronger than ionic bonds.





Correct Answer d

Explanation Covalent bonds are stronger so covalent solids are having highest melting point ionic crystal are having ionic bond not covalent Reason is true but are reason is not

Q. No. 16 Match the types of solid with their examples / properties.

No. Column A Column B Column C Id of Additional Answer

1 Molecular solid Dry ice P,r,s

2 Covalent solid Coppe R

3 Metallic solid Generally behave as insulators

Q

4 Ionic solid Generally have low r

melting points

Explanation Molecular solid constituents are molecules

generally behaves as insulator

generally have lowmelting point

e.g. Dry ice

Covalent solid constituents are atoms

generally behaves as insulator C(graphite) is conductor

excep( tion)

generally have very highmelting point

Metallic solid constituents are positive ions in esser

Positive ions in a sea of delocalized is

metallic bonding is involved

e.g. Ni

Ionic solid constituents are ions

bonding involved is electrostatic

Insulator in solid state but conductor inmolten and aq.

state

Q. No. 17 Which of the following does not represent a type of crystal system

Option 1 Triclinic

Option 2 Monoclinic

Option 3 Rhombohedral

Option 4 Isotropical

Correct Answer 4

Explanation Isotropical is not type of crystal system

Q. No. 18 Example of unit cell with crystallographic dimensions , = =90 , 90 isa b c

Option 1 Calcite

Option 2 Graphite

Option 3 Rhombic sulphur

Option 4 Monoclinic sulphur

Correct Answer 4

Explanation i Calcite CaCO3

Axial angle

= = 90 0

Axial distance a = b = c

Crystal system Rhombohedral

ii Graphite = =90 ; 120 0 0 =a b c Hexagonal

iii Rhombic Sulphur

= = =90 0 a b c Orthorhombic

iv Monoclinic sulphur

= =90 = 90 0 0 a b c Monoclinic

Q. No. 19 Crystals can be classified into ____ basic crystal habits

Option 1 3

Option 2 7

Option 3 14

Option 4 4

Correct Answer 2

Explanation There are 7 types of crystal system i) cubic ii) tetragonal iii) orthorhombic iv) monoclinic v)hexagonal vi) rhombohedral vii) triclinic

Q. No. 20 Tetragonal crystal system has the following unit cell dimensions

Option 1 = = and = = =90a b c

Option 2 = and = = =90a b c

Option 3 a b c and = = =90

Option 4 a=b c and = =90 , =120

Correct Answer 2

Explanation Tetragonal crystal system has = and = = =90a b c

Q. No. 21 Monoclinic crystal has dimensions

Option 1 , = =90 90a b c 0 0

Option 2 = = , = = =90a b c 0

Option 3 = , = = =90a b c 0

Option 4 , 90a b c 0

Correct Answer 1

Explanation Monoclinic crystal has dimensions

, - =90 90a b c 0 0

Q. No. 22 The unit cell with the structure below refer to ……crystal system

Option 1 Cubic

Option 2 Orthorhombic

Option 3 Tetragonal

Option 4 Trigonal

Correct Answer 2

Explanation Orthorhombic crystal has dimensions

= = =90a b c 0

Q. No. 23 What is the coordination number of sodium in Na2O

Option 1 6

Option 2 4

Option 3 8

Option 4 2

Correct Answer 2

Explanation 4

Q. No. 24 If an atom is present in the centre of the cube, the participation of that atom per unit cell is

Option 1 1

4

Option 2 1

Option 3 1

2

Option 4 1

8

Correct Answer 2

Explanation 1

Q. No. 25 Crystal systems in which no two axial lengths are equal are

Option 1 Triclinic ☒

Option 2 Orthorhombic ☒

Option 3 Monoclinic ☒

Option 4 Tetragonal ☐

Explanation Triclinic, orthorhombic, monoclinic

Q. No. 26 Which of the following systems do not give correct description of axial lengths and axial angles?

Option 1 Hexagonal : = , = =90 , =120a b c ☐

Option 2 Trigonal : = , = =90 , =90a b c ☒

Option 3 Monoclinic : , = = 90a b c ☒

Option 4 Cubic : = , = = =90a b c ☒

Explanation Crystal system Axial distance Axial angle

Cubic a = b = c = = =90 0

Tetragonal =a b c = = =90 0

Orthorhombic a b c = = =90 0

Monoclinic a b c = =90 ; =90 0 0

Hexagonal a = b + c = =90 ; =120 0 0

Rhombohedral a = b = c = = 90 0

Triclinic a b c =90 0

Q. No. 27 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) A : Triclinic system is the most unsymmetrical system. R : All axial lengths are different in a triclinic system.





Correct Answer 2

Explanation Both assertion and reason are true in triclinic ; =90a b c 0

Q. No. 28 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) A : Graphite is an example of tetragonal crystal system .

R : For a tetragonal system, = , = = =90 .a b c





Correct Answer 4

Explanation Graphite is example of hexagonal tetragonal system

= = = =90a b c 0

Density of a unit cell is the same as the density of the substances. So, if the density of the substance is known, we can calculate the number of atoms or dimesions of the unit cell. The density of the unit cell is related to its mass (M), number of atoms per

unit cell (Z), edge length (a in cm), and Avogadro’s constant NA as :Z M

=a N

3A

Q. No. 29 An element X crystallizes in a structure having an fcc unit cell of an edge 100 pm. If 24

g of the element contains 24 10 atoms, the density is23

Option 1 2.40 g cm -3

Option 2 40 g cm -3

Option 3 4 g cm -3

Option 4 24 g cm -3

Correct Answer 2

Explanation Z = 4 (fcc) 1pm = 10-10 cm

=100 10 cma -10

Na=6.023 1023 For M = ? Given

24 10 atom 24g 23

241 atom 6.023 10

24 10

2323

6.023 10 =6.023g 23

4 6.023=

(100 10 ) 6.023 10f

-10 3 23

4

=10 10 10 6 -30 23

=4 10 =40gcm 30-29 -3

Q. No. 30 The number of atoms present in 100 g of a bcc crystal (density = 12.5 g cm-3) having cell edge 200 pm is

Option 1 1 10 25

Option 2 1 10 24

Option 3 2 10 24

Option 4 2 10 26

Correct Answer 3

Explanation 2 M12.5gcm =

(200 10 ) 6.023 10

-3-10 3 23

From substituting Given values Μ=30.15g

30.15 g 6.023×10 atoms 23

6.023×101 g

30.15

23

6.023100 g ×100

30.5

=1.9×1024

=2 10 24

Q. No. 31 A metal A (atomic mass = 60) has a body -/ centered cubic crystal structure. The density of the metal is 4.2 g cm-3. The volume of unit cell is

