1.1 – Why are solids rigid?
Ans: Solids are rigid because of the intermolecular forces of attraction present in the solid. The intermolecular forces existing among the constituent particles of a solid are very strong. That’s why the particles of solid cannot move from their positions; they only vibrate about their mean positions.
1.2 – Why do solids have a definite volume?
Ans. Solids have strong intermolecular forces of attraction. The constituent particles of solids cannot move from their positions they can only vibrate from their mean positive. That is why do solids have a definite volume.
1.3 – Classify the following as amorphous or crystalline solids:
Polyurethane, naphthalene, benzoic acid, Teflon, potassium nitrate, cellophane, polyvinyl chloride, fiber glass, copper.
Ans. AMORPHOUS SOLIDS: : Polyurethane, Teflon, Cellophane, Polyvinyl chloride, Fiber Glass
CRYSTALLINE SOLIDS: Naphthalene, Benzoic acid, Potassium nitrate, Copper
1.4 – 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. Isotropic solid has same value of reflective index in all the direction. All isotropic solid are amorphous solid.Amorphous solid cuts into two pieces with irregular surfaces Therefore,.Amorphous solid does not show the cleavage property.
1.5 – 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. 1. Potassium sulphate – Ionic 2. Tin – Metallic 3. Benzene – Molecular (non polar) 4. Urea – Molecular (polar) 5. Ammonia – Molecular (hydrogen bonded) 6. Water – Molecular (hydrogen bonded) 7. Zinc Sulphide – Ionic 8. Graphite – Covalent or Network 9. Rubidium – Metallic 10. Argon – Molecular (non polar) 11. Silicon carbide – Covalent
1.6 – 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. 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)
1.7 – Ionic solids conduct electricity in molten state but not in solid state. Explain.
Ans. The constituent particles of ionic solid are not free to move in solid state. Therefore ionic solid are insulator in solid phase. Ions becomes free to move in molten state so ionic solid can conduct electricity in molten state
1.8. What type of solids are electrical conductors, malleable and ductile?
Ans. Metallic solids are hard but malleable and ductile. Also, they are conductors in solid as well as in the molten state.
1.9. Give the significance of a ‘lattice point’.
Ans. The significance of a lattice point is that each lattice point represents one constituent particle of a solid which may be an atom, a molecule (group of atom), or an ion.
1.10 – Name the parameters that characterise a unit cell.
Ans. The six parameters that characterise a unit cell are as follows.
(i) Its dimensions along the three edges, a, b, and c
These edges may or may not be equal.
(ii) Angles between the edges
These are the angle ∝ (between edges b and c), β (between edges a and c), and γ (between edges a and b).
1.11 – Distinguish between (i) Hexagonal and monoclinic unit cells (ii) Face-centred and end-centred unit cells.
Ans
(i) Hexagonal unit cell — For a hexagonal unit cell,
Monoclinic unit cell
For a monoclinic cell
(ii) Face−centred unit cell
A face−centred unit cell has one constituent particle which is present at each corner and one at the centre of each face.
End−centred unit cell
A end−centred unit cell has one particles at each corners and one at the centre of any
two opposite faces.
1.12 – Explain how much portion of an atom located at (i) corner and (ii) bodycentre of a cubic unit cell is part of its neighbouring unit cell.
Ans.
(i) An atom located at the corner of a cubic unit cell is shared by eight adjacent unit cells.
Therefore, 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 neighboring 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.
1.13 – What is the two dimensional coordination number of a molecule in square close-packed layer?
Ans. In two dimensional square close−packed layer, a molecule touches four neighbor atoms. Therefore, 4 is the two dimensional coordination number of a molecule in square close packed layer.
1.14 – 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.
Number of close-packed particles = 0.5 × 6.022 × 1023 = 3.011 × 1023
Therefore, number of octahedral voids = 3.011 × 1023
And, number of tetrahedral voids = 2 × 3.011 × 1023 = 6.022 ×1023
Therefore, total number of voids = 3.011 × 1023 + 6.022 × 1023 = 9.033 × 1023
1.15 – 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?
Ans. M2N3
the number of atoms of M is equal to 2 1/3 = 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 M2 N3.
1.16 Which of the following lattices has the highest packing efficiency (i) simple cubic (ii) body-centred cubic and (iii) hexagonal close-packed lattice?
Ans.
The packing efficiency of simple cubic lattice is 52.4%, body centered is 68% and that of hexagonal close packed lattice is 74%.
Therefore, (iii) hexagonal close packed lattice has highest packing efficiency, i.e. 74%.
1.17 An element with molar mass 2.7×10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7×103 kg m-3, what is the nature of the cubic unit cell?
Ans.
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).
1.18 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 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.
1.19 What type of stoichiometric defect is shown by: (i) ZnS (ii) AgBr
Ans.
(i) ZnS – Since there is large difference in the size of ions, thus it shows Frenkel defect.
(ii) AgBr – AgBr shows Frenkel defects and Schottky defects both.
1.20 Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.
Ans. When a cation of higher valence is added as impurity in an ionic solid . two or more than two cations of lower valances is replace by less than 2 ions of higher valances . To maintain the electrical neutrality few sites become vacant.
For example,
Sr2+ is added to NaCl, each Sr2+ ion replaces two Na+ ions. Because the charge on 1 ions of Sr2+ is equal to charge on two ions of Na+. and One Na+ ions leave there site
1.21 Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.
Ans.
Metal excess defects are seen because of missing of anions from regular site leaving a hole which is occupied by electron to maintain the neutrality of the compound. Hole occupied by electron is called F-centre and responsible for showing colour by the compound.
This defect is common in NaCl, KCl, LiCl, etc. Sodium atoms get deposited on the surface of crystal when sodium chloride is heated in an atmosphere of sodium vapour. In this process, the chloride ions get diffused with sodium ion to form sodium chloride and sodium atom releases electron to form sodium ion. This released electron gets diffused and occupies the anionic sites in the crystal of sodium chloride; creating anionic vacancies and resulting in the excess of sodium metal.
1.22 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. An n−type semiconductor conducts the electricity due to presence of extra electrons. So that, an element of group 14th can be converted to n−type semiconductor by doping it with the element of group 15 like P and As .
1.23 What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic. Justify your answer.
Ans.
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 unmagnetised 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 domains persists even after the removal of the magnetic field. Thus, the ferromagnetic substance becomes a permanent magnet.
