Chapter 1 The Solid State

From this chapter, you will be able to describe general characteristics of solid state. You will also be able to distinguish between amorphous and crystalline solids along with their classification on the basis of the nature of binding forces. Definition of crystal lattice and unit cell is also given. Close packing of particles is also explained. Description of different types of voids and close packed structures is also given. You will also be able to calculate the packing efficiency of different types of cubic unit cells. Correlation of the density of a substance with its unit cell properties is also given. Description of the imperfections in solids and their effect on properties is also given. Correlation of the electrical and magnetic properties of solids and their structures is also done.

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Exercise 1 ( Page No. : 29 )

  • Q1 Why are solids rigid?

    Solids have a definite volume & shape. They are rigid & lack the ability to flow.In solids, the atoms,ions & molecules are held together by relatively strong chemicalmforces-ionic bond, covalent bond, or by intermolecular Van der waals forces. They do not translate although they vibrate to some extent in their fixed positions.The interparticle spaces or voids are so small that their movement is almost negligible.

    Q2 Why do solids have a definite volume?

    Gases & liquids can flow & take up the shape of their container because they have weak intermolecular forces between them & the ion are not arranged in a regular method. As a result, when certain pressure or temperature is applied to them,their ions or molecules rearranges themselves & show flow properties. But in case of solids, the ions or molecules are arranged in a regular & repeated three dimensional pattern. So, whenever pressure or heat is applied to solids ,their ions or molecules do not rearranges themselves, and hence do not show flow properties & have a definite volume.

    Q3 Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.

    Solids are classified into 2 types:

    Crystalline solids

    Amorphous solids

    Crystalline solids-exists as small crystals,each crystal having a characterstic geometrical shape.The atoms,ions or molecules are arranged in a regular, repeating three dimensional pattern called the crystal lattice.

    Amorphous solids( Greek term amorphous-no form)- has atoms, molecules or ions arranged at random & lacks the ordered crystalline lattice.


    Naphthalene, benzoic acid, potassium nitrate, copper

    Polyurethane, teflon, cellophane, polyvinyl chloride, fibre glass


    Q4 Why is glass considered a super cooled liquid?

    Amorphous solids( Greek term amorphous-no form)- has atoms, molecules or ions arranged at random & lacks the ordered crystalline lattice. For eg glass, plastics. In their disordered structure,amorphous solids resemble liquids.Being a amorphous solid, glass has atoms arranged in random order with weak intermolecular forces, when ever heat is applied in gradual pattern the glass behave as a super cooled liquid or highly viscous liquids. The liquid nature of glass is sometimes apparent in very old window panes that have become slightly thicker at the bottom due to gradual downward flow.

    Q5 Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property?

    The solids are classified into two types:

    a) Crystalline /true solids

    b) Amorphous solids

    To distinguish between crystalline and amorphous solids, isotropy and anisotropy plays a vital role.

    Isotropy – The substances which have the same physical properties such as electrical conductivity, thermal conductivity, mechanical strength, refractive index etc. in all the directions in the space are called isotropic substances and this property is called as isotropy.

    Anisotropy – The substances which have physical properties such as electrical conductivity, thermal conductivity, mechanical strength, refractive index etc. different in different directions in the space are called anisotropic substances and the property is called as anisotropy.

    So a isotropic solid will have same refractive index in all the directions. Hence, amorphous solids are isotropic in nature. The reason for this is that in amorphous solids the arrangement of particles is random and disordered, therefore all directions are equivalent and properties are independent of direction .On the other hand, the particles in a crystalline solids are arranged & well ordered. Thus the arrangement of particles may be different in different directions.

     CLEAVAGE-A crystalline solid when cut with a sharp edged tool such as knife gives a clean cleavage but an amorphous solid undergoes an irregular cleavage i.e. when an amorphous solid is cut with a sharp edged tool, it cuts into two pieces with irregular surfaces.

    Q6 Classify the following solids in different categories based on the nature of intermolecular forces operating in them: Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide.

    Potassium sulphate → Ionic bond is formed, wherein potassium loses electrons & sulphate gains electrons.

    Tin → Tin can form two ions Sn 2+ or Sn 4+ now depending on what form it is in and what other element it is bonding with then only the type of bonding will be known.

    Benzene → It comprises of sp2 hybridized carbon atoms which makes covalent bond with each other by mutual sharing and pairing of electrons.

