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KVS REGIONAL OFFICE, RAIPUR
STUDY MODULE FOR CHEMISTRY
CLASS - XII
(MINIMUM LEARNING CONTENTS)
UNIT -1
SOLID STATE
Marks=4
Terminology and Definitions:Solid : The substances having definite shape, Mass and volume are called solid.
Characteristic properties of solid :
[1] They have definite mass, volume and shape.
[2] The inter molecular distances are short.
[3] The inter molecular forces are strong.
[4] Their constituent particles have fixed positions and can only oscillate about their mean
positions.
[5] They are incompressible and rigid.
Types of Solids : On the basis of the arrangement of the constituents particles they are of
2 types :Crystalline Solid
2. Amorphous solids
1. Crystalline solid
[1] In a crystalline solid the particles (atoms, molecules or ions) are arranged in a regular
and repetitive
three dimensional arrangements
[2] These solids have sharp melting point.
[3] These solids are Anisotropic.
Note : Anisotropic substances/solids are the substances have different values of
physical properties (such as electrical conductivity, refractive index, thermal expansion
etc.) in different directions. (Phenomenon of showing different values of all physical
properties in different directions is known as Anisotropism.)
[4] These solids can undergo a clean cleavage.
[5] These solids are generally incompressible.
Examples: All the metallic elements like iron, copper and silver;
Non – metallic elements like sulphur, phosphorus and iodine and
Compounds like sodium chloride, zinc sulphide and naphthalene
2. Amorphous solid
[1] In amorphous solid the particles (atoms, molecules or ions) are arranged in an irregular
and non-repetitive three dimensional arrangements.
[2] Rapidly solidified liquids are amorphous substances, e.g. Glass, rubber etc.
[3] These solids are generally Isotropic.
Note : Inotropic substances/Solids are the substances have the same values of
physical properties (such as electrical conductivity, refractive index, thermal expansion
etc.) in all directions.
(Phenomenon of showing same values of all physical properties in all direction is known
as Isotropism.)
[4] These solids on cleavage form smaller pieces with non-planar faces.
[5] These solids do not have sharp melting point and boiling point i.e. they melt gradually
over a temperature range.
[6] These solids are compressible.
Que: What makes a glass different from a solid such as Quartz? Under what conditions
quartz could be converted into glass?
Ans: In glass, amorphous silica ( SiO2 ) is present. SiO4 tetrahedral have an irregular
arrangement. In quartz, crystalline silica ( SiO2 ) is present. SiO4 tetrahedral have a regular
arrangement. When quartz (SiO2 ) is melted and the melt is cooled very rapidly, quartz
converted into glass.
Que. Why glass is considered as a super cooled liquid?
Ans : Amorphous solids have a tendency to flow. Since glass is an amorphous solid , so it is
called super cooled liquid or pseudo solid.
Que. Why some glass objects from ancient civilizations are found to become milky in
appearance?
Ans : Glass becomes crystalline at some temperature. For which glass objects from ancient
civilizations become milky in appearance because of some crystallization.
# Fluid
Liquids and gases are called fluids because of their ability to flow. The fluidity is due to the
fact that the molecules are free to move about.
Que: Why glass pans fixed to windows or doors of old building are invariably found to be
slightly thicker at the bottom than at the top?
Ans : Due to fluidity property, the glasses flows down very slowly and make the bottom
portion slightly thicker.
Que: Why solids are rigid in nature?
Ans: The constituent particles in solid have fixed position and can only oscillate
about their mean position, for which solids are rigid.
Que : Name the factors which determine the stability of a substance.
Ans: [1] Intermolecular forces tend to keep the molecules or constituent particles closer.
[2] Thermal energy tends to keep them apart by making them move faster.
# Crystalline Solids can be classified in to the following categories on the basis of the
intermolecular forces acting between the constituent particles :
Types of solid
[1] Molecular solids
(a) Non-polar
Constituent
particles
Attractive forces
Examples
Physical
nature
Electrical conductivity
Melting
point
Insulator
Very low
Molecules
Dispersion or London forces Ar, CCl4 , H2 , I2 , CO2 Soft
(b) Polar
Dipole-dipole interactions
HCl , SO2
Soft
Insulator
Low
(c)Hydrogen bonded
Hydrogen bonding
H2O ( ice)
Hard
Insulator
Low
[2] Ionic solids
Ions
coulombic or electrostatic
NaCl, MgO,ZnS, CaF2 Hard but Insulator
brittle
[3] Metallic solids
Positive ions
in a sea of
delocalised
electrons
Metallic bonding
Fe, Cu, Ag, Mg
Covalent bonding
SiO2 ( quartz),
SiC, Diamond ,
AlN
[4] Covalent or
network solid
Atoms
Graphite
in solid state but
conductors in molten state and in
aqueous solutions
High
Hard but
malleable Conductors in solid state Fairly high
as well as in molten state
and
ductile
Hard
Insulator
Soft
Conductor
Very high
#Lattice point : This is a constituent particles (i.e. spheres, ions, atoms, molecules) of a
crystal.
# Space lattice or crystal lattice
A regular three dimensional arrangement of points in space is called space lattice or crystal
lattice. The points represent the constituent particles of the crystal.
# Unit cell
An unit cell is the smallest portion of the crystal lattice. When it is moved repeatedly a
distance equal to its own dimension along each direction, a three dimensional crystal lattice
is generated.
# Types of unit cell : The unit cells are of 4 types : -
1. Simple/Primitive Unit cell : It has lattice points (i.e. spheres, ions, atoms,
molecules) at all the corner of a cube. (Therefore, it has total 8 points)
2. Body Centred Unit cell : It has one lattice point at the centre of the cube along with
all the points at the corner of a cube. (Therefore, it has total 9 points)
3. Face Centred Unit cell : It has lattice points at the centre of each face along with all
the points at the corner of a cube. (Therefore, it has total 14 points)
Rank of unit cell ( Z )
The rank of unit cell of :-
The number of particles as points in a unit cell is known as rank.
To calculate rank, these points should be noted:- A point at each corner of unit cell is counted as 1
8
- The points on an edge are counted as 1
4
- The points at each face are counted as
[2] Body-centred cubic =
[3] Face-centred cubic =
1
2
1
8
1
8
1
X 8
X 8
= 1
X 8
8
+
+
1
=2
1
X 6
2
=
4
[4] In hexagonal system ,each corner atom shared by 6
hexagonal unit cell.,each face shared by 2 unit cell and
inside thebody there are three atoms
- The points within the unit cell is conted as 1
Z= (
#
[1] Primitive / simple cubic =
1
6
X 12 corners) + (
1
2
X 2 faces ) + ( 3 inside the body ) = 6
Packing in metallic crystal:-
The identical solid spheres can be packed in a number of ways.
[1] In the first layer the spheres are arranged in a hexagonal manner in which each sphere
is in contact with six other spheres.
[2] In the second layer the spheres will fit into the depression of the first layer.
[3] For the third layer, there are two possibilities:(a) The spheres can be placed in the depression of the second layer i.e. the third layer is
directly above the first and the fourth layer is directly above the second. This leads to the
arrangement ABABAB……
This type of arrangement is known as hexagonal closed packed (HCP) structure.
e.g. Zinc, magnesium crystallizes in this type of structure.
[b] Alternatively, the sphere can be placed in the depressions, of the second layer; do not
lie directly above the atoms of the first layer i.e. the spheres in fourth layer lie exactly
above the first, fifth above the second, sixth above the third and so on. This leads to the
arrangement ABCABCABC……….
This type of arrangement is cubic closed packed or face centered cubic arrangement.
e.g. Cu,Ag.Au crystallizes in this type of structure.
# Writing formula of a compound based on the position of its constituent atoms/ions :
# Co-ordination number
It is the number of atoms or spheres that surrounds the single sphere / atom in a crystal.
C.N of tetragonal arrangement = 3
C.N of tetrahedral arrangement = 4
C.N of octahedral arrangement = 6
C.N of body centered cubic arrangement = 8
Any close ( tight ) packing having C.N = 12 i.e. hcp and ccp i.e. fcc having C.N = 12
# Void / Hole/ Interstices
The space which is left in between the closest pack arrangement is called void. In
close packing two types of voids are created.
Relationship between edge length (a) and radius of the sphere ( r ) in unit cell :
For
FCC , a = 2 2 r
For BCC ,
.
.
a =
.
4
3
.
r
For simple cubic ,
a=2r
(Note : Sphere = Atom, molecule, ion or any constituent.)
# Relationship between ‘d’ & ‘a’ and ‘d’ & ‘r’
S.
No
1
2
3
Type of
unit cell
Simple
BCC
FCC
d (nearest neighbor distance)
and a(edge length)
d=a
d = a × 0.866
d = a × 0.707
a (Edge length)
and r (atomic radius)
r = a/2
r = a × 0.433
r = a × 0.3535
# EFFICIENCY OF PACKING : This is the space occupied by the sphere in a particular unit
cell. It is
generally shown in %
1. The % of packing fraction in Simple Cubic = 52.36%
2. The % of packing fraction in Body Centred Cubic = 68%
3. The % of packing fraction in Face Centred Cubic = 74%
# CRYSTAL DEFECT/Crystal IMPERFECTION/IRREGULARITY
(The following defects are found in crystals)
Crystal defect
Internal irregularities of crystals is known as crystal defect.
Line defect
Stoichiometric
defect
Crystal defect
Vacancy
defect
Schottky
defect
Interstitial Frenkel
defect
defect
Metal excess
defect
Point defect
Non-Stoichiometric
defect
Impurity defect
Colour centre
or F-centre
or Anion vacancy
Meta-deficiency
defect
Cation vacancy
Single crystals are formed when the process of crystallization occurs at extremely slow rate.
# Point defects are the irregularities or deviations from ideal arrangement around a point or
an atom in a crystalline substance.
# Line defects are the irregularities or deviations from ideal arrangement in entire rows of
lattice points.
#
# Point defects do not disturb the stoichiometric of the solid is known as stoichiometric
defect or intrinsic defect or thermodynamic defect.
# Vacancy defect – When some of the lattice sides are vacant, the crystal is said to have
vacancy defect. This results the decrease in density of the substance. This defect develops
when a substance is heated.
