NCERT Solutions For Class 12 Chemistry Chapter 1 The Solid State

1.1.Why are solids rigid?
Ans: The constituent particles in solids have fixed positions and can oscillate about their mean positions. Hence, they are rigid.

1.2.Why do solids have a definite volume?
Ans: The constituent particles of a solid have fixed positions and are not free to move about, i.e., they possess rigidity. That is why they have definite volume.

1.3.Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, Teflon, potassium nitrate, cellophane, polyvinyl chloride, fibreglass, copper
Ans: Crystalline solids: Benzoic acid, potassium nitrate, copper Amorphous solids: Polyurethane, Teflon, cellophane, polyvinyl chloride, fibreglass

1.4.Why is glass considered a supercooled liquid?
Ans: Glass is an amorphous solid. Like liquids, it has a tendency to flow, though very slowly. This can be seen from the glass panes of windows or doors of very old buildings which are thickerat the bottom than at the top. Therefore, glass is considered as a supercooled liquid.

1.5.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?
Ans: As the solid has same value of refractive index along all directions, it is isotropic in nature and hence amorphous. Being amorphous solid, it will not show a clean cleavage and when cut, it will break into pieces with irregular surfaces.

1.6.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
Ans: Potassium sulphate = Ionic Tin=Metallic.
Benzene = Molecular (non-polar)
Urea=Molecular (polar).
Ammonia=Molecular (H-bonded)
Water = Molecular (H-bonded)
Zinc sulphide = Ionic Graphite=Covalent Rubidipm Metallic Argon = Molecular (non-polar)
Silicon Carbide=Covalent

1.7.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?
Ans: It is a covalent or network solid.

1.8.Ionic solids conduct electricity in molten state but not in solid state. Explain
Ans: In solid state, the ions cannot move, they are held by strong electrostatic forces of attraction. So, ionic solids do not conduct electricity in solid state. However, in the molten state, they dissociate to give tree ions and hence conduct electricity.

1.9.What type of solids are electrical conductors, malleable and ductile?
Ans: Metallic solids

1.10.Give the significance of a ‘lattice point’.
Ans: Each lattice point represents one constituent particle of the solid. This constituent particle may be an atom, a molecule or an ion.

1.11. Name the parameters that characterise a unit cell.
Ans: A unit cell is characterised by following parameters:
(i)the dimensions of unit cell along three edges: a, b and c.
(ii)the angles between the edges: α (between b and c); β (between a and c) and γ (between a and b)

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

Ans:
(i)

Hexagonal unit cell	Monoclinic unit cell
a=b≠c	a≠b≠c
α = β = 90°	α = γ = 90°
γ = 120°	β ≠ 90°

(ii)

Face-centred unit cell	End-centred unit cell
A Face-centred unit cell the constituent particles are present at the corners and one at the centre of each face.	An End-centred unit cell contains particles at the corners and one at the centre of any two opposite faces.
Total no of particles in a face centered unit cell= 4	Total no. of particles in an end centered unit cell = 2

1.13. 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.
Ans: (i) An atom at the comer is shared by eight adjacent unit cells. Hence, portion of the atom at the comer that belongs to one unit cell=1/8.
(ii)An atom at the body centre is not shared by any other unit cell. Hence, it belongs fully io unit cell.

1.14. What is the two-dimensional coordination number of a molecule in square close-packed layer?
Ans: In 2D, square close-packed layer, an atom touches 4 nearest neighbouring atoms. Hence, its CN=4

1.15. 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?
Ans:
No. of atoms in close packings 0.5 mol =0.5 x 6.022 x 10²³ =3.011 x 10²³
No. of octahedral voids = No. of atoms in packing =3.011 x 10²³
No. of tetrahedral voids = 2 x No. of atoms in packing
= 2 x 3.011 x 10²³ = 6.022 x 10²³
Total no. of voids = 3.011 x 10²³ + 6.022 x 10²³
= 9.033 x 10²³

1.16. A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy l/3rd of tetrahedral voids. What is the formula of the compound?
Ans: Atoms of N from ccp, therefore, if the lattice points are n, then
No . of atoms of N = n
No. of oct voids = n
No. of td voids = 2n= 2 x 1n/3 = 2n/3
∴ Formula of compound is: M : N
2/3 n : n
2n: 3n
2: 3
i.e., M₂N₃

1.17. Wh ich of the following lattices has the highest packing efficiency (i) simple cubic (ii) body- centred cubic and (iii) hexagonal close-packed lattice?
Ans: Packing efficiency of:
Simple cubic = 52.4% bcc = 68% hcp = 74%
hcp lattice has the highest packing efficiency.

