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SOLID STATE
General characterishics of solid states
The following are the characteristic propeties of the solid state 1. They hare definite mass, volume and shape.
2. Intermolecular distance are short. 3. Intermolecular forces are strong.
4. Their constituent particles (atoms/molecules/ions) have fixed positions and can only oscillate about their mean position .
5. They are incompressible and rigid.
Such solids which hare difinite volume and structure and do not lose their shapes are called true solids
eg. NaCl,KCl, sugar, Fe, Ag, Cu etc.
Whereas solids which lose shapes on long standing and flows under its own weight and get distorted are called pseudo solids super cooled liquids. e.g. glass.
Ordered arrangement :
In solids there is ordered arrangement of the particles of the solid (atoms/molecules/ions). This ordered arrangement is of two types
1. Long range order, and 2. Short range order. “Long range order”
In long range order there is a regular pattem of arrangement of particle which represents itself periodically over the entire solid.
“Short range order :”
In short range order the regular arrangement is periodically repeated over short distances only. On the basis of the nature of the order, solids can be classified as crystalline solids and amorphous solids. Crystalline solids - In such solids there is a definite arrangement of particles throughout the entire there dimentional network of a crystal. These solids thus have ‘long range order’. This three dimentional arrangement of particle is called crystal lattice (or) space laltice eg. NaCl, KCl, Sugar, quartz etc.
Amorphous solids - In such solids, particles do not have definite arrangement and thus are ‘short range order’ i.e. portions of regular arrangement are scaltered and in between them arrangement is disordered. eg. glass
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Difference between crystalline and Amorphous solids
1. Characteristic geometry- In the crystalline solids the partides are definitely and orderly arranged
thus these have characteristic geometry while Amorphous solids do not have definite arrangement of particles thus do not posses characteristic geometry thus called “Amorphous” (A Not, morphous structure)
2. Melting points- As crystalline solids posses a definite arrangement of constitutent particles thus
there is fixed value of bond length or bond energy between to neighbouring particles, thus a crystalline solid have a sharp melting point value i.e. it changes into liquid state at a define temeperature. On the contrary an amorphous solid have irregular arrangement of particles thus various values of bond length or bond energy therefore shows a range of temperatures where it melts. For example, when glass is heated, it softens and then starts flowing without under going any abrupt change from solid to liquid state. crystalline solids are thus classified as true solids while amorphous solids are pseudo solids.
3. Isotropy and Anisotropy :- Amorphous solids differe from crystalline solids and resemble liquids
in many respects. The properties of amorphous solids, such as electrical condutivity, thermal
conductivity, mechanical strength, refractive index, coefficient of thermal expansion etc. are same in all directions. Such solids are known as Isotropic solids. Gases and liquids are also isotropic. On the other hand, crystalline solids show these physical properties different in different direction . Therefore crystalline solids are called “Anisotropic” in nature.
Anisotropy in crystolline solid is a consequence of their orderly particle arrangement in crystals. AB Consist of one type of particles.
CO Consist of two type of particles. Due to this variation, properties are different along different directions in crystalline solids.
While in amorphous solids there is “short range order” i.e. regularity is upto small distances that one randomly scattered throughout volume of the solid, thus the resultant value of any physical property is same along all directions. Classification of crystalline solids :
Crystalline solids can be classified, on the basis of nature of intermolecular forces operating in them, into four categories.
Ionic solids, covalent solids, metallic solids and Molecular solids.
1. Ionic solids :- Ionic solids consist of ions (cation & anion) that are arranged in three dimensional space bounded by strong electrostatic forces.
A B
C
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Ionic solids are :- Hard, briltle
Possesing high melting point and boiling point.
Bad conductor of electricity in solid state while furnishes the ions in aqueous (or) molten state thus become conducting.
e.g. NaCl, KCl, KF, CsCl, Na2SO4, Zns etc.
2. Covalent solids :- These are formed by sharing of valence electrons between two atoms resulting in the formation of a covalent bond. The covalent bonds extend in two and three dimensions forming a giant interlocking structure called Network. Thus covalent solids are also called Network solids. As the covalent bonds are directonal thus the atoms are bounded tightly, this imparts rigidity and hardness to covalent solids. These solids also have high melting point.
As all the valence electrons take part in bond formation thus covalent solids are insulators and do not conduct electricity. eg. diamond and silicon carbide (SiC).
In graphite, the carbon atoms are arranged is layers, one over other. In these layers each carbon atom is covalently bonded to three other neighbouring carbon atoms. The fourth valence electron of each carbon atom is present between the layers and is free to move about. These electrons are responsible for conducting nature of graphite.
3. Metallic solids :- Metallic solids consist of large number of positive metal ions embedded in sea of free electrons. These electrons are the valence electrons of the metal atom that are contributed by each metal atom in course of formation of metallic crystals. These electrons are responsible for high electrical and thermal conductivity of metals. When an electric field is applied these electrons flows through the network of massive positive metal ions.When heat is supplied to one portion of a metal, the thermal energy is uniformally spread through out the solid by free electrons. Metallic lusture is also due to these free electrons.When light rays are incident on the metal surface these free elctrons absorb energy and get excited to higher energy levels, as the electron jumps back to ground state it emits energy in the form of light and the surfce shines.
4. Molecular solids :- Molecules are the basic constitutient particles of molecular solids. There are three types of molecular solids.
(i) Non-polar molecular solids :- These are solids of either atoms [eg. solidified Noble gases] (or)
of non polar molecules [eg. solid H2, N2, O2, CO2, I2] etc.
In these solids atoms (or) moleculars are held together by weak varder weals forces.
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(iii) Hydrogen-bonded molecular solids :- Solids like H2O (ice), C6H5COOH etc which shows polar covalent bonds between H and strong electronegative elements like F, O, N are the hydrogen bonded, molecular solids. These solids are also non-conductor of electricity.
Crystalline state
:-“A crystal is a solid comporsed of atoms/ions (or) molecules arranged in an orderly repetitive array” The crystallinity of a crystal may be defined as a condition of matter resulting from an orderly, cohesive, three dimensional arrangement of its constituting particles in space.
This three dimentional arrangement is called crystal laltice or space crystal laltice and the constituting atoms are called lattice sites or lattice points. The lattices are bound by surface that are usually planner and known as faces of the crystal.
Unit Cell :- The unit cell of a crystal is the basic structural unit possesing the symmetry of the crystal and which when repeated in all directions will develop the crystal lattice. Thus a unit cell is the smallest picture of the whole crystal.
Law of constancy of interfacial angles or Law of crystallography.
:-The angle between the two perpendiculars to the two intersecting faces is formed as the interfacial angle. It is same as the angle between the unit cell adges. Goniometer is used to measure the interfacial angle. The interfacial angle of a substance remains the same although its shape may be different due to conditions of formation. This is known as law of constancy of interfacial angle.
Characteristics of a unit cell
:-1. A unit cell is characterized by the edge lengths a, b and c along the
three edges of the unit cell. These edges may or may not be mutually perpendicular.
2. The angles between the edges are as follows. Between edges b and c,
Between edges a and c, and Between edges a and b.
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Types of unit cells
Primitive or simple cubic (PS/SC) unit cell: Spheres in one layer sitting directly on top of those in previous layer, so that all layers are identical. Each sphere is touched by six other, hence coordination number is 6. 52% of available space occupied by spheres.
Example: Polonium crystallises in simple cubic arrangement.
Z = 1 ; C.N. = 6
1.2 Body Centered cubic (BCC) unit cell: Spheres in one layer sit in the depression made by first layer in a-b-a-b manner. Coordination number is 8, and 68% of available space is occupied by atoms.
