12c NCERT Solidstate

www.studyadda.com CBSE Board - Chemistry - 12th NCERT Exercise with Solutions The Solid State

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The Solid State

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Section - I

1. Why are solids rigid ?

Ans.This is because of close packing of the constituent particles and the very strong forces of attraction.

2. Why do solids have a definite volume ?

Ans.This is because in solids intermolecular distances are short and thus, solids keep their volume due to rigid structures.

3. Classify the following as amorphous or crystalline solids : Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.

Ans. Amorphous : Polyurethane, teflon, cellophane, polyvinyl chloride, fibre glass.

Crystalline : Naphthalene, benzoic acid, potassium nitrate, copper.

4. Why is glass considered a super-cooled liquid ?

Ans.Glass has characteristics similar to liquids. It can flow extremely slowly, so it is considered to be a super-cooled liquid.

5. Refractive index of a solid is observed to have to same value along all directions. Comment on the nature of this solid. Would it show cleavage property ?

Ans. The solid is amorphous. It would not show clean cleavage. Such a solid when cut with a sharp edged tool, it gives two pieces with irregular surfaces.

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. Ionic solids : Potassium sulphate, zinc sulphide. Covalent solids : Graphite, silicon carbide.

Molecular solids : Urea, ammonia, water, argon.

Metallic solids : Tin, rubidium.

7. Solid A is 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 network solid.

8. Ionic solids conduct electricity in molten state but not in solid state. Explain.

Ans. In solid state, ionic solids have fixed arrangement of constituent ions. These can not carry current. In molten state, ions become free to move and carry current.

9. What type of solids are electrical conductors, malleable and ductile ?

Ans. Metallic solids are electrical conductors, malleable and ductile in nature.

10. Give the significance of a lattice point ? Ans. Lattice point signifies the position of constituent particles of the crystal.

11. Name the parameters that characterise a unit cell. Ans. A unit cell is characterised by its edge lengths a, b and c and the angles ( , ,α β γ ) between them. Thus, a unit cell

is characterised by six parameters.

12. Distinguish between

(i) Hexagonal and monoclinic crystal system.

(ii) Face-centred and end-centred unit cell.



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Ans. (i) (a) In hexagonal crystal system, the axial distances a b c= ≠ but in monoclinic crystal system, the three axial distances are altogether different i.e. a b c≠ ≠ .

(b) In hexagonal crystal system, axial angles are o o90 , 120α =β= γ = whereas in monoclinic type, these are o o90 , 90 .α = γ = β≠

(c) Graphite, ZnO are the examples of hexagonal crystal systems while monoclinic sulphur, Na2SO4 . 10H2O are the examples of monoclinic crystal system.

(ii) (a) In addition to the atom at the corner, in face centred unit cell, an atom is present at the centre of each of the six faces of the unit cell whereas in end-centred unit cell two particles are located at the centre of any two opposite faces. (b) Number of atoms per unit cell in face-centred unit cell is 4 but in end-centred unit cell it is 2.

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. In a cubic unit cell, an atom at the corner is shared by eight other unit cells. Thus, 18

th part of the atom at the

corner is part of its neighbouring unit cell.

On the other hand, atom located at body-centre of a cubic unit cell is not shared by neighbouring unit cells.

14. What is the two dimensional coordination number of a molecule in square close packed layer ?

Ans. 4.

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 the close packing = 0.5 mot

= 0.5 x 6.022 x1023 = 3.011 x 1023

No. of octahedral voids is equal to the no. of atoms in the packing = 3.011 x 1023

No. of tetrahedral voids = 2 x No. of octahedral voids

= 2 x 3.011 X 1023 = 6.022 x 1023

Total number of voids

= 3.011 x 1023 + 6.022 x 1023 = 9.33 x 1023

16. A compound is formed by two elements M and N. The element N forms cep and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound ?

