NCERT CBSE 12th Class XII Standard 12 Chemistry Solved Exemplary Problems BookBank Publication

268 Pages·2019·8.253 MB·English

by  BookBank Teachers

1 NCERT 12th Chemistry Solution
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Table of Content
1. The Solid State
2. Solutions
3. Electrochemistry
4. Chemical Kinetics
5. Surface Chemistry
6. Principles and Processes of Elements
Isolation
7. The p-Block Elements
8. The d - and f -Block Elements
9. Coordination Compounds
10. Haloalkanes and Haloarenes
11. Alcohols, Phenols and Ethers
12. Aldehydes, Ketones and Carboxylic Acids
13. Amines
14. Biomolecules
15. Polymers
16. Chemistry in Daily Life
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Chapter 1
The Solid State
SSSSSuuuuummmmmmmmmmaaaaarrrrryyyyy
Solids have definite mass, volume and shape. This is due to the fixed position of
their constituent particles, short distances and strong interactions between them.
In amorphous solids, the arrangement of constituent particles has only short
range order and consequently they behave like super cooled liquids, do not have
sharp melting points and are isotropic in nature. In crystalline solids there is long
range order in the arrangement of their constituent particles. They have sharp
melting points, are anisotropic in nature and their particles have characteristic
shapes. Properties of crystalline solids depend upon the nature of interactions
between their constituent particles. On this basis, they can be divided into four
categories, namely: molecular, ionic, metallic and covalent solids. They differ
widely in their properties.
The constituent particles in crystalline solids are arranged in a regular pattern
which extends throughout the crystal. This arrangement is often depicted in the
form of a three dimensional array of points which is called crystal lattice. Each
lattice point gives the location of one particle in space. In all, fourteen different
types of lattices are possible which are called Bravais lattices. Each lattice can be
generated by repeating its small characteristic portion called unit cell. A unit cell
is characterised by its edge lengths and three angles between these edges. Unit
cells can be either primitive which have particles only at their corner positions or
centred. The centred unit cells have additional particles at their body centre (body-
centred), at the centre of each face (face-centred) or at the centre of two opposite
faces (end-centred). There are seven types of primitive unit cells. Taking centred
unit cells also into account, there are fourteen types of unit cells in all, which
result in fourteen Bravais lattices.
Close-packing of particles result in two highly efficient lattices, hexagonal
close-packed (hcp) and cubic close-packed (ccp). The latter is also called face-
centred cubic (fcc) lattice. In both of these packings 74% space is filled. The
remaining space is present in the form of two types of voids-octahedral voids and
tetrahedral voids. Other types of packing are not close-packings and have less
efficient packing of particles. While in body-centred cubic lattice (bcc) 68% space
is filled, in simple cubic lattice only 52.4 % space is filled.
Solids are not perfect in structure. There are different types of imperfections
or defects in them. Point defects and line defects are common types of defects.
Point defects are of three types - stoichiometric defects, impurity defects and
non-stoichiometric defects. Vacancy defects and interstitial defects are the
two basic types of stoichiometric point defects. In ionic solids, these defects are
present as Frenkel and Schottky defects. Impurity defects are caused by the
presence of an impurity in the crystal. In ionic solids, when the ionic impurity has
a different valence than the main compound, some vacancies are created. Non-
stoichiometric defects are of metal excess type and metal deficient type.
Sometimes calculated amounts of impurities are introduced by doping in
semiconductors that change their electrical properties. Such materials are widely
used in electronics industry. Solids show many types of magnetic properties like
paramagnetism, diamagnetism, ferromagnetism, antiferromagnetism and
ferrimagnetism. These properties are used in audio, video and other recording
devices. All these properties can be correlated with their electronic configurations
or structures.
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1. Define the term ‘amorphous’. Give a few examples of amorphous
solids.
Soln. A solid is said to be amorphous if the constituent particles are not
arranged in any regular fashion. They may have only short range order.
Amorphous solids are generally obtained when the melts are rapidly
cooled, e.g., glass, plastics, amorphous silica, etc.
2. What makes a glass different from a solid such as quartz? Under what
conditions could quartz be converted into glass?
Soln. Glass is an amorphous solid in which the constituent particles (SiO
4
tetrahedra) have only a short range order and there is no long range order.
