CBSE Class 12 Chemistry Solid State Questions Answers

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This CBSE Class 12 Chemistry question bank on Chapter 1 Solid State covers 40 one mark questions, 10 two mark questions and 10 three mark questions. Each question has a full answer with working where needed, and a Teacher's Note that points out the common mistakes, for board exam practice.

CHAPTER 1: SOLID STATE

1 MARK QUESTIONS

1. What type of solids are electrical conductors, malleable and ductile? [1 Mark]

Answer: Metallic solids are electrical conductors, malleable and ductile.

Teacher's Note:
a) Free electrons move through the metal lattice, which is why metals conduct electricity.
b) Malleability and ductility come from layers of metal ions that can slide over each other.

2. Give the significance of lattice point. [1 Mark]

Answer: A lattice point shows the position of a constituent particle (atom, ion or molecule) in the crystal lattice.

Teacher's Note:
a) Every lattice point in a crystal is occupied by one constituent particle.
b) Lattice points are not the same as the particles themselves, which are only represented by the points.

3. Name the parameters that characterize the unit cell. [1 Mark]

Answer: The parameters are the three edge lengths a, b and c, and the three angles alpha (α), beta (β) and gamma (γ) between the edges.

Teacher's Note:
a) These six parameters are needed to describe the shape and size of a unit cell.
b) They are the basis for the seven crystal systems.

4. What is the two dimensional coordination number of a molecule in square close packed structure. [1 Mark]

Answer: The two dimensional coordination number is 4.

Teacher's Note:
a) In square close packing, each sphere touches four neighbours in the same layer.
b) Do not confuse this with the three dimensional coordination number of 12 in close packed structures.

5. Which of the following lattice has the highest packing efficiency, (I) simple cubic (II) BCC (III) HCP [1 Mark]

Answer: (III) HCP

Teacher's Note:
a) HCP and ccp both have a packing efficiency of 74 per cent.
b) Simple cubic has about 52 per cent and BCC about 68 per cent, so both are lower.

6. What type of stoichiometric defect is shown by (i) ZnS (ii) AgBr [1 Mark]

Answer:
(i) ZnS shows Frenkel defect.
(ii) AgBr shows both Schottky and Frenkel defects.

Teacher's Note:
a) Frenkel defect is common when the cation is small, as in ZnS and AgBr.
b) Schottky defect appears in ionic solids with cations and anions of similar size.

7. What type of substances makes better permanent magnets? Justify your answer. [1 Mark]

Answer: Ferromagnetic substances make better permanent magnets. In these substances the domains are aligned in the same direction, so they stay strongly magnetised even after the external magnetic field is removed.

Teacher's Note:
a) Iron, cobalt and nickel are common ferromagnetic substances.
b) Paramagnetic substances lose their magnetism as soon as the field is removed.

8. Why glass is considered as super cooled liquid? [1 Mark]

Answer: Glass has no long range order and no sharp melting point, and it flows slowly under the influence of gravity. Hence, it is regarded as a super cooled liquid, or a pseudo solid.

Teacher's Note:
a) Glass is an amorphous solid, so its particles are arranged randomly.
b) A true solid would keep its shape, while glass slowly changes shape over time.

9. Why KCl appears pink when heated in K vapors? [1 Mark]

Answer: When KCl is heated in potassium vapour, electrons are trapped at the anionic vacancies. These trapped electrons are called F-centres, and they absorb visible light, so the crystal appears pink.

Teacher's Note:
a) F-centres are formed when the metal atoms add extra electrons to the crystal.
b) The colour is a result of metal excess defect.

10. Give the relationship between density and edge length in an unit cell. [1 Mark]

Answer: Density is given by d = Z × M / (a³ × N_A), where Z is the number of formula units per unit cell, M is the molar mass, a is the edge length and N_A is Avogadro's number.

Teacher's Note:
a) The edge length must be in cm to get density in g/cm³.
b) Density is inversely proportional to the cube of the edge length.

11. What is F-centre? [1 Mark]

Answer: An F-centre is a vacant anionic site in an ionic crystal that is occupied by an electron.

Teacher's Note:
a) The word F comes from the German word Farbe, meaning colour.
b) F-centres give colour to the crystal by absorbing visible light.