Ans. Let, no. of atoms of N = n

Then, no. of tetrahedral voids = 2n

∴ No. of atoms of M =13

x tetrahedral voids

1 22n n3 3

= × =

∴ Ratio of M : N is N is 23

n: In = 2:3

i.e. formula of the compound is M2 N3.

17. Which of the following lattices has the highest packing efficiency (i) simple cubic (ii) body centred cubic and (iii) hexagonal close packed lattice '?

Ans. Out of these three, hexagonal close packed lattice (hcp) has maximum packing efficiency of 74%.



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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. a = 405 pm = 405 x 10–12 m,

M = 2.7 x 10–2 kg mol–1

NA = 6.022 x 1023 mot–1,d = 2.7 x 103 kg m–3

Now, 3

A3

A

d N az Md zMN a

× ××= ∴ =

×

3 3 23 1 12

2 1

(2.7 10 kgm ) (6.022 10 mol ) (405 10 m)(2.7 10 kg mol )

− − −

− −

× × × × ×=

×= 4

Since, No. of atoms per unit cell are 4, so it must be a face centred cubic unit cell.

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 can be created in which certain constituents may leave the lattice site. This leads to lowering of the density of the solid.

20. What type of stoichiometric defect is shown by :

(i) ZnS (ii) AgBr

Ans. (i) ZnS shows Frenkel defect.

(ii) AgBr shows both Frenkel and Schottky defects.

21. Explain how vacancies are introduced in an ionic solid when a cation of higher valence in added as an impurity in it.

Ans. When molten NaCl (Na+ is monovalent) containing a little amount of SrCl2 (Sr2+ is divalent) as impurity is crystallised, some of the sites of Na+ ions are occupied by Sr2+ . Each Sr2+

replaces two Na+ ions. One site is occupied by Sr2+ and the other site remains vacant. The cationic vacancies thus

produced are equal to the number of Sr2+ ions.

22. Ionic solids, which have anionic vacancies due to metal- excess defect, develop colour. Explain with the help of a suitable example.

Ans. Alkali halides like NaCI and KCI show this type of defect. When crystals of NaCI are heated in an atmosphere of sodium vapour, the sodium



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Section – II Chapter End Exercise

1. Define the term 'amorphous'. Give a few examples of amorphous solids.

Ans. Amorphous is that in which the constituent particles do not possess the regular arrangement throughout. e.g., Glass, Plastics, Rubber, Cellulose etc.

2. What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass ?

Ans. Glass is a super cooled liquid and an amorphous substance. In glass, SiO4 tetrahedral are joined in a random manner. Quartz is the crystalline form of silica (SiIO2) in which tetrahedral units SiO4 are linked with each other in such a way atoms are deposited on the surface of the crystal.

The Cl– ions diffuse to the surface of the crystal and combine with Na atoms to give NaCI. This happens by loss of electron by sodium atoms to form Na+ ions. The released electrons diffuse into the crystal and occupy anionic sites (Fig.). As a result, the crystal now has an excess of sodium. 'Thus, it is metal excess defect. The anionic sites occupied by unpaired electrons are called F-centres (from the German word Earbenzenter for colour centre). They impart yellow colour to the crystals of NaCI. The colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals. Similarly, excess of lithium makes LiCl crystals pink and excess of potassium makes KCI crystals violet (or lilac).

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. Group 15 because it has five valence electrons. One electron is left for electrical conductivity.

24. What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic. Justify your answer.

Ans.Ferromagnetic substances are better permanent magnets than ferrimagnetic substances because in the first case, the spins due to all unpaired electrons are alligned in the same direction. On the other hand, in ferrimagnetic substances, the electronic spins are alligned both in the parallel and antiparallel directions in unequal number that the oxygen atom of one tetrahedron is shared with another Si atom.

Quartz can be converted into glass by melting it and cooling the melt very rapidly.

3. Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.