1.18. An element with molar mass 2.7 x 10^-2 kg 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?
Ans:

1.19. What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?
Ans: When a solid is heated, vacancy defect is produced in the crystal. On heating, some atoms or ions leave the lattice site completely, i.e., lattice sites become vacant. As a result-of this defect, density of the substances decreases.

1.20.What type of stoichiometric defect is shown by:
(i)ZnS (ii)AgBr
Ans: (i) ZnS shows Frenkel defect
(ii)AgBr shows Frenkel as well as Schottky defect.

1.21. Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.
Ans: Let us take an example NaCl doped with SrCl, impurity when SrCl2 is added to NaCl solid as an impurity, two Na+ ions will be replaced and one of their sites will be occupied by Sr21- while the other will remain vacant. Thus, we can say that when a cation of higher valence is added as an impurity to an ionic solid, two or more cations of lower valency are replaced by a cation of higher valency to maintain electrical neutrality. Hence, some cationic vacancies are created.

1.22.Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.
Ans: Let us take an example of NaCl. When NaCl crystal is heated in presence of Na vapour, some Cl^–ions leave their lattice sites to combine with Na to form NaCl. The e^-1 s lost by Na to form Na⁺ (Na⁺ + Cl^–—> NaCl) then diffuse into the crystal to occupy the anion vacancies. These sites are called F-centres. These e-s absorb energy from visible light, get excited to higher energy level and when they fall back to ground state, they impart yellow colour to NaCl crystal.

1.23.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?
Ans: Impurity from group 15 should be added to get n-type semiconductor.

1.24.What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic. Justify your answer.
Ans: Ferromagnetic substances make better permanent magnets. This is because when placed in magnetic field, their domains get oriented in the directions of magnetic field and a strong magnetic field is produced. This ordering of domains persists even when external magnetic field is removed. Hence, the ferromagnetic substance becomes a permanent magnet.

NCERT EXERCISES

1.1. Define the term ‘amorphous’. Give a few examples of amorphous solids.
Sol. Amorphous solids are those substances, in which there is no regular arrangement of its constituent particles, (i.e., ions, atoms or molecules). The arrangement of the constituting particles has only short range order, i.e., a regular and periodically repeating pattern is observed over short distances only, e.g., glass, rubber and plastics.

1.2. What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
Sol. Glass is made up of Si04 tetrahedral units. These constituent particles have short range order only. Quartz is also made up of Si04 tetrahedral units. On heating it softens and melts over a wide range of temperature. It is a crystalline solid having long-range ordered structure. It has a sharp melting point. Quartz can be converted into glass by first melting and then rapidly cooling it.

1.3 Classify each of the following solids as ionic, metallic, modular, network (covalent) or amorphous:
(i) Tetra phosphorus decoxide (P₄O₁₀) (ii) Ammonium phosphate, (NH₄)₃P0₄ (iii) SiC (iv) I₂ (v) P₄ (vii) Graphite (viii), Brass (ix) Rb (x) LiBr (xi) Si
Sol.

1.4 (i) What is meant by the term ‘coordination number’?
(ii) What is the coordination number of atom
(a) in a cubic close-packed structure?
(b) in a body centred cubic structure?
Sol. (i) The number of nearest neighbours of a particle are called its coordination number.
(ii) (a) 12 (b) 8

1.5 How can you determine the atomic mass of an unknown metal if you know its density and the dimensions of its unit cell? Explain.
Sol. Let the edge length of a unit cell = a
Density = d
Molar mass = M
Volume of the unit cell = a3
Mass of the unit cell
= No. of atoms in unit cell x Mass of each atom = Z × m

1.6 ‘Stability of a crystal is reflected in the magnitude of its melting points’. 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?
Sol. Higher the melting point, greater are the forces holding the constituent particles together and thus greater is the stability of a crystal. Melting points of given substances are following. Water = 273 K, Ethyl alcohol = 155.7 K, Diethylether = 156.8 K, Methane = 90.5 K.
The intermoleoilar forces present in case of water and ethyl alcohol are mainly due to the hydrogen bonding which is responsible for their high melting points. Hydrogen bonding is stronger in case of water than ethyl alcohol and hence water has higher melting point then ethyl alcohol. Dipole-dipole interactions are present in case of diethylether. The only forces present in case of methane is the weak van der Waal’s forces (or London dispersion forces).