12. Give the coordination number of fcc structure. [1 Mark]

Answer: The coordination number of fcc structure is 12.

Teacher's Note:
a) Each atom in fcc touches 12 nearest neighbours.
b) The same value of 12 applies to hcp.

13. Give an example of a solid that shows both schottky and frenkel defect. [1 Mark]

Answer: AgBr shows both Schottky and Frenkel defects.

Teacher's Note:
a) In AgBr the silver ion is small enough to move into an interstitial site.
b) Both defects together keep the crystal electrically neutral.

14. What defect is observed when CdCl₂ is doped with AgCl? [1 Mark]

Answer: An impurity defect is observed.

Teacher's Note:
a) Doping introduces foreign ions that do not fit the host lattice exactly.
b) When a divalent ion such as Cd²⁺ replaces an Ag⁺ ion, a cation vacancy is created to keep the crystal neutral.

15. Give an example of a network solid. [1 Mark]

Answer: Silicon carbide (SiC) is an example of a network solid. Diamond and quartz (SiO₂) are other examples.

Teacher's Note:
a) Network solids have covalent bonds that extend throughout the whole crystal.
b) They are generally hard and have very high melting points.

16. Give an example each of 13-15 and 12-16 compounds. [1 Mark]

Answer: AlP is an example of a 13-15 compound, and ZnS is an example of a 12-16 compound.

Teacher's Note:
a) Gallium arsenide (GaAs) is another 13-15 compound used in electronics.
b) Cadmium sulphide (CdS) is another 12-16 compound.

17. How is quartz different from quartz glass? [1 Mark]

Answer: Quartz has a long range order of its constituent particles, so it is crystalline. Quartz glass has only a short range order, so it is amorphous.

Teacher's Note:
a) Both have the same chemical composition, SiO₂.
b) Quartz has a sharp melting point, but quartz glass softens over a range of temperatures.

18. What is a semiconductor? [1 Mark]

Answer: A semiconductor is a solid whose electrical conductivity lies between that of conductors and insulators, that is between 10^-6 and 10⁴ ohm^-1 m^-1. It has a small energy gap between the valence band and the conduction band.

Teacher's Note:
a) Silicon and germanium are the common elemental semiconductors.
b) The conductivity of a semiconductor increases with rise in temperature.

19. FeO is mostly found with a composition of Fe_0.95O_1.00. Why? [1 Mark]

Answer: FeO shows metal deficiency defect. Some Fe²⁺ ions are missing from their lattice sites, and some of the remaining iron is present as Fe³⁺ to keep the crystal neutral.

Teacher's Note:
a) Metal deficiency defect is common for transition metal oxides.
b) Because of this defect, the composition is slightly less than FeO exactly.

20. What is a n-type semiconductor? [1 Mark]

Answer: An n-type semiconductor is formed when a group 14 element such as silicon is doped with a group 15 element such as phosphorus. The extra electron of the group 15 element becomes a free electron that carries the current.

Teacher's Note:
a) The n in n-type stands for negative, which refers to the electrons.
b) Extra electrons increase the conductivity of the semiconductor.

21. What is photovoltaic cell? [1 Mark]

Answer: A photovoltaic cell is a device that converts sunlight into electrical energy.

Teacher's Note:
a) Solar cells are the common example of photovoltaic cells.
b) They work on the principle of a p-n junction semiconductor.

22. What do you mean by the term Doping? [1 Mark]

Answer: Doping is the introduction of a small amount of foreign particles (impurity) into a pure solid to change its electrical properties, by creating defects in the crystal.

Teacher's Note:
a) Doping is done in very small amounts, usually in parts per million.
b) Doped silicon is the basis of modern electronic devices.

23. What the term 12-16 compounds stand for? [1 Mark]

Answer: 12-16 compounds are formed by combination of group 12 elements (such as zinc or cadmium) with group 16 elements (such as sulphur or selenium). For example, ZnS.

Teacher's Note:
a) The numbers refer to the groups of the periodic table.
b) ZnS is an important phosphor used in screens.

24. What is the number of tetrahedral voids in a ccp structure of solid? [1 Mark]

Answer: A ccp structure with N atoms has 2N tetrahedral voids. For a unit cell of ccp, which has 4 atoms, there are 8 tetrahedral voids.