Option 1 8.2 10 cm -23 3

Option 2 4.72 10 cm -23 3

Option 3 3.86 10 cm -23 3

Option 4 None of these

Correct Answer 2

Explanation 2 60v=

4.2 6 10

23=4.72 10 cm -23 3

Q. No. 32 A metal X crystallizes in a unit cell in which the radius of atoms (r) is related to edge of unit cell (a) as r = 0.3535a. The total number of atoms present per unit cell is _____.

Correct Answer 0004

Is Integer Type ☐

Explanation For bcc, r = 0.3535a No.ofatom

unitcell = 0004

Q. No. 33 A cubic unit cell has one atom on each corner and one atom on each body diagonal. The number of atoms in the unit cell is _______.

Correct Answer 0005

Is Integer Type ☐

Explanation At corners contribution is

1×5=2

4

At only diagonal contribution is 1

×6=32

Total no. of atoms present in the unit cell = 2 + 3 = 5

Q. No. 34 Match the type of packing with the metal possessing it / space occupied.


1 Body - centered cubic (bcc)

Iron r

2 Hexagonal cubic 52% s

3 Cubic close packing 68% P,s

4 Simple cubic 74% q

Explanation Relation between edge length (a) of cube and radius of sphere A = 29

Atomic radius=2

a

42× r

3Pf =93

3

For simple cubic unit cell

Z = 1 A = 22

41× r

3Pf = = =52.3668r

3

3%

No. of effective atoms in a unit cell

1 1=12× +2× +3×1=6

6 2

46 r

3P.f =base area height

3

Area of equilateral triangle 3

a4

2

46 r

3=3

6 a 2 / 34

3

2

46 r

3=3

6 2r 4r 2 / 34

3

2

=0.74=74 No. of effective atoms in a unit cell

1=8× +1×1=2

8

Body diagonal of cube =4 = 39r

d = 2r

3= or

9r a

3d= a

2

Volume occupied by the spheres in the unit cell =2×volume of sphere

4=2× r

33 [2 = 2 for bcc]

42× r

3= =68.024r

3

3

3%

Effective no. of atoms

= no. of corner atoms 1 1

+no.of face centred atoms of8 2

nc

1 1=8× +6× =1+3=4

8 2

volume occupied by the spheres in the unit cellP.f =

volume of unit cell

44× r

3= =7406(2 2r)

3

3

=74.06 So, fcc is also known as ccp.

Q. No. 35 The formula for determination of density of unit cell is

Option 1 a Ng cm

N M

3-30

Option 2 N Mg cm

a N-3

30

Option 3 a Mg cm

N N

3-3

0

Option 4 M Ng cm

a N-30

3

Correct Answer 2

Explanation Molar mass of solid×rank of unit cellDensity =

edge length of unit cell×Avagadro no.

Q. No. 36 Potassium has a bcc structure with nearest neighbor distance 4.52 A

0. Its density (in kg

m-3) will be

Option 1 454

Option 2 804

Option 3 852

Option 4 911

Correct Answer 4

Explanation For bcc 72 = 2

and 3a=4r

Nearest neighbour d3a

= =4.52A2

istance0

4.52 2= A

1.732

0a

= 5.22A0

zm 2×39density =

N ,a 6.023×10 ×(5.22×10 )3 23 -8 3A

78=

6.023×142.23×10-1

78=

6.02×14.223

= 0.904 g/cm3

Q. No. 37 A metallic element has a cubic lattice. Each edge of the unit cell is 2A

0.The density of

the metal is 2.5 g cm-3. The unit cells in 200 g of the metal are

Option 1 1 10 25

Option 2 1 10 24

Option 3 1 10 22

Option 4 1 10 20

Correct Answer 1

Explanation mass of unit cellDensity of unit cell=

volume

density volume=mass of unit cell

=2.5g/cm (2 10 ) =mass(in gm) 3 -8 3

= mass o=2.5 8 1 f unit0 cell -24

No. of unit cell in 1

= ×20020×

010

20 g-24

=10 unit cellls25

mass of unit cellDensity of unit cell=

volume

density×volume=mass of unit cell

2.5×8×10 gm=massofunitcell-24

=20×10 gm-24

No. of unit cell in 21

= ×20020×

10

00gm-24

=10 unitcells25

Q. No. 38 The density of KBr is 2.75 g cm-3. The length of the unit cell is 654 pm. Atomic mass of K = 39, Br =80. Then the solid is

Option 1 Face centred cubic

Option 2 Simple cubic system

Option 3 Body centred cubic system


Correct Answer 1

Explanation kBr unit cell

Z×Mdensity =

N ×aA 3

Z×1192.75=

6.023×10 ×(6.54×10 )23 -10 3

Here Z+ represents the no. of KBr molecule Per unit cell so the actual value of Z = 2Z*