    Urea → Polar molecular solid and they mostly forms hydrogen bond with other atoms.

    Ammonia → Covalently bonded polar compound formed by mutual sharing and pairing of electrons between nitrogen and three hydrogen atoms.

    Water → Covalently bonded polar compound formed by mutual sharing of electrons between oxygen and hydrogen atoms.

    Zinc sulphide → It exist in the form of crystals and makes coordination bond.

    Graphite → It consist of carbon atoms and have covalent bond between them forming a 2-D (flat, planar etc.) network, which in itself is strong. These planes are then held together by London Forces, which are very weak.

    Rubidium → Metallic bond is formed by rubidium because it is a metal. Metallic bond is hydride of covalent bond and ionic bond.

    Argon → Non-polar molecular solid

    Silicon carbide → Covalent or network solid

    Q7 Solid A is a very hard electrical insulator in solid as well as in molten state and melts at extremely high temperature. What type of solid is it?

    The solids which are very hard have their atoms bind to each other by covalent forces which are very strong, as a result they have high melting and boiling points and do not conduct heat are called as covalent crystals (crystalline solids). Hence, solid A shows the properties of an covalent crystalline solids.

    For example  diamond (C), quartz (SiO2), &  silicon carbide SiC)

    Q8 Ionic solids conduct electricity in molten state but not in solid state. Explain.

    In an ionic solids the lattice is made of positive and negative ions. These are held together by ionic bonds-the strong electrostatic attractions between oppositely charged ions. Consequently, the cations and anions attract each other and pack together in an arrangement so that the attractive forces maximize. The ionic solids are insulators in the solid state because the ions are entrapped in fixed places in the crystal lattice and cannot move when electric field is applied. However, in molten state, they become good conductors of electricity. This is due to the fact that in molten state, the well- ordered arrangement of ions in the solids is destroyed and the ions are in a position to move about in the liquid medium when an electric field is applied. For example NaCl.

    Q9 What type of solids are electrical conductors, malleable and ductile?

    Metallic solids are electrical conductors, malleable, and ductile. They consists of an assemblage of positive ions (kernels) immersed in a ‘sea’ of mobile electrons. Thus, each electron belongs to a number of positive ions and each positive ion belongs to a number of electrons. The force that binds a metal ion to a number of electrons within its sphere of influence is known as metallic bond. This force of attraction is sufficiently strong and is thus responsible for compact solid structure of metals and therefore these solids are good conductors, malleable and ductile in nature. For example metals like copper, nickel etc.

    Q10 Give the significance of a lattice point.

    In a crystalline solid the constituent particles (atoms, ions) are arranged in a definite and regular order in space and the relative positions of such particles in a crystal are shown by points (.) The arrangement of these points in a crystal is called space lattice. The location of the points in the space lattice are called lattice points or lattice sites. These points are linked by lines to depict the picture of a space lattice. These lattice points arranges in repeated pattern in different directions to form the complete lattice known as the unit cell. These unit cells are of different types and constitutes of different solids.

    Q11 Name the parameters that characterize a unit cell.


    A unit cell is the smallest repeating unit in space lattice which when repeated over and over again produces the complete space lattice. It represents the shape of entire crystal. The parameters which characterizes a unit cell are as follows:

    1 Relative lengths (distance) of the edges along the three axis (a, b, c). These edges may or may not be equal.

    2 Angle between the edges (α, β, γ).The angle α is between the edges b & c, angle β is between the edges c & a, and angle γ is between the edges a & b.

    Q12 Distinguish between (i)Hexagonal and monoclinic unit cells (ii) Face-centred and end-centred unit cells.

    (i) Hexagonal unit cell

    For a hexagonal unit cell,


    Monoclinic unit cell

    For a monoclinic cell,


    (ii) Face-centred unit cell

    In a face-centred unit cell, the constituent particles are present at the corners and one at the centre of each face.

    End-centred unit cell

    Anend-centred unit cell contains particles at the corners and one at the centre of any two opposite faces.

    Q13 Explain how much portion of an atom located at (i) corner and (ii) body-centre of a cubic unit cell is part of its neighbouring unit cell.

    (i) An atom located at the corner of a cubic unit cell is shared by eight adjacent unit cells.

    Therefore, 1/8 portion of the atom is shared by one unit cell.