# Interstitial defect- When some constituent particles occupy an interstitial site, the
crystal is said to have interstitial defect. This defect increases the density of the substance.
# Schottky defect
[1] It occurs when a pair of ions of opposite charge are missing from the ideal lattice.
[2] The presence of a large number of Schottky defect in a crystal lowers its density.
[3] This defect occurs if cation and anion having similar size with high coordination number.
[4] It is found in NaCl (there is one schottky defect for 1016 ions. One c.c of sodium chloride
contains 1022 ) ions. Therefore, one cubic centimetre (c.c.) of NaCl possesses 106
Schottky pair of ions.
# Frenkel defect
[1] When an ion leaves its position in the lattice and occupy interstitial site leaving a gap in
the crystal i.e. it creates a vacancy defect and interstitial defect.
[2] This defect will occur if size of cation is smaller than anion, with low coordination
number.
[3] Frenkel defects are not found in pure alkali halide. Due to larger size of cations, ions
can not accommodate in interstitial site.
[4] Frenkel defect are found silver halide (AgX), because silver ions are smaller in size and
can get into the interstitial site.
[5] The Frenkel defect does not change the density of the solid.
[6] In AgBr, AgCl both Schottky and Frenkel defects are found.
# Differences between Schottky and Frenkel defects : ( These defects are
classified under stoichiometric defects i.e. ionic ratio of compound remains same.)
Schottky defect
Schottky defects
Ions leave the lattice structure
Electrical neutrality is maintained
Density of crystal decreases
Found in compounds with higher
coordination no.
Ionic size should be same e.g. NaCl,
CsCl, AgBr, AgCl etc.
Frenkel defect
Frenkel defects
Ions slip into voids in the lattice structure
Electrical neutrality is maintained
Density of crystal remains same
Found in compounds with lower coordination no.
Ionic size difference is more, Frenkel defect is
found silver halide (AgX), because silver
ions are smaller in size and can get into
the interstitial site. e.g. AgBr, AgCl etc.
N.B.- In AgBr, AgCl both Schottky and Frenkel defects are found.
# Metal excess defect :
[a] F-Centers ( here F means Farbe (i.e. colour ) : When there is an excess of
metal ions in non-stoichiometric compounds, the crystal lattice has vacant anion
site. The anion sites occupied by electrons are called F-centre.
The F-centers are associated with the colour of the compounds. Excess of K
in KCl makes the crystalviolet. Excess of Li in LiCl makes the crystal pink.
Solid containing F-centre are paramagnetic, because the electrons occupying
the F-centers are unpaired.
When the crystal having F-centers are exposed to light, they become
photoconductor.
# F- centres & its importance
‘F’ centres : F-centres are sites in an
imperfect crystal where electrons are trapped in
the anion vacancies.
F-centre
Importance of ‘F’ centres: F-centres are
responsible for imparting colour to the crystals
[b] Metal excess defect due to presence of extra cation at the interstitial site
Zinc oxide is white in colour at room temperature on heating, it loses oxygen and
turns yellow.
Now there is excess of zinc in the crystal and its formula
becomes Zn 1+ x O the excess zinc ions move to interstitial site and the electrons to
neighboring interstitial site.
# Metal deficiency defect
FeO, mostly found with a composition of Fe 0.95 O i.e. range from Fe0.93O – Fe0.96O.
In crystals of FeO, some Fe2+ ions are missing and the loss of positive charge is
made up by the presence of required number of Fe3+ ions.
# Impurity defect
This defect arises due to cation vacancies, in some ionic compounds presence of
impurity produce some defects they are called impurity defects.
e.g. 1. Presence of Cd++ ion as impurity in AgCl.
2. Presence of Sr++ ion as impurity in NaCl.
# Doping :
The process of adding impurities to a crystal so as to change its electrical
properties is called Doping.
e.g. Mixing of SrCl2 in NaCl solid.
Note : 1. Doping increases the conductivity of crystal.
2. Doping is kind of impurity defect.
For example, if we mix Strontium chloride ( SrCl2 ) with Sodium chloride, some
strontium ( Sr2+ ) ions occupy the lattice sites of sodium ions ( Na +) and equal
number of sodium ( Na+ ) sites remain vacant. Such vacancies in the crystal
increase the electrical conductivity because certain ions from the neighboring sites
can move into these vacant holes. In this defect the number of positive ions are less
as compared to negative ions. Crystals with such defects also act as semiconductor.
Since the conductivity is due to holes, these are known as p-type semiconductors.
Cation Vacancy
Fig. Doping of SrCl2 in NaCl
Que: If NaCl is doped with 10-3 mol % of SrCl2. What is the concentration of
cation vacancies?
Ans : The addition of SrCl2 to NaCl produces cation vacancies equal in number to
that of Sr2+ ions.
No. of moles of SrCl2 added to one mol of NaCl = 10-3 / 100 = 10-5 mol.
No. of holes created in one mole of NaCl = 10-5 X 6.023 X 1023 = 6.023 x 1018
# Semiconductor
These are solids whose conductivity lies in between those of conductors and insulators. The
conductivity of semiconductors increases with increase of temperature.
# Intrinsic semiconductor
An insulator capable of conducting electric current at higher temperature or when
irradiated with electromagnetic radiations, are known as intrinsic semiconductor.
This happens because certain covalent bonds are broken and the released electrons are in a
position to conduct electric current. e.g. Silicon , Germanium.
# Extrinsic semiconductor
These are formed when impurities of certain elements are added (doped) to insulator.
# n-type semiconductors
It is obtained by doping group – 14 elements with group – 15 elements.
Suppose Si (having 4 valence electrons) is doped with P (with 5 valence electrons), out of 5
valence electrons of P, only 4 valence electrons are involved in bond formation. The 5th
electron is not bound any where and can be easily promoted to the conduction band. The
conduction is thus mainly caused by the movement of electrons.
# p-type semiconductors
It is obtained by doping group – 14 elements with group – 13 elements.
Suppose Si (having 4 valence electrons is doped with Ga (which has 3 valence electrons), 3
valence electrons are involved in bond formation with neighboring Si atom.
A vacancy is left which can be filled by the transfer of a valence electron from a neighboring
Si atom. The movement of electron into the vacancy leaves behind a hole which carries
positive charge. Another electron from a neighboring Si atom can move into the hole leaving
behind another hole. It appears as if the hole has moved through the lattice.
The movement of positively charged hole is responsible for the conduction of charge.
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?
Que.
Ans : Since Cu2O is non-stoichiometric oxide, it contains Cu in two oxidation states, +1 and
+2. Cu2+ provides an excess of positive charge. As a result an electron from a neighboring
Cu+ is transferred to Cu2+. The transfer of electron leaves behind a hole, which carries an
extra positive charge and a negative hole is created. It appears that the positive hole moves
through the lattice, hence it appears as P-type semiconductor.
# 12 – 16 and 13 – 15 compounds
Combination of elements of Gr – 13 and Gr – 15 or Gr – 12 and Gr – 16 produce
compounds which stimulate average valence of four as in Ge or Si .
12 – 16 compounds –> ZnS, CdS, CdSe, HgTe
13 – 15 compounds –> InSn, AlP, GaAs
# The Band theory to explain electrical properties :
The band model of metal is based on molecular orbital theory. When a large no. of orbital
overlap in metal, it results a continuous energy level produced by a large number of
molecular orbital is called Energy Band
2p0
2p band ( conduction band )
2s2
2s- band ( valence band )
Valence band
111-63
Kj/mol
511 Kj/mol
Coduction band
[ conductor ]
1s2
[ Semi conductor ]
[ Insulator ]
1s- band
Conductivity of Insulator
10 -20 - 10-10 ohm -1 m -1
Semiconductor
10 - 6 - 104 ohm -1 m -1
Conductor
10 4 - 107 ohm -1 m -1
The lowest unoccupied energy band is known as conduction band
The highest occupied energy band is known as valence band
The energy difference between the top of valence band and the bottom of the conduction band is known as energy gap.
or forbidden zone.
# Electrical properties of solids- There
are three kinds of electrical properties (i.e.
solids may be classified into three categories on the basis of their electrical properties:)
1. Conductors
2. Insulators
3. Semiconductors
Conductors- Substances which
allow electricity to pass through
them easily
eg- Metals
Insulators- Substances which do
not allow electricity to pass
through them easily.
eg- Rubber, wood, glass
Semiconductors- Substances
which allow electricity to pass
through them partially.
eg- Doped Silicon
Conductors: The solids having
no gap (there is no forbidden
zone) between conduction band
and filled band.
Since energy gap is not there
therefore electrons can
easily/freely move from one
band to another band.
Insulators: There is large gap
(large forbidden zone)
between conduction band and
filled band.
Electrons can not move
easily/freely
Semiconductors : There is
small gap (Small forbidden
zone) between conduction
band and filled band.
Electrons can move/jump
when energy is supplied.
# Magnetic properties :
[1] Diamagnetic Property/Substance/ Solids :
Diamagnetic substances are the substances which are weakly repelled by a magnetic
field. The electrons in diamagnetic substances are all paired. They do not contain unpaired
electrons. e.g. TiO2 , NaCl , C6H6 ,N2 , Zn
[2] Paramagnetic Property/Substance/ Solids
Paramagnetic substances are those which are attracted by a magnetic field but they
lose their magnetism in the absence of magnetic field. These substances have permanent
magnetic dipole, due to presence of atoms, molecules or ions containing unpaired
electrons. e.g. Cu2+, Fe3+, O2 , NO, CuO, etc.
Substances containing unpaired electrons are further classified as:
(a)
Ferromagnetic Property/Substance/ Solids –> Ferromagnetic substances are those
substances which are strongly attracted by a magnetic field and can be
made into permanent magnets. These substances show magnetism even in
the absence of a magnetic field. The large magnetism in these substances
is due to the spontaneous alignment of magnetic moment,i.e Unpaired electron in the
same direction. These substances lose ferromagnetism and become paramagnetic on
certain temperature on heating. (This particular Temp. is called Curie Temperature)
Example: Iron, cobalt, nickel, gadolinium and CrO2
CrO2 is used to make magnetic tapes for audio recording.
