PHYSICAL CHEMISTRY

CLASS-XII

SOLID STATE

CONTENTS

" A SPECIALLY DESIGNED KIT FOR LEARNING."

THE KEY  Basic principles of subjects. An outline of the topics to be
discussed in class lectures.
THE ATLAS  Basic layout of subject. A route map correlating different
subtopics in coherent manner.

SHORT NOTES  Important Formula (Summary)

EXERCISE I  Introductory problems based on JEE to get first hand
experience of problem solving.
EXERCISE II  A collection of good problems.
QUESTION BANK  Test your objective skill.
EXERCISE III  A collection of previous 15 years JEE problems.
 THE KEY
Crystalline solids :
Crystalline solids are those whose atom, molecules or ions have an ordered arrangement extending over a
Long Range. example-(Rock salt, NaCl).

Amorphous solids :
Amorphous solids are those whose constitutent particles are randomly arrange and have no ordered long
range structure. example: Rubber, Glass ect.

TYPES OF CRYSTALLINE SOLIDS :

Type of Solid Intermolecular forces Properties Examples

Ionic Ion-Ion forces Brittle, hard high Melting NaCl, KCl, MgCl2
Dispersion forces/Dipole-
Molecular Soft, low melting non-conducting H2O, Br2, CO2, CH4
Dipole /H-bond
Covalent network Covalent bonds Hard: High melting C-Diamond SiO2
Variable hardness and melting
Metallic Metallic bonds Na, Zn, Cu, Fe
point conducting

TYPES OF UNIT CELL :

Collection of lattice points, whose repetition produce whole lattice is called a unit cell. The whole lattice can
be considered to be made by repetition of unit cell.

1. Unit Cell Space lattice

Unit Cell Parameters

Crystal Systems Bravais Lattice
Intercepts Crystal Angles
Primitive, Face Centered,
1 Cubic a=b=c  =  =  = 90°
Body Centered
Primitive, Face Centered,
2 Orthorhombic abc  =  =  = 90°
Body Centered, End Centered
3 Rhombohedral Primitive a=b=c  =  =   90°
4 Monoclinic Primitive, End Centered abc  =  = 90°,   90°
5 Triclinic Primitive abc     90°
6 Tetragonal Primitive, Body Centered a=bc  =  =  = 90°
7 Hexagonal Primitive a=bc  =  = 90°,  = 120°

== c=a

b=a
a
Rhombohedral

[2]
1.1 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.

Z = 4 ; C.N. = 12

2. Density of cubic crystals:

TYPE OF PACKING:
3. Closest packing of atoms: This is the most efficient way of packing 74% of available space is occupied by
spheres and coordination number is 12.
(i) Hexagonal close pack (A-B-A-B) type packing : Each layer has hexagonal arrangement of touching
sphere and 3rd layer is similar (exactly on top) of first layer.
(ii) Cubic close pack (A-B-C-A-B-C): AB layers are similar to HCP arrangement but third layer is offset from
both A and B layers. The fourth layer is exactly on top of first layer.

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 Hexagonal primitive unit cell

4. Types of voids
4.1 Tetrahedral void

4.2 Octahedral void

5. Radius ratio

5.1 Radius ratio for co-ordination number 3

2 r 2  3
(Triangular Arrangement): r+ + r – = 3 r– ; = = 0.155
3 r 3

5.2 Radius ratio for coordination number 4

3a 3 r–
(Tetrahedral arrangement): r+ + r– = ; 4r– = 2a=
4 2
[4]
 r 3 2
= = 0.225
r 2

5.3 Radius ratio for coordination number 6: r+ + r – = 2 r–

r
(Octahedral Arrangement) or = 2 –1 = 0.414
r
Radius ratio for coordination number 4
(Square plannar arrangement)

3
5.4 Radius ratio for coordination number 8 : r+ + r – = a
2
(Body centered cubic crystal) r+ + r – = 3 r–

r
= 3 –1 = 0.732
r

6. Types of ionic structures

6.1 Rock salt structure : (NaCl) Larger atom forming ccp

arrangement and smaller atom filling all octahedral voids.

6.2 Zinc blend (sphalerite) structure : (ZnS) Larger atom

forming ccp arrangement and smaller atom filling
half of alternate tetrahedral voids

6.3 Fluorite structure : (CaF2) Ca2+ forming ccp

arrangement and F– filling all tetrahedral voids.
6.4 Antifluorite structure :(Li2O) O2– ion forming ccp

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 and Li+ taking all tetrahedral voids.

6.5 Cesium halide structure: (CsCl) Cl– at the corners

of cube and Cs+ at the center of cube.