    (ii) An atom located at the body centre of a cubic unit cell is not shared by its neighbouring unit cell. Therefore, the atom belongs only to the unit cell in which it is present i.e., its contribution to the unit cell is 1.

    Q14 What is the two dimensional coordination number of a molecule in square close packed layer?

    Square Close Packing: In this type, the spheres are packed in such a way that the centres of all of them are in straight line. This means that the different rows have horizontal as well as vertical alignment. Since each sphere is in contact with four other spheres in the same plane, this packing is called square close packing.

    Coordination Number: The number of spheres which are touching a given sphere is called the coordination number. In square close-packed layer, a molecule is in contact with four of its neighbours. Therefore, the two dimensional coordination number of a molecule in square close-packed layer is 4.

    Q15 A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?

    Number of close-packed particles = 0.5 x 6.022 x 1023 = 3.011 x 1023

    Therefore, number of octahedral voids = 3.011 x 1023

    And, number of tetrahedral voids = 2 x 3.011 x 1023 = 6.022 x 1023

    Therefore, total number of voids = 3.011 x 1023 + 6.022 x 1023 = 9.033 x 1023

    Q16 A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound?

    The ccp lattice is formed by the atoms of the element N.

    Here, the number of tetrahedral voids generated is equal to twice the number of atoms of the element N.

    According to the question, the atoms of element M occupy 1/3rd of the tetrahedral voids.

    Therefore, the number of atoms of M is equal to 2x1/3 = 2/3rd of the number of atoms of N.

    Therefore, ratio of the number of atoms of M to that of N is M: N =2/3 :1 = 2 : 3

    Thus, the formula of the compound is M2N3.

    Q17 Which of the following lattices has the highest packing efficiency (i) simple cubic (ii) body-centred cubic and (iii) hexagonal close-packed lattice?

    Hexagonal close-packed lattice has the highest packing efficiency of 74%. The packing efficiencies of simple cubic and body-centred cubic lattices are 52.4% and 68% respectively.

    Q18 An element with molar mass 2.7 x 10-2kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 x 103 kg m-3, what is the nature of the cubic unit cell?

    It is given that density of the element, d = 2.7 ×103 kg m-3

    Molar mass, M = 2.7 ×10-2 kg mol-1

    Edge length, a= 405 pm = 405 ×10-12 m = 4.05 ×10-10 m

    It is known that, Avogadro's number, NA= 6.022 ×1023 mol-1

    Applying the relation, This implies that four atoms of the element are present per unit cell. Hence, the unit cell is face-centred cubic (fcc) or cubic close-packed (ccp).

    Q19 What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?

    When a solid is heated, vacancy defect can arise. A solid crystal is said to have vacancy defect when some of the lattice sites are vacant.

    Vacancy defect leads to a decrease in the density of the solid.

    Q20 What type of stoichiometric defect is shown by: (i) ZnS (ii) AgBr

    (i) ZnS shows Frenkel defect.

    (ii) AgBr shows Frenkel defect as well as Schottky defect.

    Q21 Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.

    When a cation of higher valence is added to an ionic solid as an impurity to it, the cation of higher valence replaces more than one cation of lower valence so as to keep the crystal electrically neutral. As a result, some sites become vacant. For example, when Sr2+ is added to NaCl, each Sr2+ ion replaces two Na+ ions. However, one Sr2+ ion occupies the site of one Na+ ion and the other site remains vacant. Hence, vacancies are introduced.

    Q22 Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.

    The colour develops because of the presence of electrons in the anionic sites. These electrons absorb energy from the visible part of radiation and gets excited.

    For example, when crystals of NaCl are heated in an atmosphere of sodium vapours, the sodium atoms get deposited on the surface of the crystal and the chloride ions from the crystal diffuse to the surface to form NaCl with the deposited Na atoms. During this process, the Na atoms on the surface lose electrons to form Na+ ions and the released electrons diffuse into the crystal to occupy the vacant anionic sites. These electrons get excited by absorbing energy from the visible light and impart yellow colour to the crystals.

    Q23 A group 14 element is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong?

    An n-type semiconductor conducts because of the presence of extra electrons. Therefore, a group 14 element can be converted to n-type semiconductor by doping it with a group 15 element.

    Q24 What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic. Justify your answer.

    Ferromagnetic substances would make better permanent magnets.

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

    The ordering of the domain persists even after the removal of the magnetic field. Thus, the ferromagnetic substance becomes a permanent magnet.