(b) Anti-ferromagnetic Property/Substance/ Solids –> Antiferromagnetic substances are those substances in which equal
number of magnetic moments are aligned in opposite directions so as
to give zero net moment. Example: MnO, MnO2 and Mn2O3
(c) Ferrimagnetic substances
Ferrimagnetism is observed when the magnetic moments
of the domains in the substance are aligned in parallel
and anti-parallel directions in unequal numbers They are weakly
attracted by magnetic field as compared to ferromagnetic substances.
eg : Fe3O4 (magnetite) and ferrites like MgFe2O4 and ZnFe2O4
These substances lose ferrimagnetism and become paramagnetic on certain
temperature(called Curie Temperature). on heating.(This particular Temp. is called
Curie Temperature)
A Few Numericals
:
Que: Gold (At.radius= 0.144nm) crystallizes in aFCC unit cell. What is the length of
the side of the unit cell?
Ans : Atomic radius= 0.144 nm
So, length of side a = 2 √ 2 r = 2 √ 2 x 0.144 = 0.406 nm
Que: Aluminum crystallizes in a cubic closed packed structure. Its metallic radius is 125 pm.
[1] What is the length of the side of the unit cell ?
[2] How many unit cells are there in one c.c of aluminium?
Ans : [1] For FCC , a = 2 √ 2 r = 2 √ 2 x 125 = 354 pm, So the edge length of the unit cell
= 354 x 10 -10 cm
[2] Volume of unit cell = a3 = ( 354 x 10 -10 ) 3 cm 3
Therefore, number of unit cells in 1 cc of aluminum = 1/ ( 354 x 10 -10 ) 3 cm 3
= 2.254 x 10 22 Unit cells
Que: Ferric oxide crystallizes in a hexagonal closed pack array of oxide ions with two
out of every three octahedral holes occupied by ferric ions. Deduce the formula of the
ferric oxide.
Ans : No. of atoms in one unit cell of hcp structure = 6
No. of oxide ions per unit cell = 6
No. of octahedral voids = 6
Since ferric ions occupy only two out of every three octahedral voids, therefore, no. of
octahedral voids occupied by ferric ions = ( 2/3 ) x 6 = 4
Stoichiometric ratio of Fe 3+ and O 2- is 4 : 6 = 2 : 3
Hence the formula of ferric oxide is Fe2O3
Que : Analysis shows that nickel oxide has the formula Ni 0.98 O 1.00. What fraction of
nickel exist as Ni2+ and Ni3+ ions?
Ans : Let amount of Ni3+ be x mol.
Then amount of Ni2+ is (0.98 – x)
Total oxidation number of Ni in the compound is 3x + 2 (0.98 – x)
Oxidation number of oxygen is -2
Since the sum of the oxidation number of all the constituents in a compound is zero.
=> 3x + 2 (0.98 – x) – 2 = 0
=> 3x + 1.96 – 2x – 2 = 0
=> x = 0.04
Hence % of Ni3+ = ( 0.04 / 0.98 ) x 100 = 4.08 %
% of Ni2+ = 100 – 4.08 = 95.92 %
Que : Relationship between density (d) and the dimension of unit cells.
Ans : Let the edge length of unit cell be a
Therefore volume of unit cell = a 3
Let no. of atoms in unit cell = Z
Gm. Atomic mass = M
Therefore mass of one atom = M / NA
Where NA = Avogadro’s number i.e. 6.023 x 1023
Mass of Z atoms = ZM
NA
Density of unit cell ( d ) =
Mass of unit cell
Volume of unit cell
=
ZM
NA
d=
a3
Silver crystallizes in FCC lattice. If edge length of the cell is 4.05 x 10
density is 10.5 gm/cm3 . Calculate atomic mass of silver.
Que :
Ans : Given data-
Edge length ( a ) = 4.07 X 10
Density ( d ) = 10.5 gm/ cm3
–8
ZM
a3 NA
cm and
cm
–8
Since silver crystallizes in fcc lattice, so rank , i.e. the no. of silver atoms per unit cell ( z ) = 4
NA = 6.023 X 10
So d =
23
ZM
a3 NA
M=
d a3NA
z
10.5 X ( 4.07 X 10 -8 ) 3 X 6.023 X 10 23
4
= 107.1 gm / mol
M =
So atomic mass of silver = 107.1 amu
A quick Revision
TERMS
:
EXPLANATION
Amorphous and Crystalline
Solids
Amorphous- short range order, Irregular shape eg-glass
Crystalline Solids- long range order, regular shape eg : NaCl
Molecular solids
Ar, CCl4, H2O (ice)
Covalent or Network solid
SiO2
No of atoms per unit cell (z )
Simple cubic -1, BCC- 2, FCC – 4 , End-Centred- 2
Coordination Number
FCC- 6:6
Calculation of number of voids
Let the number of close packed spheres be N, then:
The number of octahedral voids generated = N
The number of tetrahedral voids generated = 2N
Relation between r and a
Simple Cubic a = 2r ,
BCC 4r = a√3
Packing Efficiency
Simple Cubic52.4% ,
BCC 68% ,
diamond
BCC- 8:8
FCC 4r = a√2
FCC 74%
Calculations Involving Unit Cell
Dimensions
M=molar mass (g/mol)
a = edge length in
cm ,
NA = 6.023× 1023
Frenkel Defect:
This is a defect in which Cation is dislocated to an interstitial site.
It does cause change in the density of the solid. Frenkel defect is
shown by ionic substance in which there is a large difference in
the size of ions, for example, ZnS, AgCl, AgBr and AgI due to
small size of Zn2+ and Ag+ ions.
Schottky Defect
It is a vacancy defect. In this defect the number of missing
cations and anions are equal. This defect causes decreases in
density of the crystal. For example, NaCl, KCl, CsCl and AgBr.
Metal excess defect due to
anionic vacancies (F-centres )
When NaCl heated in an atmosphere of Na vapour, the Na atoms
deposited on the surface of the solid. The Cl– ions diffuse to the
surface of the crystal and combine with Na atoms to give NaCl.
The released electrons diffuse into the crystal and occupy anionic
sites The anionic sites occupied by unpaired electrons are called
F-centres They impart yellow colour to the crystals of NaCl.
Similarly, excess of lithium makes LiCl crystals pink and excess
of potassium makes KCl crystals violet (or lilac).
Doping
The conductivity of intrinsic semiconductors is increased by
adding an appropriate amount of suitable impurity. This process is
called doping
n / p -type semiconductors
n- type : Si + As or Sb or Bi
Tl
13 –15 compounds
13 – 15 compounds: InSb, AlP and GaAs.
12–16 compounds
12 – 16 compounds :ZnS, CdS, CdSe and HgTe
Paramagnetic substances
Weakly attracted by a magnetic field. Examples: O2, Cu2+, Fe3+, Cr3+
Diamagnetic substances
Weakly repelled by a magnetic field. Example: H2O, NaCl and C6H6
Ferromagnetism:
A few substances like iron, cobalt, nickel, gadolinium and CrO2
are attracted very strongly by a magnetic field.
Antiferromagnetism
Domains oppositely oriented and cancel out magnetic moment e.g.
MnO.
Ferrimagnetism:
domains aligned in parallel and anti-parallel directions in unequal
numbers . Example: Fe3O4 (magnetite)
p-type: Si + B or Ga or In or
(important questions with answer)
Q.1) What are the stoichiometric defects found in ionic crystals?
(1 Mark)
(Ans) The point defects in which the ratio of cations to anions remains the same as shown by the
molecular formula of the compound are known as stoichiometric defects.
Q.2) Write two examples of amorphous solids.
(Ans) Rubber and quartz glass are examples of amorphous solids.
(Q.3) Name two metals which have cubic close packed structure.
(1 Mark)
(1 Mark)
(Ans) Silver (Ag) and copper (Cu) have cubic close packed structure.
(Q.4) Name the type of solids which are malleable, ductile and electrical conductors.
(1 Mark)
(Ans) Metallic solids are malleable, ductile and electrical conductors.
(Q.5) Sometimes, crystals of common salt (NaCl) are yellow instead of being pure
white. Why?
(1 Mark)
(Ans) Sometimes, crystals of common salt (NaCl)are yellow instead of being pure white due to
the presence of electrons in some lattice sites in place of anions. These sites act as F-centers and
impart yellow colour to crystal of common salt.
(Q.6) What are voids?
(1 Mark)
(Ans) Voids are the empty spaces present between atoms or ions, when they are packed within
the crystal.
(Q.7) How many atoms are present in the unit cell of bcc and fcc?
(1 Mark)
(Ans) Number of atoms in unit cell of bcc is2and in fcc it is 4.
(Q.8) Fe3O4 is ferrimagnetic at room temperature. What happens to its magnetic
properties when it is heated to 850 K?
(1 Mark)
(Ans) When Fe3O4 is heated to 850 K it loses ferrimagnetism and becomes paramagnetic.
(Q.9) Name the point defect which lowers the density of a crystal.
(1 Mark)
(Ans) Schottky defect lowers the density of a crystal.
(Q.10) Write the unit in which the magnitude of magnetic moment is measured.
(1 Mark)
(Ans) The magnetic moment is measured in Bohr magneton (
(Q.11) What do you mean by F-center?
B).
(1 Mark)
(Ans) The anion sites which are occupied by unpaired electron are called F-centres.
Q.(12) Name the type of solids which have long range orders.
(1 Mark)
(Ans) crystalline solids have long range order.
(Q.13) Name one solid which has both Frenkel and Schottky defects?
(1 Mark)
(Ans) Silver bromide (AgBr) has both Schottky and Frenkel defects.
(Q.14) Why is the window glass of the old building thick at the bottom?
(1 Mark)
(Ans) Glass is a pseudo solid, that is, it is a supercooled liquid of high viscosity. It flows down
very slowly and makes the bottom portion of window glass of old building slightly thicker .
(Q.15) What causes the conduction of electricity by semiconductors?
(1 Mark)
(Ans) Electrons and holes produced by defects cause the conduction of electricity by
semiconductors.
(Q.16) Name the salt which can be added to AgCl so as to produce cationic vacancies.
(1 Mark)
(Ans) SrCl2orCdCl2 is added to AgCl to produce cationic vacancies.
(Q.17) What is a diode?
(1 Mark)
(Ans) Diode is a combination of n- type and p- type semiconductors. It is used as a rectifier.
