CBSE Class 11 Chemistry -Chapter 5 States of Matter- Study Materials

Subtopics of Class 11 Chemistry Chapter 5: States of Matter

  1. Intermolecular Forces
    • Dispersion Forces Or London Forces
    • Dipole-dipole Forces
    • Dipole-induced Dipole Forces
    • Hydrogen Bond
  2. Thermal Energy
  3. Intermolecular Forces Vs Thermal Interactions
  4. The Gaseous State Ex
  5. The Gas Laws Ex
    • Boyle’s Law (Pressure-volume Relationship)
    • Charles’ Law (Temperature-volume Relationship)
    • Gay Lussac’s Law (Pressure-temperature Relationship)
    • Avogadro Law (Volume – Amount Relationship)
  6. Ideal Gas Equation
    • Density And Molar Mass Of A Gaseous Substance
    • Dalton’s Law Of Partial Pressures
  7. Kinetic Molecular Theory Of Gases
  8. Behaviour Of Real Gases: Deviation From Ideal Gas Behaviour
  9. Liquefaction Of Gases
  10. Liquid State
    • Vapour Pressure
    • Surface Tension and Viscosity.

States of Matter Class 11 Notes Chemistry Chapter 5

• Intermolecular Forces
Intermolecular forces are the forces of attraction and repulsion between interacting particles
have permanent dipole moments. This interaction is stronger than the London forces but is weaker than ion-ion interaction because only partial charges are involved.
The attractive forces decrease with the increase of distance between dipoles. The interaction energy is proportional to 1/r6 where r is the distance between polar molecules.
Ion-Dipole Interaction: This is the force of attraction which exists between the ions (cations or anions) and polar molecules. The ion is attracted towards the oppositely charged end of dipolar molecules.
The strength of attraction depends upon the charge and size of the ion and the dipole moment and the size of the polar molecule.
For example: Solubility of common salt (NaCl) in water.
• Ion-induced Dipolar Interactions
In this type of interaction permanent dipole of the polar molecule induces dipole on the electrically neutral molecule by deforming its electronic cloud. Interaction energy is proportional to 1/r6 where r is the distance between two molecules.
• London Forces or Dispersion Forces
As we know that in non-polar molecules, there is no dipole moment because their electronic . charge cloud is symmetrically distributed. But, it is believed that at any instant of time, the electron cloud of the molecule may be distorted so that an instantaneous dipole or momentary dipole is produced in which one part of the molecule is slightly more negative than the other part. This momentary dipole induces dipoles in the neighbouring molecules. Thus, the force of attraction exists between them and are exactly same as between permanent dipoles. This force of attraction is known as London forces or Dispersion forces. These forces are always attractive and the interaction energy is inversely proportional to the sixth power of the
distance between two interacting particles, (i.e. 1/r6 where r is the distance between two particles).
This can be shown by fig. given below.
Hydrogen bonding: When hydrogen atom is attached to highly electronegative element by covalent bond, electrons are shifted towards the more electronegative atom. Thus a partial positive charge develops on the hydrogen atom. Now, the positively charged hydrogen atom of one molecule may attract the negatively charged atom of some other molecule and the two molecules can be linked together through a weak force of attraction.
Thermal Energy: The energy arising due to molecular motion of the body is known as thermal energy. Since motion of the molecules is directly related to kinetic energy and kinetic energy is directly proportional to the temperature.
• The Gaseous State
Physical Properties of Gaseous State
(i) ases have no definite volume and they do not have specific shape,
(ii) Gases mix evenly and completely in all proportions without any mechanical aid.
(iii) Their density is much lower than solids and liquids. :
(iv) They are highly compressible and exert pressure equally in all directions.
• Boyle’s Law (Pressure-Volume Relationship)
At constant temperature, the volume of a given mass of gas is inversely proportional to its pressure.
Charles’ law: At constant pressure, the volume of a given mass of a gas is directly proportional to its absolute temperature.
• Gay Lussac’s Law (Pressure-Temperature Relationship)
At constant volume, pressure of a given mass of a gas is directly proportional to the temperature.
• Avogadro Law (Volume-Amount Relationship)
Avogadro’s law states that equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.
V α n
Where n is the number of moles of the gas.
Avogadro constant: The number of molecules in one mole of a gas
= 6.022 x 1023
Ideal Gas: A gas that follows Boyle’s law, Charles’ law and Avogadro law strictly, is called an ideal gas.
Real gases follow these laws only under certain specific conditions. When forces of interaction are practically negligible.
• Ideal Gas Equation
This is the combined gas equation of three laws and is known as ideal gas equation.
• Dalton’s Law of Partial Pressure
When two or more non-reactive gases are enclosed in a vessel, the total pressure exerted by the gaseous mixture is equal to the sum of the partial pressure of individual gases.
Let P1 ,P2, and P3 be the pressure of three non reactive gases A, B, and C. When enclosed separately in the same volume and under same condition.
PTotal = P1+ P2 + P3
Where, PTotal = P is the total pressure exerted by the mixture of gases.
• Aqueous Tension
Pressure of non reacting gases are generally collected over water and therefore are moist. Pressure of dry gas can be calculated by substracting vapour pressure of water from total pressure of moist gas.
P2Dry gas = PTotal – Aqueous Tension
• Partial Pressure in terms of Mole Fraction
Let at the temperature T, three gases enclosed in the volume V, exert partial pressure P1 , Pand Prespectively, then
• Kinetic Molecular Theory of Gases
(i) Gases consist of large number of very small identical particles (atoms or molecules),
(ii) Actual volume occupied by the gas molecule is negligible in comparison to empty space between them.
(iii) Gases can occupy all the space available to them. This means they do not have any force of attraction between their particles.
(iv) Particles of a gas are always in constant random motion.
(v) When the particles of a gas are in random motion, pressure is exerted by the gas due to collision of the particles with the walls of the container.
(vi) Collision of the gas molecules are perfectly elastic. This means there is no loss of energy after collision. There may be only exchange of energy between colliding molecules.
(vii) At a particular temperature distribution of speed between gaseous particles remains constant.
(viii) Average kinetic energy of the gaseous molecule is directly proportional to the absolute temperature.
• Deviation From Ideal Gas Behaviour
Real Gas: A gas which does not follow ideal gas behaviour under all conditions of temperature and pressure, is called real gas.
Deviation with respect to pressure can be studied by plotting pressure Vs volume curve at a given temperature. (Boyle’s law)
Compressibility factor (Z): Deviation from ideal behaviour can be measured in terms of compressibility factor, Z.
• van der Waals Equation
Where V is a constant for molecular attraction while ‘V is a constant for molecular volume.
(a) There is no force of attraction between the molecules of a gas.
(b) Volume occupied by the gas molecule is negligible in comparison to the total volume of the gas.
Above two assumptions of the kinetic theory of gas was found to be wrong at very high pressure and low temperature.
• Liquifaction of Gases
Liquifaction of gases can be achieved either by lowering the temperature or increasing the pressure of the gas simultaneously.
Thomas Andrews plotted isotherms of C02 at various temperatures shown in figure.
Critical Temperature (Tc): It is defined as that temperature above which a gas cannot be liquified however high pressure may be applied on the gas.
Tc = 8a/27bR
(Where a and b are van der Waals constants)
Critical Pressure (Pc): It is the pressure required to Liquify the gas at the critical temperature.
Pc = a/27b2
The volume occupied by one mole of the gas at the critical temperature and the critical pressure is called the critical volume (Vc).
For Example. For C02 to Liquify.
Tc = 30.98°C
Pc = 73,9 atm.
Vc = 95-6 cm3/mole
All the three are collectively called critical constants.
• Liquid State
Characteristics of Liquid State
(i) In liquid, intermolecular forces are strong in comparison to gas.
(ii) They have definite volume but irregular shapes or we can say that they can take the shape of the container.
(iii) Molecules of liquids are held together by attractive intermolecular forces.
Vapour Pressure: The pressure exerted by the vapour of a liquid, at a particular temperature in a state of dynamic equilibrium, is called the vapour pressure of that liquid at that temperature.
Vapour Pressure depends upon two factors:
(i) Nature of Liquid (ii) Temperature
• Surface Tension
It is defined as the force acting per unit length perpendicular to the line drawn on the surface of liquid.
S.I. unit of Surface Tension = Nm-1
Surface Tension decreases with increase in temperature, because force acting per unit length decreases due to increase in kinetic energy of molecules.
• Viscosity
It is defined as the internal resistance to flow possessed by a liquid.
The liquids which flow slowly have very high internal resistance, which is due to strong intermolecular forces and hence are said to be more viscous.
When liquid flows, the layer immediately below it tries to retard its flow while the one above tries to accelerate.
Thus, force is required to maintain the flow of layers.
Effect of Temp, on Viscosity: Viscosity of liquids decreases as the temperature rises because at high temperature, molecules have high kinetic energy and can overcome the intermolecular forces to slip past one another.
• Boyle’s Law: It states that, under isothermal conditions pressure of a given mass of a gas is inversely proportional to its volume.