Exercise 2 ( Page No. : 32 )

  • Q1 Define the term 'amorphous'. Give a few examples of amorphous solids.

    The substance whose constituents are arranged in an irregular manner are called as amorphous solids. They lack shape or form unlike crystalline solids. These substances are incompressible and rigid. For example glass, rubber, plastic etc.

    Q2 What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?

    The first difference between them is that quartz contains almost 99% silicon dioxide while glass contains only 80% silicon dioxide.

    Secondly, glass has a irregular molecular structure while quartz has a symmetrical molecular structure.

    Thirdly, quartz can withstand high temperature and pressure, whereas glass is unstable at high temperature and pressure.

    Fourthly, glass act as an insulator and do not conduct electricity, while quartz conducts electricity.

    For the conversion of quartz into glass, the quartz has to be melted and cooled rapidly.

    Q3 Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous. (i) Tetra phosphorus decoxide (P4O10)           (vi) Plastic (ii) Ammonium phosphate (NH4)3PO4            (vii) Graphite (iii) SiC                                                        (viii) Brass   (iv) I2                                                           (ix) Rb (v) P4                                                          (x) LiBr                                             (xi) Si

    (i) Tetra phosphdecoxide (P4O10)  → Molecular

    (ii) Ammonium phosphate (NH4)3PO4  → ionic

    (iii) SiC → Network (Covalent)

    (iv) I2 → molecular

    (v) P4 → Molecular

    (vi) Plastic → Amorphous

    (vii) Graphite → Network (Covalent)

    (viii) Brass → Metallic

    (ix) Rb → Metallic

    (x) LiBR → Ionic

    (xi) Si → Network (Covalent)

    Q4 (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?

    (i) It is defined as the number of nearest neighbours of particle in a close packed structure.

    (ii) a) In a cubic close packed structure the coordination number is 12.

    b) In a body centered cubic close structure the coordination number is 8.

    Q5 How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.

    If we know the density and the dimenstions of unit cell of an unknown metal, the atomic mass of the metal can be determined easily.

    Let us assume a cubic crystal,then Its volume of unit cell will be = a3 


    No of atom in each cell = z

    Also molar mass of the metal =M,

    Mass of each atom = m

    Now we know, mass of unit cell = no of atom in each cell x Mass of each atom -------------------(1)

    Also Mass of each atom = molar mass / avagadro number ,i.e M/NA ------------(2)

    Putting the values of 2 in 1 we get,

    Mass of unit cell = Mz / NA ------------------(3)

    Now density of unit cell (d) = mass of unit cell / volume of unit cell

    Therefore from 1,2,3 we,get

    d = Mz / a3 NA---------------------------(4)

    From 4th equation we can calculate the atomic mass of unknown metal.

    Q6 'Stability of a crystal is reflected in themagnitude 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?

    In context with the solid state of a crystal, the stability of a crystal is dependent upon the packaging of atoms in it. The stronger the intermolecular forces of attraction present in the structure, the higher will be the stability of the crystal and vice versa.

    Melting point of:

    Solid water→ 273 K → has strong intermolecular hydrogen bonding

    Ethyl alcohol → 155 K → has intermolecular hydrogen bonding

    Diethyl Ether→156 K → has dipole-dipole interaction

    Methane → 91 K → has weak Vander waals forces

    Q7 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?
    S No Hexagonal close packing Cubic close packing
    1 It is 2dimensional packing It is a 3 dimensional packing
    2 Here the sphere of 2nd row sit into the depressions in between the spheres of the 1st row

    Here the sphere in the third layer is placed over hollows voids of the 1st layer

    3 It is abbrebiated as hcp

    It is abbrebiated as ccp

    1. 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.

    Case 1: 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.











    S. No crystal lattice unit cell
    1 It a regular 3 dimensional arrangement of the points in a crystal It is smallest but complete unit in the space lattice which is when repeated over & again in 3 dimensions



    S. No Tetrahedral void Octahedral void
    1 It is a vacant space among four spheres having tetrahedral arrangement It is avoid formed by 2 equilateral triangles with apices in opposite directions
    2 There are 2 tetrahedral sites for each sphere There is only 1 octahedral site for each sphere




    Q8 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

    (i) No. of corner atoms per unit cell = 8 corner atoms x 1/8 atom per unit cell

                     = 8 x 1/8 = 1 atom

        No. of face centred atoms per unit cell = 6 face centred atoms x 1/2 atom per unit cell

                     = 6 x 1/2 = 3 atoms

        :. Total no. of atoms or labtice points = 1+3= 4 atoms.