(Q.18) Name a transition metal oxide which has appearance and conductivity like that
of Cu?
(1 Mark)
(Ans) Rhenium oxide (ReO3)
(Q.19) What is the co-ordination number of an atom present in octahedral void? (1 Mark)
(Ans) The co-ordination number of atom present in octahedral void is 6.
(Q.20) Write one property which is caused due to the presence of F-center in a solid.
(1 Mark)
(Ans) The colours and paramagnetic behaviour of the solid is due to the presence of F-center in a solid.
(Q.21) Draw body centred and face centred cubic unit cells.
(2 Marks)
(Ans)
(Q.22) Define: (i) Crystal lattice
(ii) Packing efficiency
(Ans) (i)Crystal lattice is the well-defined regular arrangement
(2 Marks)
of atoms, molecules or
ions in three-dimensional space.
(ii) Packing efficiency is the percentage of the total space filled by the particles.
(Q.23) Explain the following with suitable example:
(i) Molecular solids
(ii) Metallic solids
(2 Marks)
(Ans) (i) In molecular solids, the constituents particles are molecules which are held
together byvan der Waals forces of attraction. e.g. I2.
(ii)Metallic solids consist of positive ions surrounded by and held together by a sea of
free electrons. These free electrons are mobile and are responsible for high thermal and
electrical conductivity of metallic solids. e.g. Ag, Cu.
(Q.24) What is rank? Find rank of face centered cubic unit cell.
(Ans)
(2 Marks)
Rank is the number of atoms per unit cell of a crystal.
In f c c Contribution of atoms present at the corners=
Contribution of atoms present at faces =
Rank = 1 + 3 = 4.
(Q.25) What is doping? How does n-type and p-type semiconductors formed? (2 Marks)
The process of introducing atoms of other elements as impurity into an insulator to
make it semiconductor is called doping. Doping of silicon or germanium with electron
rich impurities like P, As, Sb results in formation of n-type semiconductors whereas ptype semiconductors are formed by adding elements of group 13 like B, Al,Ga.
(Ans)
(Q.26) Explain the following with one example each:
(i)Ferrimagnetism
(ii)Antiferromagnetic substances
(2 Marks)
) (i) When magnetic moments are aligned in parallel and antiparallel directions in
unequal numbers it results in net moment. It is called ferrimagnetism. These substances are
weakly attracted by magnetic field as compared to ferromagnetic substances.
e.g.: Magnetite (Fe3O4)
(Ans
(ii) Antiferromagnetic substances are expected to posses paramagnetism or
ferromagnetism but actually they possess zero net magnetic moment. It is due to the
presence of equal number of domains in opposite direction. e.g.: MnO.
(Q.27)
(Ans)
Write two differences between isotropy and anisotropy.
(2 Marks)
Isotropy
Anisotropy
(i)These substances show identical values of These substances show different values of
physical properties in all directions.
physical properties in different directions.
(ii) Amorphous solids show isotropy.
Crystalline solids show anisotropy.
(Q.28) Potassium metal crystallizes in bcc. The edge length of unit cell is 4.3 A0. Find the radius
of potassium atom.
(2 Marks)
(Ans)
(Q.29) A solid is made of two elements A and B. Atoms of element A occupy all the tetrahedral
sites while atoms of element B are in ccp arrangement. From this data find the formula of the
compound.
(2 Marks)
There are 2 tetrahedral sites per atom of B because atoms of element B have ccp arrangement.
(Ans)
There are 2 atoms of element A for each atom of element B because all tetrahedral sites are occupied by
atoms of element A. Therefore, the formula of the compound is A2B.
(Q.30) A solid A+ B– has NaCl closed packed structure. The radius of anion is 245 pm. Find radius of anion.
(2 Marks)
(Ans) The co-ordination number of A+ B– = 6 (
It has NaCl type structure.)
For this,
(Q.31) The edge length of unit cell of NaCl crystal lattice is 5.6A0. The density of NaCl is 2.2g cm–3. Find the number of
formula units of NaCl per unit cell.
(2 Marks)
(Ans)
(Q.32)
(Ans)
Find the number of NaCl molecules in a unit cell of its crystal.
(2 Marks)
No. of NaCl molecules in a unit cell of NaCl= 4
(Q.33) A compound is made of two atoms X and Y. Atom X is arranged in ccp and atom Y occupies
tetrahedral site. Find the formula of compound.
(2 Marks)
(Ans)
No. of atoms of X = 8
No. of atoms of Y =(8/8 ) + (6/2) = 4
Ratio of X : Y is 2: 1
Formula of compound is X2Y.
(Q.34) The unit cell of metallic silver is face-centred cubic. What is the mass of a silver unit cell?
(Ans)
(2Marks)
(Q.35) A metal crystallizes as face centered cubic lattice with edge length of 450pm. Molar mass of metal is
50g mol–1. Find the density of metal.
(2 Marks)
(Ans)
(Q.36) A solid has bcc structure. Distance of closest approach between two atoms is 1.73A 0. Find edge length
of cell.
(2 Marks)
(Ans)
Or
In bcc, distance of closest approach =[( 3)/2] edge length
Edge length = [1.73/ 3] 2A
= 200pm
(Q.37) Classify the following solids as metallic; molecular, amorphous, covalent or ionic.
(i) SO2
(ii) Diamond
(iv) Ag
(iii)
(v) Quartz
I2
(iv)
MgO
(vi) Ar
(3 Marks)
Metallic solid - Ag
Covalent solid - Quartz
Molecular solids - I2, Ar, SO2
Ionic solids - MgO
(Ans)
(Q.38) (i) What are voids?
(ii) How a tetrahedral void is different from octahedral void?
(iii) Draw structure of tetrahedral and octahedral void.
(3 Marks)
(i) Atoms and ions are spherical in shape. A crystal is formed by close packing of atoms or
ions. Since, spheres touch each other only at points, some empty space is left between them.
This space is called void or hole.
(ii)A tetrahedral void is surrounded by four spheres(atoms), which lie at vertices of regular
tetrahedron whereas an octahedral void is surrounded by six spheres(atoms).
(iii)
(Ans)
(Q.) The density of an atom is 7.2g cm–3. It has bcc structure. The edge length is 288 pm. How many atoms
of element does 208g of element has?
(3 Marks)
(Ans)
(Q.39)
Find the type of lattice for cube having edge length of 400 pm, atomic wt. = 60 and
density = 6.25 g/cc.
(3 Marks)
(Ans)
Let the no. of atoms in a unit cell = x
Mass of one unit cell =
=
3
Volume of unit cell = (edge length) =
-24
3
= 64 x 10 cm
Density =
or, Mass =
=
=4
The unit cell has 4 atoms
It is face centered cubic lattice.
(Q.40) A mineral contains Ca, O and Ti. In its unit cell oxygen atoms are present at face centres, calcium
atoms at corners and titanium atoms at centre of cube. Find the oxidation number of titanium in the mineral.
(3 Marks)
(Ans)
No. of Ca atoms =
No. of O atoms =
No. of Ti atoms =
Formula of mineral is CaTiO3
Let oxidation number of Ti = x
In CaTiO3
+2 + x + (-2 3) = 0
x = +4
Oxidation state of titanium is + 4 in this mineral.
(Q.41) A metal oxide crystallizes in a hexagonal close packed array of oxide ions with two out of every three
octahedral holes occupied by metal. Find formula of metal oxide.
(3 Marks)
(Ans)
(Q.42) A metal has cubic lattice. Edge length of lattice cell is 2A0. The density of metal is 2.4g cm–3.How many
units cell are present in 200g of metal.
(3 Marks)
(Ans)
(Q.43) The density of NaCl crystal is 2.155g cm–3 and distance between Na+ and Cl– is 280 pm. Find value of
Avogadro’s number.
(3 Marks)
(Ans) NaCl has fcc structure.
In fcc, a (edge length) =
=
= 560 pm
pm
For fcc , Z = 4, Na = Avogadro number =?
(Q.44) In a face centered cubic lattice atoms of A occupy corner of cell and that of B occupy face centers. One
of the A atoms is missing from one corner of a unit cell. Find the simplest formula of compound.
(3 Marks)
(Ans)
UNIT-2
SOLUTIONS
Marks : 5
Solid Solutions
Gas in solid  Solution of hydrogen in palladium
Liquid in solid  Amalgam of mercury with sodium
Temp. Vs Conc.
Mass %, ppm, mole fraction and molality are independent of temperature,
whereas molarity depends on temperature. This is because volume depends
on temperature.
Henry’s law.
At a constant temperature, the solubility of a gas in a liquid is directly
proportional to the pressure of the gas. p = KH . x
KH = Henry’s law constant ( greater the KH value means lower the
solubility.)
Application of Henry’s
law.
1.To increase the solubility of CO2 in soft drinks, the bottle is sealed under
high pressure.
2. To avoid bends, the tanks used by scuba divers are filled with air diluted
with helium
Temp and Solubility of
gas
Solubility of gas increases with decrease of temperature. It is due to this
reason that aquatic species are more comfortable in cold waters rather than in
warm waters.
Raoult’s law for volatile
liquids
The partial vapour pressure of each component in the solution is directly
proportional to its mole fraction. p1 α x1 ,
p1 = p10 x1
Ideal Solutions
The solutions which obey Raoult’s law over the entire range of concentration
are known as ideal solutions. ( For ideal solution ΔmixH = 0, ΔmixV =
0) Example : Solution of n-hexane and n-heptane,
Non-ideal Solutions
Positive deviation : A-B interactions are weaker than those between A-A
or B-B,
Example - Mixtures of ethanol and acetone
Negative deviations : Forces between A-A and B-B are weaker than those
between A-B
Example- mixture of phenol + aniline. a mixture of chloroform +acetone
Azeotropes
Mixtures have same composition in liquid and vapour phase and boil at a
constant temp.
minimum boiling azeotrope (positive deviation) eg- 95% aq ethanol
maximum boiling azeotrope (negative deviation) eg- 68% aq nitric acid
Colligative properties
Depend on the number of solute particles not upon their nature.
Osmosis
Molecules of Solvent flows through the semi permeable membrane from
pure solvent to the solution.
