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1. Define the term ‘amorphous’. Give a few examples of amorphous solids.
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Solution:
(Greek-Amorphous means ‘no form’)
A category of solids which possess properties of incompressibility and rigidity to a certain extent but
do not have definite geometrical forms. Such substances are called amorphous solids. Example:
glass, rubber and plastics.
2. What makes a glass different from a solid such as quartz? Under what
conditions quartz could be converted into glass?
•
Solution:
Glass is a super cooled liquid and is an example for an amorphous solid. Quartz is good example for
crystalline solid. The arrangement of constituent particles in quartz is ordered and has a long range
order. Whereas quartz glass has a short range order arrangement of its constituent particles. Glass
can soften over a range of temperature and can be moulded and blown into various shapes. On
heating it becomes crystalline (quartz) at some temperature.
•
Solution:
(a) Tetra Phosphorous decoxide P4O10 - Molecular solid
in
3. Classify each of the solids as ionic, metallic, molecular, net work
(covalent) or amorphous.
(a) Tetra phosphorous decoxide P4O10
(b) Graphite
(c) Brass
(d) Ammonium Phosphate
(e) SiC
(f) Rb
(g) I2
(h) LiBr
(i) P4
(j) Si
(k) Plastic
(b) Graphite - Crystalline solid with network of covalent bonds
(c) Brass - Metallic solid.
(d) Ammonium Phosphate - Ionic solid
(e) SiC - Network/Covalent solid
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(f) Rb - Metallic solid
(g) I2 - Molecular solid
(h) LiBr - Ionic solid
(i) P4 - Molecular solid
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(j) Si - Covalent solid
(k) Plastic - Amorphous
4. (a) What is meant by the term Co-ordination number?
(b) What is the co-ordination number of atom
(i) in a cubic close packed structure
(ii) In a body centred cubic structure
•
Solution:
(a) Co-ordination number is defined as the number of nearest neighbouring atoms that a given atom
is surrounded within a crystal lattice.
(b) (i) The co-ordination number of atoms in a cubic close placed structure is 12.
(ii) The co-ordination number of atoms in a body centred cubic structure is 8.
5. How can you determine the atomic mass of an unknown metal if you
know its density and dimensions of its unit cell? Explain.
•
Solution:
The density of the unit cell is given by the expression
P=
x g/cm3
P = density of the unit cell
Z = Number of atoms per unit cell
M = Atomic mass of the element.
in
Volume of the unit cell = (a pm)3 = a3 × 10-30 cm3
Mass of the unit cell = Number of atoms in the unit Cell × Mass of each atom
=Z×m
Where m = mass of each atom
M=
=
Knowing all the terms in the density parameter expression the Atomic mass of the element can be
calculated
The Atomic mass (M) =
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6. ‘Stability of a crystal is reflected in the magnitude of its melting
point’. Comment. Collect melting point of water, ethyl alcohol, diethyl ether
and methane from a data book. What can you say about the intermolecular
forces between these molecules?
Solution:
Crystals posses perfect geometrical arrangements of atoms or ions or molecules in the three
dimensional region. If the atoms or ions or molecules are compactly packed in the crystal lattice the
melting point is generally high.
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Substance
Water
Ethyl alcohol
Diethyl Ether
Methane
Melting Point
273 K
156 K
157 K
91 K
Polar molecule with H-Bonds
H Bonds
Dipole – Dipole interaction
Vander waals forces.
7. How will you distinguish between the following pairs of terms:
Hexagonal close packing and cubic close packing
•
Solution:
Hexagonal close packing
(i) This represents a three dimensional
arrangement of spheres. The spheres of the
base layers is marked as A. The two types of
voids between the spheres are marked as ‘a’
and ‘b’ as shown below.
Cubic close packing
This is also a three dimensional arrangement of
spheres. The third layer of spheres are arranged in
such a manner that they cover the void ‘b’ as shown
in the figure.
The third layer of spheres are placed on the
void ‘a’ of base layer. This type of arrangement
given a stacking of layers A B – A B C A B ..
This arrangement packing is called the
Hexagonal Close packing.
e.g. Molybdenum, Magnesium and Beryllium
Crystalline in hcp Structures.