6.6 Corundum Structure: (Al2O3) O2– forming hcp and Al3+ filling 2/3 octahedral voids.

6.7 Rutile structure: (TiO2) O2– forming hcp while Ti4+ ions occupy half of the octahedral voids.

6.8 Pervoskite structure:(CaTiO3) Ca2+ in the corner of

cube O2– at the face center and Ti4+ at the centre of cube.

6.9 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.

7. Crystal defects:
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
pair, the defect is called schottky defect.

 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

[6]
 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.

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, NO - 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.

Metal deficency defect

8. PROPERTIES OF SOLIDS (CRYSTALS)
The imperfections (defects) developed in crystals gives some characteristic changes in the properties of
ionic compound. These properties of solids are directly related to their composition, their lattice structure
and the nature of bonds. We shall now study some of the properties of solids e.g.
1. Electrical properties
2. Magnetic properties
3. Dielectric properties

[7]
1. Electrical properties
Electrical conductivity of solids may arise through the motion of electrons or positive holes or through the
motion of ions. Conduction through ions or positive holes is due to electronic imperfection. The conduction
through electrons is called n-type conduction and through positive holes is called p-type conduction. Pure
ionic solids where conduction take place only through motion of ions are Ionic solids can conduct electric
current in the molten or solution state.
Types of conductors : (i) good conductors, (ii) insulators, (iii) semi-conductors.
(i) Good conductors. These allow the maximum portion of the applied electric field to flow through them.
The electrical conductivity of good conductors is of the order of 108 ohm–1 cm–1. ex. Metals.
(ii) Insulators. They do not practically allow the electric current to flow through them, the electrical conductivity
is of the order of 10–22 ohm–1 cm–1. ex. : Organic solid and inorganic solids
(iii) Semi-conductors. At room temperature, semi-conductors allow a portion of electric current to flow
through them. Actually, electrical conductivity of a semi-conductor at normal temperature lies between that
of a good conductor and an insulator; it is in the range of 10–9 to 102 ohm–1 cm–1.
Two types : (a) intrinsic (b) extrinsic.
(a) Intrinsic semi-conductors. (Semi-conductors due to thermal defects). At 0 K, pure silicon and germanium
act as insulators because electrons fixed in covalcnt bonds are not available for conduction. However, at
higher temperatures some of the covalcnt bonds are broken and the electrons so released become free to
move in the crystal and thus conduct electric current. This type of conduction is known as intrinsic conduction,
as it can be introduced in the crystal without adding an external substance.
(b) Extrinsic semi-conductors. (Semi-conductors due to impurity defects). Silicon and germanium (group
14 elements), in pure state, have very low electrical conductivity. However, the electrical conductivity of these
elements is greatly enhanced by the addition of traces of an element belonging to group 13 (III) or group 15
(V), to the crystals of group 14 (IV) elements, i.e. silicon or germanium. Introduction of group 15 and group 13
elements to the crystal lattice of group 14 elements (Si or Ge) produces n-type semi-conductors and p-type
semi-conductors respectively.
(i) n-Type semi-conductors. This type of semi-conductor is produced (X) due to metal excess defect and
(b) by adding trace amounts of group 15 element (P, As) to extremely pure germanium or silicon by a process
called doping.
(ii) p-Type semi-conductors. This type of semi-conductors are produced (a) due to metal deficiency defects,
and (b) by adding impurity atoms containing less electrons (i.e. atoms of group 13) than the parent insulator
to the insulator lattice.
Applications of semi conductors. A large variety of semi-conductors have been prepared/ by the following
types of combinations.
(i) Elements of group 14 (Si, Ge) and group 15 (P, As, Sb).
(ii) Elements of group 13 (B, Ga) and group 14 (Si, Ge)
(iii) Elcments of group 13 and group 15, e.g. InSb, AIP
(iv) Elements of group 12 and group 16, e.g. ZnS, CdS, CdSe, HgTc
Properties of a semi-conductor are considerably changed depending upon the nature of the impurity.
Semiconductors are used in transistors and in exposure meters as photoelectric devices. Combination of
p- and n- type of semi-conductors (known as p-n junction) allows electric current from outside to flow through
it in one direction. This type of junction is known as a rectifier and is used for converting alternating current
to direct current.
Super-conductivity. The electrical resistance of metals depends upon temperature. Electrical resistance
decreases with decrease in temperature and becomes almost zero near the absolute temperature. Materials
in this state are said to possess superconductivity. Thus super-conductivity may be defined as a phenomenon
in which metals, alloys and chemical compunds become perfect conductors with zero resistivity at temperatures
approaching absolute zero. Superconductors are diamagnetic. The phenomenon was first discovered by
Kammerlingh Onnes in 1913 when he found that mercury becomes superconducting at 4 K. The temperature
at which a substance starts behaving as super conductor is called transition temperature which lies between