(i) Tetra phosphorous decoxide (P4O10)

(ii) Ammonium phosphate (NH4)3PO4

(iii) siC (iv) I2

(v) P4 (vi) Plastic

(ii) Graphite (viii) Brass

(ix) Rb (x) LiBr



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(xi) Si

Ionic solids : (NH4)3PO4, LiBr

Metallic solids : Rb, Brass

Molecular solids : P4O10,I2, P4

Covalent solids : Graphite, Si, SiC

Amorphous solids : Plastic

4. (a) What is meant by the term 'coordination number' ?

(b) What is the coordination number of atoms

(i) in a cubic close packed structure ?

(ii) in a body centred cubic structure ?

Ans. (a) Coordination number is the number of spheres which are touching a particular sphere. In ionic crystals, it is the number of oppositely charged ions surroundings a particular ion.

(b)(i) Twelve (ii) Eight.

5. How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell ? Explain.

Ans. Edge of the cubic crystal in the unit cell = a pm

Volume of the unit cell

a3 (pm)3 = a3 x 10-30 cm3

Density of the unit cellMass of the unit cell

Volume of the unit cell=

Mass of the unit cell = No. Volume of the unit cell

× mass of each atom or

Mass of each atom,

A

Atomic mass M(m)Avogadro number N

= =

∴ Density of unit cell (d)

33 30

A

z M g / cma 10 N−

×=

× ×

The density of unit cell is equal to the density of the substance. If the density of the substance is known by other methods, then we can calculate the atomic mass of unknown metal by the formula

3 30

Ad a 10 NM g/ molz

−× × ×=

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 the data book. What can you say about the inter-molecular forces between these molecules ?

Ans.(a) The stability of a crystal depends upon the magnitude of forces of interaction in the constituting particles. Greater the force of attraction present, more will be the stability of the crystal. For example, ionic salts such as NaCI, KCI etc. have very high melting and boiling points while the molecular crystals such as naphthalene. iodine etc. have low values of melting and boiling points.



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(b) The melting points of different substances are '.

(i) Water = 273 K (ii) Ethyl alcohol = 155.7 K

(iii) Diethyl ether = 156.8 K (iv) Methane 90.5 K.

The intermolecular forces in molecules of water and ethyl alcohol are mainly hydrogen bonding. The magnitude is more in water than in alcohol as is clear from the values of melting points.

The forces in the molecules of diethyl ether are dipolar forces while in methane these are mainly the van der Waal's forces which is quite evident from its melting point (least among the compounds listed).

7. How will you distinguish between the following pairs of terms :

(a) Hexagonal close packing and cubic close packing ?

(b) Crystal lattice and unit cell ?

(c) Tetrahedral void and octahedral void ?

Ans. (a) In a hexagonal close packing (hcp), the spheres of the third layer are vertically above the spheres of the first layer (ABABAB …… type). On the other hand, in cubic close packing (ccp), the spheres of the fourth layer are present above the spheres of the first layer (ABCAB( …….. type).

(b) The three dimensional arrangement of atoms/ions/molecules is called a crystal lattice. On the other hand, the smallest repeating pattern in this which when repeated in three dimensions generates the crystal is called a unit cell.

(c) Tetrahedral void is a simple triangular void in a crystal and is surrounded by four spheres arranged tetrahedrally around it. On the other hand, an octahedral void is a double triangular void with one triangle vertex upwards and the other triangle tertex downwards and is surrounded by six spheres.

8. How many lattice points are there in one unit cell of each of the following lattices ?

(a) face centred cubic

(b) face centred tetragonal

(c) body centred cubic

Ans. (a) In face centred cubic arrangement,

Lattice points located at the corners of the cube = 8

Lattice points located at the centre of each face = 6

Total no. of lattice points = 8 + 6 = 14

(b) In face centred tetragonal, the number of lattice points is also the same i.e., 8 + 6 = 14.