Example: Iron, sodium and 14 other metal crystallises in this manner.
Z = 2 ; C.N. = 8
1.3 Face centered cubic (FCC) unit cell:
Examples : Al, Ni, Fe, Pd all solid noble gases etc.
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Crystal systems
:-These are seven crystal systems viz cubic, tetragonal, orthorhombic, hexagonal, rhombohedral, Monoclinic and triclinic.
A crtystal system may exist with more than one type of unit cells. In nature there exist 14 different types of crystal lattices also called Bravias lattices these crystal laltices are orises due to different possible combination of crystal systems and unit cells.
Number of atoms in a unit cell :
1. Primitive cubic unit cell
:-(i) Number of atoms in an isolated unit cell = 8
(ii) Number of unit cells that share a atom in un-isolated condition = 8 contribution of a atom in each unit cell = 1
8
(iii) Number of atoms in an un-isolated unit cell =8 1 8 = 1
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2. Body-centred cubic unit cell
:-(i) Number of atoms in an isolated unit cell = 8 + 1 = 9 (ii) Number of unit cells sharing
an atom at any one corner = 8 an atom at body centre = 1
(iii) Number of atoms in an un-isolated unit cell = 8 1 8
+ (1 × 1) = 1 + 1 = 2 3. Face-centred cubic unit cell
:-(i) Number of atoms in an isolated unit cell = 8 + 6 = 14 (ii) Number of unit cells sharing
an atom at any one corner = 8 an atom at any face centre = 2 (iii) Number of atoms in an un-isolated unit cell
8 1 8 + 6 × 2 1 = 1 + 3 = 4 Packing efficiency
:-P. E. Volume Ocupied by atoms in crystal Total volume of the crystal
1. For face centred cubic structure
:-r 2r r A B C InABC B2 = AC2 = BC2 + AB2 = 2a2 b = a 2 Radivs of sphere r = b a 4 2 4 9 2 2
As number of particles in a FCC is four therefore total volume occupied by particles =4 4
3 3 r P. E. 4 4 3 3 3 r a = 3 2 = 0.74
Therefore packing efficiency of FCC is 74% 2. For body centred cubic
:-InABC
B2 = AB2 = AC2 + BC2
a2 + a2(2)
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Packing efficiency = 3
6 = 0.68
Therefore packing efficiency of BCC is 68% 3. For simple cubic structure
:-Hence b = a and as r = b a
2 2
a
Therefore P.E. = 0.524
Therefore packing efficiency of simple cubic is 52.4%
Calculation of density of a unit cells
:-(i) Volume of a cubic unit cell = a3 [
a = edge length]
(ii) Mass of a unit cell = no. of atoms in × mass of each atom unit cell = z × m. as m = Molar massN
A
[NA Avogadro Number] or m = NM A
(iii) Density of the unit cell =
mass of unit cell
volume of unit cell= 3
a m Z = Z M3 a NA
(iv) Density of unit cell of a solid compound is same as density of any amount of mass of that compound.
Example1- The density of crystalline sodium chloride is 2.165 g cm-3 . What is the egde length of the unit
cell? What would be the dimensions of cube containing one mole of NaCl. Ans- Since density d = Z M3
a NA a3 = A N d M Z = 23 10 023 . 6 165 . 2 5 . 58 4 = 1.794 × 10-22 a = 5.64 × 10-8cm
Molar volume = DensityMass =
165 . 2 5 . 58
Therefore, Edge length (a) = 3
1 165 . 2 5 . 58 = 3 cm
Coordination Number:- It is defined as the number of nearest neighbours that an atom has in a unit cell. (a) Simple Cubic Struct)ure- C.N. = 6
(b) Face Centered Cubic structure - C.N. = 12 (c) Body Centered Cubic structure- C.N. = 8
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Closed packed structures
:-(i) Arrangment in One dimension:- The number of neasest neighbours of a particle is called its “coordination number”. When th particles are arranged along a line i.e. in 1D arrangement the coordination number of the particles is two.
(10 arrangement) [C.N. = 2]
(ii) Arrangment in Two dimensions:-Now for 2D arrangement the second line of the particle can be arranged in two manners.
I
st Particles of second line align as it is in first line
IInd Particles of second line align in depressions between the particles of first line.
Ist Case IInd Case A A A A A B A B
It is called “AAA” type arrangement joining the It is callled “ABAB” arrangement Joining the centres of four nearest particles gives a square, centre of six nearest particles gives a hexagon, thus it is also called “Square, close packing”. In thus it is also called “Hexagonal close packing” square close packing the coordination number of In “Hexagonal close packing” 2D structure the particles is 4. coordination number of particle is 6.
The triangular shaped vacant space (reated due to adjacent placement of three spheses in this case) iscalled “ Triangular void”. The adjacent triangular void in ABAB arrangement are pointing in opposite directions (i.e. up and down).
(iii) Arrangment in 3 dimension:- To build up the third layer of spheres, there are two ways . 1. ABAB Type or HCP Type- In this arrangment the spheres are
placed in the voids between the spheres forming the bottom layer, now the third layer is placed on the second layer in such a way that the third layer lies exactly above the spheres of the first layer. When this arrangment is continued it is found to have hexagonal symmetry and is called hexagonal close packing.
2. ABCABC.. type or Cubic close packing- In this type of arrangment the spheres of the second layers are placed on the hollow spaces between the spheres of the first layer. Now the third layer is placed on the second layer in such a way that the third layer occupies the hollow positions between the spheres of the second layer. Now the spheres of the fouth layer will be exactly above the spheres of the first layer. This type of arrangment is thus called ABCABC.. arrangment. When this arrangment is
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Note- FCC and CCP are the same arrangment.The Coordination number of each sphere in the above two arrangment is 12.
Interstitial sites or voids:- In close packing of spheres there is always some vacant space between the spheres, thesevacant spaces are called interstitial sites or holes or voids. Voids are of two
types-(i) Tetrahedral Voids- When a sphere is placed on the space between three spheres which are touching each other , the centres of these four spheres are at the corners of a regular tetrahedron and the void is called tetrahedral void.
In HCP and CCP each sphere is in contact with three spheres above and three spheres below. Thus there are two tetrahedral voids associated with each sphere.
So if there are Z atoms in a unit cell than-Number of tetrahedral voids = 2Z
Locating tetrahedral voids-In FCC structure, there are eight spheres in the corners of the unit cell and each sphere is in contact with three groups giving rise to eight tetrahedral voids .
Circles labelled T are the tetrahedral voids. Radius Ratio or size of tetrahedral voids
sphere void
r r
= 0.225
(ii) Octahedral voids- This type of void is formed by the combination of two triangular voids. Each octahedral void is created by superimposing two equilateral triangles with apices in opposite directions. There is one octahedral site associated with each sphere(atom).
So if there are Z atoms in a unit cell than-Number of octahedral voids = Z
Locating octahedral voids- In closed packed structures the octahedral voids are present at the centres of the edges of the cube.
Radius Ratio or Size of octahedral
voids-sphere void
r r
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Circles labelled as O represents octahedral voids.
Example- In a solid, oxide ions are arranged in ccp. Cations A occupy one – sixth of the
tetrahedral voids and cations B occupy one third of the octahedral voids. What is the formula of the compound?