[8]
 2 and 5 K in most of the metals exhibiting this phenomenon.
Efforts are going on to find materials that behave as a superconductor at room temperature because attaining
low temperature with liquid helium is highly expensive. The highest temperature at which superconductivity
has been observed is 23 K for alloys of niobium (Nb3Ge). Since 1987, many complex metal oxides have been
found to possess superconductivity at fairly high temperatures, e.g.
Y Ba2Cu3O7 90 K
Bi2Ca2Sr2Cu3O10 105 K
Tl2Ca2Ba2Cu3O10 125 K
2. Magnetic properties.
Solid substances are classified into following different types depending upon their behaviour towards magnetic
field.
(i) Paramagnetic substances
(ii) Diamagnetic substances
(iii) Ferromagnetic substances
(i) Paramagnetic substances : The substances which are weakly attracted when placed in the external
magnetic field are termed as paramagnetic substances. The paramagnetic property of the solid is due to the
presence of ions, atoms or molecules containing unpaired electrons. The unpaired electrons have magnetic
moment (electron spin) but their randomly orientation cancel each other’s effect. However, in the presence of
external magnetic field these unpaired electrons are aligned and shows the temporary magnetism.
Molecules NO and O2
Transition metals Cr, Mn, Ni, Co, Fe, etc
Metal ions Cu2+, Ni2+, Fe3+, etc.
Metal oxide VO2 and CuO
(ii) Diamagnetic substances : The diamagnetic substances arc those which arc weakly repelled by the
external magnetic field. It is because of the fact that all electrons arc paired in ions, atoms or molecules
showing diamagnetism. Being paired, the magnetic moment of one electron is compensated by the equal
and opposite magnetic moment of the other electron. ex. N2, NaCl, TiO2, Zn, Cd, Cu+ etc.
(iii) Ferromagnetic substances : The substances possessing unpaired electrons arc further classified in
three different groups based on the alignment of magnetic moments of unpaired electrons.
(a) Ferromagnetic substances
(b) Antiferromagnetic substances
(c) Ferrimagnetic substances
(a) Ferromagnetic substances : The substances which are strongly attracted by magnetic field are termed
as ferromagnetic substances. This type of substances have alignment of all the unpaired electrons in the
same direction (orientation). These substances arc permanently magnetised i.e. Examples are Ni, Fe, Co
and CrO2.
(b) Antiferromagnetic substances : When equal number of unpaired electrons are aligned in opposite
directions, their magnetic moment (electron spin) will compensate each other’s magnetic moment. Such
substances are termed as antiferromagnetic substances. For example, MnO, Mn2O3 and MnO3.
(c) Ferrimagnetic substances : When unequal number of unpaired electrons are aligned in opposite directions,
the net magnetic moment is not zero. Such substances arc termed as ferrimagnetic substances e.g. ferrite
Fe2O3.
It has been observed that the ferro-magnetic, ferrimagnetic and antiferromagnetic substances show
paramagnetic nature at higher temperatures. This arises due to randomisation of spins at higher temperature
e.g. Fe3O4 is ferrimagnetic at room temperature but becomes paramagnetic at 850 K.

[9]
3. Dielectric properties
A substance through which there is no net flow of electric charge when placed under an applied electrical
field is defined as dielectric material. This is due to the fact that the electrons in the dielectric material are
tightly held by individual atoms or ions. However, the applied field polarises the dielectric material and thus
results in the creation of dipoles. The electrical diploes are developed in the polar bonds with two equal and
opposite charges. Such electrical dipoles interact with the applied external field in the following manners.
(i) These dipoles may align themselves in such a manner that the dipole moments cancel each other completely
giving no net dipole moment in the crystal.
(ii) The dipoles are so orderly oriented that these result in a net dipole moment in the crystal.
The crystals of second type, having some net dipole moment show the undermentioned electrical properties.
(a) Piezoelectricity : When the crystals, in which the dipoles are oriented in an order ed manner, are
subjected to mechanical stress or pressure produces piezoelectricity. Such crystals are termed as
piezoelectric crystals. In piezoelectric cystals electricity is produced due to the displacement of ions from
their orderly arrangement by the application of mechanical stress.
Such crystals are used as pick ups in record players where they produce electrical signals by application of
pressure.
(b) Pyroelectricity : Upon heating, if the orderly arrangement of ions or atoms in the crystal get displaced
and produces electricity then the property of the crystal is termed as pyroelectricity.
(c) Ferroelectricity : Certain piezoelectric crytals show permanent alignment of dipoles even in the absence
of electric field. When the electric field is applied on such crystals, then the direction of polarisation of ions
is altered. Such solids are called ferroelectric substances and the phenomenon is termed as ferroelectricity.
Sodium potassium tartarate (Rochelle salt), potassium dihydrogen phosphate (KH2PO4) and barium titanate
(BaTiO3) are examples of ferroelectric solids.
(d) Antiferroelectricity : If the crystal is having the alternate electrical dipoles pointing in opposite directions
then it shall not exhibit ferroelectric properties. The net dipole moment in such solids is zero and are termed
as antiferroelectric crystals. Lead zirconate (PbZrO3) is a typical antiferroelectric solid.