    (ii) Face centered tetragonal

                 8 lattice points at corners and 6 lattice points at face centres

             :. 8 x 1/8 + 6 x 1/2 = 4


    (iii) In bcl unit cell, no. of corner atoms per unit cell 

                = 8 x 1/8 per corner atom

                = 1 atom

          No. of atoms at body centre = 1 atom

         :. Total no. of atoms (lattice points) = 1 + 1 = 2

    Q9 Explain (i) The basis of similarities and differences between metallic and ionic crystals. (ii) Ionic solids are hard and brittle.

    (i) Both metallic and ionic crystals has strong forces of attraction between their atoms or ions. In metallic crystals there is a strong metallic bond present between electrons and positively charged ions. In ionic crystals there is a strong ionic bonds between anions and cations. Both of them conduct electricity, but metallic crystals conduct electricity in all three states of matter while ionic crystals conduct electricity only in molten state.

    (ii) The ionic solids are hard and brittle because they have strong electrostatic forces of attraction between the anions and cations .As a result the anions and cations are tightly held by these forces and due to which they are unable to move and they are fixed at one position. So when force is applied to break them, they appear to be hard.

    Q10 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).

    (i) Simple cubic:

    Suppose the edge length of the unit cell = a


    Radius of the sphere = r

    Then,since the sphere are touching each other along the edge,therefore a = 2r

    Now there are 8 spheres at the corners of the cube & each sphere at the corner is shared by 8 unit cells & the contribution per unit cell is 1/8 so that

    Number of spheres per unit cell is 8 x 1/8 = 1

    Volume of sphere =4/3πr3 & volume of cube = a3 = (2r)3 = 8r3

    Now packing efficiency = (volume of one sphere / total volume of cubic unit cell) x 100


    (4/3 πr3 / 8r3) x 100 = 52.4%

    Therefore the volume occupied in simple cubic arrangement = 52.4%


    (ii) Body centered cubic:

    Let us suppose the edge leght = a & radius of each sphere = r then there are 8 spheres at the corners & 1 in the body of unit cell

    Therefore number of spheres per unit cell = (8 x1/8) + 1 = 2

    Now volume of unit cell = a3 = (4r / √3)3

    and volume of a sphere = 4 /r3

    Total volume of two spheres = 2 x 4/3πr3

    Packing efficiency = (volume of two spheres in unit cell/total volume of unit cell ) x 100

    = (2 x 4/3πr3 / (4r/√3)3 ) x 100 = 68%

    Therefore volume occupied in bcc arrangement = 68%


    (iii) Face centered:

    let us suppose the edge length of the unit cell = a

    Radius of each sphere = r

    Now there are 8 spheres at the corner & 6 at the faces

    Therefore number of spheres in unit cell = (8 x 1/8 + 6 x1/2) = 4

    From the arrangement of fcc, we get a = 2√2r

    Now volume of a unit cell = a3 = (2√2r)3 = 16√2r3

    Total volume of 4 spheres = 4 x 4/3 πr3 = 16/3 πr3

    Packing efficiency = (volume of 4 spheres in the unit cell/total volume of unit cell) x 100

    = (16/3 πr3 /16√2r3) x 100 = 74%

    Therefore volume occupied in fcc = 74%

    Q11 Silver crystallises in fcc lattice. If edge length of the cell is 4.07 x 10-8cm and density is 10.5 g cm-3, calculate the atomic mass of silver.

    From the equation, density = ZxM/a3 x N0, we get

    M = ρ x a3 x N0/Z …………………………………1

    Placing the values in equation 1,

    WE GET M= 10.5 x (4.07 x10-8)3 x 6.022 x 1023/4

    = 106.6 gmol-1

    = Therefore, atomic mass of silver = 106.6 gmol-1

    Q12 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?

    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 = 8 x 1/8 = 1

    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.

    Q13 Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm-3, calculate atomic radius of niobium using its atomic mass 93 u.

    Given Density = 8.55 gcm-3

    Let the length of the edge = a cm

    Number of atoms per unit cell, Z = 2

    Atomic mass M = 93 u

    From the formula , density = ZxM/a3 x N0, we get

    8.55 = 2 x 93/a3 x 6.022 x 1023 = 36.12 x 10-24 cm3

    Now edge length,a = (36.12 x 10-24)1/3 = 3.306 x 10-8 cm

    Now radius in body centered cubic, r = √3/4 a

    Putting the value of a, we get .r = 1.431 x 10-10 m

    Q14 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.