Solution: Let there are N atoms in CCP arrangment than
No. of tetrahedral voids = 2N and No. of Octahedral voids = N
No of A atoms = 6 1 2N = 3 N
and no. of B atoms =
3 1
N
So the formula of oxide will be
3 1 3 1B A = AB
Example- In a crystalline solid, having formula AB2O4, oxide ions are arranged in cubic close packed lattice while cations A are present in tetrahedral voids and cations B are present in octahedral voids
(i) What percentage of the tetrahedral voids is occupied by A? (ii) What percentage of the octahedral voids is occupied by B?
Solution: In a cubic close packed lattice of oxide ions there would be two tetrahedral voids and one octahedral
void for each oxide ion.
For four oxide ions there would be 8 tetrahedral and four octahedral voids two are occupied by B. Percentage of tetrahedral voids occupied by A = 1/8 × 100 = 12.5%
Percentage of tetrahedral voids occupied by B = 2/4 × 100 = 50% STRUCTURE OF CRYSTALS
1. Ionic Crystals:- In ionic crystals, the coordination number depends on the relative sizes of the ions. The ratio of the radii of cation and anion is called radius ratio. The coordination number varies with radius ratio.
Radius ratio Coordination Number Arrangment of Example
(r+/r-) anions round the cation
0.155 – 0.225 3 Planer triangular B2O3
0.225 – 0.414 4 Tetrahedral ZnS
0.414 - 0.732 6 Octahedral NaCl
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Structure of Ionic Crystals
NaCl type :- In NaCl each Na+ion is surrounded by six Cl-ions. Cl- ions form the FCC lattice and Na+ ions
occupy the octahedral holes. There are six octahedral holes around each Cl- ion. The coordination number of each ion is 6 :6.
Radius ratio = r r Na Cl = 0.524
No. of sodium ions = 12(at the edge centres) ×1
4 + 1(at body centre) = 4 No. of chloride ions = 8(at the corners)× 1
8+ 6(at face centres)× 1 2 = 4
Example- most alkali metal halides like KCl, NaI, RbF, RbI, AgF,AgCl, AgBr, NH4Br and FeO.
CsCl Type
In CsCl, Cl- ions are at the corners of a cube whereas Cs+ at the centre of the cube and
vice versa. Thus it has BCC structure. The cordination number is 8:8. Radius ratio = 0.933
examples- CsBr, CsI.
Note- At 760K CsCl structure changes into NaCl structure and coordination number changes from 8:8 to 6:6.
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ZnS or ZnOType structure
ZnS crystal has two type of structures(i) Zinc Blende structure:- It has FCC or CCP structure. The S2-ions are at the corners of the cube
and at the centre of the each face . Zn ions occupy half of the tetrahedral voids. Each Zn ions is surrounded by four sulphide ions tetrahedrally and vice versa. The coordination number is 4:4. Radius ratio = 0.4
examples- CuCl, CuBr, CuI, AgI. Diamond also posses the same structure.
Zinc blende structure
S2– Zn+2
(ii) Wurtzite structure:- It has hexagonal close packed structure. S2-ions adopt HCP arrangment and
Zn2+ ions occupy half of the tetrahedral sites. The coordination number is 4:4.
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Antifluorite structure :(Li2O)O2– ion forming ccp and Li+ taking all tetrahedral voids.
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Rutile structure: (TiO2) O2– forming hcp while Ti4+ ions occupy half of the octahedral voids.
Pervoskite structure:(CaTiO3) Ca2+ in the corner of cube O2– at the face center and Ti4+ at the centre of cube.
Spinel and inverse spinel structure: (MgAl2O4)O2– forming fcc, Mg2+ filling 1/8 of tetrahedral voids and Al3+ taking half of octahedral voids.In an inverse spinel structure, O2– ion form FCC lattice, A2+ ions occupy 1/8 of the tetrahedral voids and trivalent cation occupies 1/8 of the tetrahedral voids and 1/4 of the octahedral voids.
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CRYSTAL DEFECTS:
Real crystals are never perfect: they always contain a considerable density of defects and imperfections that affect their physical, chemical, mechanical and electronic properties. The existence of defects also plays an important role in various technological processes and phenomena such as annealing, precipitation, diffusion,sintering, oxidation and others. It should be noted that defects do not necessarily have adverse effects on the properties of materials. There are many situations in which a judicious control of the types and amounts of imperfections can bring about specific characteristics desired in a system.
The electrical behavior of semiconductors, for example, is largely controlled by crystal imperfections. The conductivity of silicon can thus be altered intype (n or p) and by over eight orders of magnitude through the addition of minute amounts of electrically active dopant elements. In this case, each atom of dopant,substitutionally incorporated, represents a point defect in the silicon lattice. The fact that such small amounts of impurity atoms can significantly alter the electrical properties of semiconductors is responsible for the development of the transistor and has opened up the entire field of solid state device technology.
POINT DEFECTS : When some ion's are missing from ionic crystals from their theoretical lattice point, the
crystal is defected structure.
Defect due to missing of ions from theoretical lattice point is called point defect. Point defect increases with increase in temperature. At absolute zero temperature, ionic crystal may not have any defect.
POINT DEFECTS ARE OF TWO TYPES :
(i) Stoichiometric defect : Defects due to which overall formula of ionic compound do not change is called stoichiometric defect.
(ii) Non- Stoichometric defects are those due to which overall formula of compound changes. Stoichiometric defect :(i) Schottky defect (ii) Frenkel defect Schottky :- When pair of holes exist in crystal lattice due to missing of positive and negative ions in
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Schottky defect
This defect is common in ionic compounds with high coordination number. It absolute zero temperate crystals tends to have perfectly order arrangement. As temperature is increased, some vacancies are always created in crystal lattice.
Ex. of crystals showing schottky defect: NaCl, KCl, NaBr etc.
Frenkel defect : In this type of defect, holes are created due to transfering of an ion from usual lattice site to a interstitial site. This type of defect very common in compounds in which there is large difference between size of cation and anion.
Frenkel defect e.g.ZnS ; AgBr; etc Consequence of defects:
- Due to schottky defect density of crystal decreases - Crystal can conduct electricity to a small extent.
Non-stoichiometric defects :
Non stoichiometric compounds are those in which the ratio of positive and negative ions present in the compounds differ from that indicated by their chemical formula. eg. Fe0.95O, Cu1.97S, etc.
These defects arise due to excess of metal or non-metal atoms–
(i) Metal excess defect (ii) Metal deficiency defect Metal excess defect arise due to
(i) Missing of a negative ion from lattice site and position taken by an electron. This defect is similar to schotty defect and also found in crystals showing schotty defect. Ex. When sodium vapours passed over NaCl crystal a yellow non-stoichiometric form of NaCl is obtained. Vaccant lattice site occupied by electron's is called F-centre (Farbe colour). Which is responsible for colour of crystal.
(ii) An extra metal occupy interstitial site and to maintain electrical neutrality, electrons occupy another interstitial site. This type of defect is very close to Frenkel defect and found in ZnO.
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When ZnO is heated, it turns yellow as it's loses some oxygen. The Zn2+ ion move to an interstitial site. Note: In this defect there is no hole in the crystal.
— Crystals with metal excess defect contain free electrons and if these migrate, they conduct an electric current.
— As amount of current carried is very small, they behave like semiconductor's. (n-type semiconductor)
Metal deficiency defect:–
(i) A metal ion is missing from it's normal lattice point the electrical neutrality is maintained by extra positive charge of same of the remaining metal ions.
FeO, FeS, NiO - exhibit this type of defect.
(ii) An extra negative ion is present in the interstial position and electrical neutrality is maintained by extra– positive charge on remaining metal ions. This type of defect is not known.