The third layer of spheres possess different
arrangement of spheres labelled as ‘c’. This pattern
of stacking of spheres gives A B C. arrangement
called cubic close packing.
in
e.g. Iron, Nickel, copper, Silver gold, aluminium
pack in ccp structures.
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ABABA ….or hcp arrangement of spheres
(b) Crystal lattice and unit cell
Crystal Lattice
A regular arrangement of the constituent
particles (i.e atoms, ions or molecules) of a
crystal in a three dimensional space is called
crystal lattice or space lattice.
(c) Tetrahedral void and Octahedral void
Tetrahedral Void
Unit Cell
The smallest three dimensional portion of a
complete space lattice which when repeated over
and again in different directions produces the
complete space lattice is called the unit cell.
Octahedral Void
In a close packed arrange ments of spheres,
The close packed arrangement gives two types of
voids or holes are created amongst the spheres voids. The void ‘X’ created by six spheres in contact
as shown in the Fig.
is called the Octahedral hole.
This void that is formed by
in
•
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8. How many lattice points are there in one unit cell of each of the
following lattice?
(a) Face centered cubic
(b) Face centered tetragonal
(c) Body centered
•
Solution:
In the case of cubic close packed or face centered cubic structure there are 8 atoms at each corner of
each unit cell. An atom at the corner is being shared among eight unit cells. Its contribution to the
unit cell is only
. 8 corner atoms x
atoms per unit cell =
= 1 atom. Similarly an atom at the
centre of the face is shared two unit cells. There are 6 face atoms. 6 face atoms ×€
cell =
atom per unit
= 3 atoms. The total number of atoms per unit cell I f.c.c. cubic structure is = 4 atoms.
(b) Face centered tetragonal structure has the number of atoms as that of f.c.c. cubic structure.
(c) Body centered cubic structure is the one which consists of atoms at the corner as well as at the
centre of the cube. Therefore the number of atoms per unit cell is = 2 atoms.
•
in
9. Explain:
(a) The basis of similarities and differences between metallic and ionic
crystals.
(b) Unit cell is not simply a cube of four sodium ions and four chloride ions.
(c) Can a cube consisting of Na+ and Cl- ions at alternate corners serve as a
satisfactory unit cell for the sodium chloride lattice.
(d) Ionic solids are hard and brittle.
Solution:
Metallic crystals are made up of atoms of one kind either crystallizing in h.c.p or c.c.p or body
centered cubic structure.
(a) In ionic crystals there are two types of ions. Cations and anions. Both may have different sizes
and charges. There is possibility that opposite charges attract each other and therefore there is a
tendency for ions of one type of charge to group around ions of the opposite charge.
(b) The unit cell of sodium chloride can be explained as follows: Chloride ions being bigger in size
adopt cubic close packed arrangement. Thus Cl-ions are present in all corners and at face centres.
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The smaller Na+ ions occupy the octahedral sites in the c.c.p arrangement . As the octahedral voids
are present at edge centres as well as at the centre of the body, Na atoms are present at the edge
centre as well as centre of the body.
The number of Sodium ions = 12 (At the edge centres) x
Number of chloride ions = 8 (at corners) x
+ 1 (at body centre) x 1 = 3 + 1 = 4
+6
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(at face centres) x
=1+3=4
Thus a unit cell of Na+ Cl- consists of four sodium ions and four chloride ions.
(c) It is not possible to get satisfactory unit cell for NaCl consisting of Na+ and Cl- ions at alternate
corners as it does not satisfy the condition that an ion occupies similar lattice points in the crystal
lattice.
(d) Ionic solids are made up of positive and negative ions arranged in a regular fashion. The forces of
attraction between the ions are electrostatic in nature. There is no free space between the ions. Their
position in the space is rigid. Therefore they are hard and brittle.
10. Calculate the efficiency of packing in case of a metal crystal for
(a) Simple cubic
(b) face centred cubic (with the assumptions that atoms are touching each
other)structure the atoms are at the corners and at the centre of the face.
(c) Body centred cubic structure
•
Solution:
(a)
No. of spheres per unit cell =
in
In this arrangement the atoms are present only at the corners.