Teacher's Note:
a) Each atom in a close packed structure creates one octahedral void and two tetrahedral voids.
b) Octahedral voids are equal in number to the atoms, so they are N in number.

25. Crystalline solids are anisotropic in nature. Comment on this statement. [1 Mark]

Answer: Crystalline solids are anisotropic because their particles are arranged in a regular order, so a physical property such as refractive index or electrical conductivity differs in different directions.

Teacher's Note:
a) Isotropic substances, like glass, show the same property in all directions.
b) Graphite conducts electricity much better along its layers than across them.

26. What is the percentage efficiency of a solid whose particles are having ccp structure? [1 Mark]

Answer: The packing efficiency of a ccp structure is 74 per cent.

Teacher's Note:
a) The 26 per cent unoccupied space is the empty space in the structure.
b) hcp also has a packing efficiency of 74 per cent.

27. LiCl crystal appears to be pink in colour. Why? [1 Mark]

Answer: The pink colour is due to the presence of F-centres. Extra electrons are trapped in the anionic vacancies of the crystal, which give it colour.

Teacher's Note:
a) Pure LiCl is colourless, so the colour comes only from the defect.
b) This is the same type of defect that causes colour in KCl.

28. When Ge is doped with slight amount of In, what type of semiconductor will be formed? [1 Mark]

Answer: A p-type semiconductor is formed.

Teacher's Note:
a) Indium is a group 13 element with three valence electrons, one fewer than germanium.
b) The missing electron creates a hole, which acts as a positive charge carrier.

29. Metal deficiency defect with extra anions in interstitial sites can't be produced in solids. Why? [1 Mark]

Answer: Anions are usually larger in size than cations, so they cannot fit into the small interstitial sites.

Teacher's Note:
a) Metal deficiency defect involves missing cations, not extra anions.
b) Interstitial defects are shown mainly by small cations.

30. The electrical conductivity of semiconductors increases with increase in temperature but in case of metals conductivity decreases with rise in temp. Why? [1 Mark]

Answer:
1. In semiconductors, a rise in temperature gives electrons enough energy to cross the small band gap, so more charge carriers are available.
2. In metals, a rise in temperature makes the metal ions vibrate more strongly, which obstructs the flow of free electrons.

Teacher's Note:
a) Increased vibration of ions is the main reason for higher resistance in metals.
b) In semiconductors, the number of carriers matters more than the obstruction by vibrations.

31. The window panes seem to be thicker at the bottom in case of old buildings. What can be the reason? [1 Mark]

Answer: Glass is a super cooled liquid, so it flows very slowly under the influence of gravity. Over a long time, the glass has flowed downwards, making the bottom of the pane thicker.

Teacher's Note:
a) The flow is very slow and is noticed only over many years.
b) This is the reason given in textbooks, and it is related to the fact that glass has no sharp melting point.

32. Assign the relationship between edge length (a, b, c) and values of α, β and γ for tetragonal unit cell. [1 Mark]

Answer: For a tetragonal unit cell, a = b ≠ c and α = β = γ = 90°.

Teacher's Note:
a) Two edges are equal and the third is different in a tetragonal cell.
b) All the angles are right angles, as in a cube.

33. Give an example of solid which can exhibit schottky defect as well as frenkel defect. [1 Mark]

Answer: AgBr can exhibit both Schottky and Frenkel defects.

Teacher's Note:
a) This is the same example as in question 13.
b) Silver halides are the common examples of this behaviour.

34. Atoms of element A are occupying all the octahedral voids while atoms of B are in 1/4th tetrahedral voids and atoms of C are arranged in hexagonal arrays. What will be the formula of the compound? [1 Mark]

Answer:
1. Let there be N atoms of C, arranged in hexagonal close packing.
2. Octahedral voids = N, so there are N atoms of A.
3. Tetrahedral voids = 2N, so 1/4th of them is N/2, and there are N/2 atoms of B.
4. The ratio A : B : C = N : N/2 : N = 2 : 1 : 2, so the formula is A₂BC₂.

Teacher's Note:
a) Always count octahedral voids as equal to the number of atoms.
b) Tetrahedral voids are twice the number of atoms.