CBSE Class 11 Chemistry Chapter-5 Important Questions

1 Marks Questions

1.Define Van der waal’s forces.

Ans. Attractive intermolecular forces between molecules is known as Van der waals forces.

2.Give an example to show dipole-dipole forces.

Ans. Dipole-dipole forces act between the molecules possessing permanent dipole. Ends of the dipoles possess partial charges which are responsible for the interaction eg. The interaction between two HCl molecules.

3.What type of bond exists between ,HF, NH3, C2H5 OH molecule.?

Ans. In H2O, HF, NH3, C2H5OH molecule, hydrogen bond exists between hydrogen and the other electronegative atom attached to it.

4.Define Boyle’s law.

Ans. At constant temperature, the pressure of a fixed amount (i.e; number of moles n) of gas varies inversely with its volume.


(at constant T and n)



5.Why helium and hydrogen gases not liquefied at room temperature by applying very high pressure?

Ans. Because their critical temperature is lower than room temperature and gases cannot be liquefied above the critical temperature even by applying very high pressure.

6.How is the pressure of a given sample of a gas related to temperature at volume?

Ans. Pressure is directly proportional to the temperature , i.e; PT.

7.Define absolute zero temperature.

Ans. The lowest hypothetical or imaginary temperature at which gases are supposed to occupy zero volume is called Absolute zero.

8.Define an ideal gas.

Ans.A gas which obeys the ideal gas equation (PV = nRT) at all temperature and pressure is called an ideal gas.

9.What is aqueous tension?

Ans.Pressure exerted by saturated water vapors is called aqueous tension.

10.What is the value of R at STP?

Ans. At S. T. P, R = 8.20578 x 10-2 L at m k-1 mol-1.

11.Molecule A is twice as heavy as the molecule B. which of these has higher kinetic energy at any temperature?

Ans. Kinetic energy of a molecule is directly proportional to temperature and independent of mass so both the molecules will have the same kinetic energy.

12.Write Van der waal’s equation for n moles of a gas.

Ans. is Van der waal’s equation for n moles of a gas.

13.Out of NH3 and N2, which will have (i) larger value of ‘a’ and (ii) larger value of ‘b’?

Ans. (i) NH3 will have larger value of ‘a’ because of hydrogen bonding.

(ii)N2 should have large value ‘b’ because of larger molecular size.

14.What property of molecules of real gases is indicated by van der waal’s constant ‘a’?

Ans. Intermolecular attraction

15.Under what conditions do real gases tend to show ideal gas behaivour?

Ans. When the pressure of the gas is very low and the temperature is very high.

16.How are Van der waal’s constants ‘a’ and ‘b’ related to the tendency to liquefy?

Ans.The Van der waal’s constant ‘a’ is a measure of intermolecular attractions. Therefore, the value of ‘a’ reflects the tendency of the gas to liquefy. The gas having larger value of ‘a’ will liquefy more easily.

17.When does a gas show ideal behaivour in terms of volume?

Ans. The gases show ideal behaviors when the volume of the gas occupied is large so that the volume of the molecules can be neglected in comparison to it.

18.Define Boyle point.

Ans. The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called Boyle temperature or Boyle point.

19.Define standard boiling point.

Ans.If the pressure is 1 bar then the boiling point is called standard boiling point of the liquid.

20.What is surface energy?

Ans. The energy required to increase the surface area of the liquid by one unit is defined as surface energy.

21.What is surface tension? What is its S.I unit?