Suppose the edge length of a unit cell = a
and radius of the sphere = r
As the spheres are touching each other a = 2 r
×8=1
Volume of the sphere =
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Volume of the cube = a3 = (2r)3= 8r3
∴ Fraction occupied =
% occupied = 52.4%
= 0.524
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(b)
The spheres on the face are touching the length AC = 4 r
From the right angled triangle A B C
∴
or a =
Volume of the unit cell = a3 =
=
No. of spheres per unit cell = 8 ×
Volume of 4 spheres = 4 ×
Fraction occupied =
% occupied = 74%
in
(c)
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The sphere at the centre touches the spheres at the corners. The body diagonal AD = 4r. Further the
face diagonal AC=
and body diagonal AD =
a.
a = 4 r or a =
Volume of the unit cell = a3 =
No. of spheres per unit cell = 8 ×
Volume of 2 spheres = 2 ×
Fraction occupied =
% occupied = 68%.
11. Silver crystallises in fcc lattice. If edge length of the cell is 4.077 × 108
cm and density is 10.5 gm cm-3. Calculate the atomic mass of silver.
•
Solution:
in
The density of silver unit cell is calculated using the formula ρ =
For f c.c. unit cell the number of Ag atoms per unit cell = 4
M is the atomic mass of silver
NA is the Avogadro’s number
10.5 =
M=
M = 107.4 gm mol-1
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12. A cubic solid is made of two elements P & 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 co-ordination numbers of P & Q?
•
Solution:
An atom at the corner is shared by eight other cubes. The number of atoms per unit cell
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is
Therefore there is one atom of Q per unit cell. Since atom P is at the body center, it contributes to
the unit cell wholly. The formula of the compound is PQ. The co-ordination number of P and Q are
8:8 respectively.
13. Niobium crystallizes in body centered cubic structure. If density is
8.55 gm cm-3. Calculate atomic radius of niobium using its atomic mass 93
u.
•
Solution:
Relationship between edge length and the density of the crystal is
x=
a3
Z for b.c.c. = 2
a3 =
a3 = 3.606 × 10-23
a3 = 36.06 × 10-24
a=
=
=
[1.5570 -24.0000
= 8.5190
The edge length is related to the radius by the relationship
r=
in
a =330.4 pm
a
r=
= × 330.4
r = 142.9 pm
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14. If the radius of the octahedral void is r and the radius of the atoms
in close packing is R, derive the relation between r and R.
Solution:
An octahedral void is created by six spheres touching each other. A sphere fitting in to the octahedral
void is known by shaded circle. A sphere above and a sphere below this small sphere are not shown
in the fig.
•
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BC2 = AB2 + AC2
(2R)2 = (R+r)2 + (R+r)2 = 2 (R+r)2
or
= (R+r)2
or
2
or
= (R+r)2
R = R+r
r=
R- R = R
r = R (1.414-1) = 0.414 R
r = 0.414 R
15. Copper crystallises into a f.c.c. lattice with edge length 3.61 x 10cm.Show that the calculated density is in agreement with its measured
value of 8.92 g cm-3.
8
•
Solution:
P=
P=
P = 8.97 gm/cc
The calculated and the observed value of densities of Cu are in agreement.
in
16. Analysis shows that Nickel oxide has formula NiO98O1.00. What
fractions of the nickel exists as Ni2+ and Ni3+ ions?
•
Solution:
Let the number of O2- ion in Ni0.98O1.00 = 100
The total number of Ni2+ and Ni3+ ions = 98
The total negative charged carried by
O2- ions = 200
Total positive charge carried by Ni2+ ions = 2x
Total positive charge carried by Ni3+ ions = 3 y
By the principle of electrical neutrality = 2x +3y = 200
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‘x’ is the total number of Ni2+ ions
‘y’ is the total number of Ni3+ ions
x + y = 98 (i)
2x + 3y = 200 (ii)
Solving eq (i) an (ii) the value of x = 94 and y = 4
The percentage of Ni2+ ions =
× 100 = 96%
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The percentage of Ni3+ ions =
∴ The % of Ni2+ ions = 96%
% of Ni3+ ions = 4%.
× 100 = 4%
17. What is a Semi conductor? Describe the two main types of semi
conductors and contrast their conduction mechanism.