35. What is the co-ordination number of a particle in ccp structure? [1 Mark]

Answer: The coordination number of a particle in ccp structure is 12.

Teacher's Note:
a) The coordination number of ccp is the same as that of fcc, because they are the same structure.
b) Each particle has 6 neighbours in its own layer and 3 in each layer above and below.

36. What happens when Na vapour is passed over crystalline NaCl? [1 Mark]

Answer: The crystal turns yellow. Sodium atoms from the vapour are deposited on the surface, and the electrons released by them move into the anionic vacancies, forming F-centres.

Teacher's Note:
a) The yellow colour is due to the F-centres, which absorb light from the visible region.
b) This is an example of metal excess defect.

37. Ferromagnetic and ferri magnetic solids on heating converted into paramagnetic in nature. Justify the reason. [1 Mark]

Answer: On heating, the thermal energy randomises the alignment of the spins or domains. Above the Curie temperature, the magnetic domains are destroyed, so the solids behave as paramagnetic substances.

Teacher's Note:
a) The temperature at which this change happens is called the Curie temperature.
b) Paramagnetic substances are only weakly attracted by a magnetic field.

38. Glass is assumed to be super cooled liquid. Why? [1 Mark]

Answer: Glass has no sharp melting point and it has a tendency to flow under the influence of gravity, which is the behaviour of a liquid. Its particles are randomly arranged, like those in a liquid.

Teacher's Note:
a) A true solid has a sharp melting point and a regular arrangement of particles.
b) This is the same reasoning as in question 8.

39. Give an example of solid which has its structure as fluorite type structure. [1 Mark]

Answer: Calcium fluoride (CaF₂) has the fluorite type structure.

Teacher's Note:
a) In CaF₂, the Ca²⁺ ions form a ccp lattice and F^- ions occupy all the tetrahedral voids.
b) Be careful about the ratio of ions, which is 1 : 2 in CaF₂.

40. Copper is conducting as such while copper sulphate is conducting only in aqueous solution. Why? [1 Mark]

Answer: Copper is an electronic conductor because it has free electrons. Copper sulphate is an ionic conductor, and its ions are free to move only in the molten state or in aqueous solution.

Teacher's Note:
a) Solid copper sulphate has ions fixed in the lattice, so it does not conduct.
b) The conduction of ions is by movement of the ions themselves, while metals conduct by movement of electrons.

2 MARKS QUESTIONS

1. The edge length of a unit cell having molecular weight 75 g/mol is 5 Å which crystallizes in bcc lattice. If the density is 2 g/cc then find the radius of the metal atom. [2 Marks]

Answer:
1. For bcc, Z = 2. Using d = Z × M / (a³ × N_A), a³ = 2 × 75 / (2 × 6.022 × 10²³) cm³ = 1.245 × 10^-22 cm³, so a = 5 × 10^-8 cm.
2. In bcc, the atoms touch along the body diagonal, so √3 a = 4r.
3. r = √3 × 5 Å / 4 = 2.165 Å = 216.5 pm.

Teacher's Note:
a) In bcc the body diagonal equals four times the atomic radius.
b) Convert Å to pm by multiplying by 100 to get 216.5 pm.
c) Check the edge length by substituting it back into the density formula, which should give 2 g/cc.

2. Potssium crystallizes in bcc lattice. Calculate the approximate number of unit cell in 1 gm of potassium. (atomic mass of K = 39) [2 Marks]

Answer:
1. Moles of potassium in 1 g = 1 / 39 = 0.02564 mol.
2. Atoms in 1 g = 0.02564 × 6.022 × 10²³ = 1.544 × 10²².
3. A bcc unit cell has 2 atoms, so number of unit cells = 1.544 × 10²² / 2 = 7.72 × 10²¹.

Teacher's Note:
a) Divide the number of atoms by the number of atoms per unit cell, which is 2 for bcc.
b) Keep the molar mass and Avogadro's number in the working, so the steps can be followed.

3. A compound is formed by two elements M & N. The element N forms ccp & atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound? [2 Marks]

Answer:
1. N atoms form ccp, so there are N atoms of N and 2N tetrahedral voids.
2. M occupies 1/3rd of the tetrahedral voids, so number of M atoms = 1/3 × 2N = 2N / 3.
3. Ratio M : N = 2N/3 : N = 2 : 3, so the formula is M₂N₃.