Ans. Surface tension is defined as the force acting per unit length perpendicular to the line drawn on the surface of liquid S.I unit is expressed as Nm-1.

22.How does surface tension change when temperature is raised?

Ans. Surface tension decreases as the temperature is raised.

23.Why is glycerol highly viscous?

Ans. Glycerol has three hydrogen bonds which results in an extensive hydrogen bonding. That is why glycerol is highly viscous.

24.Some tiny light hollow spheres are placed in a flask. What would happen to these spheres, if temperature is raised?

Ans. The spheres would start moving faster randomly and colliding with each other.

25.The boiling points of a liquid rises on increasing pressure. Give reason.

Ans. A liquid boils when its vapors pressure becomes equal to the atmospheric pressure. An increase in pressure on liquid, causes a rise in the boiling temperature of the Liquid.

26.  What are fluids? Give examples.

Ans. Substances which flow are fluids e.g. liquids and gases.

27.  Solids are rigid why?

Ans. Rigidity in solids is due to fixed positions of the constituent particles and their oscillations about their mean positions

28.  How are solids classified?

Ans. Solids may be classified into two categories – crystalline and amorphous.

29. Why do crystalline solids are anisotropic nature and amorphous solid are isotropic in nature. 

Ans. Anisotropy in crystals is due to different arrangement of particles along different directions. Isotropic in amorphous solid is due to its long range order in them and arrangement is irregular along all the directions.

30. Define the term: Crystal lattice

Ans. A regular three dimensional arrangement of points in space is called crystal lattice.

31. What is a unit cell?

Ans. The smallest portion of a crystal lattice which, when, repeated in different directions, generates the entire lattice, is called its unit cell.

32. What are the axial angles and edge length in a cubic crystal system?

Ans. Axial angles,  and edge lengths a = b = c.

33. Give one example of each – Tetragonal and hexagonal crystal system.

Ans. Tetragonal crystal system – white tin,.

Hexagonal crystal system – Graphite, ZnO.

34. What is square close packing?

Ans. The close packing of spheres – atoms or ions, in which each sphere is in contact with four of is neighbors, whose centres, if joined, form a square, is called square close packing.

35. What is the coordination number in:-

(a) Square close packing        

(b) Hexagonal close packing.

Ans. (a) in square close packing, the coordination no. is 4.

(b) In hexagonal close packing, the coordination no. is 12.

36. Define – (a) void (b) coordination Number

Ans. (a) void – the empty space left between close packed spheres are voids.

(b) Coordination number – the number of spheres or atoms surrounding a single sphere or atom in a crystal is called coordination number.

37. What is the packing efficiency in

Ans. (a) hcp packing efficiency is 74%.

(b) bcc packing efficiency is 68%.

(c) packing efficiency in simple cubic structure is 52.4%.

38. What is the meaning of term ‘defect’ with reference to crystal

Ans. The defects are irregularities in the arrangement of constituent particles in a crystal.

39. Name the types of point defect.

Ans. point defects are of three types – stoichiometric defect, impurity defect & non – stoichiometric defect.

40. What are F centres?

Ans. F – centre is the position of an anion in an ionic crystal which is occupied by a trapped electron

41. Why do the heating of NaCl in an atmosphere of sodium vapour impact yellow colour ?

Ans. In an atmosphere of sodium vapour the Cl  diffuses to the surface of the crystal and combine with Na atom to form NaCl. During this process Na atom loses an electron –cl for. As a result the crystal has an excess of sodium which import yellow colour to the crystal.

42. Give an example which shows both frenkel and Schottky defect.

Ans. AgBr.

43. Define the term – doping.

Ans. Doping – The process of introduction of impurity atoms into an insulator to make it a semiconductor is called doping.

44. What is the meaning of 13 – 15 compounds?  

Ans. A semiconductor formed by combination of 13 groups & 15 groups elements is 13 – 15 compound.

45. Name an element which can be added to silicon to give a –  

(i) p – type semiconductor

(ii) n – type semiconductor.

Ans. (i) For p – type semiconductor Borom can be added.

(ii) For n – type semiconductor, Phosphorous can be added.

46. What is the difference between ferromagnetic and paramagnetic substances?

Ans. A ferromagnetic substance has permanent magnetic behaviour whereas a paramagnetic substance acts as a magnet only in the presence of an external magnetic field.

2 Marks Questions

1.Ice has lower density than water. Give reason.

Ans. Hydrogen bonding affect the physical properties of compounds. Ice is H-bonded molecular solid having open type structure with wide holes whereas liquid water has H-bonding having closed type structure that is why ice has lower density than water.

2.Water has maximum density at 40C. Give reason.

Ans. Water has maximum density at 40C because when temperature is increased from o to 40C, some of the H-bonds break and molecules come closer and density increases till 40C because volume decreases. But, above 40C, the kinetic energy of molecules increases which leads to increase in volume and density decreases.

3.Define thermal energy

Ans. Thermal energy is the energy of a body arising from motion of its atoms or molecules. It is directly proportional to the temperature of the substance.

4.What are the factors responsible for the strength of hydrogen bonds?

Ans. Strength of the hydrogen bond is determined by the coulombic interaction between the lone-pair electrons of the electronegative atom of one molecule and the hydrogen atom of other molecule.

5.At what temperature will the volume of a gas at 00 c double itself, pressure remaining constant?

Ans. Let the volume of the gas at 00C = Vml.


V1 = Vml, V2 = 2Vml

T1 = o + 273 T2 = ?

= 273k

By applying Charle’s law

= > 

= > T2 = 

T2 = 546 – 273 = 2730C

6. 50 cm3 of hydrogen gas enclosed in a vessel maintained under a pressure of 1400 Tor, is allowed to expand to 125 cm3 under constant temperature conditions. What would be its pressure?

Ans. Since, temperature is constant, we have

PV = constant

P1 V1 = P2 V2

P1 = 1400 ; P2 = ?

V= 50cm3 ; V2 = 125 cm3

 P2 = 

The final pressure of the gas after expansion would be 560 Torr.

7.State the law depicting the volume-temperature relationship.

Ans. The law is known Charle’s law.

“Pressure remaining constant the volume of a given mass of a gas increases or decreases by 1/273 of its volume at 00C for every one degree centigrade or fall in temperature.


Where Vt is the volume of the gas at t0C and Vo is its volume at 00C.