2.75×6.023×10 ×(654×10 )Z*=

119

23 -10 3

4.6331×10 ×10 ×10=

119

23 -30 9

=0.03893×102 =3.8 4 fcc

Q. No. 39 A metal has bcc structure and the edge length of its unit cell is 3.04A

0. The volume of

the unit cell in cm3 will be

Option 1 1.6 10 cm -21 3

Option 2 2.81 10 cm -23 3

Option 3 6.02 10 cm -23 3

Option 4 6.6 10 cm -24 3

Correct Answer 2

Explanation = 3. A04

0a

=3.04 10 cm -8

Volume 2.81 1 cm= 0= 3 -23 3a

Q. No. 40 A face - centred cubic element (atomic mass 60) has a cell edge of 400 pm. What is its density?

Option 1 6.2 g cm-3

Option 2 24.8g cm-3

Option 3 12.4g cm-3

Option 4 3.1g cm-3

Correct Answer 1

Explanation 4×60F=

(400×10 cm) ×N-10 3A

4×60=

(400×10 cm) ×6.02-10 3

= 6.2 g cm-3

Q. No. 41 For the structure given below the site marked as S is a

Option 1 Tetrahedral void

Option 2 Cubic void

Option 3 Octahedral void


Correct Answer 3

Explanation Octahedral void at edge centre

Q. No. 42 For a solid with the following structure, the co-ordination number of the point B is

Option 1 3

Option 2 4

Option 3 5

Option 4 6

Correct Answer 4

Explanation The mild of the edge as fcc surrounded by 4 face centered atoms and 2 corner atom.

Q. No. 43 The coordination number of a cation occupying a tetrahedral hole is

Option 1 6

Option 2 8

Option 3 12

Option 4 4

Correct Answer 4

Explanation Tetrahedral void , The vacant position in a closed packed unit cell where a foreign atom is in contract with four host atom in the form of a tetrahedral void.

Q. No. 44 In hcp arrangement, the number of nearest neighbours are

Option 1 8

Option 2 6

Option 3 4

Option 4 12

Correct Answer 4

Explanation Factual

Q. No. 45 The coordination number of a metal crystallizing in a hexagonal close packed structure is

Option 1 4

Option 2 12

Option 3 8

Option 4 6

Correct Answer 2

Explanation Factual

Q. No. 46 The ratio of close - packed atoms to tetrahedral holes in cubic close packing is

Option 1 1:1

Option 2 1:2

Option 3 1:3

Option 4 2:1

Correct Answer 2

Explanation Factual

Q. No. 47 The number of octahedral voids in a unit cell a cubical closest packed structure is

Option 1 1

Option 2 2

Option 3 4

Option 4 8

Correct Answer 3

Explanation

The position in a closed packed unit cell where a foreign atom is in contact with six host atoms in the form of an octahedral is called octahedral void. No. of effective void = no. of effective host atom = 4

Q. No. 48 In octahedral holes (voids)

Option 1 A simple triangular void surrounded by four spheres

Option 2 A bi-triangular void surrounded by four spheres

Option 3 A bi-triangular void surrounded by six spheres

Option 4 A bi-triangular void surrounded by eight spheres

Correct Answer 3

Explanation

Q. No. 49 The number of tetrahedral voids in a unit cell of cubical closest packed structure is

Option 1 1

Option 2 2

Option 3 4

Option 4 8

Correct Answer 4

Explanation Number of effective tetrahedral void =2 no. of effective atoms

=2 84 =

Q. No. 50 Which of the following are not true about hexagonal close packing?

Option 1 It has a coordination number of 6. ☒

Option 2 It has 26% empty space. ☐

Option 3 It has ABAB …… type of arrangement ☐

Option 4 It is as closely packed as body-centered cubic packing. ☒

Explanation

In this geometry of hexagonal unit cell only primitive cell is found to be symmetrical. It is closest packing like fcc where 74% space is occupied. Has co-ordination no. 12., contain 6 atoms per unit cell.

Q. No. 51 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) A : A hexagonal close packed structure is more closely packed than cubic - close packed structure. R : Hexagonal close packing and cubic packing have coordination number of 12.





Correct Answer 4

Explanation PfofHcp=74%

Pfofccp=74%

Ccp and hcp has coordination no. = 12

Q. No. 52 In the following questions, a statement of assertion (A) is followed by a statement of

reason (R) A : The number of tetrahedral voids is double the number of octahedral voids. R : The size of the tetrahedral void is half of that of the octahedral void.





Correct Answer 3

Explanation Location T void is such has no. of T void is double the no. of a void

Q. No. 53 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) A : Size of tetrahedral void is much larger than an octahedral void. R : The cations may occupy more space than anions in crystal packing.





Correct Answer 4

Explanation Assertion False

Reason True

Passage Text X-ray studies show that the packing of atoms in a crystal of a metal is found to be in layer such that starting from any layer, every fourth layer is exactly identical. The density of the metal is found to be 19.4 g cm-3 and its atomic mass is 197 u.

Q. No. 54 The coordination number of metal atom in the crystal is

Option 1 4

Option 2 6

Option 3 8

Option 4 12

Correct Answer 4

Explanation ABC A type arrangement so coordination no. will be 12.

Q. No. 55 The fraction occupied by metal atoms in the crystal is

Option 1 0.52

Option 2 0.68

Option 3 0.74

Option 4 1.0

Correct Answer 3

Explanation 74%

Q. No. 56 The approximate number of unit cells present in 1 g of metal is

Option 1 3.06 10 21

Option 2 1.53 10 21

Option 3 3.82 10 20

Option 4 7.64 10 20

Correct Answer 4

Explanation Z M= for this arrangement Z =4

N

3 A

fa

Z M 4 197= =

N 19.4 6.023 10

323

A

af

=6.74 103 -23a

Massdensity =

volume

density volume=mass of 1 unit cell

19.4 6.74 10 =mass of 1 unit cell-23

130.83 10 g=mass of 1 unit cell-23

1g 7.64 10 unit cell 20

Q. No. 57 The length of the edge of the unit cell will be

Option 1 407 pm.

Option 2 189 pm.

Option 3 814 pm.

Option 4 204 pm.