•
Solution:
Semi conductors are the substances which posses conductivities ranging between 102 to 10-9ohm1
cm-1. In the case of semi conductors, the gap between the valence band and conduction band is
small. Therefore some of the electrons may jump from valence band to conduction band and some
conductivity is observed. The electrical conductivity increases with increasing temperature for semi
conductors. This is due to the fact that large number of electrons from the valence band can jump to
the conduction band. Pure substances that exhibit conducting behaviour like that of silicon and
germanium are called intrinsic semi conductors. In pure crystalline silicon and germanium four
valence electrons of each of pure atom forms covalent bonds with adjacent atoms at room
temperature. When Si crystal is doped with a group–15 element, such as P, As, Sb or Bi the
structure of the crystal lattice is left unchanged, but an occasional atom with five valence electrons
occupies a site that would normally be occupied by a silicon atom. The foreign atom has an extra
electron which is free to contribute to electrical conduction. Silicon that has been doped with a
group–15 element is called an ‘n’ type semi conductor.
Doping a silicon crystal with a group 13 element, such as B, Al, Ga or In produces a silicon crystal
structure in which an occasional dopant atom is present with only three valence electrons. The place
where the fourth valence electron is missing is called an electron vacancy or a hole. Such type of
semi conductors are called ‘p’ type semi conductors. The direction of motion of the holes in an
electric field is opposite to that of the electrons.
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 semi
conductor?
in
•
Solution:
In the non stoichiometric Cu2O the oxide ions form covalent linkages with neighbouring atoms by
contributing two valence electrons. Each oxide ion is bonded to two Cu2+ ions as shown in the fig:
As the ratio of Cu :O is not exactly equal to 2:1, electrical conduction arises. Such substances are
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semiconductors and the electrical transport is mainly by ‘holes’. These are called p-type
semiconductors.
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.
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•
Solution:
n ferric oxide the oxide ions forms hexagonal close packing with the occupation at corners and
Fe3+ ions occupy 2/3rds of the octahedral voids.
The number of oxide present per
Unit cell =
× 8 =1
In hexagonal close packing of atoms for every one atom the number of octahedral void =1.
∴ the formula of ferric oxide =
O1.
The Stoichiometric formula is Fe2 O3
20. Classify each of the following as being either a p- type or an n-type
semi conductor.
(i) Ge doped with In (ii) B doped with Si.
•
Solution:
Ge belongs to group-14. If Ge is doped with Indium of Group-13 it form p-type semi conductors. In
germanium crystal structure, Indium atoms with three valence electrons will be present. The place
where the 4th valence electron is missing is called an electron vacancy or a hole. Such holes (positive
in charge) appear to be responsible for electrical conductivity.
When boron is doped with silicon, n-type semi conductor is formed. The occasional atom of silicon
with four valence electrons occupies a site that would normally be occupied by a B atom with three
valence electrons. Since one extra electron is not needed for normal bonding, it is delocalized and it
contributes to conducting property of that element.
21. Gold (atomic radius = 0.414 nm) crystallises in a face centered unit
cell. What is the length of a side of the cell.
•
r=
a
0.414 =
a=
in
Solution:
For a face centered unit cell the relation between radius and edge length is
a
= 1.171 nm.
22. In terms of band theory, what is the difference
(i) between a Conductor and an insulator
(ii) between a Conductor and a semi conductor
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•
Solution:
(i) In a conductor the partially filled band overlaps with conduction band as shown in the fig
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In an insulation there is large energy gap between conduction band
(ii) In the case of semi conductors, the gap between valence band and the conduction band is small
and therefore some of the electrons may jump from valence band to the conduction band and some
conductivity is observed.
23. Explain the following terms with suitable examples. (i) Schottky
defect (ii) frenkel defect (iii) Interstitials (iv) F- centres.
in
•
Solution:
(i) Schottky defect
When an atom or anion is missing from its normal lattice site, a lattice vacancy is created. In
stoichiometric ionic crystal, a vacancy of one ion has to be accompanied by the vacancy of the
oppositely charged ion in order to maintain electrical neutrality. In NaCl, there are approximately
106schottky pairs per cm3 at room temperature. The presence of large number of schottky defects in
a crystal lowers its density markedly.