(In both the cases, particles per unit cell = 8

x 1/8 + 6 x 1/2 = 4)

(c) In body centred cubic arrangement,

Lattice points located at the corners of the cube = 8

Lattice points located in the centre of the body = 1

Total no. of lattice points = 8 + 1 = 9

The total number of particles per unit cell

= 8 x 1/8 + I = 2.

9. Explain :

(a) the basis of similarities and differences between metallic and ionic crystals.



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(b) Ionic solids are hard and brittle.

Ans. (a) The basis of similarities between the metallic and ionic crystals are the electrostatic forces of attraction. These arc present among the ions in the ionic crystals and among the kernels and valence electrons in the metallic crystals. That is why both metals and ionic compounds are good conductors of electricity and have high melting points.

The basis of difference is the absence of mobility of ions in the ionic crystals while the same is present in the valence electrons in case of metallic crystals. Asa consequence, the ionic compounds conducts electricity only in the the molten state while the metals can do so even in the solid state.

(b) The ionic solids are hard and brittle because of strong electrostatic forces of attraction which are present in the oppositely charged ions.

10. Calculate the efficiency of packing (or packing fraction) in case of a metal crystal for

(a) simple cubic

(b) body centered cubic

(c) face centered cubic

(with the assumptions that atoms are touching each other).

Ans. Packing efficiency is the percentage of the total space which is occupied by the particles in a certain packing. The fraction of the total space filled by spheres is called packing fraction.

Packing fraction.

No. of spheres per unit cell

Volume occupied by atoms in the unit cell

Volume of unit cell

(a) In a simple cubic unit cell,

Suppose the edge length of the unit cell = a and radius of the sphere = r

As spheres are touching each other so, a = 2r

1 8 18

= × =

Volume of 1 sphere 34 r3

= π

Volume of the cube = a3 = (2r)3 = 8r3

∴ Fraction occupied i.e. packing fraction

3

3

4 r3 0.5248r

π= =

or % of space occupied = 52.4%

Percentage of unoccupied space

= 100 – 52 .4 = 47 .6%

(b) In body centered cubic structure,



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As the sphere at the centre touches the spheres at the corner, body diagonal AD = 4r

Further face diagonal,

AC 2 2 2 2AC AB BC a a= + +

and body diagonal,

2 2 2 2AD AC CD 2a a 3a= + = + =

4r3a 4r or a3

= =

∴ Volume of the unit cell 3 3

3 4r 64 ra3 3 3

= =

No. of spheres per unit cell = 8 18 1 28

= + =

Volume of 2 spheres 3 34 82 r r3 3

= × π = π

∴ Fraction occupied i.e. packing fraction

3

3

8 r3 0.6864r3 3

π= =

or % of space occupied = 68%


=100 – 68 = 32%

(c) In face-centred cubic structure

As spheres on the face are touching, so AC = 4r But from the right angled triangle ABC,

2 2 2 2AC AB BC a a 2a= + = + =

2 a 4r∴ =

or 4a r2

=

∴ Volume of the unit cell



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3

3 34 32a r r2 2

= = =

No. of spheres in the unit cell

1 18 6 48 2

= × + × =

Volume of 4 spheres 3 34 164 r r3 3

= × π = π

Fraction occupied i.e. packing fraction or efficiency of packing

3

3

16 r / 3 0.7432r / 2

π= = or

% of space occupied = 74%


= 100 – 74 = 26%

11. Silver crystallizes in fcc lattice. If edge length of the cell is 4.077 x 10–8 cm and density is 10.5 g em–3, calculate the atomic mass of silver.

Ans. We know that 3A

z Mda N

×=

×

or 3

Ad a NMz

× ×=

According to available data :

Edge length (a) = 4.07 x 10–8 cm

No. of atoms per unit cell (z) = 4

( of fcc structure)

Density of silver = 10.50 g cm–3

Avogadro's number (NA) = 6.022 x 1023 mol–1

∴ Atomic mass of the element (M)

3 3 23 1 8 3(10.50gcm ) (4.077) (6.222 10 mol )(10 cm)

4

− − −× × ×=

= 107.12 g mol –1.