[10]
THE ATLAS

[11]
 SUMMARY
1. Some Formula :
ZM Z  Vatom
(i) d  ; (ii) Packing Fraction = ; (iii) Void Fraction = 1 – Packing Fraction ;
N0a 3 Vunit cell
(iv) C.No. = No. of First Neighbours where d = density ; M = Atomic Weight of element ;
N0 = Avogadro's number ; a = Edge length of cube.
2. Cubic System :

S.No. Pr operty Simple Cubic Lattice BCC Lattice FCC Lattice

Atomic radius (r ) a a 3 a 2
1. r r r
a  edge length of cube 2 4 4
2. No of atoms per unit cell (Z ) z 1 z2 z4
3. Co  ordination No. (C.N.) 6 8 12
4. Packing fraction (P.F.) 0.52 0.68 0.74

3. Packing :

S.No. Pr operty Hexagonal close packing (HCP) Cubic close packing (CCP )
1. Co  ordination No. (C.N.) 12 12
2. No. of atoms per unit cell (z) z6 z4
3. Packing fraction (P.F.) P.F.  0.74 P.F.  0.74
4. Type of packing ABAB........ ABCABC.......

4. VOID :

S.No. Name of Void rvoid / rsphere Co  ordination Number (C.N.)

1. Triangular void 0.155 3
2. Tetrahedra l void 0.225 4
3. Octahedral void 0.414 6
4. Cubic void 0.732 8

5. TYPES OF IONIC STRUCTURE :

S.No. Name of Structure Location of Particle

1. Rock salt ( AB) NaCl B – : ccp lattice ; A  : Octahedral void
2. Zinc blende ( ZnS) S – 2 : ccp lattice ; Zn  2 : half of alternate tetrahedra l void
3. CsCl Cs  : Cube centre ; Cl – : Corner of cube
4. Fluorite structure (CaF2 ) Ca  2 : FCC lattice ; F – : Tetrahedral void
5. Antifluorite structure Na 2 O O – 2 : FCC lattice ; Na  : Tetrahedra l void

6. DEFECTS : (i) Point defects (ii) Line defects

Point Defects

Stoichiometric Non-stoichiometric

Schottky Frenkel Metal Metal

(ion-pair missing) (Dislocation of ions) excess defficent

Anionic Vacency Extra caton in Cationic vacency Extra anion in

(electron present in anionic vacency) interstitial site defect interstitial site (Not found)

[12]
 EXERCISE-I

Formula of ionic solid from unit cell description

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.

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.

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.

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?

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 formula of spinel.

Edge length, density and number of atoms per unit cell

6. 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.

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

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 4.0 Å.

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.

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?

11. Gold crystallizes in a face centered cubic lattice. If the length of the edge of the unit cell is 400 pm, calculate
the density of gold as well as its atomic radius assuming it to be spherical. Atomic mass of gold = 197 amu.

12. 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.

13. The effective radius of the iron atom is 1.414 Å. It has FCC structure. Calculate its density (Fe = 56 amu)

14. Xenon crystallises in the face-centred cubic lattice and the edge of the unit cell is 565.6 pm. What is the
nearest neighbour distance and what is the radius of xenon atom?

[13]
15. 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+.

16. 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.

17. CsCl has the bcc arrangement and its unit cell edge length is 400 pm. Calculate the interionic distance in CsCl.

18. 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)

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 volume
of unit cell.

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?

21. 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?

22. 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.

23. 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.

24. If the radius of Mg2+ ion, Cs+ ion, O2– ion, S2– ion and Cl– ion are 0.65Å, 1.69Å, 1.40Å, 1.84Å and 1.81Å
respectively. Calculate the co-ordination numbers of the cations in the crystals of MgS, MgO and CsCl.

25. Assume End centered cubic (ECC) unit cell exists in nature then calculate total effective number of atoms in
FCC unit cell & ECC unit cell

[14]
 EXERCISE-II

1. The figures given below show the location of atoms in three crystallographic planes in FCC lattice. Draw the
unit cell for the corresponding structure and identify these planes in your diagram.

2. KCl crystallizes in the same type of lattice as does NaCl. Given that rNa  0.5 rCl– and rNa  0.7 rK 
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.

3. 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.

4. Prove that the void space percentage in zinc blende structure is 25%.

5. The composition of a sample of wurzite is Fe0.93O1.0. What percentage of iron is present in the form of Fe(III)?

6. 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.

7. 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.