8.State Avogadro’s Law. Is the converse of Avogadro’s law true?

Ans. Avogadro’s Hypothesis: This law was given by Avogadro in 1811. According to this law, “Equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules.”

Volume = a constant X Number of Moles (Temperature and pressure constant).

The converse of Avogadro’s law is also true. Equal number of molecules of all gases occupy equal volume’s under the same conditions of temperature and pressure. It follows that one gram molecular mass of any gas (containing 6.023×1023 molecules) will occupy the same volume under the same conditions. The volume occupied by one gram molecular mass of any gas at 00C and 760 mm of Hg is 22.4 dm3 (liters) is called the gram molecular volume or simply molar volume.

9.Deduce the relation pv = nRT where R is a constant called universal.

Ans. According to Boyle’s law,

(at constant T and n)

According to charle’s law,

(at constant p and n)

According to Avogadro’s law,

(at constant T and p)

By combining the three laws,


Where R is a constant called universal gas constant.

10.At 250C and 760 mm of Hg pressure a gas occupies 600ml volume. What will be its pressure at a height where temperature is 100C and volume of the gas is 640mL. Calculate the volume occupied by 5.0 g of acetylene gas at 500C and 740mm pressure.

Ans. P1 = 760mm Hg, V1 = 600mL

T1 = 25 + 273 = 298 k

V2 = 640 mL and T2 = 10 = 273 = 283 k

According to combined gas law,


P2 = 

11.Explain how the function pv/RT can be used to show gases behave non-ideally at high pressure.

Ans. The ratio pv/RT is equal to the number of moles of an ideal gas in the sample. This number should be constant for all pressure, volume and temperature conditions. If the value of this ratio changes with increasing pressure, the gas sample is not behaving ideally.

12.Mention the two assumptions of kinetic theory of gases that do not hold good.

Ans. The two assumptions of the kinetic theory that do not hold good are –
(i)There is no force of attraction between the molecules of a gas.
(ii)Volume of the molecules of a gas is negligibly small in comparison to the space occupied by the gas.

13.Calculate the pressure exerted by one mole of CO2 at 273 k if the Van der waal’s constant a = 3.592 dm6 at m mol-1. Assume that the volume occupied by CO2 molecules is negligible.

Ans. According to Van der waal’s equation

 (for 1 mole of the gas)

Since, the volume occupied by CO2 molecules is negligible, therefore, b =o

a = 3.592 dm6 at m mol-1
V = 22.4 dm3
R = o.o82 L at m k mol-1
T = 273 k.

= 0.9993 – 0.0071

14.Why does viscosity of liquids decrease as the temperature is raised?

Ans. Viscosity of liquids decreases as the temperature rises because at high temperatures molecules have high kinetic energy and can overcome the intermolecular forces to slip over one another between the layers.

15.What is the effect of temperature on
(i) density
(ii) vapors pressure of a liquid?

Ans.(i) The volume increases, with the increase of temperature. Therefore, density decreases with the rise of temperature.
(ii) As the temperature of a liquid is increased, the vapors pressure increases.

16. The window panes of the old buildings are thick at the bottom. Why?  

Ans. Glass panes of old buildings are thicker at the bottom than at the top as from is an amorphous solid and flows down very slowly and makes the bottom portion thicker.

17. The stability of a crystal is reflected in the magnitude of its melting point. Explain

Ans. Melting point of a solid gives an idea about the intermolecular forces acting between particles. When these forces are strong, the melting point is higher and when these forces are weak, low melting point is observed. Higher is the melting point, more stable the solid is.

18. Graphite is soft and good conductor of electricity. Explain.   

Ans. Graphite is soft and good conductor due to its typical structure here carbon atoms are arranged in different layers and each atom a covalently bonded to three of its neighbouring atoms in the same layer. The fourth electron of each atom is free to move about due to which it conducts electricity. Different layers can slide over the other which makes it a soft solid.

19. Ionic solids are good conductors in molten state and in aqueous solutions but not in solid state. Why?

Ans. In the solid state, the ions in the ionic solids are not free to move about due to their rigid structure & strong electrostatic forces. Therefore they cannot conduct electricity whereas in molten state and aqueous solution, the ions become free to move about and they conduct electricity.

20. Name three types of cubic unit cells?

Ans. (a) Simple cubic

(b) Face – centred cubic

(c) Body centred cubic

21.  How many atoms are there in a unit cell of a metal crystallizing in a:

  (a) FCC structure

  (b) BCC structure

Ans. (a) FCC = 4     (b) BCC = 2

22.  What is the contribution of an atom per unit cell if the atom is:

  (a) At the corner of the cube.

  (b) On the face of the cube.

  (c) In the centre of the cube.

Ans. (a) When atom is at the corner of the cube, the contribution is atom.

(b) When the atom is on the face of the cube, its contribution is  atom.

(c) If the atom is in the centre of the cube, its contribution is 1 atom.

23.  A compound formed by A & B crystallizes in the cubic structure where ‘A’ are at the corners of the cube and B are at the face centre. What is the formula of the compound?

Ans. Contribution of atom A per unit cell = atom

Contribution of atom B per unit cell =  atom

Ratio of A & B = 1:3               Formula =.

24.  Calculate the no. of atoms in a cubic based unit – cell having one atom on each corner and two atoms on each body diagonal. 

Ans. No. of atoms contributed by 8 corners per unit cell = atom.

No. of atoms contributed by one diagonal = 2

No. of diagonal = 4

Total contribution by diagonal = 4 2 = 8

Total no. of atoms = 8 + 1 = 9 atoms

25. What is the no. of octahedral and tetrahedral voids present in a lattice?

Ans. No. of octahedral voids present in a lattice is equal to the no. of close packed particles and the number of tetrahedral voids is twice the no. of close packed particles.

26.  Give the relationship between density and edge length of a cubic crystal.

Ans. Density, d of a cubic cell is given by –

Where Z = no. of atoms per unit cell

M = molar mass

= Avogadro number

a = edge length

27.  Copper which crystallizes as a face – centred cubic lattice has a density of at. calculate the length of the unit cell.

Ans. 63.1 u

28.  An element crystallizes in BCC structure. The edge of its unit cell is 288 pm. If the density is, calculate the atomic mass of the element.