(ii) Frenkel defect
In a Frenkel defect the ion leaves its position in the lattice and occupies an interstitial void. This
defect is commonly found in ionic solids and does no change the density of the solids. This defect
arises mainly when there is large difference in size between cation and anion and also when co-
13
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ordination number is low.
(iii) Interstitials
Atoms or ions which occupy the normally vacant interstitial position in a crystal are
called Interstitials.
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(iv) F-Centres
Anion vacancies in alkali halides are produced by heating the allkali halide crystals in an atmosphere
of the alkali metal vapour. This is because the metal atoms deposit on the surface of the crystal of
alkali halide, and the halides, which migrate from the bulk to the surface, combine with the alkali
metals. The electrons set free here are moved into the bulk of the crystal and occupy the anionic
vacancies created because of the movement of anions. These electrons trapped in anionic vacancies
are referred to as F-centres. Excess K in KCl makes the crystal appear violet and excess Li in LiCl
makes it pink.
24. Aluminium crystallises in 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 cm3 of aluminium?
•
Solution:
For fcc, the relationship between radius and edge length is
r=
a
a=
= 353.60 pm
Volume of the unit cell = a3 = (353.60)3 × 10-30 cm3
Volume of the unit cell = 4.42 × 10-23 cm3
The number of unit cells in 1 cm3 of Al= 2.26 × 1022 units cells.
25. If NaCl is doped with 10-3 mol% of SrCl2, What is the concentration of
cation vacancies?
•
Solution:
A cation of Sr 2+ would create one cation vacancy in NaCl. Therefore, the number of cation vacancies
created in the lattice of NaCl is equal to the number of divalent Sr2+ ions added.
Therefore, concentration of cation vacancy on being doped with
in
10-3 mol% of SrCl2 =
= 10-5 mol
No. of Sr2+ ions in 10-5mol = 10-5 x 6.023 x 1023
= 6.023 x 1018 Sr2+ ions.
Therefore No. of cation vacancies= 6.023 x 1018.
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26. Explain the following with suitable example.
(a) Ferromagnetism
(b) Para magnetism
(c) Ferri magnetism
(d) Piezo electric effect
(e) Antifluorite Structure
(f) 12-16 and 13-15 compounds.
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Solution:
Ferro magnetism
Ferro magnetic substances show permanent magnetism even when the magnetic field is removed
(e.g. Fe, CrO2, Fe, Co, NiCl)
(b) Paramagnetism
These are the solids which possess permanent dipoles due to the presence of atoms, ions or
molecules with unpaired electrons (e.g. O2, Cu2+, Fe3+) and these materials are attracted by a
magnetic field. They lose their magnetism in the absence of a magnetic field.
(c) Ferri magnetism
In these materials the magnetic moments are aligned in parallel and anti-parallel directions in
unequal numbers resulting in a net moment (e.g. Fe3O4)
(d) Piezoelectric effect
These are the substances which convert mechanical strains to electrical energy and vice versa e.g.
Titanates or Ba and Pb. Lead Zirconate (PbZrO3). Ammonium dihydrogen phosphate (NH4H2PO4) and
quartz. We find Piezoelectric applications in microphones , photograph, pickups, etc.
(e) Antifluorite Structure
In the antifluorite structure the position of cations and anions as obtained in the fluorite structure get
reversed. In this structure the smaller cations occupy the positions of fluoride ions, and the larger
anions occupy the position of the Calcium ions in the fluorite structure.
For e.g. in Na2O the oxide ions form a ccp arrangement and Na+ ions occupy all the tetrahedral
voids. The other examples are Li2O, K2O, Rb2O, and Rb2S.
(f) 12-16 and 13-15 compounds
These are the solids formed by the combination of Zn, Cd and Hg with group 16 elements S, Se and
Te, e.g. ZnS, CdS, CdSe and HgTe. In these compounds the bonds are not perfectly covalent and
ionicity depends on the electronegativities of the two elements. These compound act as semi
conductors possessing an average valency of four as Ge or Si.
in
(g) 13-15 compounds
Typlical examples of 13-15 compounds InSb, AlP and GaAs. These substances are semi conductors
similar to Ge or Si. The bonds formed as the result of sharing of electrons.