8. If NaCl is dopped with 10–3 mol % of SrCl2, what is the numbers of cation vacancies?

9. 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.

10. NaH crystallizes in the same structure as that of NaCl. The edge length of the cubic unit cell of NaH is 4.88Å.
(a) Calculate the ionic radius of H–, provided the ionic radius of Na+ is 0.95 Å.
(b) Calculate the density of NaH.

11. 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.

12. 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). How many molecules are contained in a given
unit cell? [density of ice = 0.92 gm/cm3]

[15]
13. 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
14. 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?

15. Prove that void space in fluorite structure per unit volume of unit cell is 0.243.

16. 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 are 60 and 90 respectively.

17. 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 1/4th of octahedral voids and divalent ions occupying 1/4th 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.

18. 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.

19. Diamond structure can be considered as ZnS (Zinc blende) structure in which each Zn2+ in alternate tetrahedral
void and S2– in cubic close pack arrangement is replaced by one carbon atom. If C–C covalent bond length
in diamond is 1.5Å, what is the edge length of diamond unit cell (z = 8).

20. 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. Atomic wt. Ge = 72.6, B = 11

[16]
 QUESTION BANK
1. A solid has a structure in which W atoms are located at the corners of a cubic lattice, O atom at the centre
of the edges and Na atom at centre of the cubic. The formula for the compound is
(A) NaWO2 (B) NaWO3 (C) Na2WO3 (D) NaWO4
2. The density of CaF2 (fluorite structure) is 3.18 g/cm3. The length of the side of the unit cell is
(A) 253 pm (B) 344 pm (C) 546 pm (D) 273 pm
3. An element has FCC structure with edge length 200 pm. Calculate density if 200 g of this element contains
24 × 1023 atoms.
(A) 4.16 g cm–3 (B) 41.6 g cm–3 (C) 4.16 kg m–3 (D) 41.6 kg m–3
4. The coordination number of cation and anion in Fluorite CaF2 and CsCl are respectively
(A) 8:4 and 6:3 (B) 6:3 and 4:4 (C) 8:4 and 8:8 (D) 4:2 and 2:4
5. The interstitial hole is called tetrahedral because
(A) It is formed by four spheres.
(B) Partly same and partly different.
(C) It is formed by four spheres the centres of which form a regular tetrahedron.
(D) None of the above
6. The tetrahedral voids formed by ccp arrangement of Cl– ions in rock salt structure are
(A) Occupied by Na+ ions (B) Occupied by Cl– ions
+ –
(C) Occupied by either Na or Cl ions (D) Vacant
7. The number of nearest neighbours around each particle in a face-centred cubic lattice is
(A) 4 (B) 6 (C) 8 (D) 12
8. In FCC unit cell, what fraction of edge is not covered by atoms?
(A) 0.134 (B) 0.293 (C) 0.26 (D) 0.32
9. If the anions (A) form hexagonal closest packing and cations (C) occupy only 2/3 octahedral voids in it, then
the general formula of the compound is
(A) CA (B) CA2 (C) C2A3 (D) C3A2
10. A solid is formed and it has three types of atoms X, Y, Z. X forms a FCC lattice with Y atoms occupying all
the tetrahedral voids and Z atoms occupying half the octrahedral voids. The formula of the solid is:
(A) X2Y4Z (B) XY2Z4 (C) X4Y2Z (D) X4YZ2
11. The intermetallic compound LiAg crystallizes in cubic lattice in which both lithium and silver have coordination
number of eight. The crystal class is
(A) Simple cubic (B) Body centred cubic
(C) Face centred cubic (D) None
12. A compound XY crystallizes in BCC lattice with unit cell edge lenght of 480 pm. If the radius of Y– is 225 pm,
then the radius of X+ is
(A) 127.5 pm (B) 190.68 pm (C) 225 pm (D) 255 pm
13. In a solid “AB” having NaCl structure “A” atoms occupy the corners of the cubic unit cell. If all the
face-centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid is
(A) AB2 (B) A2B (C) A4B3 (D) A3B4