Ans. 52 g/ mol

29.  The compound CuCl has ZnS structure and the edge length of the unit cell in 500 pm. Calculate the density. (Atomic masses: Cu = 63, Cl = 35.5, Avogadro no =)


30.  In a compound, B ions form a close – packed structure & A ions occupy all the tetrahedral voids. What is the formula of the compound?

Ans. Let the no. of B ions = 100

No. of A ions = no. of tetrahedral voids

= 2 no. of B = 2 100 = 200

Ratio of A & B = 200 : 100 = 2 : 1

Formula =.

31.  Define two main types of defects.

Ans.  Defects are of two types –

Point defects: – Irregularities or deviations from ideal arrangement around a point or on atom in a crystal.

Line defect: – Irregularities or deviation from ideal arrangement in entire row of lattice points.

 32.  (a) Identify the defect in figure below :     

  (b) How does it affect the density of crystal?

  (c) Give an example of crystal where this defect can be found.

  (d) What is its effect on electrical neutrality of crystal?

Ans.  (a) Schottky defect.

(b) It decreases the density of crystal.

(c) NaCl, KCl

(d) The crystal remains electrically neutral.

33.  Which defect is observed in a solid solution of and AgCl? Explain.

Ans.  In a solid solution of and AgCl, impurity defect is observed. In the crystal of AgCl, some of the sites of are occupied by, each  replacing two . It replaces the site of one ion and other site remains vacant.

34. Excess of lithium makes LiCl crystal pink. Explain.

Ans. When crystals of LiCl are heated in an atmosphere of Li vapour the Lithium atoms are deposited on the surface of crystal. The Cl– ions diffuse to the surface of crystal & combine with Li atoms to from LiCl which happens by loss of electrons by Li atoms to form ions. These released elements diffuse into the crystal & electrons get excited after absorbing light from visible region & emit pink colour.

 35. In the fig –


  (a) Identify the magnetic behaviour of substance.

  (b) How are these substances different from diamagnetic substances?

Ans. (a) Antiferromagnetic field.

(b)In Antiferromagnetic substance the domain structure is similar to ferromagnetic substance but the domains are oppositely oriented and cancel out each others magnetic moment whereas in diamagnetic substance all individual electron are paired which cancels the individual magnetic moments.

36. Define the terms – Intrinsic semiconductor and extrinsic semiconductor.

Ans. Intrinsic semiconductor – An insulator which conducts electricity when heated.

Extrinsic semiconductor – An insulator which conducts electricity on addition of an impurity.

37. Classify solids on the basis of their conductivities.   

Ans. Solids can be classified into three types on the basis of their conductivities –

Solids     Conductivities

1. Conductor    

2. Insulators    

3. Semi – conductor   

38. Give two examples of each.        

  (a) Ferromagnetic substances           

  (b) Ferrimagnetic substances

Ans. (i) Ferromagnetic substances –Fe, Co, Ni, , Gadolinium.

(ii) Ferrimagnetic substance –, Magnetite and Ferrites.

39. Give two application of p – type and n – type semiconductors.

Ans. Application –

(i) A combination of both p – type and n – type semiconductors is used in diode which is used as rectifiers.

(ii) npn & pnp types of transistors are used as amplifiers.

40. Non-stoichiometric cuprous oxide, can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?

Ans. In the cuprous oxide prepared in the laboratory, copper to oxygen ratio is slightly less than 2:1. This means that the number of ions is slightly less than twice the number of  −ions. This is because some ions have been replaced by ions. Every ion replaces two ions, thereby creating holes. As a result, the substance conducts electricity with the help of these positive holes. Hence, the substance is a p-type semiconductor.

41. Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.

Ans. Let the number of oxide ions be x.

So, number of octahedral voids = x

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

So, number of ferric ions=23x

Therefore, ratio of the number of Fe3+ions to the number of O2− ions, Fe3+: O2−=23 x:x


= 2 : 3

Hence, the formula of the ferric oxide isFe2O3.

42. Classify each of the following as being either a p-type or an n-type semiconductor:

(i) Ge doped with In (ii) B doped with Si.

Ans. (i) Ge (a group 14 element) is doped with In (a group 13 element). Therefore, a hole will be created and the semiconductor generated will be a p-type semiconductor.

(ii) B (a group 13 element) is doped with Si (a group 14 element). Thus, a hole will be created and the semiconductor generated will be a p-type semiconductor.

43. Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?

Ans. For a face-centred unit cell:a=

It is given that the atomic radius, r= 0.144 nm


= 0.407 nm

Hence, length of a side of the cell = 0.407 nm

44. In terms of band theory, what is the difference

(i) Between a conductor and an insulator

(ii) Between a conductor and a semiconductor

Ans. (i) The valence band of a conductor is partially-filled or it overlaps with a higher energy, unoccupied conduction band.

On the other hand, in the case of an insulator, the valence band is fully- filled and there is a large gap between the valence band and the conduction band.

(ii) In the case of a conductor, the valence band is partially-filled or it overlaps with a higher energy, unoccupied conduction band. So, the electrons can flow easily under an applied electric field.

On the other hand, the valence band of a semiconductor is filled and there is a small gap between the valence band and the next higher conduction band. Therefore, some electrons can jump from the valence band to the conduction band and conduct electricity.

45. How many lattice points are there in one unit cell of each of the following lattice?

(i) Face-centred cubic

(ii) Face-centred tetragonal

(iii) Body-centred

Ans. (i) There are 14 (8 from the corners + 6 from the faces) lattice points in face-centred cubic.

(ii) There are 14 (8 from the corners + 6 from the faces) lattice points in face-centred tetragonal.

(iii) There are 9 (1 from the centre + 8 from the corners) lattice points in body-centred cubic.

46. Explain

(i) The basis of similarities and differences between metallic and ionic crystals.

(ii) Ionic solids are hard and brittle.

Ans. (i) The basis of similarities between metallic and ionic crystals is that both these crystal types are held by the electrostatic force of attraction. In metallic crystals, the electrostatic force acts between the positive ions and the electrons. In ionic crystals, it acts between the oppositely-charged ions. Hence, both have high melting points.

The basis of differences between metallic and ionic crystals is that in metallic crystals, the electrons are free to move and so, metallic crystals can conduct electricity. However, in ionic crystals, the ions are not free to move. As a result, they cannot conduct electricity. However, in molten state or in aqueous solution, they do conduct electricity.