Correct Answer 1

Explanation 4 19719.4=

a 6.023 10

3 -23

4 197 10a =

19.4 6.023

-233

A=67.4 10 -24

A = 407 pm.

Q. No. 58 Assuming the metal atom to be spherical, its radius will be

Option 1 103.5 pm.

Option 2 143.9 pm.

Option 3 146.5 pm.

Option 4 267.8 pm.

Correct Answer 2

Explanation a=r

2 2

143.r= 9 pm.

Q. No. 59 In hexagonal close packing, the difference in the number of tetrahedral and octahedral voids in a unit cell is ______.

Correct Answer 0006

Is Integer Type ☐

Explanation 12 tetrahedral voids and 6 octahedral voids. 12-6 = 0006

Q. No. 60 Atoms of element A from hcp arrangement and those of element B occupy 2/3rd of tetrahedral voids. The total number of A and B per formula unit is _____.

Correct Answer 0007

Is Integer Type ☐

Explanation No of atoms of hcp = 6

2

3rd of tetrahedral voids

2= 12=8

3

A6B8 = A3B4 0007

Q. No. 61 The following structure drawn is of

Option 1 Fluorite

Option 2 Caesium chloride

Option 3 Wurtzite

Option 4 Zinc blende

Correct Answer 4

Explanation S2- ions occupy main position and Zn2+ ions are present in alternate tetra-hedral voids in fcc explaination Effective formula = Zn4S4 Co-ordination of Zn2+ = 4 Co-ordination No. of S2- = 4

a 3r +r =

42+ 2-zn s

This slr. Is zinc blende slr

Q. No. 62 The number of formula unit in unit cell of CsCl,type is

Option 1 1

Option 2 2

Option 3 3

Option 4 4

Correct Answer 1

Explanation Co-ordination no. of Cs+ = 8 Co-ordination no. of Cl- = 8 Effective formula = cscl

Q. No. 63 The interionic distance for caesium chloride crystal will be

Option 1 a

Option 2 a

2

Option 3 3a

2

Option 4 2a

3

Correct Answer 3

Explanation a 3r +r =

2+ -cs cl

Q. No. 64 In CsCl lattice coordination number of Cl- ion is

Option 1 2

Option 2 6

Option 3 8

Option 4 12

Correct Answer 3

Explanation 8

Q. No. 65 If the coordination number of Ca2+ in CaF2 is 8, then the coordination number of F- ion would be

Option 1 3

Option 2 4

Option 3 6

Option 4 8

Correct Answer 2

Explanation Co-ordination no. of Ca+2 = 8 Co-ordination no. of f- = 4

Q. No. 66 How many chloride ions are there around sodium ion in sodium chloride crystal .

Option 1 3

Option 2 8

Option 3 4

Option 4 6

Correct Answer 4

Explanation Co-ordination no. of Na+ = 6 Cl- = 6

Q. No. 67 The structure of MgO is similar to NaCl. What would be the coordination number of magnesium.

Option 1 2

Option 2 4

Option 3 6

Option 4 8

Correct Answer 3

Explanation CNO will be 6.

Q. No. 68 In NaCl lattice the coordination number of Cl- ion is

Option 1 2

Option 2 4

Option 3 6

Option 4 8

Correct Answer 3

Explanation CNO will be 6

Q. No. 69 In the crystals, which of the following ionic compounds would you expect maximum distance between centres of cations and anions

Option 1 LiF

Option 2 CsF

Option 3 CsI

Option 4 LiI

Correct Answer 3

Explanation CSI has maximum distance between centres of cation and anion.

Q. No. 70 If the radius ratio is in the range of 0.732 - 1, then the coordination number will be

Option 1 2

Option 2 4

Option 3 6

Option 4 8

Correct Answer 4

Explanation Co-ordination no. - 8 (BCC s.b. arrangement)(Cascium chloride type slr.)

Q. No. 71 Which of the following contains zinc blende structure

Option 1 CuCl

Option 2 CuBr

Option 3 AgI

Option 4 All of these

Correct Answer 4

Explanation Cucl, CuBr, AgI all has zinc blend slr.

Q. No. 72 The number of unit cells in 58.5 g of NaCl is nearly

Option 1 6 10 20

Option 2 3 10 22

Option 3 1.5 10 23

Option 4 0.5 10 24

Correct Answer 3

Explanation In NaCl structure, Na+ ions occupy corners and face-centres and Cl- ions occupy octahedral voids. So 1 unit cell has Na4cl4.

No. of molecule in 58.5 g =6.023 1023

No. of unit cells in6.02

58.4

5 g3 10

= 23

=1.25 1023

Q. No. 73 Potassium fluoride has NaCl type structure. What is the distance between K+ and F-

ions if cell edge is a cm

Option 1 2 a cm

Option 2 a/2 cm

Option 3 4a cm

Option 4 a/4 cm

Correct Answer 2

Explanation In Nacl structure

- a2 r +r r +r =

2+ - +

Q. No. 74 The intermetallic compound LiAg crystallizes in cubic lattice in which both lithium and silver have coordination number of eight. The crystal class is

Option 1 Simple cube

Option 2 Body-centred cube

Option 3 Face-centred cube


Correct Answer 2

Explanation If in a crystal lattice, the coordination no. is 8, then this cubic cell is body-centered.

Q. No. 75 Ferrous oxide has a cubic structure and each edge of the unit cell is 5.0A

0

. assuming density of the oxide as 4.0 g cm-3,then the number of Fe2+ and O2- ions present in each unit cell will be

Option 1 Four Fe2+ and four O2-

Option 2 Two Fe2+ and four O2-

Option 3 Four Fe2+ and two O2-

Option 4 Three Fe2+ and three O2-

Correct Answer 1

Explanation Edge length = 5=5A =5 10 cm

0-8

Density = 4g/cm3 Density Volume=Mass

4 (5 10 ) =Mass-8 3

500 10 g=Mass-24

Mass of 1 unit cell5 10 g= -22

Mass of 1 molecule of Fe56+16 72g

= =N 6.023 10

O 23

A

No. of molecule in one unit cellMass of unit cell

=Mass of 1 molecule

5 10 300 3006.023 10

72 88 72

-2223

So the no. of Fe+2 ions = 4 = no. of O2- ions.