Teacher's Note:
a) Tetrahedral voids are always twice the number of atoms in ccp.
b) Write the formula with the smallest whole number ratio.

4. A unit cell consists of a cube in which there are atoms A at the corners and atoms B at the face centers. Two A atoms are missing from the two corners of a unit cell. What is the formula of the compound? [2 Marks]

Answer:
1. Atoms A at the remaining corners: (8 - 2) × 1/8 = 0.75.
2. Atoms B at face centres: 6 × 1/2 = 3.
3. Ratio A : B = 0.75 : 3 = 1 : 4, so the formula is AB₄.

Teacher's Note:
a) A corner atom is shared by eight unit cells, so it contributes 1/8.
b) A face centre atom is shared by two unit cells, so it contributes 1/2.

5. Analysis shows that nickel oxide has formula Ni_0.98O_1.00. What fraction of nickel exist as Ni²⁺ and Ni³⁺. [2 Marks]

Answer:
1. Let the number of Ni³⁺ ions be x. Then the number of Ni²⁺ ions is 0.98 - x.
2. Charge balance: 2(0.98 - x) + 3x = 2, so x = 0.04.
3. Ni³⁺ = 0.04 / 0.98 = 0.041, which is about 4 per cent; Ni²⁺ = 0.94 / 0.98 = 0.959, which is about 96 per cent.

Teacher's Note:
a) The total charge on the oxide must be zero, so the oxygen charge of minus 2 is balanced by the nickel ions.
b) The fractions must add up to 1 when expressed as parts of the total nickel.

6. If NaCl is doped with 10^-3 mol % of SrCl₂. What is the concentration of cation vacancy? [2 Marks]

Answer:
1. 10^-3 mol % means 10^-3 mol of SrCl₂ in 100 mol of NaCl, which is 10^-5 mol of SrCl₂ per mole of NaCl.
2. Each Sr²⁺ ion creates one cation vacancy, so the concentration of cation vacancies is 10^-5 mol per mole of NaCl.
3. Number of cation vacancies per mole of NaCl = 10^-5 × 6.022 × 10²³ = 6.022 × 10¹⁸.

Teacher's Note:
a) Each divalent cation added in place of a monovalent cation creates one vacancy to keep the crystal neutral.
b) Be careful to convert mol % into a mole fraction before multiplying.

7. Under what conditions will sodium chloride conduct electricity? [2 Marks]

Answer: Sodium chloride conducts electricity when it is in the molten state or when it is dissolved in water (aqueous solution). In both states its ions are free to move and carry the current.

Teacher's Note:
a) In the solid state, the ions are held in fixed positions in the lattice.
b) Electrolysis of molten NaCl is the basis for the industrial production of sodium.

8. Name the binding force in each of the following. (a) Molecular (b) Ionic (c) Covalent (d) Metallic. [2 Marks]

Answer:
(a) Molecular solids are held by van der Waals forces.
(b) Ionic solids are held by electrostatic forces of attraction between ions.
(c) Covalent solids are held by covalent bonds, formed by sharing of electrons.
(d) Metallic solids are held by metallic bonds, in which electrons are delocalised over the lattice.

Teacher's Note:
a) Van der Waals forces are the weakest of these forces.
b) Covalent and ionic forces are strong, which explains the high melting points of these solids.

9. Differentiate between anisotropy & isotropy by giving examples. [2 Marks]

Answer:
1. Anisotropy is the property of a substance in which physical properties differ in different directions. Example: crystals such as graphite and mica.
2. Isotropy is the property of a substance in which physical properties are the same in all directions. Example: glass and other amorphous solids.

Teacher's Note:
a) Crystalline solids are anisotropic, while amorphous solids are isotropic.
b) Refractive index and electrical conductivity are good examples of properties that can be checked for these differences.

10. Derive the relationship between edge length and radius of atom in fcc unit cell. [2 Marks]

Answer:
1. In fcc, atoms touch along the face diagonal, so the face diagonal = 4r.
2. For a cube of edge a, face diagonal = √2 a.
3. So √2 a = 4r, which gives a = 4r / √2 = 2√2 r.