(ii) The constituent particles of ionic crystals are ions. These ions are held together in three-dimensional arrangements by the electrostatic force of attraction. Since the electrostatic force of attraction is very strong, the charged ions are held in fixed positions. This is the reason why ionic crystals are hard and brittle.

47.  Define the term ‘amorphous’. Give a few examples of amorphous solids.

Ans. Amorphous solids are the solids whose constituent particles are of irregular shapes and have short range order. These solids are isotropic in nature and melt over a range of temperature. Therefore, amorphous solids are sometimes called pseudo solids or super cooled liquids. They do not have definite heat of fusion. When cut with a sharp-edged tool, they cut into two pieces with irregular surfaces. Examples of amorphous solids include glass, rubber, and plastic.

48.  What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?

Ans. The arrangement of the constituent particles makes glass different from quartz. In glass, the constituent particles have short-range order, but in quartz, the constituent particles have both long range and short range orders.

Quartz can be converted into glass by heating and then cooling it rapidly.

49.   Classify the following as amorphous or crystalline solids:

Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.

AnsAmorphous solids: Polyurethane, teflon, cellophane, polyvinyl chloride, fibre glass

Crystalline solids: Naphthalene, benzoic acid, potassium nitrate, copper

50. (i) What is meant by the term ‘coordination number’?

(ii) What is the coordination number of atoms:

(a) in a cubic close-packed structure?

(b) in a body-centred cubic structure?

Ans. (i) The number of nearest neighbours of any constituent particle present in the crystal lattice is called its coordination number.

(ii) The coordination number of atoms

(a) in a cubic close-packed structure is 12, and

(b) in a body-centred cubic structure is 8

3 Marks Questions

1. Differentiate between amorphous and crystalline solids with reference to

(1) Melting point

(2) Cleavage property

(3) Nature


PropertyCrystalline solidsAmorphous solids
1. Melting pointThey have sharp melting pointThey have a range of melting point.
2. Cleavage propertyThey split into pieces of plain and smooth surfaces when cut with a sharp edged tool.When cut with a sharp edged tool, pieces of irregular surfaces are obtained.
3. NatureThey are true solids.They are pseudo solids or super cooled liquids.

2. How are crystalline solids classified on the basis of nature of bonding? Explain with examples.

Ans. Classification of crystalline solids.

1. Molecular solids: The forces operating between molecules are dispersion or London forces, dipole – dipole interactions, hydrogen bounding e.g., HCl, ice etc.

2. Ionic solids:The intermolecular forces are coulomibic or electrostatic forces, e.g. NaCl, MgO etc.

3. Metallic solids: The forces operating is metallic bonding e.g. Fe, Cu, Ag etc.

4. Covalent or network solids: The attractive forces are covalent bonding e.g. Diamond, Quartz etc.

3. In crystalline solid, anions C are arranged in cubic close – packing, cations A occupy 50% of tetrahedral voids & cations B occupy 50% of octahedral voids. What is the formula of solid?

Ans. Suppose no. of anions, C = 100

Suppose no. of cations, A =  no. of tetrahedral voids

No. of cations, B = no.   of octahedral voids

Ratio of ions A : B : C = 100 : 50 : 100

= 2 : 1 : 2

Formula = 

4. Which type of ionic substances show?                                                                   

(a) Schottky defect

(b) Frenkel defect

Ans. (a) Schottky defect – ionic substances in which the cation and anion are of almost similar sizes eg. NaCl, KCl, CrCl.

(b) Feenkel Defect – Ionic substances in which there is large difference in size of ions eg. ZnS, AgCl, AgBr.

5. Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm.

(i) What is the length of the side of the unit cell?

(ii) How many unit cells are there in 1.00 of aluminium?

Ans. (i) For cubic close-packed structure:

= 353.55 pm

= 354 pm (approximately)

(ii) Volume of one unit cell = (354 pm)3

= 4.4×107pm3


= 4.4×

Therefore, number of unit cells in 1.00 =1.00cm 

= 2.27×1022

6. If NaCl is doped with of, what is the concentration of cation vacancies?

Ans. It is given that NaCl is doped with % of .

This means that 100 mol of NaCl is doped with 10−3mol of .

Therefore, 1mol of NaCl is doped with103100mol of 

mol of 

Cation vacancies produced by one ion = 1

Therefore, Concentration of the cation vacancies

Produced by mol of ions = 

Hence, the concentration of cation vacancies created by is of NaCl.

7. Silver crystallizes in fcc lattice. If edge length of the cell is and density is 10.5 g , calculate the atomic mass of silver.

Ans. It is given that the edge length, 


As the lattice is fcc type, the number of atoms per unit cell, z= 4

We also know that, 

Using the relation:

Therefore, atomic mass of silver = 107.13 u

8. A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q?

Ans. It is given that the atoms of Q are present at the corners of the cube.

Therefore, number of atoms of Q in one unit cell = 

It is also given that the atoms of P are present at the body-centre.

Therefore, number of atoms of P in one unit cell = 1

This means that the ratio of the number of P atoms to the number of Q atoms, P:Q = 1:1

Hence, the formula of the compound is PQ.

The coordination number of both P and Q is 8.

9. Niobium crystallises in body–centred cubic structure. If density is 8.55 g , calculate atomic radius of niobium using its atomic mass 93 u.

Ans. It is given that the density of niobium, d= 8.55 g cm – 3

Atomic mass, M = 93 g 

As the lattice is bcc type, the number of atoms per unit cell, z= 2

We also know that, 

Applying the relation:


For body-centred cubic unit cell:

= 14.32 nm

10. If the radius of the octachedral void is r and radius of the atoms in close packing is R, derive relation between r and R.

Ans. A sphere with centre O, is fitted into the octahedral void as shown in the above figure. It can be observed from the figure that is right-angled

Now, applying Pythagoras theorem, we can write:

11. Copper crystallises into a fcc lattice with edge length . Show that the calculated density is in agreement with its measured value of 8.92 g .

Ans. Edge length, 

As the lattice is fcc type, the number of atoms per unit cell, z= 4

Atomic mass, M = 63.5 g 

We also know that, 

Applying the relation:

The measured value of density is given as . Hence, the calculated density is in agreement with its measured value.