Crystal with metal deficiency defect behaves like p-type semiconductors.
A
B
A
B
B
B
B
B
A
A
A
+ + + + 2+e
(a) Ex. FeO, NiO, FeS
Metal deficency defect
PROPERTIES OF SOLIDS 1. Electrical properties of solids :
On the basis of electrical conductivity, solids are classified into three types. i. metals ii. semi-conductors iii. insulators
Electrical conductivity of metals is very high and is of the order of 106 – 108 ohm–1 cm–1.
Electrical conductivity of solids is due to the movement of electrons and positive holes or through the motion of ions.
The conduction through electrons is called n-type conduction and through positive ions is called p-type conduction.
Pure ionic solids, where conduction takes place only through motion of ions, are insulators. The presence of defects in the crystal structure increases their conductivity.
The solids whose conductivity lies between those of metallic conductors and insulators are called semi-conductors.
Semi - conductors have conductivity which lies in the range of 102 – 10–9 ohm–1 cm–1.
The solids which do not allow the.passage.of electric current through them are called insulators. Eg: Wood, rubber, sulphur, phosphorus etc..
The conductivity of semi conductors and insulators. is mainly due to the presence of interstial electrons and positive holes in the solids due 10 imperfections.
The conductivity of semi conductors and insulators increases with increase in temperature while that of metals decreases.
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2. Magnetic properties of solids :
The substances which are weakly repelled by magnetic field are called diamagnetic substances. Eg : TiO2, NaCI, benzene etc..
Diamagnetism arises when all the electrons are-paired.
The substances which are weakly attracted by magnetic field are called paramagnetic substances. Atoms,-ions or molecules with unpaired electrons exhibit paramagnetism.
Paramagnetic substances lose their magnetism in the absence of magnetic field. Eg : TiO, VO2' CuO etc.
The substances which are strongly attracted by magnetic field are called ferromagnetic substances. Ferromagnetic substances show permanent magnetism even in the absence of magnetic field.
Eg : Iron, cobalt, nickel, CrO2 etc.
Ferromagnetism arises due to spontoneous alignment of magnetic moments in the same direction. Anti ferromagnetism arises due to the alignment- of magnetic moments in opposite direction in a com-pensatory manner and resulting in a zero magnetic moment.
Eg : MnO, MnO2, Mn2O3.
Ferrimagnetism arises due to alignment of magnetic moments in opposite directions resulting in a net magnetic moment.
Eg : Fe3O4, M2+Fe
2O4 ; (M = Mg, Cu, Zn etc.)
Ferromagnetic and ferrimagnetic substances change into paramagnetic substances at higher tempera-tures due to randomisation of spins.
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EXERCISE I Formula of ionic solid from unit cell description
Q.1 A cubic solid is made up of two elements A and B. Atoms B are at the corners of the cube and A at the body centre. What is the formula of compound.
Q.2 A compound alloy of gold and copper crystallizes in a cubic lattice in which gold occupy that lattice point at corners of the cube and copper atom occupy the centres of each of the cube faces. What is the formula of this compound.
Q.3 A cubic solid is made by atoms A forming close pack arrangement, B occupying one. Fourth of tetrahedral void and C occupying half of the octahedral voids. What is the formula of compound.
Q.4 What is the percent by mass of titanium in rutile, a mineral that contain Titanium and oxygen, if structure can be described as a closet packed array of oxide ions, with titanium in one half of the octahedral holes. What is the oxidation number of titanium?
Q.5 Spinel is a important class of oxides consisting of two types of metal ions with the oxide ions arranged in CCP pattern. The normal spinel has one-eight of the tetrahedral holes occupied by one type of metal ion and one half of the octahedral hole occupied by another type of metal ion. Such a spinel is formed by Zn2+, Al3+ and O2–, with Zn2+ in the tetrahedral holes. Give the formulae of spinel.
Edge length, density and number of atoms per unit cell
Q.6 KF crystallizes in the NaCl type structure. If the radius of K+ ions 132 pm and that of F– ion is 135 pm,
what is the shortest K–F distance? What is the edge length of the unit cell? What is the closet K–K distance?
Q.7 A closed packed structure of uniform spheres has the edge length of 534 pm. Calculate the radius of sphere, if it exist in
(a) simple cubic lattice (b) BCC lattice (c) FCC lattice
Q.8 Calculate the density of diamond from the fact that it has face centered cubic structure with two atoms per lattice point and unit cell edge length of 3.569 Å.
Q.9 An element crystallizes into a structure which may be described by a cubic type of unit cell having one atom on each corner of the cube and two atoms on one of its body diagonals. If the volume of this unit cell is 24×10–24 cm3 and density of element is 7.2 g cm–3, calculate the number of atoms present in
200 g of element.
Q.10 Silver has an atomic radius of 144 pm and the density of silver is 10.6 g cm–3. To which type of cubic
crystal, silver belongs?
Q.11 AgCl has the same structure as that of NaCl. The edge length of unit cell of AgCl is found to be 555 pm and the density of AgCl is 5.561 g cm–3. Find the percentage of sites that are unoccupied.
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Q.12 Xenon crystallises in the face-centred cubic lattice and the edge of the unit cell is 620 pm. What is the nearest neighbour distance and what is the radius of xenon atom?
Q.13 The two ions A+ and B– have radii 88 and 200 pm respectively. In the closed packed crystal of compound
AB, predict the co-ordination number of A+.
Q.14 CsCl has the bcc arrangement and its unit cell edge length is 400 pm. Calculate the interionic distance in CsCl. Q.15 Gold crystallizes in a face centered cubic lattice. If the length of the edge of the unit cell is 407 pm, calculate the density of gold as well as its atomic radius assuming it to be spherical. Atomic mass of gold = 197 amu.
Q.16 The density of KBr is 2.75 g cm–3 . The length of the edge of the unit cell is 654 pm. Show that KBr has
face centered cubic structure.
(N = 6.023 ×1023 mol–1 , At. mass : K = 39, Br = 80)
Q.17 An element crystallizes in a structure having FCC unit cell of an edge 200 pm. Calculate the density, if 200 g of this element contains 24×1023 atoms.
Q.18 The effective radius of the iron atom is 1.42 Å. It has FCC structure. Calculate its density (Fe = 56 amu)
Q.19 A crystal of lead(II) sulphide has NaCl structure. In this crystal the shortest distance between Pb+2 ion and S2– ion is 297 pm. What is the length of the edge of the unit cell in lead sulphide? Also calculate the
unit cell volume.
Q.20 If the length of the body diagonal for CsCl which crystallises into a cubic structure with Cl– ions at the
corners and Cs+ ions at the centre of the unit cells is 7 Å and the radius of the Cs+ ion is 1.69 Å, what is the radii of Cl– ion?
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EXERCISE II
Q.1 Iron has body centered cubic lattice structure. The edge length of the unit cell is found to be 286 pm. What is the radius of an iron atom?
Q.2 Cesium chloride forms a body centered cubic lattice. Cesium and chloride ions are in contact along the body diagonal of the unit cell. The length of the side of the unit cell is 412 pm and Cl– ion has a radius of
181 pm. Calculate the radius of Cs+ ion.
Q.3 In a cubic closed packed structure of mixed oxides the lattice is made up of oxide ions, one eighth of tetrahedral voids are occupied by divalent ions (A2+) while one half of the octahedral voids occupied trivalent ions (B3+). What is the formula of the oxide?