14. In the closest packing of atoms A (radius : ra), the radius of atom B that can be fitted into tetrahedral voids is
(A) 0.155 ra (B) 0.225 ra (C) 0.414 ra (D) 0.732 ra
15. Which one of the following schemes of ordering closed packed sheets of equal sized spheres do not generate
closeset packed lattice.
(A) ABCABC (B) ABACABAC (C) ABBAABBA (D) ABCBCABCBC
16. In the closest packing of atoms,
(A) the size of tetrahedral void is greater than that of octahedral void
(B) the size of tetrahedral void is smaller than that of octahedral void
(C) the size of tetrahedral void is equal to that of octahedral void
(D) the size of tetrahedral void may be greater or smaller or equal to that of octahedral void depending upon
the size of atoms.
[17]
17. The coordination number of hexagonal closest packed (hcp) structure is
(A) 12 (B) 10 (C) 8 (D) 6
18. The number of nearest neighbours around each particle in a face-centred cubic lattice is
(A) 4 (B) 6 (C) 8 (D) 12
19. An ionic compound AB has ZnS type structure. If the radius A+ is 22.5 pm, then the ideal radius of B– would
be
(A) 54.35 pm (B) 100 pm (C) 145.16 pm (D) none of these
20. NH4Cl crystallizes in a body-centered cubic type lattice with a unit cell edge length of 387 pm. The distance
between the oppositively charged ions in the lattice is
(A) 335.1 pm (B) 83.77 pm (C) 274.46 pm (D) 137.23 pm
21. Which of the following is incorrect about NaCl structure?
(A) Nearest neighbours of Na+ ion is six Cl¯ ion
(B) Na+ ion make fcc lattice
(C) Cl¯ ion pack in cubic close packed arrangement
(D) There are eight next nearest neighbours of Na+ ion
22. Which of the following will show schottky defect
(A) CaF2 (B) ZnS (C) AgCl (D) CsCl
23. Give the correct order of initials T (true) or F (false) for following statements.
I. In an anti-fluorite structure anions form FCC lattice and cations occupy all tetrahedral voids.
II. If the radius of cations and anions are 0.2Å and 0.95Å then coordinate number of cation in the crystal is 4.
III. An atom/ion is transferred from a lattice site to an interstitial position in Frenkel defect.
IV. Density of crystal always increases due to substitutinal impurity defect.
(A) TFFF (B) FTTF (C) TFFT (D) TFTF
24. If the positions of Na+ and Cl– are interchanged in NaCl, the crystal lattice with respect to Na+ and Cl– is :
(A) unchanged (B) changes to 8:8 coordination from 6:6
(C) additivity of ionic radii for “a” is lost (D) none
25. The effective number of Na+ ions fitting all octahedral voids in NaCl structures are
(A) 4 (B) 6 (C) 8 (D) 13
26. How many nearest neighbours Cs+ are present in CsCl structure
(A) 6 (B) 8 (C) 12 (D) 4
27. For an ionic crystal of the general formula AX and the coordination number 6, the values of radius ratio will be
(A) Greater than 0.73 (B) In between 0.73 and 0.41
(C) In between 0.41 and 0.22 (D) Less than 0.22
28. A mineral having the formula AB2 crystallises in the cubic close-packed lattice, with the A atoms occupying
the lattice points. The co-ordination number of the A atoms, that of B atoms and the fraction of the tetrahedral
sites occupied by B atoms are
(A) 8, 4, 100% (B) 2, 6, 75% (C) 3, 1, 25% (D) 6, 6, 50%
Question Number 29 to 33 (5 questions)
In haxagonal close packing second row of spheres are placed in depressions of first row whereas third row is
vertically aligned with first, 2nd with fourth forming a pattern AB.AB...... The number of nearest atoms is called
the coordination number. It has six fold axis of symmetry.
29. What is the coordination number of a central sphere in hcp
(A) 8 (B) 12 (C) 6 (D) 4
r
30. If lies between 0.225 to 0.414, cation occupies
r–
(A) tetrahedral void (B) octahedral void (C) trigonal void (D) cubic void
31. The coordination number of octahedral void is
(A) 8 (B) 4 (C) 6 (D) 3
32. Which of the following packing are most efficient
(A) hcp (B) fccp (C) A & B both (D) bccp
33. ABC.ABC....... is called
(A) hcp (B) fccp (C) bccp (D) simple cubic