12. How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.

Ans. By knowing the density of an unknown metal and the dimension of its unit cell, the atomic mass of the metal can be determined.

Let ‘a’ be the edge length of a unit cell of a crystal, ‘d’ be the density of the metal, ‘m’ be the mass of one atom of the metal and ‘z’ be the number of atoms in the unit cell.

Now, density of the unit cell =Mass of the unit cell Volume of the unit cell


[Since mass of the unit cell = Number of atoms in the unit of one atom]

[Volume of the unit cell = (Edge length of the cubic unit cell)3]

From equation (i), we have:


Now, mass of one atom of metal (m) = Atomic mass (M) Avogadro’s number (NA)

Therefore, M = da3NAz………………………(iii)

If the edge lengths are different (say a, b and c), then equation (ii) becomes:

m = d(abc)NAz……………………………..(iv)

13. ‘Stability of a crystal is reflected in the magnitude of its melting point’. Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?

Ans. Higher the melting point, greater is the intermolecular force of attraction and greater is the stability. A substance with higher melting point is more stable than a substance with lower melting point.

The melting points of the given substances are:

Solid water = 273 K

Ethyl alcohol = 158.8 K

Diethyl ether = 156.85 K

Methane = 89.34 K

Now, on observing the values of the melting points, it can be said that among the given substances, the intermolecular force in solid water is the strongest and that in methane is the weakest.

5 Marks Questions

1. Explain conduction of electricity on the basis of band theory.

Ans. The conductivity of metals depend upon the no. of valance electrons available per atom. The atomic orbitals of metal atoms form molecular orbitals which are so close in energy to each other as to form a band. If this band is partially filled & overlaps with a higher energy unoccupied, conduction band, then electrons can flow easily under an applied electric field & metal is a conductor. If the gap between filled valance band and conduction band is large, electron cannot jump to it & the substance is an insulator where as if the gap is small & on heating, can be overcome, the substance acts as semiconductor

2. How will you distinguish between the following pairs of terms:

(i) Hexagonal close-packing and cubic close-packing?

(ii) Crystal lattice and unit cell?

(iii) Tetrahedral void and octahedral void?

Ans. A 2-D hexagonal close-packing contains two types of triangular voids (a and b) as shown in figure 1. Let us call this 2-D structure as layer A. Now, particles are kept in the voids present in layer A (it can be easily observed from figures 2 and 3 that only one of the voids will be occupied in the process, i.e., either a or b). Let us call the particles or spheres present in the voids of layer A as layer B. Now, two types of voids are present in layer B (c and d). Unlike the voids present in layer A, the two types of voids present in layer B are not similar. Void c is surrounded by 4 spheres and is called the tetrahedral void. Void d is surrounded by 6 spheres and is called the octahedral void.

Now, the next layer can be placed over layer B in 2 ways.

Case1: When the third layer (layer C) is placed over the second one (layer B) in such a manner that the spheres of layer C occupy the tetrahedral voids c.

In this case we get hexagonal close-packing. This is shown in figure 4. In figure 4.1, layer B is present over the voids a and layer C is present over the voids c. In figure 4.2, layer B is present over the voids b and layer C is present over the voids c. It can be observed from the figure that in this arrangement, the spheres present in layer C are present directly above the spheres of layer A. Hence, we can say that the layers in hexagonal close-packing are arranged in an ABAB…..pattern.

Case 2: When the third layer (layer C) is placed over layer B in such a manner that the spheres of layer C occupy the octahedral voids d.

In this case we get cubic close-packing. In figure 5.1, layer B is present over the voids a and layer C is present over the voids d. In figure 5.2, layer B is present over the voids b and layer C is present over the voids d. It can be observed from the figure that the arrangement of particles in layer C is completely different from that in layers A or B. When the fourth layer is kept over the third layer, the arrangement of particles in this layer is similar to that in layer A. Hence, we can say that the layers in cubic close-packing are arranged in an ABCABC…..pattern.

3. Calculate the efficiency of packing in case of a metal crystal for

(i) simple cubic

(ii) body-centred cubic

(iii) face-centred cubic (with the assumptions that atoms are touching each other).

Ans. (i)Simple cubic

In a simple cubic lattice, the particles are located only at the corners of the cube and touch each other along the edge.

Let the edge length of the cube be ‘a’ and the radius of each particle be r.

So, we can write: = 2r

Now, volume of the cubic unit cell = 

We know that the number of particles per unit cell is 1.

Therefore, volume of the occupied unit cell =

Hence, packing efficiency = Volume of one particle Volume of cubic unit 




(ii) Body-centred cubic

It can be observed from the above figure that the atom at the centre is in contact with the other two atoms diagonally arranged.

From ΔFED, we have:


Again, from , we have:

c2=a2+b2⇒c2=a2+2a2⇒c2=3a2 (Since b2=2a2)


Let the radius of the atom be r.

Length of the body diagonal, c=




Volume of the cube, 

A body-centred cubic lattice contains 2 atoms.

So, volume of the occupied cubic lattice = 

Therefore, Packing efficiency = Volume occupied by two spheres in the unit cell Total volume of the unit cell

(iii) Face-centred cubic

Let the edge length of the unit cell be ‘a’ and the length of the face diagonal AC be b.

From ΔABC, we have:

4.  Analysis shows that nickel oxide has the formulaNi0.98O1.00. What fractions of nickel exist as and  ions?

Ans. The formula of nickel oxide is Ni0.98O1.00.

Therefore, the ratio of the number of Ni atoms to the number of O atoms,
Ni : O = 0.98 : 1.00 = 98 : 100

Now, total charge on 100 -ions = 

= -200

Let the number of ions be x.

So, the number ofions is 98 –x.

Now, total charge on ions = x(+2)

= +2x

And, total charge on ions = (98 –x)(+3)

= 294 –3x

Since, the compound is neutral, we can write:

2x+ (294 –3x) + (– 200) = 0

⇒–x+ 94 = 0

⇒x= 94

Therefore, number of ions = 94

And, number of ions = 98 – 94 = 4

Hence, fraction of nickel that exists as =9498

= 0.959

And, fraction of nickel that exists as =498

= 0.041

Alternatively, fraction of nickel that exists as = 1 – 0.959

= 0.041

5. What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.

Ans. Semiconductors are substances having conductance in the intermediate range of 10^–6to10^4ohm^–1m^–1.