12. A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-center. What is the formula of the compound ? What are the coordination numbers of P and Q ?

Ans. Number of P atoms per unit cell = 1 (within the body)

x 1 = 1

Number of Q atoms per unit cell

= 8 (at corners) 1 18

× =

∴ Formula of the compound is PQ.

For body centred cubic, coordination numbers of both P and Q are 8.



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13. Niobium crystallizes in body centered cubic structure. If density is 8.55 g cm-3, calculate atomic radius of niobium using its atomic mass 93 u.

Ans. d = 8.55 g cm–3, z = 2 (bcc), M = 93 g mol–1

3A

z.Mda .N

=

or 3

A

z.Mad.N

=

1

3 23 1

2 93g mol8.55 g cm 6.022 10 mol

−

− −

×=

× ×

= 36.13 x 10–24 cm3

a = [36.13 x 10–24 cm3]1/3 = 3.31 x 10–8 cm

In bee, atomic radius (r) 3 a4×

=

8

71.732 3.31 10 cm 14.29 10 cm4

−−× ×

= = ×

= 14.29 nm.

14. If the radius of the octahedral void is r and radius of the atoms in close packing is R, derive relation between r and R.

Ans. An octahedral site is created by the arrangement of four spheres (atoms) in one plane, one sphere (atom) on top and one sphere (atom) below this plane. The shaded portion in the figure represent

an octahedral void. For clarity, the spheres present above and below the void are not shown.

Let, length of each side of the square is 'a' and the radii of the void and the sphere (atom) are r and R respectively. Consider the right angled triangle ABC,

2 2 2 2AC AB BC a a 2 a= + = + =

As AC = 2r + 2R

2 a 2r 2R∴ = + ….(i)

Now AB = 2R

or a = 2R

Dividing Eqn. (i) by Eqn. (ii), we get

2 a 2r 2Ra 2R

+=

r2 1R

= +



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r 2 1 1.414 1 0.414.R= − = − =

Thus, for an octahedral void, void

sphere

r 0.414r

=

∴ For octahedral void, the radius ratio = 0.414

15. Copper crystallizes into a fcc 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 g cm–3.

Ans. a = 3.61 x 10–8 cm, z = 4 (fcc)

M = 63.5 g mol –1

3A

z Mda .N×

=

1

8 3 23 1

4 63.5 g mol(3.61 10 cm) 6.022 10 mol

−

− −

×=

× × ×

d = 8.96 g cm–3

Thus, the calculated value of density is in agreement with the measured value (8.92 g cm–3).

16. Analysis shows that nickel oxide has formula Nio.98 O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?

Ans. Formula of nickel oxide : Ni0.98 O1.00

or Ni98 O100

Let no. of Ni2+ ions = x

∴ no. of Ni3+ ions = 98 – x

Total –ve charge on 100, O2– ion = 200

Total +ve charge x, Ni2+ ions = 2x

Total +ve charge on (98 – x), Ni3+ ions

= 3 (98 – x)

Total +vc charge on Ni ions = 2x + 3(98 – x)

For the substance to be electrically neutral

Total +ve charge = Total –ve charge

2x + 3(98 – x) = 200

2x – 3x = 200 – 294

x= 94

Percentage of Ni2+ ions x 100

98= ×

94 100 96%98

= × =

∴ Percentage of Ni3+ ions = 100–96 = 4%

17. What is a semiconductor ? Describe the two main types of semiconductors and contrast their conduction mechanism.

Ans. Semiconductor is the solid which is perfect insulator at 0 K but conduct some electricity at room temperature. e.g., Silicon and Germanium. Two main types of semiconductors are n-type and p-type semiconductors.



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(i) n-type semiconductors. Silicon and germanium (Group 14) have very low electrical conductivity in the pure state.