Osmotic pressure
The extra pressure applied on the solution that just stops the flow of solvent
is called osmotic pressure of the solution
Isotonic solutions
Two solutions having same osmotic pressure
Hypertonic
Higher osmotic pressure than a particular soln.
Hypotonic
Lower osmotic pressure than a particular soln.
Reverse Osmosis
The direction of osmosis can be reversed if a pressure larger than the osmotic
pressure is applied to the solution side. That is, now the pure solvent flows
out of the solution
Application : Desalination of sea water
van’t Hoff factor (i)
Ratio of normal molar mass to experimentally determined molar mass or as
the ratio of observed colligative property to the calculated colligative
property.
Questions for Practice :
(Q1) Define binary solution?
(Ans)Binary solution is a solution containing only
(Q2) What is molarity?
(Ans) The number of moles of solute dissolved in
as molarity.
(1 Mark)
one solute dissolved in a solvent.
3
(1 Mark)
one litre or 1dm of solution is known
(Q3) What do you understand by saying that molality of a solution is 0.2?
(1 Mark)
(Ans) This means that 0.2 mol of the solute is dissolved in 1Kg of the solvent.
(Q4) Why is the vapour pressure of a liquid remains constant at constant temperature?
(1 Mark)
(Ans) At equilibrium, the rate of evaporation = rate of condensation. Hence the vapour
pressure of a liquid is constant at constant temperature.
(Q5) Define Azeotropes?
(1 Mark)
(Ans) Constant boiling mixtures are called Azeotropes.
(Q6) Which substance is usually added into water in the car radiator to act as antifreeze?
(1 Mark)
(Ans) Ethylene glycol is usually added into water in the car radiator to act as antifreeze.
(Q7) Which liquids form ideal solution?
(1 Mark)
(Ans) Liquids having similar structure and polarities.
(Q8) Which property of solution depend only upon the number of moles of solute dissolved and
not on the nature of the solute?
(1 Mark)
(Ans) Colligative properties.
(Q9) Write one example each of solid in gas and liquid in gas solution?
(1 Mark)
(Ans) Solid in gas e.g. Camphor in nitrogen gas.
Liquid in gas – e.g. Chloroform mixed with N2 gas
(Q10) What is molal elevation constant or ebullioscopic constant?
(Ans) The elevation in boiling point which takes place when molality
(1 Mark)
of the solution is
unity, is known as ebullioscopic or molal elevation constant.
(Q11) Define van’t Hoff factor.
(Ans) The ratio of the observed
(1 Mark)
colligative property to the theoretical value is called van’t
Hoff factor.
(Q11 A) Two liquids A and B boil at 1200c and 1600c respectively. Which of them has higher
vapour pressure at 700 c?
(1 Mark)
(Ans)Lower the boiling point, more volatile it is .So liquid A will have higher vapour
pressure at 700c.
(Q12) What happens when blood cells are placed in pure water?
(Ans) Water molecules move into blood cells through the cell
(1 Mark)
walls. So, blood cells swell
and may even burst.
(Q13) What is the effect of temperature on the molality of a solution?
(1 Mark)
(Ans) No effect.
(Q14) Write Henry’s law.
(1 Mark)
(Ans) The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas at
a given temperature.
(Q15) What is the relation between normality and molarity of a given solution.
(1 Mark)
(Ans) Normality = 2 x molarity.
(Q16) What is an antifreeze?
(1 Mark)
(Ans) An antifreeze is a substance which is added to water to lower its freezing point. e.g. Ethylene glycol
(Q17) Why cutting onions taken from the fridge is more comfortable than cutting onions lying at
room temperature?
(1 Mark)
(Ans) The vapour pressure is low at lower temperature. So, less vapours of tear – producing chemicals are
produced
(Q18) What will be the van’t Hoff factor for O.1 M ideal solution?
(1 Mark)
(Ans) Van't Hoff factor = 1, because ideal solution does not undergo dissociation or association.
(Q19) Name the substances which are used by deep sea divers to neutralize
the toxic effects of nitrogen dissolved in the blood.
(1 Mark)
(Ans) They use mixture of helium and oxygen to neutralize the harmful effects of nitrogen.
(Q20) Why is Anoxia disease very common at higher altitudes?
(1 Mark)
(Ans) Anoxia is very common at higher altitudes because of lower partial pressure of oxygen at
higher altitudes.
(Q21) What is the optimum concentration of fluoride ions for cleaning of tooth?
(1 Mark)
(Ans)
The optimum concentration of fluoride ions for the cleaning of tooth is 1.5 ppm.
[If it is more than 1.5 ppm it can be poisonous and if less than 1.5 ppm it is in effective.]
(Q22) What happens to blood cells when they are placed in water containing less than
0.99% (mass/volume) salt?
(1 Mark)
(Ans)
The blood cells collapse due to loss of water by osmosis when placed in water containing less than
0.9% (mass/volume) salt.
(Q23) Why is molality generally preferred over molarity as unit for expressing
concentration of solutions?
(1 Mark)
(Ans) Molality is preferred over molarity as unit for expressing concentration of solutions because
it involves mass term which is not affected by change in temperature while molarity involves
volume term which changes with temperature.
(Q24) Calculate the van't Hoff factor of K4[Fe(CN)6].
(1 Mark)
+
4-
(Ans) K4[Fe(CN)6] 4 K + [Fe(CN)6]
Since one molecule of K4[Fe(CN)6] dissociates to produce 5 ions, the value of van't Hofff factor is 5.
(Q25) Name the law which explains the relationship between solubility of a gas in a
liquid and pressure above the liquid surface. Also write the name of the variable which
is kept constant for this law.
(1 Mark)
(Ans)
This is Henry’s law and temperature is kept constant for this law.
[According to Henry's law the partial pressure of gas in vapour phase is proportional to the mole
fraction of gas in the solution at constant temperature.]
(Q26) How does Henry constant vary with the solubility of gas in the solution at a given pressure?
(1 Mark)
(Ans) Higher the value of “Kh” (Henry constant), lower is the solubility of gas in solution.
(Q27) What care should be taken while preparing intravenous injection?
(1 Mark)
(Ans) They should be prepared in aquatic medium and salt concentrations should be same as
blood plasma levels.
(Q28) Why osmotic pressure is better technique for determination of molar mass of biomolecules?
(1 Mark)
(Ans)
Osmotic pressure is better technique for the determination of molar mass of biomolecules because
they are generally not stable at higher temperatures and have poor solubility. Therefore, other techniques like
elevation in boiling point and depression in freezing point cannot be used to determine molecular masses of
biomolecules.
(Q29) What will be the molecular mass of acetic acid when it is dissolved in benzene?
(1 Mark)
(Ans) Normal molecular mass of acetic acid (CH3COOH) is 60. It dimerises when dissolved in benzene.
Therefore, molecular mass of acetic acid will be 120 when it is dissolved in benzene.
(Q30) How does increase in temperature affects the solubility in endothermic and
exothermic dissolution of substances?
(1 Mark)
(Ans) The solubility in endothermic dissolution increases while in exothermic it decreases.
(Q31) What will be the shape of graph which is obtained by plotting partial pressure of
a gas against mole fraction of gas in solution?
(1 Mark)
(Ans) It is a straight line since partial pressure of a gas and mole fraction of gas are directly
proportional to each other.
(Q32) Aquatic animals are more comfortable in cold water than in warm water. Explain?
(1 Mark)
(Ans) This is because “Kh” (Henry constant) values for both N 2 and O2 increase with increase in
temperature indicating that the solubility of gases increases with decrease in temperature.
(Q33) Deep sea divers are advised not to come to surface immediately from deep waters. Why?
(1 Mark)
(Ans) Deep sea divers are advised not to come to surface immidiately from deep waters because
sudden change in outside pressure can be fatal for divers because N2 will bubble out of blood
vessels causing severe pain and can be dangerous.
(Q34) What are colligative properties? Name them.
(2 Marks)
(Ans) The properties of solute which depend upon the number of particles present in definite amount of
solvent but not on the chemical nature of solute are called colligative properties. They are-
(i) Relative lowering of vapour pressure
(ii) Elevation in boiling point
(iii) Depression in freezing point
(iv) Osmotic pressure
(Q35) Which is better method for expressing concentration of solution – Molarity or Molality?
(2 Marks)
(Ans)
Volume changes with temperature whereas mass does not change with temperature. So molality, which
does not have volume term in it is better method for expressing concentration.
(Q36) Define- (i)
Mole fraction
(ii) Molality
(2 Marks)
(Ans) ( i) The mole fraction of a particular component in a solution is the ratio of the number of moles of that
component to the total number of moles of all the components present in the solution.
(ii) Molality of a solution is defined as a number of moles of solute dissolved per 1000g of solvent.
(Q37) The osmotic pressure of human blood is 7.65 atm at 37°C. For injecting glucose solution it
is necessary the glucose solution has same osmotic pressure as of human blood. Find the
molarity of glucose solution having same osmotic pressure as of human blood.
(2 Marks)
(Ans)
=
Or 7.65 =
=
= Molarity = 0.30
(Q38) Vapour pressure of two liquid A & B are 120 and 180mm Hg at a given temp. If 2 mole of A and 3 mole
of B are mixed to form an ideal soln, calculate the vapour pressure of solution at same temperature.
(2 Marks)
(Ans) Total moles = 2 + 3 = 5
P solution =
=
= 48 + 108
= 156mm.
(Q39) Density of 1 M soln of glucose 1.18g/cm3. Kf for H2O is 1.86 Km–1. Find freezing point of solution.
(2 Marks)
(Ans) Mass of solution =volume x density
=1000 x 1.18
=1180g
Mass of water =1180 - 180
=1000g
=1m
=
=
=
= 0 – 1.86 = –1.86°C
(Q40) An aqueous solution freezes at –0.186°C.
Kf = 1.86°, Kb = 0.512. Find elevation in boiling point.
(Ans)
(2 Marks)
= 0 – (0.186) = 0.186°C
=
Or m =
=
=
(Q41) Vapour pressure of dilute solution of glucose is 750 mm of Hg at 373K. Calculate the mole fraction of
solute.
(2 Marks)
(Ans) 373K = 100°C
Vapour pressure of pure water = 760 mm Hg.