27. Why are solids rigid?
•
Solution:
The constituent particles in solids have fixed positions and can only oscillate about their positions.
Hence solids are rigid in nature.
28. Why do solids have a definite volume?
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•
Solution:
Molecules, atoms or ions that make up a solid are closely packed; they are held together by strong
cohesive forces and cannot move at random. Hence solids have a definite volume.
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29. Classify the following as amorphous or crystalline solids:
Polyurethane, naphthalene, benzoic acid, potassium nitrate, cellophane,
PVC, fibre glass, copper.
•
Solution:
Amorphous solids
PVC, Polyurethane, cellophane, fibre glass
Crystalline solids
Copper, Potassium nitrate, fibre glass, naphthalene, benzoic acid.
30. Why is glass considered a super cooled liquid?
•
Solution:
If glass is heated gradually, it softens and starts to flow without under going a definite and abrupt
change into liquid state. Hence glass is considered as a super cooled liquid.
31. 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.
•
Solution:
If a particular physical property like refractive index has the same value along all directions, then
such solid is said to be amorphous. Because amorphous solids are isotropic in nature (possess same
physical properties in all directions). They won’t show cleavage property.
32. Classify the following solids in different categories based on the
nature of inter-molecular forces operating in them. Potassium sulphate,
tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium,
argon, silicon carbide.
•
Solution:
Electrostatic or coulombic inter molecular force exists in: Potassium sulphate, Zinc sulphide.
Force of attraction due to metallic bonding exists in: Tin, Rubidium.
in
Force of attraction due to covalent bonding exists in: Graphite, Silicon carbide.
Urea: Hydrogen bonding exists in: Ammonia, Water.
Benzene: Dispersion or London forces exist in: Argon.
33. Solid A is a very hard electrical insulator in solid as well as in molten
state and melts extremely at a high temperature. What type of solid is it?
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•
Solution:
Solid A must be a covalent or network solid. They are crystalline solids of non-metals result from the
formation of covalent bonds between adjacent atoms throughout the crystal. Covalent bonds are
strong and directional and hence the atoms in a covalent solids are held very strongly in their
positions. They have extremely high melting points and are extremely high melting points and are
insulators even in the molten state.
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34. Ionic solids conduct electricity in the molten state but not in solid
state. – Why?
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Solution:
Ionic solids are formed by the three dimensional arrangements of cations and anions bound by
strong electrostatic forces. These ions are not free to move about; hence they are electric insulators
in the solid state. However in the molten state, the ions are free to move and hence they conduct
electricity.
35. What type of solids are electrical conductors, malleable and ductile?
•
Solution:
Metallic solids are electric conductors, malleable and ductile (They have positive ions in a sea of
delocalized electrons).
36. Give the significance of a ‘lattice point’.
•
Solution:
Each point in a crystal lattice is called a lattice point. Lattice points are joined by straight lines to
bring out the geometry of the lattice.
37. Name the parameters that characterise a unit cell.
•
Solution:
Crystallographic axes, Interfacial angles and primitives are the parameters that characterize a unit
cell.
38. Distinguish between (i) Hexagonal and monoclinic unit cells.
(ii) Face centered and end centered unit cells
Solution:
(i) Hexagonal and monoclinic unit cells:
in
•
Hexagonal unit cell
The constituent particles of the unit cell are present only on the corner positions of a unit cell
(primitive).
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a= b ≠ c , α = β = 90˚ γ = 120˚
Monoclinic unit cell
Apart from the corners, one constituent particle is present at the centre of any two opposite faces
beside the ones present at its corners.
a≠ b≠ c, α = γ = 90˚, β ≠ 120˚
(ii) Face centered and end centered unit cells:
Face centered unit cell
Such a unit cell contains one constituent particle present at the centre of each face, besides the ones
that are at its corners.
in
End centered unit cell
In such a unit cell, one constituent particle is present at the centre of any two opposite faces besides
the ones present at its corners.
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39. What is the two dimensional co ordination number of a molecule in a
square close-packed layer?
•
Solution:
The two dimensional coordination number of a molecule in a square close-packed layer is four.