If we add certain elements like phosphorus (P) or arsenic (As) of group 15 to these covalent crystals, their atoms will also get linked with those of group 14 elements by covalent bonds but will have the extra electron which is not involved in the bonding (atoms have five valence electrons). These extra electrons will lead to electrical conductivity resulting in n-type semi conductors as these are conducting due to the movement of electrons.

(ii) p-type semiconductors. If in the covalent crystals of group 14 elements, the addition of small amounts of the element aluminium (Al) or gallium (Ga) belonging to group 13 is done, the atoms of such elements can share only three electron with the atoms of group 14. Thus, the holes will be created in the lattice since there is no fourth electron available for sharing. The holes will lead to electrical conductivity and the crystals thus formed are also semiconductors. These are known as p-type semiconductors.

18. Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is

slightly less than 2 : 1. Can you account for the fact that this substance is a p-type semiconductor ?

Ans. The ratio less than 2: 1 in Cu2O shows that some cuprous (Cu+) ions have been replaced by cupric (Cu2+) ions. In order to maintain the electrical neutrality, every two Cu2+ ion which results in creating cation vacancies leading to positive holes. Since the conduction is due to positive holes, it is a p-type semiconductor.

19. Ferric oxide crystallises 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.

Ans. There is one octahedral hole for each atom in hexagonal close packed arrangement.

If the number of oxide ions (O2–) per unit cell =1

Then, no. of Fe3+ ions 2 octahedral holes3

= ×

2 213 3

= × =

Thus, the formula of the compound

= Fe2/3 O1 = Fe2O3

20. Classify each of the following as being either a p-type or an n-type semiconductor :

(i) Ge doped with In (ii) B doped with Si

Ans. (i) p-type semiconductor.

(ii) n-type semiconductor.

21. Gold (atomic radius = 0.144 nm) crystallises in a face centred unit cell. What is the length of a side of the cell ?

Ans. r = 0.144 nm ; Z = 4 (fee)

For fee unit cell, ar

2 2=



14

or a = r x 2 x 2

0.144 nm x 2 x 1 × 414 = 0.407 nm

22. In terms of a hand theory, what is the difference

(i) between a conductor and an insulator

(ii) between a conductor and a semiconductor ?

Ans. (i) In a conductor, the energy gap between the valence band and the conduction band is very small and electrons easily flow into the conduction band. Thus, it conducts electricity.

On the other hand, in an insulator, the energy gap between the top of the valence band and the bottom of the conduction band is so large that enough energy is not available under normal conditions to promote the electrons from valence band to conduction band. The in-between region is called the forbidden zone.

(ii) In conductors, the energy gap between the valence band and conduction band is very small and electrons easily flow into the conduction band. Thus, it conducts electricity. On the other hand, in a semi-conductor, the energy gap between the valence band and the conduction band is smaller than in insulators but larger than in conductors. The thermal energy available under ordinary conditions is sufficient to excite some electrons from valence band to conduction band and thus a small amount of current can flow through them.

23. Explain the following terms with suitable examples :

(i) Schottky defect (ii) Frenkel defect

(iii) Interstitials (iv) F-centres.

Ans. (i) Schottky defect.

This type of defect is created when equal number of positive and negative ions are missing from their respective positions leaving behind holes. Since the number of missing positive ions is equal to the number of missing negative ions the crystal as a whole is electrically neutral. This defect is more common in ionic compounds with high coordination number and where the ions (positive and negative) are of similar size. For example NaCI, KCI, CsCI and KBr.

Since the number of ions decreases, therefore as a result of large number of Schottky defects in solid, the density of solid decreases.



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(ii) Frenkel defect

This type of defect is created when an ion leaves its lattice site and occupies an interstitial site. In this case the crystal remains electrically neutral because the number of anions and cations remain the same. Since there is no absence of ions from the lattice, the density remains the same. Frenkel defects generally occur in ionic compounds

(i) which have low co-ordination number, and

(ii) in which anions are much larger in size than the cations.