In quartz, the constituent particles (SiO tetrahedra) have both short range
4
as well as long range orders. On melting quartz and then cooling it rapidly,
it is converted into glass.
3. Classify each of the following solids as ionic, metallic, molecular,
network (covalent) or amorphous.
(i) Tetraphosphorus decoxide, P O
4 10
(ii) Ammonium phosphate, (NH ) PO
4 3 4
(iii) SiC (iv) I (v) P
2 4
(vi) Plastic (vii) Graphite (viii) Brass
(ix) Rb (x) LiBr (xi) Si
Soln. P O - molecular, (NH ) PO - ionic, SiC - network(covalent),
4 10 4 3 4
I - molecular, P - molecular, plastic - amorphous, graphite - covalent,
2 4
brass - metallic, Rb - metallic, LiBr - ionic, Si - covalent
4. (i) What is meant by the term ‘coordination number’?
(ii) What is the coordination number of atoms :
(a) in a cubic close-packed structure?
(b) in a body-centred cubic structure?
Soln. (i) Coordination number is defined as the number of nearest
neighbours in a close packing. In ionic crystals, coordination number of
an ion in the crystal is the number of oppositely charged ions surrounding
that particular ion.
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(ii) (a) Coordination number of atoms in a cubic close-packed structure
is 12.
(b) Coordination 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 the dimension of its unit cell? Explain.
Soln. Suppose the edge of the unit cell = a pm
Number of atoms present per unit cell = Z
Atomic mass of the element = M
∴ Volume of the unit cell = (a pm)3 = (a × 10–10 cm)3
= a3 × 10–30 cm3
Mass of the unit cell a pm
Density of the unit cell =
Volume of the unit cell
Mass of the unit cell
= Number of atoms in the unit cell × Mass of each atom = Z × m
where m = mass of each atom
Atomic mass M
m= =
Avogadro's number N
0
Z×M
∴ Density of the unit cell, ρ = g/cm3
a3 ×N ×10−30
0
where edge, a is in pm and molar mass, M in g mol–1.
Atomic mass can be calculated by using the expression
ρ×a3 ×N ×10−30
M = 0 gmol−1
Z
6. ‘Stability of a crystal is reflected in the magnitude of its melting point’.
Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether
and methane from a data book. What can you say about the intermolecular
forces between these molecules?
Soln. 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
Higher the melting point, stronger are the forces holding the constituent
particles together and hence greater is the stability.
The intermolecular forces in water and ethyl alcohol are mainly hydrogen
bonding. Higher melting point of water as compared to alcohol shows
that hydrogen bonding 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
Waals forces (London dispersion forces).
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7. How will you distinguish between the following pairs of terms ?
(i) Hexagonal close-packing and cubic close-packing
(ii) Crystal lattice and unit cells
(iii) Tetrahedral void and octahedral void
Soln. (i) Hexagonal close-packing (hcp) : The first layer is formed
utilizing maximum space, thus wasting minimum space. In every
second row, the particles occupy the depressions (also called voids)
between the particles of the row (fig.). In the third row, the particles
are vertically aligned with those in the first row giving AB AB AB.......
arrangement. This structure has hexagonal symmetry and is known
as hexagonal close-packing (hcp) structure. This packing is more
efficient and leaves small space which is unoccupied by spheres. In
two dimension central sphere is in contact with six other spheres. Only
26% space is free. In three dimension, the coordination number is 12.
A single unit cell has 4 atoms.
Cubic close-packing (ccp) : Again, if we start with hexagonal layer of
spheres and second layer of spheres is arranged by placing the spheres
over the voids of the first layer, half of these holes can be filled by these
spheres. Presume that spheres in the third layer are arranged to cover
octahedral holes. This arrangement leaves third layer not resembling with
either first or second layer, but fourth layer is similar to first, fifth layer
to second, sixth to third and so on giving pattern ABC ABC ABC....... This
arrangement has cubic symmetry and is known as cubic close-packed
(ccp) arrangement. This is also called face centred cubic (fcc).
(ii) Crystal lattice : A regular arrangement of the constituent particles
(i.e., atoms, ions or molecules) of a crystal in three dimensional space is
called crystal lattice or space lattice.
Unit cells : The smallest three dimensional portion of a complete space
lattice which when repeated over and over again in different directions
produces the complete space lattice is called the unit cell.