Teacher's Note:
a) Use Pythagoras theorem on the face of the cube to get the face diagonal.
b) The relationship a = 2√2 r is used in calculating the packing efficiency of fcc.

3 MARKS QUESTIONS

1. Calcium metal crystalises in a fcc lattice with edge length of 0.556 nm calculate the density of the metal if it contains, (i) 0.5% frenkel defect (ii) 0.2% schottky defect. [3 Marks]

Answer:
1. For fcc, Z = 4, M = 40 g/mol, a = 0.556 nm = 5.56 × 10^-8 cm.
2. Ideal density d = Z × M / (a³ × N_A) = 4 × 40 / ((5.56 × 10^-8)³ × 6.022 × 10²³) = 1.5458 g/cm³.
3. (i) A Frenkel defect does not change the number of atoms or the mass, so the density remains 1.5458 g/cm³.
4. (ii) A 0.2% Schottky defect removes 0.2% of the atoms (mass) without changing the volume, so d = 1.5458 × (1 - 0.002) = 1.5427 g/cm³.

Teacher's Note:
a) A Frenkel defect moves an ion but does not remove it from the crystal.
b) Schottky defect lowers the density because mass is lost while the unit cell volume stays the same.
c) Keep the edge length in cm, because the density is in g/cm³.
d) The results show that a small Schottky defect causes only a small fall in density.

2. How is ferromagnetism different from paramagnetism & antiferromagnetism & explain what type substances show antiferromagnetism. [3 Marks]

Answer:
1. Ferromagnetic substances are strongly attracted by a magnetic field and keep their magnetism even after the field is removed.
2. Paramagnetic substances are weakly attracted by a magnetic field, and lose their magnetism when the field is removed.
3. Antiferromagnetic substances have spins aligned in opposite directions, which cancel each other, so they are not attracted. Substances such as MnO show antiferromagnetism.

Teacher's Note:
a) Ferromagnetism is due to alignment of domains in the same direction.
b) In antiferromagnetic substances the net magnetic moment is zero.
c) Paramagnetism comes from unpaired electrons.

3. What is electrical conductivity due to in (i) metals (ii) ionic solids (iii) semiconductors. [3 Marks]

Answer:
1. Metals: due to the flow of free electrons.
2. Ionic solids: due to the flow of ions in the molten state or in solution, and due to the movement of ions through defects in the solid.
3. Semiconductors: due to the movement of electrons in the conduction band and holes in the valence band.

Teacher's Note:
a) Metals conduct well even in the solid state because electrons are free to move.
b) Ionic solids do not conduct in the solid state because their ions are fixed.
c) Temperature affects semiconductors differently from metals.

4. What is the difference between schottky defect & frenkel defect. [3 Marks]

Answer:
1. Schottky defect: equal numbers of cations and anions are missing from their lattice sites, which creates vacancies. Density of the solid decreases.
2. Frenkel defect: a smaller ion (usually cation) leaves its normal site and occupies an interstitial site. Density of the solid is not affected.
3. Schottky defect is shown by ionic solids with cations and anions of similar size, and Frenkel defect is shown by ionic solids with a large difference in size.

Teacher's Note:
a) Schottky defect is a vacancy defect and Frenkel defect is an interstitial defect.
b) Frenkel defect does not change the density because no ion is removed.

5. Derive an expression for the calculation of density of the cubic crystal of an element whose edge is "a" pm & atomic mass is M. [3 Marks]

Answer:
1. Let Z be the number of atoms per unit cell. Mass of one unit cell = Z × M / N_A.
2. Volume of one unit cell = a³ pm³ = (a × 10^-10)³ cm³ = a³ × 10^-30 cm³.
3. Density d = mass / volume = \( \frac{Z \times M}{N_A \times a^{3} \times 10^{-30}} \) g/cm³, where a is in pm.

Teacher's Note:
a) Convert pm to cm before calculating the volume, otherwise the answer will be wrong by a large factor.
b) For the same expression with a in cm, the 10^-30 term is dropped.
c) Z is 1 for simple cubic, 2 for bcc and 4 for fcc.

6. How would account for the following. [3 Marks]

a. Frenkel defects are not found in alkali metal halides.

