The Solid State: CBSE Board - Chemistry - 12 NCERT Exercise With Solutions
The Solid State: CBSE Board - Chemistry - 12 NCERT Exercise With Solutions
The Solid State: CBSE Board - Chemistry - 12 NCERT Exercise With Solutions
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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,
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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 = = 90o , = 120o whereas in monoclinic type, these are
= = 90o , 90o.
(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,
1
th part of the atom at the
8
No. of octahedral voids is equal to the no. of atoms in the packing = 3.011 x 10
No. of tetrahedral voids = 2 x No. of octahedral voids
23
23
= 2 x 3.011 X 10 = 6.022 x 10
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 =
1
x tetrahedral voids
3
1
2
= 2n = n
3
3
Ratio of M : N is N is
2
n: In = 2:3
3
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18. An element with molar mass 2.7 x 102 kg mol1 forms a cubic unit cell with edge length 405 pm. If its density is
3
3
2.7 x 10 kg m , what is the nature of the cubic unit cell?
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M = 2.7 x 10 kg mol
23
m,
1
1
d NA a3
z M
z
=
M
NA a3
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
replaces two Na+ ions. One site is occupied by Sr2+ and the other site remains vacant. The cationic vacancies thus
2+
produced are equal to the number of Sr 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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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
Amorphous solids :
Plastic
=
(m)
Atomic mass
M
=
Avogadro number NA
z M
g / cm3
a 10 30 NA
3
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
M=
d a3 10 30 NA
g / mol
z
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
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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.
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 = 1
8
4 3
r
3
4 3
r
3
0.524
= =
8r 3
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 AC =
AB2 + BC2
a2 + a2
AD=
AC2 + CD2=
=
3a 4r
=
or a
2a2 + a2=
3a
4r
3
4r
64 r 3
3 3
1
+ 1= 2
8
4 3 8 3
r = r
3
3
8 3
r
= 3=
0.68
64r 3
3 3
or % of space occupied = 68%
Percentage of unoccupied space
=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,
AC =
AB2 + BC2 =
a2 + a2 =
2a
2a =
4r
or a =
4
2
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3
32 3
4
= a=
r =
r
2
2
3
1
1
= 8 + 6 = 4
8
2
Volume of 4 spheres = 4
4 3 16 3
r = r
3
3
16 r 3 / 3
= 0.74 or
32r 3 / 2
11. Silver crystallizes in fcc lattice. If edge length of the cell is 4.077 x 10
the atomic mass of silver.
Ans. We know that d =
or M =
z M
a3 NA
d a3 NA
z
3
23
1
= 107.12 g mol .
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 bodycenter. 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)
x1=1
Number of Q atoms per unit cell
= 8 (at corners)
1
=
1
8
10
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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.
3
z.M
a3 .NA
d=
or a3 =
z.M
d.NA
2 93 g mol1
8.55 g cm3 6.022 1023 mol1
a = [36.13 x 10
3 a
4
1.732 3.31 10 8 cm
= 14.29 10 7 cm
4
= 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,
AC =
AB2 + BC2 =
a2 + a2 =
2a
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 + 2R
=
a
2R
2
=
r
+1
R
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r
=
R
2 1= 1.414 1= 0.414.
rvoid
rsphere
= 0.414
d=
=
zM
a3 .NA
4 63.5 g mol1
(3.6110 8 cm)3 6.022 1023 mol1
d = 8.96 g cm3
3
Thus, the calculated value of density is in agreement with the measured value (8.92 g cm ).
2+
3+
16. Analysis shows that nickel oxide has formula Nio.98 O1.00. What fractions of nickel exist as Ni and Ni ions?
2+
ions = x
no. of Ni ions = 98 x
3+
2
Total ve charge on 100, O ion = 200
2+
ions = 2x
3+
Total +ve charge on (98 x), Ni 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 Ni
ions
=
2+
x
100
98
94
100 =96%
98
Percentage of N
i3+
ions = 10096 = 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 ?
+
2+
Ans. The ratio less than 2: 1 in Cu2O shows that some cuprous (Cu ) ions have been replaced by cupric (Cu ) ions.
2+
In order to maintain the electrical neutrality, every two Cu 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.
2
If the number of oxide ions (O ) per unit cell =1
3+
Then, no. of Fe ions=
2
octahedral holes
3
2
2
1=
3
3
a
2 2
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or a = r x 2 x
(ii) Frenkel
(iii) Interstitials
(iv) F-centres.
defect
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(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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1=
a
2 2
Now, 44.361 x 10
1
1cm3 contains =
44.36110 30
= 2.26 x 1022 unit cells
3
25. If NaCl is doped with 10 mot % of SrCl2, what is the concentration of cation vacancies ?
+
2+
Ans. 2 Na ions are replaced by 1 Sr ion,
2+
3
Each Sr ion causes one cation vacancy. Hence, concentration of cation vacancies on being doped with 10
3
mol % SrCl2 = 10 mol %
103
10 3
=
6.022 1023
mol =
100
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
2+
electrons. e.g., Cu , 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.
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.
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