    Radius of octahedral void = r       (given)

    Radius of atom = R                    (given)

    Now by Pythagoras theorem,we have

    2R2 = (R+ r)2 + (R+ r)2


    4R2 = 2(R +r)2


    2R2 = (R+ r)2


    R√2 = R +r


    R√2 – R = r


    r = R√2 – R

    Putting the value of √2, we get .

    r = R(1.414-1)

    r = 0.414R

    Q15 Copper crystallises into a fcc lattice with edge length 3.61 x 10-8cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm-3.

    In fcc lattice Z= 4

    Atomic mass M = 63.5

    a = 3.61x10-8

    Therefore from the formula density = ZxM/a3 x N0

    We get density = 4x 63.5/3.61x10-8 x 6.022 x1023

    = 8.96 gcm-3.

    The value is approximately equal to 8.92.

    Q16 Analysis shows that nickel oxide has the formula Ni0.98 O1.00. What fractions of nickel exist as Ni2+and Ni3+ions?

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

    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 O2 - ions = 100 × ( - 2) = - 200

    Let the number of Ni2+ions be x.

    So, the number of Ni3+ions is 98 - x.

    Now, total charge on Ni2+ions = x(+2) = +2x

    And, total charge on Ni3+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 Ni2+ions = 94

    And, number of Ni3+ions = 98 - 94 = 4

    Hence, fraction of nickel that exists as Ni2+ = 94/98 = 0.959

    And, fraction of nickel that exists as Ni3+ = 4/98 = 0.041

    Alternatively, fraction of nickel that exists as Ni3+= 1 - 0.959 = 0.041

    Q17 What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.

    Semi- conductors are the solids that have properties intermediate between metals and insulators.They have only small difference in energy between the filled valence band and empty conduction band.The conductivity is also intermediate between that of a metal and an insulator and depends upon the number of electrons in the conduction band.

    The semiconductors behaves in two types as

    a) Intrinsic conduction

    b) Extrinsic conduction


    (a) Intrinsic conduction = pure silicon and germanium are poor conductors because they have network of four strong covalent bonds. But when electricity is provided to the crystals, the electrons moves leaving behind a positive charge or hole at the site of missing electron, as a result of which the crystal will now conduct electricity.

    (b) Extrinsic conduction = here the conductivity of intrinsic conductors is increased by adding an appropriate amount of suitable impurity. This process is called doping. The impurities are of 2 types:

    (1) Electron rich

    (2) Electron deficient


    (1) Electron rich impurity – here atoms with five valence electrons are added to the semiconductor, as a result of which four out of five electrons are used in formation of four covalent bonds with four neighbouring silicon atoms. The fifth electron is extra and becomes delocalised. These delocalised electrons increase the conductivity of doped silicon, hence increases the conductivity. This type of conductor is called n-type semiconductor.

    (2) Electron deficient impurity – here an atom with three valence electrons are added to the semiconductor, as a result of which silicon or germanium atom are replaced by the impurity atom. The valence electrons of impurity will make bond with three atom but fourth atom remains free in the crystal of silicon or germanium, which is now available for conducting electricity. This type of conductor is called p-type semi- conductor.

    Q18 Non-stoichiometric cuprous oxide, Cu2O 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?

    From the given data it is clear that the ratio of copper to oxygen is less than 2:1. Therefore, cuprous oxide is non stoichiometric crystal. This means that some Cu+ ions have been replaced by Cu2+ ions and to maintain electrical neutrality, every 2 Cu+ ions will be replaced by one Cu2+ ions, making a hole. Since the conduction will be due to the presence of these positive holes, the given substance is a p-type semi- conductor.

    Q19 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.

    There are one octahedral hole for each atom in hexagonal close packed arrangement. If the no. of oxide ions per unit cell = 1.

       then no. of fe3+ ions = 2/3 x octahedual holes 

                = 2/3 x 1 = 2/3

       formula of the compound = fe2/3 Oor fe2O3

    Q20 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.

    (i) Ge belongs to group 14 and In belongs to group 13, In has 3 valence electrons while Ge has 4 valence electron. On replacement an electron deficient hole was created, therefore it is a p-type semi- conductor.