Q. No. 76 The ratio of cationic radius to anionic radius in an ionic crystal is greater than 0.732. Its coordination number is

Option 1 6

Option 2 8

Option 3 1

Option 4 4

Correct Answer 2

Explanation CN will be 8

Q. No. 77 A solid A+ B- has a body centred cubic structure. The distance of closest approach

between the two ions is 0.767A0

. The edge length of the unit cell is

Option 1 3pm

2

Option 2 142.2 pm

Option 3 2pm

Option 4 88.56 pm

Correct Answer 4

Explanation 3a=r +r

2 + -Na Cl

3a=767 10 pm

22

767 2 10a=

1.732

2

= 88.56 pm

Q. No. 78 A binary solid x+ y- has a zinc blende structure with y- ions forming the lattice and x+ ion occupying 25% tetrahedral sites. The formula of the solid is

Option 1 xy

Option 2 x2y

Option 3 xy2

Option 4 xy4

Correct Answer 3

Explanation Fcc slr. No. of tetrahedral voids =2 no. of effective atom =2 (say)

x- ion occupying 25.1 of

Tetrahedral si25 x

=2 =1

t00

e2

formula will be y 2

Q. No. 79 In closest packing of A type of atom (radius rA) the radius of atom B that can be fifed into octahedral void is

Option 1 0.155 rA

Option 2 0.125 rA

Option 3 0.414 rA

Option 4 0.732 rA

Correct Answer 3

Explanation 0.414 rA

Q. No. 80 A sample of electrically neutral NaCl crystal is analysed for its density which has some unoccupied sites. Two reading where taken.

1.Density of NaCl crystal assuming all sites are occupied =2.178 10 kgm 3 -3

2.Density of NaCl crystal by not considering the unoccupied sites but only the occupied

sites =2.165 10 kgm 3 -3 .

The percentage of unoccupied sites in Nacl crystal is

Option 1 5.96 10 -2

Option 2 5.96 10 -1

Option 3 5.96

Option 4 8.68

Correct Answer 2

Explanation will be 5.96 10 -1

Q. No. 81 NH4Cl crystallizes in a body centred cubic lattice, with a unit cell edge length of 387 pm. The distance between the oppositively charged ions in the lattice is :

Option 1 335 pm

Option 2 154 pm

Option 3 460 pm

Option 4 320 pm

Correct Answer 1

Explanation r + r

3=a

2cation anion

1.732 387= =335 pm

2

Q. No. 82 Which of the following statement are correct?

Option 1 The coordination number each type of ion in CsCl crystal is 8. ☒

Option 2 A metal that crystallizes in bcc structure has coordination number of 12. ☐

Option 3 A unit cell of an ionic crystal shares some of its ions with other unit cells. ☒

Option 4 The length of the edge of unit cell of NaCl is 552 pm

r =95pm, r =181pm+ -Na cl

☒

Explanation 2) Metal that crystalises in bcc slr. Has co-ordination no. 8 4) for NaCl

ar +r =

2+ -Na Cl

edge length(a) - 2(95 + 181) = 552 pm

Q. No. 83 The coordination number of anion is 4n

Option 1 Sodium chloride ☐

Option 2 Zinc chloride ☒

Option 3 Caesium chloride ☐

Option 4 Calcium fluoride ☒

Explanation For zinc chloride, calcium fluride CNO is multiple of Cl is 4n.

Q. No. 84 Which of the following statements are true?

Option 1 An element with bcc structure has two atoms per unit cell. ☒

Option 2 An ionic compound A+B- with bcc structure has one AB formula unit per unit cell.

☒

Option 3 The shape of the octahedral void is octahedral. ☐

Option 4 The edge length of the crystal A+B- is equal to the distance between A+ and B- ions.

☐

Explanation Octahedral void

Position of octahedral void = Body centre edge centre For NaCl type slr.

ar r =

2+ -Na Cl

For zns

a 3r +r =

42+ 2-2n s

For fluorite slr.

a 3r +r =

42+ -zn s

Anti fluorite slr.

a 3r +r =

4+ 2-Li O

Cscl type slr.

a 3r +r =

2+ -cs cl

Q. No. 85 In which of the following structures, the coordination number of both the ions is same?

Option 1 Caesium chloride ☒

Option 2 Sodium chloride ☒

Option 3 Zinc chloride ☐

Option 4 Sodium oxide ☐

Explanation For Nacl co. no of Na+ = co. no. of cl- For Cscl co. no of cs+ = co. no. of cl-

Q. No. 86 Which of the following are true?

Option 1 In NaCl crystals, Na+ ions are present in all the octahedral voids. ☒

Option 2 In ZnS (zinc blends), Zn2+ ions are present in alternate tetrahedral voids. ☒

Option 3 In CaF2, F- ions occupy all the octahedral voids. ☐

Option 4 In Na2O, O2- ions occupy half of the octahedral voids. ☐

Explanation For Na2O Na+ ions all tetrahedral sites O2- ionsadopt ccp arrangement O2- occupy the position of calcium ions in fluorite slr. For caf2 (fluorite slr.)

-F occupies all 8 tetrahedral voids.

Ca are located at the face centred cubic lattice point and therefore, have cubic2+

closed packed arrangement.

Q. No. 87 The coordination number of 8 for cation is found in

Option 1 CsCl ☒

Option 2 NaCl ☐

Option 3 CaF2 ☒

Option 4 Na2O ☐

Explanation C. no. of 8 is found for Cscl, Caf2.