40. 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?
•
Solution:
Number of atoms of the compound = 0.5 x 6.023 x 1023
=
N = 3.011 x 1023
Number of tetrahedral voids = 2N
= 2 x 3.011 x 1023
= 6.023 x 1023
Total number of voids in 0.5 mol of the compound = 6.023 x 1023 + 3.011 x 1023
= 9.034 x 1023
in
Number of octahedral voids = N = 3.011 x 1023
Therefore total number of voids in 0.5 mol of the compound = 9.034 x 1023.
41. 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?
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•
Solution:
The ccp lattice is formed by the element N. The number of tetrahedral voids generated is equal to
twice the number of atoms of N. Since 1/3rd of tetrahedral voids are occupied by the atoms of M, the
ratio of the number of atoms of M and N is 2x1/3:1 or 2:3 and the formula of the compound is M2N3.
42. Which of the following lattices has the highest packing efficiency?
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(i) sc (ii) bcc and (iii) hcp lattice?
•
Solution:
Packing efficiency of: sc is 52.4%, bcc is 68% and hcp is 74% Hence hcp lattice has the highest
packing efficiency.
43. What type of defect can arise when a solid is heated? Which physical
property is affected by it and in what way?
•
Solution:
‘Metal excess defect’ is the type of defect that arises when a solid is heated. The colour of the solid is
affected by metal excess defect. It can be explained by having the following example. When alkali
metal halide crystals are heated in the atmosphere of the alkali metal vapours. The metal atoms get
deposited on the surface of the alkali metal halide crystal, halide ions move onto the surface and
combine with metal atoms. Thus the excess of potassium in KCl makes the crystal appear violet,
excess of lithium in LiCl makes the crystal appear pink.
44. What type of stoichiometric defect is shown by: (i) ZnS (ii) AgBr
•
Solution:
Ionic solids having low co ordination number generally exhibit Frenkel defect and in which anions are
larger in size than cations. Eg. ZnS, AgCl, AgBr.Frenkel defect is produced when some atoms or ions
are displaced from their normal sites and occupy interstitial sites. AgBr is a solid, which exhibits both
Frenkel and Schottky defects.
45. Explain how vacancies are introduced in an ionic solid when a cation
of higher valance is added as an impurity in it.
•
in
Solution:
When a cation if higher valence is added as an impurity, it replaces the already existing cations. Very
less number of sites are occupied due to its greater valency and hence other sites remains vacant.
For eg., If molten NaCl containing a little amount of SrCl2 is crystallised, some of the sites of Na+ ions
are occupied by Sr2+ ions. Each Sr2+ ion 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.
46. Ionic solids, which have anionic vacancies due to metal excess
defect, develop colour. Explain with the help of a suitable example.
•
Solution:
Metal excess defects due to anionic vacancies
In this case, negative ions may be missing from their lattice sites leaving holes in which the electrons
remain entrapped to maintain the electrical neutrality.
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47. 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?
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Solution:
Silicon, a group 14 element which have four valence electrons can be converted into n-type
semiconductor by doping it with a group 15 element like P or As which contains five valence
electrons. These elements will occupy some of the lattice sites in silicon crystal. Four out of five
electrons are used in the formation of four covalent bonds with the four neighbouring silicon atoms.
The fifth electron is extra and becomes delocalized. These delocalized electrons increase the
conductivity of doped silicon. Here the increase in conductivity is due to the negatively charged
electron, hence silicon doped with electron-rich impurity is called n-type semiconductor.
48. What type of substances would make better permanent magnets,
ferromagnetic or ferrimagnetic? Justify your answer.
•
Solution:
Ferromagnetic substances would make better permanent magnets than ferrimagnetic substances.
Substances like iron, cobalt, nickel etc., are attracted very strongly by a magnetic field and are called
ferromagnetic substances. When the substance is placed in a magnetic field all the domains (small
regions) get oriented in the direction of the magnetic field and a strong magnetic effect is produced.
This ordering of domains persists even when the magnetic field is removed and the ferromagnetic
substance becomes a permanent magnet. Ferri magnetic substances are weakly attracted by
magnetic field as compared to ferromagnetic substances.
in
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