[18]
One or More than One
34. Select write statement(s)
(A) 8 Cs+ ions occupy the second nearest neighbour locations of a Cs+ ion
(B) Each sphere is surrounded by six voids in two dimensional hexagonal close packed layer
(C) If the radius of cations and anions are 0.3 Å and 0.4 Å then coordination number of cation in the
crystal is 6.
(D) In AgCl, the silver ion is displaced from its lattice position to an interstitial position such a defect is
called a frenkel defect

35. Which of the following statements are correct :

(A) The coordination number of each type of ion in CsCl is 8.
(B) A metal that crystallises in BCC structure has a coordination number 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. [ rNa = 95 pm ; rCl  = 181 pm ]

36. Which of the following system have two Bravais Lattice

(A) Cubic (B) Orthorhombic (C) Tetragonal (D) Monoclinic
37. Which of the following system have a = b = c and  =  = 
(A) Cubic (B) Orthorhombic (C) Rhobohedral (D) Monoclinic
38. Which of the following system have more than two Bravais Lattice
(A) Cubic (B) Orthorhombic (C) Tetragonal (D) Monoclinic
39. In a AB unit crystal of NaCl types assuming Na+ forming FCC
(A) The nearest neighbour of A+ is 6 B– ion (B) The nearest neighbour of B– is 6 A+ ion
3
(C) The second neighbour of A+ is 12 A+ (D) The packing fraction of AB crystal is
8
Match the Column
40. Column-I Column-II
(Distance in terms of edge length of cube (a))
(A) 0.866 a (P) Shortest distance between cation & anion
in CsCl structure
(B) 0.707 a (Q) Shortest distance between two cation in
CaF2 structure
(C) 0.433 a (R) Shortest distance between carbon atoms in
diamond
(S) Shortest distance between two cations in
rock salt structure
Integer Type :
41. In zinc blende structure, if Zn+2 is introduced in all vacent tetrahedral void. Calculate the co-ordination number
of S–2 in new structure.
42. Calculate distance (in Å) between A+2 and B– having Fluorite structure. Given edge length of unit cell is
4.62Å.
43. Comp. AB2 have antifluorite structure. Calculate sum of minimum distance and maximum distance between
cation (in Å). Given edge length of unit cell = 2.93 Å.
44. A molecule X2Y (molecular weight = 166.4) occupies orthorhombic lattice with a = 5Å, b = 8Å and c = 4Å. If
density of X2Y is 5.2 gm cm–3. Calculate number of molecules present in one unit cell (NA = 6 × 1023).
45. Iron occurs as bcc as well as fcc unit cell. If effective radius of an atom of iron is 124 pm. Calculate the value
of .
6 3 dfcc
Where    .
2 dbcc
dfcc = density of fcc unit cell. dbcc = density of bcc unit cell.

[19]
 EXERCISE-III
1. The packing efficiency of the two-dimensional square unit cell shown below is - [JEE 2010]

L
(A) 39.27% (B) 68.02% (C) 74.05% (D) 78.54%

2. The number of hexagonal faces that are present in a truncated octahedron is .......... . [JEE 2011]

3. A compound MP Xq has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown below.
The empirical formula of the compound is [JEE 2012]

M
X

(A) MX (B) MX2 (C) M2X (D) M5X14

4. The arrangement of X– ions around A+ ion in solid AX is given in the figure (to drawn to scale.) If the radius of
X– is 250 pm, the radius of A+ is [JEE Advance-2013]

–
X

+
A

(A) 104 pm (B) 125 pm (C) 183 pm (D) 57 pm

5. Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+ and M3+ in its
oxide. Fraction of the metal which exists as M3+ would be [JEE Mains-2013]
(A) 5.08% (B) 7.01% (C) 4.08% (D) 6.05%

6. CsCl crystallises in body centred cubic lattice. If 'a' is its edge length then which of the following expression
is correct? [JEE Mains-2014]
3a 3
(A) rCs   rCl –  (B) rCs  rCl –  a (C) rCs   rCl–  3a (D) rCs  rCl–  3a
2 2

[20]
7. Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 Å cubic lattice with a
unit cell edge of 4.29 Å. The radius of sodium atom is approximatley [JEE Mains-2015]
(A) 0.93 Å (B) 1.86 Å (C) 3.22 Å (D) 5.72 Å

8. Which of the following compounds is metallic and ferromagnetic ? [IIT JEE Main-2016]
(A) TiO2 (B) CrO2 (C) VO2 (D) MnO2

9. If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with m fraction of octahedral
holes occupied by aluminium ions and n fraction of tetrahedral holes occupied by magnesium ions, m and n,
respectively, are [JEE Advanced-2015]
1 1 1 1 1 1 1
(A) , (B) 1, (C) , (D) ,
2 8 4 2 2 4 8

10. The CORRECT statement(s) for cubic close packed (ccp) three dimensional structure is(are)
(A) The number of the nearest neighborus of an atom present in the topmost layer is 12
(B) The efficiency of atom packing is 74%
(C) The number of octahedral and tetrahedral voids per atom are 1 and 2, respectively

(D) The unit cell edge lengths is 2 2 times the radius of the atom [JEE Advanced-2016]

11. A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is 'a', the closest
approach between two atoms in metallic crystal will be : [IIT JEE Main-2017]
a
(A) (B) 2a (C) 2 2a (D) 2a
2

12. A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of 400 pm. If the
density of the substance in the crystal is 8 g cm–3, then the number of atoms present in 256 g of the crystal
is N × 1024. The value of N is : [JEE Advanced-2017]

13. Consider an ionic solid MX with NaCl structure. Construct a new structure (Z) whose unit cell is constructed
from the unit cell of MX following the sequential instructions given below. Neglect the charge balance.
(i) Remove all the anions (X) except the central one
(ii) Replace the all the face centered cations (M) by anions (X)
(iii) Remove all the corner cations (M)
(iv) Replace the central anion (X) with cation (M)
 number of anions 
The value of   in Z is [JEE Advanced-2018]
 number of cations 

14. Which type of 'defect' has the presence of cations in the interstitial sites ? [IIT Main 2018]
(A) Metal deficiency defect (B) Schottky defect
(C) Vacancy defect (D) Frenkel defect

[21]
15. The cubic unit cell structure of a compound containing cation M and anion X is shown below. When compared
to the anion, the cation has smaller ionic radius. Choose the correct statement(s) :

[JEE Advanced-2020]

(A) The empirical formula of the compound is MX.