Q. No. 88 The density of KBr is 2.75 g cm-3. The length of the unit cell is 654 pm. Atomic mass of K = 39, Br = 80. Then, which of the following statements are true?

Option 1 It has 4 K+ and Br- ions per unit cell. ☒

Option 2 It has a body-centered structure. ☐

Option 3 It has rock-salt type structure. ☒

Option 4 It can have Frenkel defects. ☐

Explanation Density of kBr = 2.75 g cm-3 a = 654 pm Atomic mass of k = 39 Br = 80

2=4 fcc It has Nacl type slr.

It has CNO = 4 for both cation/anion

Q. No. 89 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) A : CsCl has face-centered cubic arrangement. R : CsCl has one Cs+ ion and one Cl- ions in its unit cell.





Correct Answer 4

Explanation Cscl Bcc lattice arrangement

Q. No. 90 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) A : In CaF2, F

- ions occupy all the tetrahedral sites. R : The number of Ca2+ is double the number of F- ions.





Correct Answer 3

Explanation In CaF2 F- ion occupy all the tetrahedral sites (AB2 type)

Simple type of ionic crystal in which cations and anions are in ratio of 1:2 or 2:1

Ca are located at the face centered cubic lattice paint and therefore have cubic2+

closed packed arrangement. The fluoride ions (I- ion) occupy all the eight tetrahedral voids.

Q. No. 91 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) A : Zinc blende and wurtzite both have fcc arrangement of sulphide ions. R : There are four formula units of ZnS in both..





Correct Answer 4

Explanation Zinc blend ccp

Zns wurtzite Bcc

Only half or the alternate T void are occupied by zn ions in sphalerite r. sl2+

One unit cell has six, s at corner, face centre and 3 atom in body centre2-

there are four formula units of zns in both.

Q. No. 92 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) A : In NaCl crystal, all the octahedral voids are occupied by Na+ ions . R : The number of octahedral voids is equal to the number of Na+ ions in the packing .





Correct Answer 1

Explanation In Nacl all the octahedral voids are occupied by Na ions.+

The no. of octahedral voids is equal to the no. of Na ions in the packing.+

Passage Text Crystalline solids have an ordered internal arrangement of atoms due to which they exhibit symmetry. The atoms are arranged in a symmetrical fasion in the three- dimensional network known as space lattice. Each point in a crystal lattice is known as lattice point or lattice site. Each point in a crystal lattice represents a constituent particle which can be an atom, ion or a molecule. The smallest repeating subunit in the lattice is known as unit cell. The unit cells are described as simple (points at all the corners), body - centered (points at all the corners and in the center), face-centered (points at all the corners and corners and centers of two opposite end faces) unit cells. For the stable ionic crystalline structures, the radius ratio limit for a cation to fit perfectly in the lattice of anions is determined by the radius ratio rule. This also defines the number of nearest neighbors of opposite charges surrounding the ion, which is known as the coordination number. This depends upon the ratio of radii of two types of ions, r+/r-. The radius rations for coordination numbers 3, 4, 6 and 8 are 0.155 - 0225-0.414, 0.414 -0.732 and 0.732 - 1, respectively. The coordination number of ionic solids increases with

increase in pressure and decreases with increase in temperature.

Q. No. 93 The number of atoms per unit cell in primitive (sc), body-c entered (bcc), face-centered (fcc) and end-centered (ecc) unit cell decreases as

Option 1 fcc bcc ecc sc> > >

Option 2 fcc bcc=ecc sc> >

Option 3 bcc fcc sc=ecc> >

Option 4 fcc bcc ecc=sc> >

Correct Answer 2

Explanation No. of atoms per unit cell Arrangement

4 Fcc

2 Bcc

2 Ecc

1 Simple cubic

Q. No. 94 A metal crystallizes in a face-centered unit cell. Its edge length is 0.410 nm. The radius of the metal is

Option 1 0.205 nm

Option 2 0.290 nm

Option 3 0.145 nm

Option 4 0.578 nm

Correct Answer 3

Explanation 2ar =

4

2 .410r = = 2 1025=.145nm

4

Q. No. 95 In a cubic lattice of ABC, A atoms are present at all corners except one corner which is occupied by B atoms. C atoms are present at face centers. The formula of the compound is

Option 1 A8BC24

Option 2 ABC3

Option 3 A7B24C

Option 4 A7BC24

Correct Answer 4

Explanation A at all corner-1 B at one corner

C at face centres

A B C

7

8

1

8

6=3

2

Formula A B C7 1 24

Q. No. 96 The ionic radii of Na+, K+ and Br- are 137, 148 and 195 pm, respectively. The coordination number of cation in KBr and NaBr structures are, respectively,

Option 1 8,6

Option 2 6,4

Option 3 6,8

Option 4 4,6

Correct Answer 1

Explanation Limiting radius Ratio (x = r+/r-) 0.414 x 0.732 < 6[c. No.] 0.732 x 0.999 < 8[c. No.]

C. No. of cation in KBr and NaBr slr. Core 8,6

Q. No. 97 In NaCl structure, Cl- ions have ccp arrangement and Na+ ions occupy all the octahedral sites. The total number of Na+ and Cl- ions per unit cell is _______.

Correct Answer 0008

Is Integer Type ☐

Explanation Total no. of Na+ and Cl- ions per unit cell = 8

r+ 0.524=

r- C.No. 6

Q. No. 98 The radius ratio of an ionic solid r+/r- is 0.524. The coordination number of this type of structure is_______.

Correct Answer 0006

Is Integer Type ☐

Explanation NH++ Br- ion have ionic radii 13 apm 186 pm,

139r+/r = =.747

186

CN = 8

Q. No. 99 NH4+ and Br- ions have ionic radii of 139 pm and 186 pm, respectively. The

coordination number of NH4+ ion in NH4Br is ____.