(iii) The empty spaces left between closed packed spheres are called voids
or holes.
T T T T T T
O O O O O O
T T T T T
O O O O O O
(a) Octahedral voids : This void is surrounded by six spheres and
formed by a combination of two triangular voids of the first and second
layer. There is one octahedral void per atom in a crystal. The radius ratio
 
r
void
  is 0.414.
r
 sphere
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(b)Tetrahedral voids : These voids are surrounded by four spheres which
lie at the vertices of a regular tetrahedron. There are two tetrahedral voids
 r 
void
per atom in a crystal and the radius ratio   is 0.225.
r 
 sphere 
8. How many lattice points are there in one unit cell of each of the following
lattice?
(i) Face-centred cubic
(ii) Face-centred tetragonal
(iii) Body-centred
Soln. Lattice points in face-centred cubic or face-centred tetragonal
= 8 (at corners) + 6 (at face-centres) = 14
Lattice points in body-centred cube
= 8 (at corners) + 1 (at body-centre) = 9
9. Explain :
(i) The basis of similarities and differences between metallic and ionic
crystals.
(ii) Ionic solids are hard and brittle.
Soln. (i) Similarities : Both ionic and metallic crystals have electrostatic
forces of attraction. In ionic crystals, these are between the oppositely
charged ions. In metals, these are among the valence electrons and the
kernels. That is why both have non-directional bonds.
Differences : (a) 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 move. Hence, they can conduct electricity in the solid state.
(b) 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.
(ii) Ionic crystals are hard because they have strong electrostatic forces
of attraction among the oppositely charged ions. They are brittle because
ionic bond is non-directional.
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10. Calculate the efficiency of packing in case of a metal crystal for
(i) simple cubic
(ii) face-centred cubic (with the assumptions that atoms are touching each
other).
(iii) body-centred cubic
Soln. (i) In a simple cubic unit cell :
Suppose the edge length of the unit cell = a and radius of the sphere = r
a
As spheres are touching each other, evidently, a = 2r
1
No. of spheres per unit cell = ×8=1
8
4
Volume of the sphere = pr3 r r
3
Volume of the cube = a3 = (2r)3 = 8r3 a
4 
∴ Fraction occupied, i.e., packing fraction = pr3 /8r3 = 0.524
 
3 
or % occupied i.e., packing efficiency = 52.4%
(ii)In face-centred cubic structure : As sphere A
on the face-centre is touching the spheres at
the corners, evidently AC = 4r.
B C
But from right angled triangle ABC,
AC = AB2 + BC2 = a2 + a2 = 2a
4
∴ 2a= 4r or a= r
2
3
 4  32
∴ Volume of the unit cell = a3 = r = r3 a
 2  2 A B
1 1 C
No. of spheres in the unit cell = 8 × + 6 × = 4
8 2
4 16
Volume of four spheres = 4 × pr3 = pr3
3 3
16pr3 /3
∴ Fraction occupied i.e., packing fraction = = 0.74
32r3 / 2
or % occupied i.e., packing efficiency = 74%
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(iii) In body-centred cubic structure : As the
C
sphere at the body-centre touches the spheres at
the corners, body diagonal, AD = 4r. A
B
Further, face diagonal,
AC = AB2 + BC2 = a2 + a2 = 2a D
and body diagonal,
AD = AC2 +CD2 = 2a2 + a2 = 3a
4r
∴ 3a= 4r or a=
3
 4r 3 64r3
∴ Volume of the unit cell = a3 =  =
 3  3 3
1
No. of spheres per unit cell = 8 × +1= 2
8
4 8
Volume of two spheres = 2 × pr3 = pr3
3 3
8
pr3
3
∴ Fraction occupied i.e., packing fraction = = 0.68
64r3
3 3
or % occupied i.e., packing efficiency = 68%
11. Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10–8 cm
and density is 10.5 g cm–3, calculate the atomic mass of silver.
d × a3 × N 10.5×(4.07 ×10−8)3 ×(6.02 ×1023)
Soln. M = A =
Z 4
= 106.5 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-centre. What is the formula of the
compound? What are the coordination numbers of P and Q ?
Soln. As atoms Q are present at the eight corners of the cube, therefore
1
number of atoms of Q in the unit cell = ×8 =1
8
As atoms P are present at the body-centre, therefore, number of atoms of
P in the unit cell = 1
Hence, the formula of the compound is PQ.
