The Solid State NCERT Solutions Class 12 Chemistry- PDF Download
Page No. 4
1.1. Why are solids rigid?
Answer
The solids are rigid because of the strong intermolecular forces of attraction present in solids which makes the constituent particles of solids in fixed positions. However, they can oscillate about their mean positions.
1.2. Why do solids have a definite volume?
Answer
The solids have a definite volume because of the strong intermolecular forces of attraction that are present in solids which makes the constituent particles of solids in fixed positions. Hence, 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, fibre glass, copper.
Answer
Amorphous solids: Polyurethane, Teflon, cellophane, polyvinyl chloride, fibre glass.
Crystalline solids: Naphthalene, benzoic acid, potassium nitrate, copper.
1.4. Why is glass considered a super cooled liquid?
Answer
Like liquids, amorphous solids have a tendency to flow, though very slowly. Therefore, sometimes these are called pseudo solids or super cooled liquids.
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?
Answer
Nature of the solid: When measured along different directions, an isotropic solid has the same value of physical properties. Since the give solid is observed to have the same value along all directions, it is said to be isotropic in nature. Hence, the solid is an amorphous solid. Cleavage Property: When an amorphous solid is cut with a sharp edged tool, it cuts into two pieces with irregular surfaces.
Page No. 6
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.
Answer
Ionic solids: Potassium sulphate, Zinc sulphide
Metallic solids: Tin, Rubidium
Molecular (non-polar) solid: Benzene, Argon
Polar molecular solids: Urea, Ammonia
Hydrogen bonded molecular solid: Water
Covalent or network solid: Graphite, Silicon carbide
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?
Answer
The covalent solid or network solids are very hard electrical insulator in solid as well as in molten state and melts at extremely high temperature. Hence, the solid A must be covalent solid or network solid.
1.8. Ionic solids conduct electricity in molten state but not in solid state. Explain.
Answer
Ions are the constituent particles of ionic solids. In solid state, ions are held together by strong electrostatic forces and are not free to move about within the solid. Hence, ionic solids do not conduct electricity in solid state. However, in molten state or in solution form, the ions are free to move and can conduct electricity.
1.9. What type of solids are electrical conductors, malleable and ductile?
Answer
Metallic solids are electrical conductors, malleable and ductile.
Page No. 10
1.10. Give the significance of a 'lattice point'.
Answer
The significance of a lattice point is that they are joined by straight lines to bring out the geometry of the lattice.
1.11. Name the parameters that characterize a unit cell.
Answer
A unit cell is characterised by:
(i) Its dimensions along the three edges, a, b and c. These edges may or may not be mutually perpendicular.
(ii) Angles between the edges, α(between b and c), β(between a and c) and γ(between a and b). Thus, a unit cell is characterised by six parameters, a, b, c, α, β and γ.
1.12. Distinguish between:
(i) Hexagonal and monoclinic unit cells
(ii) Face-centred and end-centred unit cells.
Answer
(i) Difference between Hexagonal and Monoclinic Unit Cells
|
Hexagonal Unit Cells |
Monoclinic Unit Cells |
|
The axial distances, a≠b≠c |
The axial distances, a≠b≠c |
|
The axial angles, α=β=90°, γ=120° |
The axial angles, α=γ=90°, β≠90° |
|
Examples- Graphite, ZnO, CdS etc. |
Examples- Monoclinic sulphur, Na2SO4.10H2O |
(ii) Difference between Face-centred and End-centred Unit Cells
|
Face-centered Unit Cells |
End-centred Unit Cells |
|
The constituent particles are present at the corners and one at the centre of each face. |
The constituent particles are present at the corners and one at the centre of any two opposite faces. |
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.
Answer
(i) In a cubic unit cell, the atoms are located only at its corner. Each atom at a corner is shared between eight adjacent unit cells i.e., four unit cells in the same layer and four unit cells of the upper (or lower) layer. Therefore, only 1/8th of an atom or molecule or ion belongs to a particular 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.
Page No. 21
1.14. What is the two dimensional co-ordination number of a molecule in square close packed layer?
Answer
In square close packed layer arrangement, each sphere is in contact with four of its neighbours. Therefore, the two dimensional co-ordination number is 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?
