Chemistry 12 Solid State Chapter 1 – Exercise 2

NCERT EXERCISES Chapter – 1 (Exercise 1.2)

Q – 1.1 Describe the term ‘amorphous’. Give a few examples of amorphous solids.
Ans. Amorphous solids are those solids in which the constituent particles may have a short range order but do not have a long range order. They have irregular shapes and are isotropic in nature. They do not undergo a clean cleavage. They do not have sharp melting points or definite heats of fusion.
Example. Glass, rubber and plastics.
Q – 1.2 What makes a glass different from a solid such as quartz ? Under what conditions could quartz be converted into glass ?
Ans. Glass is an amorphous solid in which the constituent (SiO4  tetrahendra) have only a short range order and there is no long range order. In quartz, the constituent particles (SiO4 tetrahendra) have short range as well as long range order. On melting quartz and then cooling it rapidly, it is converted into glass.
Q –  1.3 Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous : (i) Tetraphosphorus decoxide (P4O10)   (ii) Ammonium phosphate, (NH4)3PO4
                (iii) SiC    (iv) I2    (v) P4    (vi) Plastics  (vii) Graphite   (viii) Brass   (ix) Rb  (x) LiBr  (xi) Si
Ans. Ionic = (NH4)3PO4 and LiBr ; Metallic = Brass, Rb ; Molecular =P4O10,  I2, P4, ; Network (covalent) = Graphite, SiC,  Si ; Amorphous = Plastics.
Q – 1.4  What is meant by the term ‘coordination number’? What is the coordination number of atoms (a) in a cubic close packed structure ?    (b) in a body-centred cubic structure ?
Ans. The coordination number of a constituent particle ( atom, ion or molecule) in a crystal is the number of constituent particles which are the immediate neighbours of that particles in the crystals. In ionic crystals coordination number of an ion in the crystal is the number of oppositely charged ions surrounding that particular ion.
(a) 12 (b) 8
Q – 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 your answer.
chemistry solid state Q 1.5


Q – 1.6 (a) ‘Stability of a crystal is reflected in the magnitude of its melting point’. Comment .
(b) The melting points of some compounds are given below :
Water = 273 K, Ethyl alcohol = 155.7 K, Diethyl ether = 156.8 K, Methane = 90.5 K.
What can you say about the intermolecular forces between these molecules ?
Ans. (a) Higher the melting point, greater are the forces holding the constituent particles
together and hence grater is the  stability.
(b) The intermolecular forces in water and ethyl alcohol are mainly the hydrogen bonding. Higher melting point of water then alcohol shows that hydrogen bounding in ethyl alcohol molecules is not as strong as in water molecules . Diethyl ether is a polar molecule. The intermolecular forces present in them are dipole-dipole attraction. Methane is a non-polar molecule. The only forces present in them are the weak van der wall’s forces (London dispersion forces).
Q – 1.7 How well 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.

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chemistry solid state Q 1.7.1png

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Q – 1.8. How many lattice points are there  in one unit cell of each of the following lattices ?
chemistry solid state Q 1.8
Q – 1.9. Explain : (a) The basis of similarities and differences between metallic and ionic crystals.  (b) Ionic crystals are hard and brittle.
Ans. (a) Similarities.   Both ionic and metallic crystals have electronic forces of attraction. In ionic crystal, these are between the oppositely charged ions. In metals, these are among the valence electrons and the kernels. Thar is why both have high melting point.
In both cases, the bond is non – directional.
Differences.  In ionic crystals, the ions are not free to move. Hence, they cannot conduct electricity in the solid state. They can do so only in the molten state or in aqueous solution. In metals, the valence electrons are free to flow. Hence, they can conduct electricity in the solid state.
Ionic bond is strong due to electrostatic forces of attraction. Metallic bond may be weak or strong depending upon the number of valence electrons and the size of the kernels.
(b) Ionic crystals are hard because there are strong electrostatic forces of attraction among the oppositely charged ions. They are brittle because ionic is non-directional.
Q – 1.10. 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 assumption that atoms are touching each other )
chemistry solid state Q 1.10
Q – 1.11. Silver crystallises in fcc lattice. If the edge length of the cell is 4.077 × 10-8 cm and density is 10.5 g cm-3 , calculate the atomic mass of silver.

chemistry solid state Q 1.11

Q – 1.12. A cubic solid is made up of two elements P and Q . Atoms Q are present at the corners of the cube and atoms P at the body centre. What is the formula of the compound ? What are the coordination numbers of P and Q ?
chemistry solid state Q 1.12
Q – 1.13. Niobium crystallizes in a body centred cubic structure. If density is 8.55 g cm-3 , calculate atomic radius of niobium , given that is atomic mass is 93 u.