Q.4 A solid A+ and B– had NaCl type closed packed structure. If the anion has a radius of 250 pm, what
should be the ideal radius of the cation? Can a cation C+ having a radius of 180 pm be slipped into the tetrahedral site of the crystal of A+B– ? Give reasons for your answer.
Q.5 Calculate the value of Avogadro’s number from the following data: Density of NaCl = 2.165 cm–3
Distance between Na+ and Cl– in NaCl = 281 pm.
Q.6 If the radius of Mg2+ ion, Cs+ ion, O2– ion, S2– ion and Cl– ion are 0.65 Å , 1.69 Å, 1.40 Å, 1.84 Å, and
1.81 Å respectively. Calculate the co-ordination numbers of the cations in the crystals of MgS, MgO and CsCl.
Q.7 Iron occurs as bcc as well as fcc unit cell. If the effective radius of an atom of iron is 124 pm. Compute the density of iron in both these structures.
Q.8 KCl crystallizes in the same type of lattice as does NaCl. Given that 0.5 Cl r Na r and 0.7 K r Na r Calculate: (a) The ratio of the sides of unit cell for KCl to that for NaCl and
(b) The ratio of densities of NaCl to that for KCl.
Q.9 An element A (Atomic weight = 100) having bcc structure has unit cell edge length 400 pm. Calculate the density of A and number of unit cells and number of atoms in 10 gm of A.
Q.10 Prove that the void space percentage in zinc blende structure is 25%.
Q.11 A unit cell of sodium chloride has four formula units. The edge of length of the unit cell is 0.564 nm. What is the density of sodium chloride.
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Q.12 In a cubic crystal of CsCl (density = 3.97 gm/cm3) the eight corners are occupied by Cl–ions with Cs+ ions at the centre. Calculate the distance between the neighbouring Cs+ and Cl– ions.
Q.13 KF has NaCl structure. What is the distance between K+and F– in KF if density of KF is 2.48 gm/cm3. Q.14 The composition of a sample of wustite is Fe0.93O1.0. What percentage of iron is present in the form of
Fe(III)?
Q.15 BaTiO3 crystallizes in the prevoskite structure. This structure may be described as a cubic lattice with barium ions occupying the corner of the unit cell, oxide ions occupying the face-centers and titanium ion occupying the center of the unit cell.
(a) If titanium is described as occupying holes in BaO lattice, what type of holes does it occupy? (b) What fraction of this type hole does it occupy?
Q.16 Rbl crystallizes in bcc structure in which each Rb+ is surrounded by eight iodide ions each of radius 2.17 Å. Find the length of one side of RbI unit cell.
Q.17 If NaCl is dopped with 10–3 mol % SrCl
2, what is the numbers of cation vacancies?
Q.18 Find the size of largest sphere that will fit in octahedral void in an ideal FCC crystal as a function of atomic radius 'r'. The insertion of this sphere into void does not distort the FCC lattice. Calculate the packing fraction of FCC lattice when all the octahedral voids are filled by this sphere.
Q.19 A cubic unit cell contains manganese ions at the corners and fluoride ions at the center of each edge. (a) What is the empirical formula of the compound?
(b) What is the co-ordination number of the Mn ion?
(c) Calculate the edge length of the unit cell, if the radius of Mn ion is 0.65 Å and that of F– ion is 1.36 Å. Q.20 NaH crystallizes in the same structure as that of NaCl. The edge length of the cubic unit cell of NaH is
4.88 Å.
(a) Calculate the ionic radius of H–, provided the ionic radius of Na+ is 0.95 Å. (b) Calculate the density of NaH.
Q.21 Metallic gold crystallises in fcc lattice. The length of the cubic unit cell is a = 4.07 Å. (a) What is the closest distance between gold atoms.
(b) How many “nearest neighbours” does each gold atom have at the distance calculated in (a). (c) What is the density of gold?
(d) Prove that the packing fraction of gold is 0.74.
Q.22 Ice crystallizes in a hexagonal lattice. At the low temperature at which the structure was determined, the lattice constants were a = 4.53 Å, and b = 7.60 Å(see figure).
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Q.23 Using the data given below, find the type of cubic lattice to which the crystal belongs.
Fe V Pd
a in pm 286 301 388
in gm cm–3 7.86 5.96 12.16
Q.24 Potassium crystallizes in a body-centered cubic lattice with edge length, a = 5.2 Å. (a) What is the distance between nearest neighbours?
(b) What is the distance between next-nearest neighbours? (c) How many nearest neighbours does each K atom have? (d) How many next-nearest neighbours does each K atom have? (e) What is the calculated density of crystalline potassium?
Q.25 Prove that void space in fluorite structure per unit volume of unit cell is 0.243.
Q.26 A compound formed by elements X & Y, Crystallizes in a cubic structure, where X is at the corners of the cube and Y is at six face centers. What is the formula of the compound? If side length is 5Å, estimate the density of the solid assuming atomic weight of X and Y as 60 and 90 respectively.
Q.27 The metal nickel crystallizes in a face centred cubic structure. Its density is 8.9 gm/cm3. Calculate (a) the length of the edge of the unit cell.
(b) the radius of the nickel atom. [Atomic weight of Ni = 58.89]
Q.28 The olivine series of minerals consists of crystals in which Fe and Mg ions may substitute for each other causing substitutional impurity defect without changing the volume of the unit cell. In olivine series of minerals, oxide ion exist as FCC with Si4+ occupying
4 1th
of octahedral voids and divalent ions occupying
4 1th
of tetrahedral voids. The density of forsterite (magnesium silicate) is 3.21 g/cc and that of fayalite (ferrous silicate )is 4.34 g/cc. Find the formula of forsterite and fayalite minerals and the percentage of fayalite in an olivine with a density of 3.88 g/cc.
Q.29 The mineral hawleyite, one form of CdS, crystallizes in one of the cubic lattices, with edge length 5.87Å. The density of hawleyite is 4.63 g cm–3.
(i) In which cubic lattice does hawleyite crystallize? (ii) Find the Schottky defect in g cm–3.
Q.30 A strong current of trivalent gaseous boron passed through a germanium crystal decreases the density of the crystal due to part replacement of germanium by boron and due to interstitial vacancies created by missing Ge atoms. In one such experiment, one gram of germanium is taken and the boron atoms are found to be 150 ppm by weight, when the density of the Ge crystal decreases by 4%. Calculate the percentage of missing vacancies due to germanium, which are filled up by boron atoms.