P = 750 mm Hg (given)
Or
Ethylene glycol solution having molality 0.5 is used as coolant in a car. Calculate the
freezing point of solution (given Kf=1.86 °C/mole)
(2 Marks)
(Q42)
(Ans)
(Q43) Calculate the molality of solution prepared by dissolving 18g of glucose 500g of water.
(Ans) Mol. Wt. of glucose
(2 Marks)
=(12 x 6) + (1 x 12) + (16 x 6)
= 180
Molality =
=
(Q44) Find the vant Hoff factor for Al2(SO4)3
(Ans)
(2 Marks)
Total ions produced = 2 + 3 = 5
(Q45) On a hill station pure water boils at 99.82°C. The Kb of water is 0.513°C Kg mol–1. Calculate the boiling
point of 0.69m solution of urea.
(Ans)
(2 Marks)
Boiling point of solution = Boiling point of water +
= 99.82 + 0.3539
= 100.17° C
(Q46) A soln of ethanol in water is 1.6 molal. How many gms of ethanol is present in 500g of solution.
(2 Marks)
Ans)
Mass of ethanol =Molality x Molecular weight
= 1.6 x 46
= 73.6 g
Total of mass of solution = 73.6 + 1000 = 1073.6g
1073.6g of solution contain 7.6g of ethanol.
Mass of ethanol in 500g of solution = (73.6 / 1073.6)x 500
(Q47)
List two conditions that ideal solutions must satisfy.
(2 Marks)
1. Hmixing and Vmixing of ideal solutions should be zero.
2.They should obey Raoult’s law over the entire range of concentrations.
(Ans)
(Q48) Explain ideal and non-ideal solutions with respect to intermolecular interactions in a binary
solution of A and B.
(2 Marks)
(Ans)
For the given binary solution of A and B, it would be ideal if A-B interactions are equal to A-A
and B-B interactions and it would be non-ideal if they are different to each other.
The deviation from ideal behaviour will be positive if A-B interactions are weaker as
compared to A-A and B-B. The deviation will be negative if A-B interactions are stronger
as compared to A-A and B-B.
(Q49) (i) What are minimum boiling and maximum boiling azeotropes?
(ii) Can azeotropes be separated by fractional distillation?
(2 Marks)
(Ans)
(i) Minimum boiling azeotropes are the non-ideal solutions showing positive
deviation while maximum boiling azeotropes are those which show negative deviation.
Because of positive deviation their vapour pressures are comparatively higher and so
they boil at lower temperatures while in case of negative deviation, the vapour pressures
are lesser and so higher temperature are required for boiling them.
(ii) No, azeotropes can’t be separated by fractional distillation.
(i) When a non-volatile solute is added to solvent,there is increase in boiling
point of solution. Explain.
(ii) Define ebullioscopic constant and give its units.
(2 Marks)
(Q50)
(i) When a non-volatile solute is added to a volatile solvent the vapour pressure of
pure solvent decreases because a part of the surface is occupied by non-volatile solute
which can’t volatilise. As a result, the vapour pressure of solution decreases and hence,
the solution requires a comparatively higher temperature to boil causing an elevation
of boiling point.
(ii) Ebullioscopic constant is defined as the elevation in boiling point of a solution of a
non-volatile solute when its molality is unity. Its units are K Kg mol-1
(Ans)
One molal solution of a given solvent is always less concentrated than one molar
solution. Explain.
(2 Marks)
(Q51)
In one molar solution one gram mole of solute is dissolved in one litre of solution
while in case of one molal solution same one gram mole of solute is dissolved in 1000
gm of solvent only, which on considering normal density parameters of water, can’t be
lesser in amount than solvent part present in one litre solution. Therefore,more amount
of solvent is present in one molal solution than in one molar solution .
(Ans)
(Q52)
State Raoult’s law. Prove that it is a special case of Henry’ law?
(2 Marks)
(Ans)
Raoult’s law states that partial pressure of a volatile component of a solution is directly
proportional to its mole fraction. It is a special case of Henry’ law because it becomes the
same when “Kh” (Henry constant) is equal to pressure of pure solvent.
(Q53) How did Van’t Hoff explain the abnormal molecular masses of electrolytes like KCl in water
and non-electrolytes like benzoic acid in benzene.
(2 Marks)
(Ans)
The molecular mass of KCl in aqueous medium has been observed to be almost half than
expected and it has been explained as dissociation of KCl into K+ ions and Cl- ions when actual no. of
particles become double and so become the colligative properties but since molecular mass is always
inversely proportional to colligative property it becomes almost half.
In case of benzoic acid in benzene, association of molecules take place when they dimerise and
their no. becomes almost half and so molecular mass doubles as a result.
(Q54) Calculate molality of an aerated drink having 2.5 gm of carbonic acid dissolved in 150 gm of
water.
(2 Marks)
(Ans)
Molality= moles of solute / Mass of solvent in Kg
= [(2.5 / 62.)/ (150 / 1000)] = 0.268.
(Q55) When 3.49 gm of a non-volatile solute was dissolved in 125 gm of benzene, its boiling
raised by 1.23K. Calculate the molecular mass of non-volatile solute. (Kb for benzene is 2.53 K kg
Mol-1)
(2 Marks)
(Ans)
(Q56) When 5.29 g of a non-volatile solute was dissolved in 400 g of water, its freezing point decreased by
1.79 K. Calculate the molecular wt of solute if Kf for water is 1.86 KkgMol-1
(Ans)
(2 Marks)
(Q57) Calculate the mass of ethanol which is present in 500g of 1.6m solution of ethanol in water. (2 Marks)
(Ans)
One litre seawater is found to contain 5.8 10-3g of dissolved oxygen. Calculated the ppm of
dissolved oxygen in seawater. (Density of seawater = 1.03 g/cc).
(2 Marks)
(Q58)
(Ans)
(Q59) When a non-volatile solute is added to pure water its vapour pressure decreases by 4 mm Hg. Calculate
molality of solution. (Vapour pressure of pure water is 40mm Hg)
(2 Marks)
(Ans)
(Q60) What are azeotropes? Give an example of maximum boiling azeotrope. (2 Marks)
(Ans) Azeotropes are binary mixtures having the same composition in liquid and vapour phase and
boil at a constant temperature. Mixture of nitric acid and water is an example of maximum boiling
azeotrope.
(Q61) A decimolar solution of NaCl exerts OP(∏) of 4.6 atm at 300K. Find the degree of
dissolution.
(3 Marks)
(Ans) NaCl
→ Na
+
+ Cl -
Initial moles 1 0 0
Moles at
Equilibrium
Total moles at equilibrium =
= 0.868 = 86.8%
An aqueous solution of a non-volatile and non-electrolyte substance boils at 100.5°C.
Calculate osmotic pressure of this solution at 27°C. Kb (for water) per 1000g = 0.50. (3 Marks)
(Q62)
(Ans)
=
= 100.5 –100 = 0.5°C
=
Molality of solution = 1
Solvent is water.
density of solution = 1
Volume of solution = volume of solvent = 1000/1 = 1000ml = 1 L
= nRT
=
The vapour pressure of benzene at certain temperature is 640mm Hg. To 39.08 of benzene,
non-volatile and non-electrolyte solid-weighing 2.175g was added. The vapour pressure of
solution was 600mm of Hg. Find the mass of the solute?
(3 Marks)
(Q63)
(Ans)
or
=
=
=
or m =
The vapour pressure of water at 296K is 19.8 mm of Hg, 0.1 mol of glucose dissolved in
178.2g of water. Calculate the vapour pressure of resultant solution.
(3 Marks)
(Q64)
(Ans) n glucose = 0.1 (given)
n H2O =
x (water) =
(Q65) A solution is prepared by dissolving 30g of non-volatile non-electrolyte solute in 90g water. The vapour
pressure of solution was 2.8 K Pa at 298K. When 18g of water was further added to it, the vapour pressure
became 2.9 k Pa at 298K. Calculate molar mass of solute.
(3 Marks)
(Ans)
(When A is for H2O) n = moles of solute
2.8 =
________ (1)
2.9 =
_______ (2)
Dividing eq. (1) and (2), we get
=
n=
Mass of solute = 30 and molecular mass = 23
(Q66) Vapour pressure of pure water is 40mm. If a non-volatile solute is added to it, vapour pressure falls by
4 mm. Calculate molality of solution.
(3 Marks)
(Ans)
The solution has 0.1 moles of solute in 0.9 moles of water.
Mass of water =
Molality of solution =
= 6.17m
(Q67) Conc. H2SO4 has a density 1.9g/ml and is .99% H2SO4 by weight. Find molarity of solution. (3 Marks)
(Ans) Mass of 1000 ml of H2SO4 = density
volume
=
= 1900 g
Mass of H2SO4 present in 1900 g (1L) of H2SO4
=
= 1881 g
Mole of H2SO4 present in 1L = Molarity =
=
= 19.197 M
(Q68) A solution contains 25% water, 25% ethanol and 50% acetic acid by mass. Find mole fraction of each
of the component.
(Ans)
(3 Marks)
Moles of water = n1 =
Moles of
= n2 =
Moles of
= n3 =
Total moles in solution =
Mole fraction of water =
Mole fraction of
Mole fraction of
=
=
(Q69) Which will have more osmotic pressure and why?
Solution prepared by dissolving 6g/L of CH3COOH or Solution
prepared by dissolving 7.45g/L of KCl
(Ans)
(3 Marks)
Moles of CH3COOH =
Moles of KCl =
Molar concentration of both the solutions is same.
KCl ionizes into K+ and Cl– where as CH3COOH does not ionize:
Osmotic pressure is colligative property.
Its value depend on number of particles.
Since, KCl produces more ions so, osmotic pressure of KCl will be more than that of CH3COOH .
(Q70) A solution is prepared by mixing 50ml of chloroform and 50ml of acetone. What will be the
resulting volume of solution? 100 ml or >100 ml or <100 ml.
(3 Marks)
(Ans)
Intermolecular H-Bonding
When chloroform and acetone are mixed, they form intermolecular hydrogen bonds. The hydrogen bonds
are strong forces of attraction. As a result, volume of the solution will be less than 100ml.