These defects can be found in silver halides (such as AgCI, AgBr or Agl) because of small size of the Ag+ ion, it can go into the interstitial sites.

(iii) Interstitials. Interstitial sites are the holes or voids in the crystals. Atoms (or ions) which occupy the vacant

interstitial positions in a crystal are called Interstitials.

(iv) F-centres (Farbe's centre or colour centres) F-centres are the free electrons trapped in the anionic vacancies which are responsible for colour and electrical conductance in non stoichiometric compounds.

e.g. When sodium chloride is heated in an atmosphere of sodium vapours, the excess of metal atoms get deposited on the surface of alkali metal crystal. Halide ions then diffuse to the surface where they combine with metal ions. The electrons so produced by the ionisation of the metal diffuse back into the crystal and occupy anion vacancy. These electrons absorb some energy of the white light, giving yellow colour to NaCl. These amionic sites occupied by unpaired electron are referred to as F-ccntres (German : Farbezenter means colour centre).

24. Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm.

(a) What is the length of the side of the unit cell ?

(b) How many unit cells are there in 1.00 cm3 of aluminium ?

Ans. (a) r = 125 pm

For cubic close packed structure,



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a12 2

=

a= r × 2 × 2 = 125 nm x 2 × 1.414 = 354 pm

(b) Volume of unit cell,

a3 = (354 x 10–12 cm)3 = 44.361 x 10–30 cm3

Now, 44.361 x 10–30CM3 contains = 1 unit cell

330

11cm contains44.361 10−∴ =

×

= 2.26 x 1022 unit cells

25. If NaCl is doped with 10–3 mot % of SrCl2, what is the concentration of cation vacancies ?

Ans. 2 Na+ ions are replaced by 1 Sr2+ ion,

∴ Each Sr2+ ion causes one cation vacancy. Hence, concentration of cation vacancies on being doped with 10–3 mol % SrCl2 = 10–3 mol %

3 3

2310 10mol 6.022 10100 100

− −

= = × ×

= 6.022 X 1018 vacancies.

26. Explain the following with suitable examples :

(i) Ferromagnetism (ii) Paramagnetism

(iii) Ferrimagnetism (iv) Antiferromagnetism

(v) 12-16 and 13-15 compounds.

Ans. (i) Ferromagnetism. A substance which is strongly attracted by the magnetic field and shows magnetism even in the absence of magnetic field is called Ferromagnetic substance. e.g., Fe, Ni, Co etc. Ferromagnetism arises due to the spontaneous alignment of magnetic moments due to unpaired electrons in the same direction. CrO2 is ferromagnetic oxide used to make magnetic tapes for cassette recorders.

(ii) Paramagnetism. A substance which is weakly attracted by the magnetic field is called paramagnetic substance and the property thus exhibited is called paramagnetism. This property is due to the presence of unpaired electrons. e.g., Cu2+, O2. These substances lose their magnetism in the ahsence of the magnetic field.

(iii) Ferrimagnetism. If the magnetic moments arc aligned in parallel and antiparallel directions in unequal number resulting in a small net magnetic moment, the substance is called ferrimagnetic substance and the property thus exhibited is known as ferrimagnetism e.g., FC3O4.

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Ferromagnetic Antifcrromagnetic Ferrimagnetic

(iv)Antiferromagnetism. Certain paramagnetic substances allign the magnetic moments due to the unpaired electrons under the influence of external magnetic field in such a way that they mutually cancel. As a result, they possess zero or no magnetic moment. Such substances are known as anti-ferromagnetic substances and this property is called anti-ferromgnetism. Managanese oxide (MnO) is anti-ferromagnetic in nature.

(v) 12-16 and 13-15 Compounds The solid binary compounds prepared by combining elements of group 12 and 16 are called 12-16 compound e.g. ZnS and CdS those prepared by combining elements O, group 13 and 15 are called 13-15 compounds e.g. AIP and GaAs. These are used as semiconductors.


12c NCERT Solidstate

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