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OBJECTIVE QUESTION BANK
Q.1. Transition metals, when they form interstitial compounds, the non-metals (H, B, C, N) are accommo dated in :
(A) Voids or holes in cubic - packed structure (B) Tetrahedral voids (C) Octahedral voids (D) All of these
Q.2. In a crystal both ions are missing from normal sites in equal number. This is an example of :
(A) F-centres (B) Interstitial defect (C) Frenkel defect (D) Schottky defect Q.3. The number of tetrahedral and octahedral voids in hexagonal primitive unit cell are :
(A) 8, 4 (B) 2, 1 (C) 12, 6 (D) 6, 12
Q.4. KBr shows, which of the following defects :
(A) Frenkel defect (B) Schottky defect (C) Metal excess defect (D) Metal deficiency Q.5. Hexagonal close packing is found in crystal lattice of
(A) Na (B) Mg (C) Al (D) None of these
Q.6. Schottky defect is noticed in :
(A) NaCl (B) KCl (C) CsCl (D) All
Q.7. Sodium metal crystallizes in bcc lattice with cell edge = 4.29A. The radius of sodium atom will be : (A) 1.50 A (B) 1.86A (C) 2.80 A (D) None of these
Q.8. The edge length of cube is 400 pm. Its body diagonal would be :
(A) 500 pm (B) 693 pm (C) 600 pm (D) 566 pm
Q.9. The unit cell dimensions of an orthorhombic lattice (with edges; a, b, c and the angle between them being,,)
(A) a = b = c ; = = = 90° (B) a b = c ; = = 90° 90° (C) a = b c ; = = 90° 90° (D) a b c ; = = = 90°
Q.10. A metallic element exists as cubic lattice. Each edge of the unit cell is 2.88A . The density of the metal is 7.20 g cm–3. How many unit cell will be present in 100 g of the metal :
(A) 6.85 × 102 (B) 5.82 × 1023 (C) 4.37 × 105 (D) 2.12 × 106
Q.11. Fraction of total volume occupied by atoms in a simple cube is : (A) 2 (B) 3 8 (C) 2 6 (D) 6
Q.12. Iron crystallizes in a b.c.c. system with a lattice parameter of 2.861A . Calculate the density of iron in the b.c.c. system (Atomic weight of Fe = 56, NA = 6.02 × 1023 mol–1)
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(A) 124.2 × 10–14 m, 5.5 g/cc (B) 3 4 × 287 pm, 7.5 g/cc (C) 2 4 × 287 pm, 6.5 g/cc (D) 143.5 pm, 4.5 g/ccQ.14. Lithium borohydride crystallizes in a orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are a = 6.8 A , b = 4.4 A and c = 7.2 A . If the molar mass is 21.76, then the density of crystals is :
(A) 0.6708 g cm–3 (B) 1.6708 g cm–3 (C) 2.6708 g cm–3 (D) None of these
Q.15. An alloy of coper, silver and gold is found to have copper constituting the ccp lattice. If silver atoms occupy the edge centres and gold is present at body centre, the alloy has a formula :
(A) Cu4Ag2Au (B) Cu4Ag4Au (C) Cu4Ag3Au (D) CuAgAu Q.16. The radius ratio, rA/rB for an atom A to fit into a simple cubic B lattice is :
(A) 0.414 (B) 0.225 (C) 0.732 (D) 0.500
Q.17. The anions A from hexagonal closest packing an atoms C occupy only 2/3 of octahedral voids in it, then the general formula of the compound is :
(A) CA (B) A2 (C) C2A3 (D) C3A2
Q.18. Number of atoms in the unit cell of KBr with 654 pm edge and 2.75 g/cc density, are :
(A) 1 (B) 2 (C) 4 (D) 6
Q.19. The lattic parameters of a crystal are a = 5.62 A, b = 7.41 A C = 9.48 A . The three coordinates are mutually perpendicular to each other. The crystal is :
(A) Tetragonal (B) Orthorhombic (C) Monoclinic (D) Trigonal Q.20. Non-stoichiometric form of NaCl is
(A) Pink (B) Yellow (C) Violet (D) Red
Q.21. The unit cell cube length of LiCl (Just like NaCl structure) is 5.14A. Assuming anion-anion contact, the ionic radius for chloride ion is :
(A) 1.815A (B) 2.8 A (C) 3.8A (D) None of these
Q.22. The ratio between packing fraction of bcc and fcc is : (A) 0 68 0 74 . . (B) 0 74 0 68 . . (C) 1:1 (D) 2 : 1
Q.23. Frenkel defect is noticed in :
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Q.24. A solid PQ have rock salt type structure in which Q atoms are at the corners of the unit cell. If the body centred atoms in all the unit cells are missing, the resulting stoichiometry will be :
(A) PQ (B) PQ2 (C) P3Q4 (D) P4Q3
Q.25. At room temperature, sodium crystallises in a body centred cubic cell with a 4.24A . The theoretical density of sodium is (Atomic mass of sodium = 23.0 g mol–1)
(A) 2.05 g cm–3 (B) 3.45 g cm–3 (C) 1.00 g cm–3 (D) 3.55 g cm–3
Q.26. Which of the following expressions is correct for a CsCl unit cell with lattice parameter a ? (A) rCs + r Cl= 2a (B) rCs + rCl = a 2 (C) rCs + rCl= 3 2 a (D) rCs + rCl = 3 2 a
Q.27. Diffraction studies of tungsten crystals using X-rays of = 1.85A showed first order reflection at 36° (sin 36° = 0.588).The distance between the successive layers in a tungsten crystal is :
(A) 3.14A (B) 1.57A (C) 5.17A (D) None of these Q.28. The limiting ratio r+ / r– for a tetrahedral system with co-ordination number 4 is :
(A) 0.155 (B) 0.225 (C) 0.414 (D) 0.732
Q.29. The crystal system in which a bc and the angles is :
(A) Triclinic (B) Monoclinic (C) Hexagonal (D) Cubic Q.30. A face centred cubic element with atomic mass 60 has a cell edge 300 pm. Its density is :
(A) 6.23 g/cc (B) 1.48 g/cc (C) 3.3 g/cc (D) 4.18 g/cc Q.31. In the unit cell of an fcc system the number of octahedral and tetrahedral holes are :
(A) 4, 4 (B) 4, 8 (C) 1, 8 (D) 4, 1
Q.32. A solid X melts silightly above 273 K and is a poor conductor of heat and electricity. To which of the following categories does it belong :
(A) Ionic solid (B) Covalent solid (C) Metallic (D) Molecular
Q.33. A binary solid A+B– has zinc blende structure with B– ions constituting the lattic and A+ ions occupying
50% tetrahedral holes formula of the solid is :
(A) A2B (B) AB (C) AB2 (D) AB4
Q.34. An internmetallic compound AB crystallises in cubic structure where A and B atoms have co-ordination number 8. The crystal unit cell belongs to :
(A) fcc (B) hcp (C) bcc (D) Antifluorite
Q.35. 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 a C atom at the body centre :
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Q.36. The most efficient packing of similar spheres is obtained in (A) The simple cubic system and the body centred cubic system (B) The simple cubic system and the hexagonal close packed system (C) The face centered cubic system and the hexagonal close packed system (D) The body centered cubic system and the face centred cubic system
Q.37. A mineral is composed of Ca, Ti and oxygen. With Ca+2 at cube corners, Ti ions in the centre and oxide
ions in the face centre of the unit cell. The formula of the mineral and the oxidation state of Ti are : (A) Ca Ti2O3; + 2 (B) Ca Ti O3; + 4 (C) Ca Ti3 O4 ; + 2 (D) Ca Ti O2 ; + 2 Q.38. Copper metal has a face-centred cubic structure with the unit cell length equal to 0.36 nm. Picturing
copper ions in contact along the face diagtonal. The apparent radius of a copper ion is :
(A) 0.128 (B) 1.42 (C) 3.22 (D) 4.22
Q.39. Choose the correct matching sequence from the possibilities given :
i. Cyrstal defect 1. AB AB AB .... type crystal
ii. hcp 2. Covalent crystal
iii. CsCl 3. Frenkel
iv. Diamond 4. Face centred in cube
v. NaCl 5. Body centered in cube
i. ii. iii. iv. v. i. ii. iii. iv. v.