Coordination number of each of P and Q = 8
13. Niobium crystallises in body-centred cubic structure. If density is
8.55 g cm–3, calculate atomic radius of niobium using its atomic mass 93 u.
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Soln. Given d = 8.55 g cm–3, M = 93 g mol–1, Z = 2 (for bcc), N = 6.022 × 1023,
A
r = ?
Using formula
M×Z 93× 2
a3 = = ⇒ a3 = 3.61 × 10–23 = 36.1 × 10–24
d×N 8.55 ×6.022×1023
A
∴ a = 3.304 × 10–8 cm = 330.4 × 10–12 m = 330.4 pm
For body-centred cubic,
3
r = a = 0.433 a = 0.433 × 330.4 pm = 143.1 pm
4
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.
Soln. R and r are the radii of the octahedral site and atoms respectively,
then from Pythagoras theorem we get
AC2 = AB2 + BC2
(2R)2 = (R + r)2 + (R + r)2
rBr
R R
or 2R= R + r ⇒ ( 2 −1)R= r
A R R C
∴ r = 0.414 R
15. Copper crystallises into a fcc lattice with edge length 3.61 × 10–8 cm.
Show that the calculated density is in agreement with its measured value of
8.92 g cm–3.
Soln. Given a = 3.61 × 10–8 cm, Z = 4 (for fcc), M = 63.5 (for Cu),
N = 6.022 × 1023, d = ?
A
Z×M 4×63.5
Using formula, d = = = 8.97 g cm–3
a3 ×N (3.61×10−8)3 ×(6.022×1023)
A
This value is in close agreement with the measured value (8.92 g cm–3).
16. Analysis shows that nickel oxide has the formula Ni O . What
0.98 1.00
fractions of nickel exist as Ni2+ and Ni3+ ions?
Soln. 98 Ni-atoms are associated with 100 O-atoms. Out of 98 Ni-atoms,
suppose Ni present as Ni2+ = x
Then Ni present as Ni3+ = 98 – x
Total charge on x Ni2+ and (98 – x) Ni3+ should be equal to charge on 100
O2– ions.
Hence, x × 2 + (98 – x) × 3 = 100 × 2
or 2x + 294 – 3x = 200 or x = 94
94
∴ Fraction of Ni present as Ni2+ = ×100 =96%
98
4
Fraction of Ni present as Ni3+ = ×100 =4%
98
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17. What is a semiconductor? Describe the two main types of semiconductors
and contrast their conduction mechanism.
Soln. Those solids which have intermediate conductivities ranging from
10–6 to 104 ohm–1 m–1 are classified as semiconductors. As the temperature
rises, there is a rise in conductivity value because electrons from the
valence band jump to conduction band.
(i) n-type semiconductor : When a silicon or germanium crystal is doped
with group 15 element like P or As, the dopant atom forms four covalent
bonds like Si or Ge atom but the fifth electron, not used in bonding, becomes
delocalised and continues its share towards electrical conduction. Thus
silicon or germanium doped with P or As is called n-type semiconductor,
n-indicative of negative since it is the electron that conducts electricity.
(ii) p-type semiconductor : When a silicon or germanium is doped with
group 13 element like B or Al, the dopant is present only with three
valence electrons. An electron vacancy or a hole is created at the place
of missing fourth electron. Here, this hole moves throughout the crystal
like a positive charge giving rise to electrical conductivity. Thus Si or Ge
doped with B or Al is called p-type semiconductor, p stands for positive
hole, since it is the positive hole that is responsible for conduction.
Positive hole
Silicon atom Mobile electron (no electron)
As B
Perfect crystal n-type p-type
18. Non-stoichiometric cuprous oxide, Cu O can be prepared in laboratory.
2
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?
Soln. The ratio less than 2 : 1 in Cu O shows cuprous (Cu+) ions have been
2
replaced by cupric (Cu2+) ions. For maintaining electrical neutrality, every
two Cu+ ions will be replaced by one Cu2+ ion thereby creating a hole. As
conduction will be due to the presence of these positive holes, hence 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.
Soln. Suppose the number of oxide ions (O2–) in the packing = 90
∴ Number of octahedral voids = 90