Correct Answer 0008

Is Integer Type ☐

Explanation

Q. No. 100 Frenkel defect is caused due to

Option 1 An ion missing from the normal lattice site creating a vacancy

Option 2 An extra positive ion occupying an interstitial position in the lattice

Option 3 An extra negative ion occupying an interstitial position in the lattice

Option 4 The shift of a positive ion from its normal lattice site to an interstitial site

Correct Answer 4

Explanation The shift of a positive ion from its normal lattice site to an interstitial site

Q. No. 101 Ionic solids, with Schottky defects, contain in their structure

Option 1 Equal number of cation and anion vacancies

Option 2 Anion vacancies and interstitial anions

Option 3 Cation vacancies only

Option 4 Cation vacancies and interstitial anions

Correct Answer 1

Explanation Equal number of cation and anion vacancies

Q. No. 102 In Ag Br crystal, the ion size lies in the order Ag Br+ -<< .The AgBr crystal should have

the following characteristics

Option 1 Defectless (perfect) crystal

Option 2 Schottky defect only

Option 3 Frenkel defect only

Option 4 Both Schottky and Frenkel defects

Correct Answer 3

Explanation Frenkel defect only

Q. No. 103 The density of KBr is 2.75 g cm-3. The length of the unit cell is 654 pm. Atomic mass of K = 39, Br = 80. Then, which of the following statements are true?

Option 1 It has 4 K+ and Br- ions per unit cell. ☒

Option 2 It has a body- centered structure. ☐

Option 3 It has rock-salt type structure. ☒

Option 4 It can have Frenkel defects. ☐

Explanation Z Mdensity = 1pm=10 cm

a N

-103

A

Z 1192.75=

(654 10 ) 6.023 10-10 3 23

Z 4 It has rock-salt type.

Q. No. 104 In which of the following defects, the cations are present in interstitial sites?

Option 1 Frenkel defect ☒

Option 2 Schottky defect ☐

Option 3 Metal deficient non-stoichometric compound ☐

Option 4 Metal excess non-stoichiometric compound. ☒

Explanation Explained already Frenkel defect

Metal excess non-stoichiometric camp.

Q. No. 105 If Al3+ ions replace Na+ ions at the edge centres of NaCl lattice, then the number of vacancies in 1 mole of NaCl will be

Option 1 3.01 10 23

Option 2 6.02 10 23

Option 3 9.03 10 23

Option 4 12.04 10 23

Correct Answer 1

Explanation 1 mole of NaCl contains 1 mole Na+ ions i.e. 6.023 1023 Na+ ions. NaCl has fcc arrangement of Cl- ions and Na+ ions are present at the edge centres and body-centre. As there are 12 edge centres and each edge centre is shared by 4 unit cells, their

contribution pe1

r = 12=34

unit

Contribution of Na+ ion at the body-centre = 1 Thus, for every and Na+ ions, the ions present at the edge centres = 3.

This means that Na ions which have be3

= 6.0e 23 10 =4.517 10n replaced4 23+ 23

1Al3+ ion will replace 3 Na+ ions which is done to maintain electrical neutrality. One vacancy will be occupied by Al3+ ion and the remaining 2 will be vacant.

This means that 2

of these positions will be occupied by Al ions and1

33

rd rd3+

will remain vacant. Hence, no. of vacancies in 1 mole of

2NaCl= 4.517 10 =3.01 10

3 23 23

Q. No. 106 Analysis of a sample of iron oxide shows that it has the formula Fe0.90O. The fraction of iron present as fe2+ will be about

Option 1 90%

Option 2 60%

Option 3 78%

Option 4 70%

Correct Answer 3

Explanation Fe.90O The ratio of Fe and O atoms in pure iron oxide [FeO] = 1 : 1 Let x atoms of Fe II be replaced by Fe III in wustite.

No. of Fe II atoms present = .90 - x

Since the oxide is neutral Total charge on ion atoms - charge on oxygen atom 2(.90 - x) + 3x = 2 1.80 - 2x + 3x = 2 x = 2 - 1.80

of iron III in wusiti.2

= 100=.22=22.90

te % %

of iron II 78% %

Q. No. 107 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) A : In ZnO, the excess Zn2+ ions are present in interstitial sites. R : Metal excess crystals have either missing cation or anion in interstitial site.





Correct Answer 3

Explanation

Q. No. 108 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) A : FeO is non-stoichiometric with Fe0.95O. R : Some Fe2+ ions are replaced by Fe3+ as 3Fe2+ = 2Fe3+ to maintain electric neutrality.





Correct Answer 1

Explanation FeO is non- stoichiometric with Fe0.9S0

Means some Fe2+ ions are replaced by Fe3+ as 3Fe3+ = 2Fe3+ to maintain electric neutrality.

Q. No. 109 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) A : Frenkel defects are shown by silver halides. R : The size of silver ions is large.





Correct Answer 3

Explanation Size of X- (halide ion) is large.

Q. No. 110 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) A : Due to Frenkel defect there is no effect on density of a solid. R : ions shift from lattice sites to interstitial sites in Frenkel defect.





Correct Answer 1

Explanation Density is not changed while frenkel defect b/c it involves the dislocation of ions not removal of ion.

Q. No. 111 Al3+ ions replace Na+ ions at the edge centers of NaCl lattice. The number of vacancies

in one mole Nacl is calculated as n 10 23 .The value of n approximately is ____.

Correct Answer 0003

Is Integer Type ☐

Explanation _____ 3 = n

Q. No. 112 Match the imperfection in solids with the characteristic features.


1 Schottky defects Some cations occupy interstitial sites

R

2 Frenkel defects Conduct electricity due to free electrons

P

3 Metal excess defects

Act as p-type semiconductors

P,q,s

4 Metal deficient defects

Are non-stoichiometric defects

R,s

Explanation Schottky defects Acts as p-type

Semiconductor holes are created in P-type semiconductor and electrons get trapped in those holes

Frankel defects Some cations occupy interstitial sites.

metal excess defects are nonstoichiometric defects due to deficiency of metal

due to excess of metal due to presence of extra metal ions

due to anion vacancies of f -centre Metal deficient defects Is non stoichiometric defect no. of cation and anion ratio changes so stoichiometry is disturbed by this defects.

vacancies are created as holes in p-type semiconductor

Q. No. 113 Which of the following statements are not true?

Option 1 The conductivity of metals increases with increase in temperature. ☒

Option 2 Electrical conductivity of semiconductors increases with increase in temperature.