(Q71) One litre sample of seawater is found to contain 5.8 x 10-3 g of dissolved oxygen. Calculate the
concentration of dissolved oxygen in seawater in ppm. (Density of seawater = 1.03 g/cc).
(Ans)
(3 Marks)
Mass = Volume x density
=
= 1030g
6
Mass of oxygen present in 10 (one million) gof seawater
=
=
(Q72) A solution is prepared by mixing 50ml of chloroform and 50ml of acetone.
(i) The volume of the resulting solution will be 100 ml or less than 100ml or more than 100 ml?
(ii) What happens to the net vapour pressure of resultant solution- it increases or decreases or remains
same? Support your answer with suitable explanation.
(3 Marks)
(i) When chloroform and acetone are mixed they exhibit negative deviation from ideal behavior because
intermolecular hydrogen bonds are formed between molecules of acetone and chloroform. Since the
hydrogen bonds are comparatively stronger forces of attraction, the volume of the solution decreases.
Therefore, the volume of the resulting solution will be less than 100ml.
(ii) The net vapour pressure of the solution decreases because newly formed hydrogen bonds are stronger
forces of attraction than the original forces of attraction existing in separate solutions.
(Ans)
(Q73) Find molarity of a conc. H2SO4 sample having ρ 1.9 g/ml & is 99% pure by weight.(3 Marks)
(Ans) Mass of 1000 ml of H2SO4 = Density Volume
= 1.9 g/ml 1000 ml
= 1900 g
Mass of H2SO4 present in 1900 g (1Litre) of H2SO4
=
= 1881 g
Molarity of sulphuric acid =
=
= 19.197
(Q74) What are colligative properties? Why are they also called democratic properties? Is osmotic pressure a
colligative property? Prove it.
(3 Marks)
These are the properties of solution that depend on the number of solute particles in the solution not at
all on the nature of solute particle. Above mentioned is also the reason as why are they sometime called as
democratic properties.
(Ans)
Yes, osmotic pressure is a colligative property because osmotic pressure of a solution is
proportional to its molarity at a given temperature.
Why semi permeable membrane is so important in the phenomenon of osmosis? What are
isotonic, hypo tonic and hyper tonic solutions? Does osmosis take place in all three types of
solutions?
(3 Marks)
(Q75)
The semi permeable membrane is very important in the phenomenon of osmosis because they
only permit the movement of solvent molecules through them.
(Ans)
Two solutions having similar osmotic pressure at a given temperature are called isotonic
solutions. If the given solution has less osmotic pressure it is called hypo tonic and it is hyper
tonic if its osmotic pressure is higher than the solution on the other side of semi permeable
membrane.
Osmosis takes place only in hypo tonic and hypertonic solutions.
(Q76)
(Ans)
List three general applications of phenomenon of osmosis?
(3 Marks)
1. Shrinking and swelling of blood cells when put into aqueous NaCl having concentration more or less
than 0.9%.
2. Water movement from soil into plant body.
3. Preservation of meat and fruits by salting against bacterial infection.
(Q77) Why is reverse osmosis of great practical utility to meet potable water requirements these days? Can it
make sea water drinkable? Do we use ordinary semi permeable membranes for this?
(3 Marks)
(Ans)
Reverse osmosis is of great practical utility since it can result in absolute purification of any water sample
because semi permeable membrane allows only solvent molecules to pass through.
It can make sea water drinkable though ordinary semi permeable membrane may not be able to sustain
high pressure observed in desalination of sea water i.e. reverse osmosis of sea water.
Semi permeable membrane of cellulose acetate is generally used for this.
-1
(Q78) The following solutions which have higher osmotic pressure is 1. Solution having 6gL of
-1
CH3COOH
2. Solution having 7.45gL of KCl Explain.
(3 Marks)
(Ans)
Calculate the boiling point of a solution of urea prepared by dissolving 10.36 g of urea in
250 g of water on a hill station where pure water boils at 99.82°C. (Kb of water is 0.513°C Kg
mol–1)
(3 Marks)
(Q79)
(Ans)
Calculate the mole fraction of each of the component in a solution containing 25% water,
25% ethanol and 50% acetic acid by mass.
(3 Marks)
(Q80)
(Ans)
(Q81) A solution has been prepared by dissolving 30g of non-volatile solute in 90g of water. The vapour
pressure of solution is 2.8 K Pa at 298K. When 18g of water was added to it, the vapour pressure rose to 2.9
kPa at same temperature. Calculate molar mass of solute.
(3 Marks)
According to Raoult’s law the vapour pressure of a component of a solution is equal to the product of its
vapour pressure in pure state and its mole fraction
(Ans)
So
(When A is for H2O) n = moles of solute
Dividing equation (1) by (2), we get
=
n = 0.882
Molecular Mass of solute = 30 x 0.882 = 26.470
Calculate van’t Hoff factor of the following solutions
Al2(SO4)3 is dissolved in water
MgCl2 is dissolved in water
Benzoic acid is dissolved in benzene
Glucose is dissolved in water
10 gm of KCl is dissolved in 1 Litre of water
(Q82)
1.
2.
3.
4.
5.
(5 Marks)
1.When Al2(SO4)3 is dissolved in water, its one molecule dissociates into 2 Al3+ ions and 3 SO42- ions, so its
van’t Hoff factor for this is 5.
2. When MgCl2 is dissolved in water, its one molecule dissociates into 1 Mg2+ ions and 2 Cl- ions, so its van’t
Hoff factoris 3.
3.When benzoic acid is dissolved in benzene, its two molecule associate to form a dimer. Therefore, its van't
Hoff factor is 1/2.
(Ans)
4. Glucose does not undergo association or dissociation, hence van’t 'Hoff factor for glucose is 1.
5.When KCl is dissolved in water, its every molecule dissociates in two ions. So its van’t Hoff factor for this is 2.
The osmotic pressure of human blood is 7.65 atm. Find the molarity of glucose solution that
can be injected into human blood. If adding some more amount of glucose into it disturbs its
molarity and now it boils at 100.50C, find its new molality at 270C. What will be the effect on
osmotic pressure of solution? Kb (for water) per 1000g = 0.50
(5 Marks)
(Ans) As we know for injecting glucose solution into human blood it is necessary that glucose solution has
(Q83)
same osmotic pressure as of blood. So osmotic pressure of solution should be 7.65 atm.
=
(Human body temperature is 370C)
7.65 =
=
= Molarity = 0.30
Now since elevation in boiling point has taken place with the addition of more glucose, it can be calculated as:
=
= 100.5 –100 = 0.5°C
=
Molality of solution = 1
The osmotic pressure of solution will increase because no of solute particles will increase.
Calculate the vapour pressure of a solution prepared by mixing 349.5 gm of component A
(Mol. Mass 134) and 416.8 gm of component B (Mol. Mass 169) at 298 K. Vapour pressures of
pure components A and B are 421mm Hg and 562mm Hg respectively. Also calculate mole
fraction of A and B in vapour phase.
(5 Marks)
(Q84)
(Ans)
As shown above component B is more volatile than component A and vapour phase is richer in component B
(Q85) 3.9 gm of a protein was dissolved in 63 gm of benzene when its freezing point was observed
to go down by 3.26 K. Kf for benzene is 4.9 K Kg Mol-1. Calculate the percentage association of
protein if it forms dimers in the solution?( molar mass of protein is 925)
(5 Marks)
(Ans)
Wt of protein= 3.9 gm
Kf = 4.9 K Kg Mol-1
Wt of benzene= 63 gm
Depression in freezing point= 3.26 K
Thus experimental molecular mass of protein in benzene is 930.470
Now suppose the degree of association of protein be X and 1-X mol of protein will be left unassociated and
corresponding X/2 as associated at equilibrium.
Therefore total no. of moles of particles at equilibrium will be
1-X+X/2 = 1-X/2
So total no moles at equilibrium equals von’t hoff factor (i)
But as we know i = Theoretical molar mass/ Experimental molar mass
Therefore, degree of association of protein in benzene is only 1.2%
(Q86) A solution of 10 gm NaCl in 1000 gm water freezes at –0.604K. Calculate the degree of
dissociation of NaCl. (Kf for water is 1.86Kkgmol-1)
(5 Marks)
(Ans)
UNIT -3
ELECTROCHEMISTRY
1.
Differences between electrochemical reaction and electrolysis.
Electrochemical reaction
Chemical reaction produce electricity
It is spontaneous, ∆G is -ve
2.
Marks =5
Electrolysis.
Electricity causes chemical reaction to
take place
It is non-spontaneous, ∆G is +ve
Galvanic cell, its representation-
Representation of Electrochemical Cell :
Anode / anodic electrolyte(M) ǁ Cathodic electrolyte(M) / cathode
Example :
Zn/ Zn++(0.1M) ǁ Cu++(0.12M)/Cu
Functions of Salt Bridge 1. Salt Bridge maintains electrical neutrality in solutions.
2. It completes circuit internally
Alphabet rule for an electrochemical cell- According to this rule we write that term first
which come first alphabetically as show belowL.H.S.
ANODE
OXIDATION
NEGATIVE
Note :Remember LOAN (Left, Oxidation,
Anode, Negative charge)as an
abbreviation
for anodic properties of Electrochemical Cells
R.H.S.
CATHODE
REDUCTION
POSITIVE
3.
SPECIFIC CONDUCTANCE & MOLAR CONDUCTANCE, And their calculation :
Specific conductance- The conductance of 1 cm3/ 1m3 of the solution of an electrolyte. It is
represented by ƙ (kappa)
Molar Conductance- It the conductance due to all the ion produced from 1 mole of an electrolyte in V
cm3 of the solution. It is represented by λcm at conc. ‘C’ and λ∞m at infinite dilution.
Relationship between them-
4.
Effect of dilution on K and λm
λm for strong electrolyte increases
constantly on dilution. λ∞m can be calculated
on extra plotting the curve.
λm for weak electrolyte steeply (sharply) on
dilution. λ∞m can not be calculated on extra
plotting the curve. It is determined using
Kohlrausch’s law.
Factors affecting electrolytic conductance of an electrolyte :-
5.
Kohlrausch law, its applicationKohlrausch Law- Each ion makes a definite contribution to the total molar conductivity of an
electrolyte at infinite dilution irrespective of the nature of other ion of the electrolyte.