The two main types of semiconductors are:

(i) n-type semiconductor

(ii) p-type semiconductor

n-type semiconductor: The semiconductor whose increased conductivity is a result of negatively-charged electrons is called an n-type semiconductor. When the crystal of a group 14 element such as Si or Ge is doped with a group 15 element such as P or As, an n-type semiconductor is generated.

Si and Ge have four valence electrons each. In their crystals, each atom forms four covalent bonds. On the other hand, P and As contain five valence electrons each. When Si or Ge is doped with P or As, the latter occupies some of the lattice sites in the crystal. Four out of five electrons are used in the formation of four covalent bonds with four neighbouring Si or Ge atoms. The remaining fifth electron becomes delocalised and increases the conductivity of the doped Si or Ge.

p-type semiconductor: The semiconductor whose increased in conductivity is a result of electron hole is called a p-type semiconductor. When a crystal of group 14 elements such as Si or Geis doped with a group 13 element such as B, Al, or Ga (which contains only three valence electrons), a p-type of semiconductor is generated.

When a crystal of Si is doped with B, the three electrons of B are used in the formation of three covalent bonds and an electron hole is created. An electron from the neighboring atom can come and fill this electron hole, but in doing so, it would leave an electron hole at its original position. The process appears as if the electron hole has moved in the direction opposite to that of the electron that filled it. Therefore, when an electric field is applied, electrons will move toward the positively-charged plate through electron holes. However, it will appear as if the electron holes are positively-charged and are moving toward the negatively- charged plate.

6.  Explain the following terms with suitable examples:

(i) Schottky defect

(ii) Frenkel defect

(iii) Interstitials and

(iv) F-centres

Ans. (i) Schottky defect: Schottky defect is basically a vacancy defect shown by ionic solids. In this defect, an equal number of cations and anions are missing to maintain electrical neutrality. It decreases the density of a substance. Significant number of Schottky defects is present in ionic solids. For example, in NaCl, there are approximately 106Schottky pairs per cm3at room temperature. Ionic substances containing similar-sized cations and anions show this type of defect. For example: NaCl, KCl, CsCl, AgBr, etc.

(ii) Frenkel defect: Ionic solids containing large differences in the sizes of ions show this type of defect. When the smaller ion (usually cation) is dislocated from its normal site to an interstitial site, Frenkel defect is created. It creates a vacancy defect as well as an interstitial defect. Frenkel defect is also known as dislocation defect. Ionic solids such as AgCl, AgBr, AgI, and ZnS show this type of defect.

(iii) Interstitials: Interstitial defect is shown by non-ionic solids. This type of defect is created when some constituent particles (atoms or molecules) occupy an interstitial site of the crystal. The density of a substance increases because of this defect.

(iv) F-centres:When the anionic sites of a crystal are occupied by unpaired electrons, the ionic sites are called F-centres. These unpaired electrons impart colour to the crystals. For example, when crystals of NaClare heated in an atmosphere of sodium vapour, the sodium atoms are deposited on the surface of the crystal. The Cl ions diffuse from the crystal to its surface and combine with Na atoms, forming NaCl. During this process, the Na atoms on the surface of the crystal lose electrons. These released electrons diffuse into the crystal and occupy the vacant anionic sites, creating F-centres.

7. Explain the following with suitable examples:

(i) Ferromagnetism

(ii) Paramagnetism

(iii) Ferrimagnetism

(iv) Antiferromagnetism

(v) 12-16 and 13-15 group compounds.

Ans. (i) Ferromagnetism: The substances that are strongly attracted by a magnetic field are called ferromagnetic substances. Ferromagnetic substances can be permanently magnetised even in the absence of a magnetic field. Some examples of ferromagnetic substances are iron, cobalt, nickel, gadolinium, and.

In solid state, the metal ions of ferromagnetic substances are grouped together into small regions called domains and each domain acts as a tiny magnet. In an un-magnetised piece of a ferromagnetic substance, the domains are randomly-oriented and so, their magnetic moments get cancelled. However, when the substance is placed in a magnetic field, all the domains get oriented in the direction of the magnetic field. As a result, a strong magnetic effect is produced. This ordering of domains persists even after the removal of the magnetic field. Thus, the ferromagnetic substance becomes a permanent magnet.

Schematic alignment of magnetic moments in ferromagnetic substances

(ii) Paramagnetism: The substances that are attracted by a magnetic field are called paramagnetic substances. Some examples of paramagnetic substances

are , and .

Paramagnetic substances get magnetised in a magnetic field in the same direction, but lose magnetism when the magnetic field is removed. To undergo paramagnetism, a substance must have one or more unpaired electrons. This is because the unpaired electrons are attracted by a magnetic field, thereby causing paramagnetism.

(iii) Ferrimagnetism: The substances in which the magnetic moments of the domains are aligned in parallel and anti-parallel directions, in unequal numbers, are said to have ferrimagnetism. Examples include (magnetite), ferrites such as and .

Ferrimagnetic substances are weakly attracted by a magnetic field as compared to ferromagnetic substances. On heating, these substances become paramagnetic.

Schematic alignment of magnetic moments in ferrimagnetic substances

(iv) Antiferromagnetism: Antiferromagnetic substanceshave domain structures similar to ferromagnetic substances, but are oppositely-oriented. The oppositely-oriented domains cancel out each other’s magnetic moments.

Schematic alignment of magnetic moments in antiferromagnetic substances

(v) 12-16 and 13-15 group compounds: The 12-16 group compounds are prepared by combining group 12 and group 16 elements and the 13-15 group compounds are prepared by combining group 13 and group15 elements. These compounds are prepared to stimulate average valence of four as in Ge or Si. Indium (III) antimonide (IrSb), aluminium phosphide (AlP), and gallium arsenide (GaAS) are typical compounds of groups 13-15. GaAs semiconductors have a very fast response time and have revolutionised the designing of semiconductor devices. Examples of group 12-16 compounds include zinc sulphide (ZnS), cadmium sulphide (CdS), cadmium selenide (CdSe), and mercury (II) telluride (HgTe). The bonds in these compounds are not perfectly covalent. The ionic character of the bonds depends on the electronegativities of the two elements.

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