(B) The cation M and anion X have different coordination geometries.
(C) The ratio of M-X bond length to the cubic unit cell edge length is 0.866.
(D) The ratio of the ionic radii of cation M to anion X is 0.414.

16. For the given close packed structure of a salt made of cation X and anion Y shown below (ions of only one

packingefficiency
face are shown for clarity), the packing fraction is approximately (packing fraction = )
100

Y
X
Y
Y
[JEE Advanced-2021]
(A) 0.74 (B) 0.63 (C) 0.52 (D) 0.48

17. Atom X occupies the fcc lattice sites as well as alternate tetrahedral voids of the same lattice. The packing
efficiency (in %) of the resultant solid is closest to : [JEE Advanced-2022]
(A) 25 (B) 35 (C) 55 (D) 75

18. Atoms of metals x, y, and z form face-centred cubic (fcc) unit cell of edge length Lx, body-centred cubic
(bcc) unit cell of edge length Ly, and simple cubic unit cell of edge length Lz, respectively.

3 8 3
If rz = ry ; ry = rx ; M z = M and Mz = 3Mx, then the correct statements(s) is(are)
2 3 2 y

[Given: Mx, My, and Mz are molar masses of metals x, y, and z, respectively.
rx, ry, and rz are atomic radii of metals x, y, and z, respectively.]
(A) Packing efficiency of unit cell of x > Packing efficiency of unit cell of y > Packing efficiency of unit cell of z
(B) Ly > Lz
(C) Lx > Ly
(D) Density of x > Density of y [JEE Advanced-2023]

[22]
 ANSWER KEY
EXERCISE-I
1. A-B 2. AuCu3 3. A4B2C2 4. 59.95%, +4

5. ZnAl2O4 6. 7.887 g/cc , 8.59 gm/cm3 7. 267 pm, 231.2 pm, 188.8 pm

8. 2.5 g cm–3 9. 3.472 × 1024 atoms 10. FCC

11. 21.41 g/cm3, 141.4 pm 12. 41.67 g cm–3 13. 5.83 g cm–3

14. 400 pm, 200 pm 15. 6 16. 0.24%

17. 346.4 pm 18. 19. a = 5.94 × 10–8cm, V = 2.096 × 10–22 cm–3

20. 1.81Å 21. AB2O4 22. 103.4 pm, No 23. 2.16 gm/cm3

24. 4, 6, 8 25. 6
EXERCISE-II
1. (i) Face plane, (ii) Face diagonal plane, (iii) diagonal plane 2. (a) 1.143, (b) 1.172

3. 5.188 gm/cm 3, 6.023 × 1022 atoms of A, 3.0115 × 1022 unit cells

5. 15.053 6. 3.57 Å 7. 4.34 Å 8. 6.02 × 1018 mol–1

9. 0.414 r, 79.3% 10. (a) 1.49 Å, (b) 1.37 g/cm3 11. (a) 2.88 Å, (b) 12 , (c) 19.4 g/cc

12. 4 molecules of H2O 13. for Fe is bcc, for V is bcc, for Pd is face centered

14. (a) 4.5 Å, (b) 5.2 Å, (c) 8, (d) 6, (e) 0.92 g/cm3 16. XY3, 4.38 g/cm3
17. Mg2SiO4, Fe2SiO4, 59% 18. (i) 3.90, (ii) 0.120 g/cc
19. 3.46 Å 20. 2.376%

QUESTION BANK
1. B 2. C 3. B 4. C 5. C 6. D 7. D
8. B 9. C 10. A 11. B 12. B 13. D 14. B
15. C 16. B 17. A 18. D 19. B 20. A 21. D
22. D 23. D 24. A 25. A 26. B 27. B 28. A
29. B 30. A 31. C 32. C 33. B 34. B, D 35. A,C,D
36. C, D 37. A, C 38. A, B 39. A, B, C 40. A  P ; B  Q, S ; C  R
41. 8 42. 2 43. 4 44. 3 45. 8

EXERCISE-III
1. D 2. 8 3. B 4. A 5. C 6. B 7. B
8. B 9. A 10. BCD 11. A 12. 2 13. 3 14. D
15. A,C 16. B 17. B 18. A,B,D

[23]