(A) 3 1 2 5 4 (B) 3 1 5 2 4
(C) 3 5 1 2 4 (D) 5 3 4 2 1
Q.40. A compound alloy of gold and copper crystallizes in a cube lattice in which the gold atoms occupy the lattice points at the corners of a cube and the copper atoms occupy the centres of each of the cube faces. The formula of this compound is :
(A) AuCu (B) AuCu2 (C) AuCu3 (D) None of these Q.41. Select the correct statement (s)
i. The C.N. of cation occupying a tetrahedral hole is 4 ii. The C.N. of cation occupying a octahedral hole is 6. iii. In schottky defects, density of the lattice decreases
(A) i, ii (B) ii, iii (C) i, ii, iii (D) i, iii
Q.42. A fcc lattice has a lattice parameter a = 40 pm. The molar volume including all empty space is :
(A) 10.8 ml (B) 96 ml (C) 8.6 ml (D) 9.6 ml
Q.43. A spinel is an important class of oxides consisting of two types of metal ions with the oxide ions arranged in ccp layers. The normal spinel has 1
8th of the tetrachedral void occupied by one type of metal and one half of the octahedral voids occupied by another type of metal ions. Such a spinel is formed by Zn2+, Al3+
and O2– with Zn2+ in tetrahedral void.Give the simplest formula of the spinel.
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Q.44. In a face centred cubic arrangement of metallic atoms, what is the relative ratio of the sizes of tetrahedral and octahedral voids ?
(A) 0.543 (B) 0.732 (C) 0.414 (D) 0.637
Q.45. Which of the following figures represented the cross-section of an octahedral site ?
(A) (B)
(C) (D)
Q.46. A body centred cubic arrangement is shown : O is the body centre; A, B, C, D, ..., H are the corners. What is the magnitude of the angle AOB ?
H E D A O B C G F (A) 120° (B) 109°28 (C) 104°31 (D) 70°32
Q.47. Lithium borohydride (LiBH4), crystallises in an orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are : a = 6.81 A , b = 4.43A , c = 7.17 A . If the molar mass of LiBH4 is 21.76 g
mol–1. The density of the crystal is :
(A) 0.668 g cm–3 (B) 0.585 g cm2 (C) 1.23 g cm–3 (D) None of these
Q.48. In an f.c.c. unit cell, atoms are numbered as shown below. The atoms not touching each other are (Atom numbered 3 is face centre of front face).
1
2
3 4
(A) 3 & 4 (B) 1 & 3 (C) 1 & 2 (D) 2 & 4 Q.49. Compute the percentage void space per unit volume of unit cell in zinc fluoride structure
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Q.50. A rhombohedral unit cell is shown. What is its volume ? Side length = aA (A) a 3 3 (B) a 3 3 2 60° 60° 60° 120° 120° 120° 60° a a (C) a 3 2 (D) a 3 3 2
Q.51. A crystal formula AB3 has A ions at the cube corners and B ions at the edge centres. The coordination numbers of A and B are respectively
(A) 6 and 6 (B) 2 and 6 (C) 6 and 2 (D) 8 and 8
Q.52. A metal crystallizes in two cubic phases, face centred cubic (fcc) and body centred cubic (bcc) whose unit cell length are 3.5 and 3.0A respectively. Calculate the ratio of density of fcc and bcc.
(A) 2.123 (B) 1.259 (C) 5.124 (D) 3.134
Q.53. Unit cell of Fe3O4 (ferrous ferrite) has 32 O2– ions in the unit cell. Then the unit cell of Fe
3O4 has :
(A) 16 Fe2+ ions (B) 8 Fe3+ ions
(C) 16 Fe3+ ions and 8 Fe2+ ions (D) None of these
Q.54. For the structure of solid given below if the littice points represent A+ ions and the B– ions occupy the
tetrahedral voids then co-ordination number of A is :
(A) 2 (B) 4 (C) 6 (D) 8
Q.55. TiO2 is well known example of :
(A) Triclinic system (B) Tetragonal system (C) Monoclinic system (D) None of these Q.56. For a solid with the following structure, the coordination number of the point B is :
B A
(A) 3 (B) 4
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Q.57. For the structure given below the site marked as S is a :
(A) Tetrahedral void (B) Cubic void S
(C) Octahedral void (D) None of these
Q.58. Vanadium crystallises with a body centred cubic structure with a unit cell of side length 3.011A. Calculate the atomic radius and density (V = 50.94) :
(A) 2.609A , 16.199 g/cm3 (B) 1.303A , 16.99 g/cm3
(C) 2.609 A , 7.200 g/cm3 (D) 1.303A , 6.199 g/cm3
Q.59. In a face centred cubic arrangement of A and B atoms whose A atoms are at the corner of the unit cell and B atoms at the face centred. One of the A atom is missing from one corner in unit cell. The simplest formula of compound is :
(A) A7B3 (B) AB3 (C) A7B24 (D) A7/8B3
Q.60. A solid A+B– has the B– ions arranged as below. If the A+ ions occupy half of the tetrahedral sites in the
structure. The formula of solid is :
(A) AB (B) AB2 (C) A2B (D) A3B4
Q.61. Which of the following statements is not correct ?
(A) The co-ordination number of each type of ion in CaCl crystal is 8 (B) A metal that crystallises in bcc structure has a coordination no. of 12 (C) A unit cell of an ionic crystal shares some of its ions with other unit cells. (D) The length of the unit cell in NaCl is 552 pm (r
Na= 95 pm, rCl=181 pm)
Q.62. The density of KBr is 2.75 g 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 centred cubic system with coordination number = 6 (B) Solid has simple cubic system with coordination number = 4 (C) Solid has face centred cubic system with co-ordination number = 1 (D) None of these
Q.63. Silver (atomic wt. 107.88) crystallises with fcc lattice for which the side length of the unit cell is 4.0774A ,
density of Ag is 10.53 g cm–3. Calculate the Avogadro’s number :
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Q.65. CsBr crystallises in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and of Br = 80 amu Avogadro’s number being 6.02 × 1023 mol–1, the density of CsBr
is :
(A) 8.50 g/cm3 (B) 4.25 g/cm3 (C) 42.5 g/cm3 (D) 0.425 g/cm3
Q.66. The densities of NaCl crystal as measured by puknometer and X-ray method are 2.165 × 103 kg. m–3
and 2.178 × 103 kg. m–3 respectively. The fraction of unoccupied sites in soldium chloride crystal is :
(A) 5.96 × 10–3 (B) 5.96 (C) 5.96 × 10–2 (D) 5.96 × 10–1
Q.67. The compound formed by elements X and Y crystallizes in cubic structure in which X atoms are at the corners of the cube and Y atoms are at the face centre. The formula of the compound is :
(A) XY2 (B) XY (C) XY3 (D) X3Y
Q.68. The density of a pure substance ‘A’ whose atoms pack in cubic close pack arrangement is 1 gm/cc. If B atoms can in gm/cc. [Atomic mass (A) = 30 gm/mol and atomic mass (B) = 50 gm/mol]
(A) 3.33 (B) 4.33 (C) 2.33 (D) 5.33
Q.69. In which type of voids magnesium and silicon is present respectively in asbestos [Mg3Si2O5] if
r
Mg/rO = 0.59 and rSi+4/rO–2 is 0.29
(A) Tetrahedral & octahedral (B) Both tetrahedral (C) Octahedral & tetrahedral (D) Both Octahedral
Q.70. The composition of a sample of Wustite is Fe0.93O. What is the percentage of iron present as Fe3+ ?