☐

Option 3 A diode is a combination of conductor and a semiconductor. ☒

Option 4 V2O5 behave as an insulator. ☐

Explanation For conductivity of are seseptible diode combination of

Q. No. 114 An oxide of transition metal that has conductivity as well as appearance like that of copper is

Option 1 Ti2O3

Option 2 V2O3

Option 3 ReO3

Option 4 Mn2O3

Correct Answer 3

Explanation ReO3

Q. No. 115 The oxide that is insulator is

Option 1 VO

Option 2 CoO

Option 3 ReO3

Option 4 Ti2O3

Correct Answer 2

Explanation CoO

Q. No. 116 Silicon is a

Option 1 Conductor

Option 2 semiconductor

Option 3 Nonconductor

Option 4 Superconductor

Correct Answer 2

Explanation semiconductor

Q. No. 117 Which of the following is ferromagnetic?

Option 1 Ni

Option 2 Co

Option 3 Fe3O4

Option 4 All are correct

Correct Answer 4

Explanation Ferromagnetic Permanent magnetism even in the absence of magnetic field. Above a temp. called curie temp, there is no alignment Eg. Ni, Co, Fe2O4

Q. No. 118 Ferromagnetic substances have

Option 1 Zero magnetic moment

Option 2 Small magnetic moment

Option 3 Large magnetic moment

Option 4 Any value of magnetic moment.

Correct Answer 2

Explanation Ferromagnetism This arrises when there is net dipole moment

eg. CO3O4, Fe3O4 ferrites.

Q. No. 119 Ferromagnetism is maximum in

Option 1 Fe

Option 2 Ni

Option 3 Co

Option 4 None

Correct Answer 1

Explanation Fe

Q. No. 120 Anti-ferromagnetic substances possess

Option 1 Low magnetic moment

Option 2 Large magnetic moment

Option 3 Zero magnetic moment

Option 4 Any value of magnetic moment

Correct Answer 3

Explanation Anti-ferromagnetic sub. This arises when net dipole alignment is zero due to equal and opposite alignment

eg. MnO, MnO3 Mn2O3, FeO, Fe2O3 NiO, Cr2O3, feO

Q. No. 121 In the following questions, a statement of assertion (A) is followed by a statement of reason (R) A : Antiferromagnetic substances become paramagnetic on heating to high temperature. R : Heating results in spins of electrons becoming random.





Correct Answer 1

Explanation Reason is self-explanatory for the assertion.

Passage Text Crystals as such are never perfect and in general have some defect, which simply

refers to a disruption in the periodic order of a crystalline material. Based on their nature, these are classified into three types: stoichiometric, impurity and non-stoichiometric defects. In non-stoichiometric defects, the ratio of the number of atoms of one kind to the number of atoms of the other kind does not correspond exactly to the ideal whole number ratio implied by the formula. For example, the ideal formula of ferrous oxide should be FeO, but in actually found to be Fe0.93 O in a sample, because some ferric ions have replaced ferrous ions in the crystal. These defects affect the properties of the crystals. Sometimes such defects are introduced to improve the properties of the crystals for a specific use, such as in the field of electronics. Elements of Group 14 are commonly doped with impurities of elements of Group 13 or 15. In ionic crystals, impurities are introduced with cations having higher valence than the cations in the pure crystal structure.

Q. No. 122 Which one of the following doping will produce p-type semi - conductor?

Option 1 Silicon doped with arsenic

Option 2 Germanium doped with phosphorus

Option 3 Germanium doped with aluminium

Option 4 Silicon doped with phosphorus

Correct Answer 3

Explanation P-type is when the semiconductor is doped with a group 13 element

Q. No. 123 Which one of the following defects does not affect the density of the crystal ?

Option 1 Schottky defect

Option 2 Interstitial defect

Option 3 Frenkel defect

Option 4 Both b and c

Correct Answer 3

Explanation Only is Schottky defect the ions are not present so density decreases.

Q. No. 124 NaCl was doped with 10 mol-3 % SrCl2. The concentration of cation vacancies is

Option 1 6.02 10 mol 18 -1

Option 2 6.02 10 mol 15 -1

Option 3 6.02 10 mol 21 -1

Option 4 6.02 10 mol 12 -1

Correct Answer 1

Explanation Dopping of NaCl with 10-3 SrCl2 means that 100 moles of NaCl are doped with 10-3 mole of SrCl2.

1 mole of NaCl is doped10

= with SrCl mole100

-3

2

= 10-5 mole As each sr2+ ion introduces one cation vacancy, therefore, concentration of cation vacancies = 10-5 mol/mol of NaCl

= 6.023 1010 mol 23-5 -1

=6.02 10 mol 18 -1

Q. No. 125 The percentage of iron present as Fe(III) in Fe0.93 O1.0 is

Option 1 17.7%

Option 2 7.84%

Option 3 11.5%

Option 4 9.6%

Correct Answer 3

Explanation The ratio of the Fe and O atoms in pure iron oxide (Feo) = 1:1

Let x atoms of fe II be replaced by fe III In wustite

No. of Fe II atoms present = (.93 - ) x

Since the oxide is neutral, Total charge on iron atoms = charyon oxygen atom = (.93 - x) + 3x = 2 1.89 - 2x + 3x = 2 x = 2 - 1.86 = .14

of iron III in wu.14

= 100.93

stite

=15.05%

Covalent Solid Censtituants are atoms → Bonding is covalent ...

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