    (ii) B belongs to group 13 with 3 valence electrons and Si belongs to group 14 with 4 valence electrons. Therefore on replacement there will be a free electron and it is n-type semiconductor.

    Q21 Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?

    We know, for a fcc unit cell,radius of atom

    r = a/2√2


    a = 2r√2

    By putting the values of r &√2 we get

    a = 2 x 0.144 x 1.414 = 0.407nm

    Q22 In terms of band theory, what is the difference (i) Between a conductor and an insulator (ii) Between a conductor and a semiconductor

    (i) In a conductor the energy gap is either very small or there is overlapping between valence band and conduction band. In case of an insulator the energy gap between valence band and conduction band is very large. Due to which, when an electric field is applied to an insulator, the electrons cannot jump from valence band to conduction band, hence poor or no conduction of electricity.

    (ii) In a conductor, the energy gap between valence band and conduction band is very small but in an semiconductor there is always a small energy gap between them.

    Q23 Explain the following terms with suitable examples: (i) Schottky defect (ii) Frenkel defect (iii) Interstitials and (iv) F-centres

    (i) Schottky defect – It was discovered by German scientist Schottky in 1930. It is basically a vacancy defect in ionic solids. In order to maintain electrical neutrality, the number of missing cations and anions are equal.

    (ii) Frenkel defect – It was discovered by a Russian scientist Frenkel in 1926. It arises when an ion is missing from its normal position and occupies an interstitial site between the lattice points. In this case the crystal remains electrically neutral because the number of cations and anions remains the same.

    (iii) Interstitials defect – When some constituent particles occupy an interstitial positions, the crystal is said to have a interstitial defect. Thus defect increases the density of the substance.

    (iv) F-centres – During an anion vacancy, the electrons trapped in anion vacancies are called as F centres. It is originated from a German word farbenzenter which means colour centre.

    Q24 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 cm3of aluminium?

    In a cubic close packed structure,the length of the side of unit cell is related to radius by an equation r = a/2√2


    a = r x 2√2

    Putting the values in the equation, we get

    a = 125 x 2 x 1.414 = 353.5 pm

    Volume of the unit cell= (353.5 x 10-10)3 = 4.42 x 10-23

    Number of unit cells in 1 cm3 = 1 / 4.42 x 10-23 = 2.26 x 1022 unit cells.


    Q25 If NaCl is doped with 10-3mol % of SrCl2, what is the concentration of cation vacancies?

    1 Cation of Sr2+ will create 1 cation vacancy in Nacl.Therefore the number of cation vacancies created in the lattice of Nacl is equal to the number of divalent Sr2+ ions added.

    Now the concentration of cation vacancy on being doped with 10-3 mol% of Srcl2.

    = 10-3 mol% = 10-3 / 100 = 10-5 mol\

    Also number of Sr2+ ion in 10-5 mol = 10-5 x 6.023 x 1023 = 6.023 x 1018

    Therefore number of cation vacancies = 6.023 x 1018

    Q26 Explain the following with suitable examples: (i) Ferromagnetism (ii)Paramagnetism (iii)Ferrimagnetism (iv)Antiferromagnetism (v)12-16 and 13-15 group compounds.

    (i) Ferromagnetism – The substances which are strongly attracted by the magnetic field and show permanent magnetism even when the magnetic field is removed are known as ferromagnetic substances. When there is spontaneous alignment of magnetic moments of domain in the same direction, ferromagnetism happens.

    (ii) Paramagnetism- The substances are weakly attracted by a magnetic field. They are magnetised in a magnetic field in the same direction. They lose their magnetism in the absence of magnetic field. Paramagnetism is due to presence of one or more unpaired electrons which are attracted by the magnetic field.

    (iii) Ferrimagnetism- When the magnetic moment of domains are aligned in parallel and antiparallel directions in unequal numbers resulting in net magnetic moment, ferrimagnetism happens.

    (iv) Antiferromagnetism- Substances showing antiferromagnetism have domain structure similar to ferromagnetic substance, but their domains are oppositely oriented and cancel out each other's magnetic moment.

    (v) 12-16 & 13-15 group - The solid binary compound prepared by combining elements of group 12 & 16 are called 12-16 compound. The compound prepared by combining elements of group 13 & 15 are called 13-15 compounds. These compounds are used as semiconductor.

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