Answer: Alkali metal halides have cations and anions of nearly similar size, so a cation cannot easily leave its site and fit into an interstitial void. Such crystals show Schottky defects instead.

Teacher's Note:
a) Frenkel defect needs a large difference in size between the cation and the anion.
b) Alkali halides are the classic examples of Schottky defect.

b. Schottky defects lower the density of related solids.

Answer: In Schottky defects, equal numbers of cations and anions are missing. The mass of the crystal decreases, but its volume remains the same, so the density falls.

Teacher's Note:
a) Density equals mass divided by volume, so a loss of mass lowers density.
b) This is the reason a Schottky defect can be detected by measuring density.

c. Impurity dopped silicon is a semiconductor.

Answer: Silicon is a group 14 element. When it is doped with a group 13 element, such as aluminium, a hole is created. When it is doped with a group 15 element, an extra electron is provided. Both create charge carriers, which make doped silicon a semiconductor.

Teacher's Note:
a) Group 13 doping gives a p-type semiconductor.
b) Group 15 doping gives an n-type semiconductor.
c) Only a very small amount of dopant is needed.

7. Define the following terms in relation to crystaline solids. [3 Marks]

a. Unit cell

Answer: A unit cell is the smallest portion of a crystal lattice which, when repeated in all directions, gives the entire lattice.

Teacher's Note:
a) Unit cells can be primitive or centred.
b) The unit cell is characterised by its six parameters.

b. co ordination number

Answer: The coordination number is the number of nearest neighbours of a particle in a crystal lattice.

Teacher's Note:
a) For a cation in NaCl, the coordination number is 6.
b) For ccp and hcp structures, the coordination number is 12.

c. P-type semiconductor

Answer: A p-type semiconductor is formed when a group 14 element such as silicon is doped with a group 13 element such as indium or aluminium. The holes created by the missing electrons act as positive charge carriers.

Teacher's Note:
a) The p stands for positive, which refers to the holes.
b) Boron and aluminium are common p-type dopants for silicon.

8. What type of defect can arise when solid is heated? Which physical property is affected by it & in what way? [3 Marks]

Answer:
1. Schottky defect can arise when a solid is heated, because some ions leave their lattice sites and create vacancies.
2. The density of the solid is affected, and it decreases because mass is lost while the volume remains the same.

Teacher's Note:
a) Frenkel defect can also be produced on heating, but it does not change density.
b) The electrical conductivity of the solid may increase because the vacancies help ions move.

9. An element has bcc structure with cell edge of 288 pm. The density of the element is 7.2 gm/cm³. How many atoms are present in 208 gm of the element? [3 Marks]

Answer:
1. Molar mass from d = Z × M / (a³ × N_A): M = d × a³ × N_A / Z = 7.2 × (2.88 × 10^-8)³ × 6.022 × 10²³ / 2 = 51.8 g/mol.
2. Moles in 208 g = 208 / 51.8 = 4.015 mol.
3. Number of atoms = 4.015 × 6.022 × 10²³ = 2.42 × 10²⁴ atoms.

Teacher's Note:
a) Convert 288 pm to 2.88 × 10^-8 cm before using the formula.
b) Z = 2 for bcc, which must be used in the calculation of the molar mass.
c) The answer is about 2.42 × 10²⁴ atoms, not 2.42 × 10²³.

10. (I) Solid A is very hard, electrical insulator in solid as well as molten state & melt at extremely high temprature what type of solid is it? (II) A compound forms hcp structure. What is the total no of voids in 0.5 mol of it? How many of these are tetrahedral voids? [3 Marks]

Answer:
(I) Solid A is a covalent (network) solid, because it is very hard, does not conduct electricity in any state and has a very high melting point.
(II) In hcp, the number of octahedral voids = N and tetrahedral voids = 2N, so total voids = 3N. For 0.5 mol of atoms (N = 0.5 mol), the total number of voids = 3 × 0.5 = 1.5 mol.
Number of tetrahedral voids = 2 × 0.5 = 1.0 mol.

Teacher's Note:
a) Network solids like diamond and silicon carbide have strong covalent bonds throughout.
b) Octahedral voids are equal in number to the atoms, and tetrahedral voids are twice that number.
c) In terms of particles, 1.5 mol of voids is 1.5 × 6.022 × 10²³ voids.