Answer
Number of atoms in close packing = 0.5 mol
1 atom in close packing = 6.022×1023 particles
Thus,
Number of close-packed particles = 0.5×6.022×1023
= 3.011×1023
Therefore, Number of octahedral voids = 3.011×1023
Now,
Number of tetrahedral voids = 2×No. of octahedral voids
= 2×3.011×1023
= 6.022×1023
Total no. of voids = Tetrahedral voids + Octahedral voids
= 6.022×1023 + 3.011×1023
= 9.033×1023
Page No. 22
1.16. 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?
Answer
Let the number of atoms of the element (N) be 'n'.
Number of tetrahedral voids = 2×Number of atoms in close packing
Therefore,
Number of tetrahedral voids = 2n
According to the questions,
The atoms of element M occupy 1/3rd tetrahedral voids. So,
The number of atoms of M
Now, the ratio of the number of atoms of M and N = M:N = 2n/3 : n
Multiplying by 3 and dividing by n, we get
M:N = 2:3
Hence, the formula of the compound = M2N3.
1.17. Which of the following lattices has the highest packing efficiency:
(i) simple cubic,
(ii) body-centred cubic and
(iii) hexagonal close-packed lattice?
Answer
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.
1.18. 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?
Answer
Molar mass, M = 2.7×10-2kg mol-1
Edge length, a = 405 pm =405×10-12m = 4.05×10-10m
Density of the element, d = 2.7×103 kg m-3
Avogadro's number, NA = 6.022× 1023 mol-1
Substituting the values, we get
= 400.04 ×10-2
= 4.004 ≈ 4
This shows 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).
Page No. 29
1.19. What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?
Answer
When a solid is heated, vacancy defect is formed. The physical property, density is affected. The vacancy defect leads to a decrease in the density of the solid.
1.20. What type of stoichiometric defect is shown by:
(i) ZnS
(ii) AgBr?
Answer
(i) ZnS [Zinc sulphide] shows Frenkel defect.
(ii) AgBr [Silver bromide] shows Frenkel defect 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?
Answer
When a cation of higher valence is added to an ionic solid as an impurity, 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, if molten NaCl containing a little amount of SrCl2 is crystallised, some of the sites of Na+ ions are occupied by Sr2+. Each Sr2+ replaces two Na+ ions. It occupies the site of one ion and the other site remains vacant. The cationic vacancies thus produced are equal in number to that of Sr2+ ions.
1.22. Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.
Answer
Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. The colour is formed due to the excitation of the unpaired electrons occupying anionic sites when they absorb energy from the visible light falling on the crystals.
For example, when crystals of NaCl are heated in an atmosphere of sodium vapour, the sodium atoms are deposited on the surface of the crystal. The CI- ions diffuse to the surface of the crystal and combine with Na atoms to give NaCl. This happens by loss of electron by sodium atoms to form Na+ ions. The released electrons diffuse into the crystal and occupy anionic sites. As a result the crystal now has an excess of sodium. The anionic sites occupied by unpaired electrons are called F- centres. These electrons get excited by absorbing energy from the visible light and impart yellow colour to the crystals. They impart yellow colour to the crystals of NaCl. Similarly, excess of lithium makes LiCl crystals pink and excess of potassium makes KCl crystals violet (or lilac).
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?
Answer
Silicon and germanium, each has four valence electrons and they belong to 14th group of periodic table. Arsenic and phosphorous belong to 15th group of periodic table and they have valence electrons equal to 5. When silicon or germanium is doped with phosphorous or arsenic, four electrons of phosphorous or arsenic out of five; make covalent bonds with four electrons of silicon or germanium leaving one electron free; which increases the electrical conductivity of silicon or germanium. Since the electrical conductivity of silicon or phosphorous is increased because of negatively charged particle (electron), this is known as n-type of semiconductor. In other words, 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.
1.24. What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic? Justify your answer.
Answer
Metal ions of ferromagnetic substances are randomly oriented in normal condition and substances do not act as a magnet. However, when metal ions are grouped together in small regions, called domains, each domains act like a tiny magnet and produce strong magnetic field, in such condition ferromagnetic substance act like a magnet. When the ordering of domains in group persists even after removal of magnetic field a ferromagnetic substance becomes a permanent magnet while domains are grouped in parallel and anti-parallel direction but in unequal number in ferromagnetic compounds. Thus, Ferromagnetic substances, such as Ni, Co, Fe would make better permanent magnets rather than ferromagnetic substances.