1.7 How will you distinguish between the following pairs of terms:
(i) Cubic close packing and hexagonal close packing?
(ii) Crystal lattice and unit cell?
(iii) Tetrahedral void and octahedral void?
Sol. (i) Cubic close packing: When the third layer is placed over the second layer in such a way that the spheres cover the octahedral voids, a layer different from first (A) and second (B) is produced. If we continue packing in this manner, then a packing is obtained where the spheres in every fourth layer will vertically aligned. This pattern of packing spheres is called ABCABC pattern or cubic close packing.
Hexagonal close packing: When a third layer is placed over the second layer in such a manner that the spheres cover the tetrahedral void, a three dimensional close packing is obtained where the spheres in every third or alternate layers are vertically aligned. If we continue packing in this manner, then the packing obtained would be called ABAB pattern or hexagonal close packing.
(ii) Crystal lattice: It is a regular arrangement of the constituent particles (i?.e., ions, atoms or molecules) of a crystal in three dimensional space.
Unit cell: The smallest three dimensional portion of a complete space lattice which when repeated over and over again in different directions produces the complete crystal lattice is called the unit cell.
(iii) Tetrahedral void: A simple) triangular void is a crystal is surrounded by four spheres and is called a tetrahedral void.
Octahedral void: A double triangular void is surrounded by six spheres and is called a octahedral void.

1.8 How many lattice points are there is one unit cell of each of the following lattices?
(i) Face centred cubic (if) Face centred tetragonal (iii) Body centred cubic
Sol.

1.9 Explain:
(i) The basis of similarities and differences between metallic and ionic crystals.
(ii) Ionic solids are hard and brittle.
Sol. (i) Metallic and ionic crystals
Similarities:
(a) There is electrostatic force of attraction in both metallic and ionic crystals.
(b) Both have high melting points.
(c) Bonds are non-directional in both the cases.
Differences:
(a) Ionic crystals are bad conductors of electricity in solids state as ions are not free to move. They can conduct electricity only in die molten state or in aqueous solution. Metallic crystals are good conductors of electricity in solid state as electrons are free to move.
(b) Ionic bond is strong due to strong electrostatic forces of attraction.
Metallic bond may be strong or weak depending upon the number of valence electrons and the size of the kernels.
(ii) Ionic solids are hard and brittle.Ionic solids are hard due to the presence of strong electrostatic forces of attraction. The brittleness in ionic crystals is due to the non- directional bonds in them.

1.10 Calculate the efficiency of packing in case of a metal crystal for (i) simple cubic, (ii) body centred cubic, and (iii) face centred cubic (with the assumptions that atoms are touching each other).
Sol. Packing efficiency: It is the percentage of total space filled by the particles.

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

1.12 A cubic solid is made up 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?
Sol. As atoms Q are present at the eight comers of the cube, therefore, the contribution of atoms of
Q in the unit cell = 1/8 x 8 = 1.
As atom P is present at the body centre, therefore, the contribution of atoms of P in the unit cell = 1.
.•. Ratio of atoms of P: Q = 1:1
Hence, the formula of the compound = PQ The coordination number of each P and Q = 8 .

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

1.14 If the radius of the octahedral void is r and radius of the atoms in close-packing is R, derive relation between rand R.
Sol. A sphere is fitted into the octahedral void as shown in the diagram.

1.15 Copper crystallises into a fee lattice with edge length 3.61 x 10^-8 cm. Show that the calculated density is in agreement with its measured value of 8.92 gcm^-3.
Sol.

This calculated value of density is closely in agreement with its measured value of 8.92 g cm³.

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