(A) 15% (B) 25% (C) 35% (D) 45%
Q.71. In a compound AB the ionic radii of A+ and B– are 88 pm and 200 pm respectively the coordination
number of A+ is
(A) 6 (B) 8 (C) 4 (D) 12
Q.72. In a solid ‘AB’ having NaCl structure atoms, occupy the corners of the unit cell. If all the face centred atoms along one of the axis are removed, then the resultant stoichiometry of the solid is :
(A) AB2 (B) A2B (C) A4B3 (D) A3B4
Q.73. A compound formed by elements A and B crystallizes in the cubic structure where A atoms are at the corners of a cube and B atoms are at the face centres. The formula of the compound is :
(A) AB3 (B) AB (C) A2B (D) A2B2
Q.74. Potassium has a bcc structure with nearest neighbour distance 4.52A. Its atomic weight is 39. Its
density will be :
(A) 454 kg m–3 (B) 804 kg m–3 (C) 852 kg m–3 (D) 908 kg m–3
Q.75. A compound formed by elements A and B has a cubic structure in which A atoms are at the corners of the cube and B atoms are at the face centres. The formula of the compound will be :
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Q.76. CsBr has BCC structure. The length of its one side is 4.3A. The minimum distance between Cs+ and Br–
ion will be
(A) 0.897A (B) 3.72A (C) 1.749A (D) None of these
Q.77. The edge of face centred unit cubic cell is 508 pm. If the radius of the cation is 110 pm , the radius of the anion is :
(A) 144 pm (B) 288 pm (C) 628 pm (D) 398 pm
Q.78. The number of atoms in 100 g of a fcc crystal with density 10.0 g/cc and edge length as 100 pm is (A) 3 × 1025 (B) 4 × 1025 (C) 1 × 1025 (D) 2 × 1025
Q.79. The unit cell of aluminium is a cube with an edge length of 405 pm. The density of aluminium is 2.7 g/cc. The type of lattice Al has is :
(A) bcc (B) fcc (C) sc (D) End centred
Q.80. The face centred unit cell of nickel has an length 352.39 pm. The density of nickel is 8.9 g/cc. Avogadro number is :
(A) 6.029 × 1023 (B) 6.023 × 1023 (C) 5.9 × 1023 (D) 6.4 × 1023
Q.81. An ionic compound AB has ZnS type of structure, if the radius A+ is 22.5 pm, then the ideal radius of B–
is :
(A) 54.35 pm (B) 100 pm (C) 145.16 pm (D) None of these
Q.82. Addition of CdCl2 to AgCl yields solid solution where the divalent cations Cd2+ occupy the Ag+ sites.
Which one of the following statements is true ?
(A) The no. of cationic vacancies is equal to that of divalent ions added.
(B) The no. of cationic vacancies is one half of the no. of that of divalent ions added. (C) The no. of anionic vacancies is equal in no. to that of divalent ions added. (D) No cationic or anionic vacancies are produced.
More than one option is correct Q.83. In the fluorite structure if the radius ratio is
2 3
- 1 , how many ions does each cation touch ? (A) 4 anions (B) 12 cations (C) 8 anions (D) No cations
Q.84. TiO2 (rutile) shows 6 : 3 coordination. Which of the following solids have a rutile-like structure ?
(A) MnO2 (B) ZnS (C) KCl (D) SnO2
Q.85. The intermetallic compound LiAg has a cubic crystalline structure in which each Li atom has 8 nearest neighbour silver atoms and vice-versa. What is the type of unit cell ?
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Q.86. Which of the following compounds represent a normal 2 : 3 spinel structure ? (A) MgIIAl 2 IIIO 4 (B) Co II(CoIII) 2O4 (C) Zn(TiZn)O4 (D) Ni(CO)4
Q.87. Which of the following statements are correct ?
(A) The crystal structure of rock salt is an fcc array of anions in which the cations occupy all the octahedral holes (or vice versa)
(B) The sphalerite crystal structure is an expanded fcc anion lattic with cations occupying one type of tetrahedral hole.
(C) In the fluorite crystal structure, cations occupy half the cubic holes of a primitive cubic array of anions
(D) None of these
Q.88. Which of the following crystals have 6 : 6 coordination ?
(A) NH4I (B) MgO (C) MnO (D) ZnS
Q.89. Which of the following statement is correct ?
(A) A rutile (TiO2) structure consists of an hcp anion lattice with cations occupying half the octahedral holes
(B) The wurtzite structure is derived from an expanded hcp anion array with cation occupying one type of octahedral holes
(C) In the fluorite structure (CaF2), anions occupy both types of tetrahedral holes in an expanded fcc lattice of cations
(D) None of these
Q.90. The hcp and ccp structure for a given element would be expected to have (A) The same coordination number (B) The same density (C) The same packing fraction (D) All of these
Q.91. Which of the following statements are correct ?
(A) The coordination number of each type of ion in a CsCl crystal is eight (B) A metal that crystallizes in a bcc structure has coordination number of twelve (C) A unit cell of an ionic crystal shares some of its ions with other unit cells (D) The length of the unit cell is NaCl is 552 pm (given thatr
Na= 95 pm and rCl = 181 pm)
Q.92. Pick up the correct statements :
(A) The ionic correct of AgBr has Schottky defect
(B) The unit cell having crystal parameters, a = b C, = = 90°, = 120° hexagonal (C) In ionic comopunds having Frenkel defect, the ratio r+/r– is high.
(D) The coordination number of Na+ ion in NaCl is 4
Q.93. Select the correct statements (s)
(A) In NaCl crystal, each ion has an octahedral coordination. (B) The C.N. of an atom in hcp structure is 12.
(C) Diamagnetic materials are attracted by the magnetic fields (D) An octahedral hole is smaller than a tetrahedral hole.
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Q.94. Which of the following are not the characteristics of crystalline solids ?
(A) They are exhibit polymorphism (B) They are isotropic
(C) They do not have thermodynamic defects (D) After melting, they become crystalline Q.95. Which of the following statements are false :
(A) The radius of a metal atom is taken as half the nearest metal metal distance in a metallic crystal (B) One tetrahedral void per atom is present in hcp structure.
(C) In the fluorite structure (CaF2), the Ca2+ ions are located at the lattice points and the fluoride ions fill
all the tetrahedral holes in the ccp crystal
(D) In the antifluorite structure (Li2O, Rb2S) the cations are located at the lattice points and anions fill the tetrahedral holes in the ccp structure.
Q.96. If the radius of Na+ ion is 95 pm and that of Cl– ion is 181 pm then :
(A) Coordination no. of Na+ is 6 (B) Coordination no. of Na+ is 8
(C) Length of the unit cell is 552 pm (D) Length of the unit cell is 380 pm Q.97. Which of the following statement (s) is (are) correct ?
(A) When the radius ratio is in the range 0.414 – 0.732, a bcc arrangement with coordination no. 8 (B) When the radius ratio is in the range 0.225 – 0.414, a tetrahedral arrangement with coordination no.
4
(C) When the radius ratio is in the range 0.155 – 0.225, an octahedral arrangement with coordination no. 6
(D) In B2O3, smaller cations occupy triangular voids and a planar trigonal arrangement with coordination no. 3
Q.98. In the crystal structure of CsCl :
(A) Cl– ions are present at the corners of a cube (B) Cs+ ions are present in the cubic voids
(C) Cl– ions are present in the cubic voids (D) The close packed structure is formed by
Cs+ ions.
Q.99. Which of the following statement (s) is (are) correct for CaF2: (A) Ca2+ ions are present only at the corners of a cube
(B) ccp type structure
(C) F– ions are present in all the octahedral voids
(D) The structure has 8 : 4 coordination Q.100. Select the correct statements :
(A) For CsCl unit cell (edge-length = a), rc + ra = 3
2 a
(B) For NaCl unit cell (edge - length = ), rc + ra = 2 (C) The void space in a bcc unit cell is 0.68