chemistry solid state Q 1.13

Q – 1.14. If the radius of the octahedral void is r and the radius of the atoms in the close packing is R, derive relationship between r and R.
Ans.  A sphere fitting into the octahedral void is shown by a small circle  ( Fig ). A sphere above and a sphere below this small sphere have not been shown in the fig .
Obviously, ABC is a right angled tringle. Applying  Pythagoras theorem,
chemistry solid state Q 1.14
Q – 1.15. Copper crystallizes into a fcc lattice with edge length 3.61 × 10-8  cm. Show that the calculated density is in agreement with its measured value if 8.92 g cm-3 .

chemistry solid state Q 1.15

Q – 1.16. The composition of a sample of wustite id Fe0.93O1.00. What percentage of the iron is present in the form of Fe (III)  ?
chemistry solid state Q 1.16
Q – 1.17. What is a ‘semiconductor’ ? describe the two main types of semiconductors and contrast their conduction mechanisms.
Ans. Substances whose conductance lies in between that of metals (conductors) and insulators are called semiconductors.
(i) n-type semiconductors These delocalized electrons increase the conductivity of silicon or germanium . as the increase in conductivity is due to negatively charged electrons , the silicon or germanium crystal doped with electron rich impurities are called.chemistry solid state n-type semiconductors(ii) An electron from the neighbouring atom can jump to fill up this electron hole but then an electron hole is created at the site from where electron has jumped. As it continues, the electron holes will move in a direction opposite to that of the flow of electrons. Hence, silicon and germanium doped with electron-deficit impurities are called p-type semiconductors.
chemistry solid state p-type semiconductors
Q – 1.18 . Non-stoichiometric cuprous oxide, Cu2O can be prepared in the laboratory. In this oxide, copper to oxygen ratio is slightly less then 2 : 1 . Can you account for the fact that this substance is a p-type semiconductor.
Ans. The ratio less then 2 : 1 in Cu2O shows that some cuprous (Cu+) ions have been replaced by cupric (Cu2+) ions. To maintain electrical neutrality, every two Cu+ ions will be replaced by one Cu2+ ion thereby creating a hole. As conduction will be due to presence of these positive holes, hence it is a p-type semiconductor.
Q – 1.19. Ferric oxide crystallizes 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.
chemistry solid state Q 1.19
Q – 1.20. Classify each of the following as being either a p-type or n-type semiconductor :
Ge doped with in                      B doped with Si.
Ans. (i)  Ge is group 14 elements and in is Group 13 element. Hence, an electron deficit hole is created and therefore, it is  p-type.
(ii) B is group 13 element and Si is group 14 element, there will be a free electron, Hence, it is n-type.
Q – 1.21. Gold ( atomic radius = 0.144 nm ) crystallises in a  face centred unit cell. What is the length of the side of the cell ?

chemistry solid state Q 1.21

Q – 1.22. In terms of band theory, what is the difference
between a conductor and an insulator ?   between a conductor and a semiconductor ?
Ans.  (i) The energy gap between the valence band and conduction band in an insulator is very large whereas in a conductor, the energy gap is very small or there is overlapping between valence band.
(ii) In a conductor, there is very small energy gap or there is overlapping between valence band and conduction band but in a semiconductor, there is always a small energy gap between them.
Q – 1.23. Explain the following terms with suitable examples :
(i) Schottky defect           (ii)  Frenkel defect         (iii)  Interstitials

Ans. (i) Schottky defect if in an ionic crystal of the type A+B , equal number of cations and anions are missing from their lattice sites so that the electrical neutrality is maintained, it is called schottky  defect.

chemistry solid state schottky defect

(ii) Frenkel defect if an ion is missing from its lattice site (causing a vacancy or a hole there) and it occupies the interstitial site, electrical neutrality as well as the stoichiometry of the compound are maintained. This type of defect is called frenkle defect.

chemistry solid state frenkle defect

(iii) Interstitials defect when some extra constituent particle are present in the interstitial site, the crystal is said to have interstitial defect.

chemistry solid state interstitial defect

Q – 1.24. 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 cm3 of aluminium  ?
chemistry solid state Q 1.24
Q – 1.25. If NaCl is doped with 10-3 mol % SrCl2, what is the concentration of cation vacancies ?
chemistry solid state Q 1.25
Q – 1.26. Explain the following with suitable examples :
(i) ferromagnetism      (ii) paramagnetism       (iii) ferrimagnetism         (iv) anti – ferromagnetism
Ans. (i)  ferromagnetism which show permanets magnetism even in the absence of the magnetic field
are called ferromagnetism .
(ii) paramagnetism which are attached by the external magnetic field are called paramagnetism .
(iii) ferrimagnetism which are expected to possess large magnetism on the basis of the magnetic
moments of the domains but actually have small net magnetic moment are called
(iv) Anti – ferromagnetism which are expected to possess paramagnetism or ferromagnetism on the
basis of magnetic moments of the domains but actually they possess zero net magnetic moment
are called  anti – ferromagnetism.
chemistry solid state ferromagnetism


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