Application of Kohlrausch Law - It helps us to calculate λ∞m of weak electrolytes (if the
λ∞m of strong electrolytes or ions is known.)
Numericals related to Kohlrausch Law :
6.
NERNST equation (to show the effect of conc. of electrolyte and temperature on E° and
E°cell), its different forms, writing Nernst equation for an electrochemical reaction
i) Nernst equation for electrode potential-
ii) Nernst equation for electrode potential-
For an electrochemical reaction at equilibrium Ecell = 0 then,
From E°cell we can calculate ∆G° using following formula
7.
Application of electrochemical seriesi) To compare the relative oxidizing and reducing powers- A chemical species with higher
value of E° will be stronger oxidizing agent and a chemical species with more –ve E° will be
stronger reducing agent
ii) To compare the relative activities of metals-A metal with more –ve E° value will be more
reactive
iii) To calculate the E°cell of given cellE°cell = E°cathode - E°anode or E°cell = E°RHS - E°LHS
N.B. - E° for SHE is taken 0 (Zero)
iv) To predict whether a metal will displace hydrogen from its compound or not- All the chemical
species which have –ve E° value will displace hydrogen from its compounds
M → Mn+ + ne- (electron lost is gained by H+ ion to undergo reduction and H2 is
produced)
2 H+ + 2e- → H2
v) To predict spontaneity of a redox reaction- For a cell E°cell/
Ecell is +ve the cell reaction will be spontaneous
8. Commercial cells- These are of 4 types
1) Dry cell (primary cell) – It is a Non rechargeable cell. Used in Torch, Wall clock etc.
The following reaction takes place at Anode and Cathode in the Dry cell :
The above reaction takes place at anode i.e. Zn plate
The above reaction takes place at cathode i.e. Carbon rod
2) Mercury cell/Button Cell – It is used in hearing aids and watches etc.
The following reaction takes place at Anode and Cathode in the
Mercury cell/Button Cell :
The above reaction takes place at anode i.e. Zn container
The above reaction takes place at cathode i.e. Carbon rod
3) Lead storage battery – It is a Rechargeable battery. Used in automobiles, Rechargeable
batteries used in house hold appliances etc.
4) Fuel cell : This cell is based on the combustion of fuel like Hydrogen by oxygen:
Advantages of Fuel Cells : 1. As works as long as we supply Hydrogen & oxygen gas
2. It causes no pollution 3. Its efficiency is 70% 4. Its By product (Water) is also useful.
The following reaction takes place at Anode and Cathode
in the fuel cell.
3. Predicting products of electrolysisThe nature of the product obtained at cathode or anode depends upon E° value of given
chemical species. The chemical species which has most –ve E° will form product at
cathode and The chemical species which has least –ve/most +ve E° will form product at
anode. Inert electrodes do not take part in chemical reactions
Faraday’s 1st Laws
The amount of chemical reaction which occurs at any electrode during
electrolysis is proportional to the quantity of electricity passed through the
electrolyte (solution or melt).
Faraday’s 2nd Laws
The amounts of different substances liberated by the same quantity of
electricity passing through the electrolytic solution are proportional to their
chemical equivalent weights (Atomic Mass of Metal ÷ Number of electrons
required to reduce the cation).
Product of Electrolysis
NaCl (molten)
NaCl (aq)
H2SO4(dil)
H2SO4(conc)
AgNO3(aq)-Ag electrodes
AgNO3(aq)- Pt electrodes
CuCl(aq)- Pt electrodes
Cathode : Na+(l) + e– → Na(s)
Anode : Cl–→ ½Cl2+e–
–
–
Cathode : H2O (l ) + e → ½H2(g) + OH Anode : Cl–→ ½Cl2+e–
Cathode : H+ + e-  ½ H2
Anode: 2H2O(l )→ O2(g) + 4H+(aq) + 4e–
Cathode : H+ + e-  ½ H2
Anode: 2SO4 2– (aq) → S2O8 2– (aq) + 2e–
+
Cathode : Ag (aq) + e  Ag(s)
Anode: Ag(s)  Ag+(aq) + eCathode : Ag+(aq) + e-  Ag(s) Anode: 2H2O(l) → O2(g)+4H+(aq)+4e–
Cathode : Cu+(aq) + e-  Cu(s) Anode: 2H2O(l)→O2(g)+4H+(aq)+4e–
Standard Hydrogen ElectrodeDefinition of STANDARD HYDROGEN ELECTRODE
(S.H.E.) : Electrode is used to determine E° value of
a given half cell by connecting this electrode to the
other half cell.
The E° value of other Half cell is determined
according to the following steps :
1.We will constitute one cell (SHE + Given Half Cell).
2. The EMF of the above cell is determined which is
also called E°cell .
3. Knowing the EMF/ E°cell the E° of given half cell
is determined asE°cell = E°cathode - E°anode
4. E° value of S.H.E. is taken Zero.
(Note:S.H.E. may work as anodic or cathodic half cell)
Corrosion- It is a process of eating away of metals on their surfaces, it is an unwanted process as it
results in loss of mass of metals.
In this process metal surface reacts with atmospheric oxygen to form a layer of oxide. It is
an electrochemical reaction
The following reaction takes place during rusting :
Anodic Reaction
Cathodic Reaction
Over all Reaction
Some Important questions :
Q1.
Ans:-
How much electricity in terms of Coulomb is required to reduce 1 mol of Cr 2O72- to Cr3+.
2Cr2O7-2-------------2Cr+3, 2Cr+6+6e----------------2Cr3+
Therefore the coulomb of electricity required =6F,
=6x96500 C= 579000 C
Q2.
What is Fuel Cell?
Ans:- Fuel cell is a device which produce the energy during the combustion of fuels like
Hydrogen , Methane, Methanol.
Q3.
A solution of CuSO4 is electrolysed using a current of 1.5 amperes for 10 minutes. What mass of Cu is
deposited at the cathode? (Atomic mass of Cu=63.7)
Ans:- The reaction is Cu+2(aq.) + 2e ------------ Cu (s)
The mass of copper deposited=ECu
X
I
X
t
= 63.7 x 1.5 x10 x 60
/ 2x96500C =0.297 g.
96500C
Q4.
Q5.
Q6.
Calculate the equilibrium constant for the reaction Cu (s) + 2Ag+ Cu +2 + 2Ag(s) Eo Cu2+/Cu =
+0.34V, EoAg+/Ag = + 0.80V.
Write the Nernst equation and emf of the following cells at 298K:
Sn/Sn2+(0.050M)//H+(0.020M)/H2(g)/Pt(s) EoSn2+/Sn= - 0.13V
Calculate the standard free energy change for the following reaction at250C,
Au(s) +Ca+2 (1 M)Au3+(1M) + Ca (s) ,The electrode potential values are Ca2+ /Ca = –2.87V,Au3+ /
Au = +1.50V. Predict whether the reaction will be spontaneous or not at 250C.
Q7.
Q8.
How do you account for conductivity of strong and weak electrolyte with concentration? Plot the graphs also.
Q9.
Resistance of conductivity cell filled with 0.1molL-1 KCl solution is 100 ohm. If the resistance of the
same cell when filled with 0.02molL-1 KCl solution is 520 ohm. Calculate the conductivity & molar
conductivity of 0.02molL-1 KCl solution. The conductivity of 0.1 molL-1 solution of KCl is 1.29Sm-1.
A Copper –silver is set up. The copper ion concentration is 0.10M. The concentration of silver is not
known. The cell potential measured 0.422V.Determine the concentration of silver ion in the cell.
Eo (Ag+/Ag) = +0.80V, Eo (Cu2+/Cu)= +0.34V.
A voltaic cell is set up at 250C with the following half cells:
Al(s)/Al3+(0.001M) and Ni2+(0.50)/Ni(s) ,Write the equation for the cell reaction that occurs when the
State Kohlrausch law . Calculate Limiting molar conductivity of NaCl, HCl and NaAc (Sodium Acetate)
are 126.4, 425.9 &91 SCm2 mol-1.Calculate Limiting molar conductivity of HAc (Acetic Acid).
Ans:- According to this law, Molar conductivity of an electrolyte, at infinite dilution can be expressed as the
sum of contributions from its individual ions. It the molar conductivity of the cation is denoted by Λo+
and that of the anions by Λo- then the law of independent migration of ions is
Λom=v+ Λo+ + v- Λo-.
Q10.
Q11.
cell
generates an electric current and determine the cell potential (given E o Ni2+/Ni = -0.25V, EoAl(s)/Al3+=-1.66V)
Q12.
Q13.
Q14.
Q15.
Write the reaction involved in the following cells: (a) Fuel Cell (b) Lead Storage Battery.
Three electrolytic cells A,B,C containing solutions ZnSO4,AgNO3,and CuSO4 respectively are
connected in series .a Steady current of 1.5 amperes was respectively are connected in series .A steady
current of 1.5 amperes was passed through them until 1.45g of silver deposited at the cathode of cell B.
How long did the current flow? What mass of copper and zinc were deposited?
Conductivity of 0.00241M acetic acid is 7.896 X 10-6 S cm-1. Calculate its molar conductivity. If Λ0 for
acetic acid is 390.5 S cm2 mol-1. What is its dissociation constant?
2+
(a) Two half cell reactions of an electrochemical cell are given below: MnO4- + 8 H+ + 5 e- 
+
o
4 H2O E =+1.51V
Sn2+ Sn4++ 2e- Eo = +0.15V.
Construct the redox reaction from the two half cell reaction and predict if the reaction favours formation
of reactants or product shown in the reaction
(b). How much electricity in terms of Faraday is required to produce (i) 20g of Ca from molten CaCl2
(ii) 40g of Al from molten AlCl3
VALUE BASED QUESTIONS :
Q.1 In news paper Yogesh read that iron pipes of sewerage system of the town will be protected by cathodic
protection method (sacrificial method). His friend Juned said in this method something will replace iron to
protect it from rusting. Explain the above method for Yogesh and also write the value associated with.
Q.2 Sanjay went to buy dry cell for wall clock. His father instructed him to check date of manufacturing before
buying the cells. Sanjay was wondering if buying new cells with seal packed what to do with manufacturing
date. If Sanjay is write? Give your opinion and the value associated with.
UNIT – 5
CHEMICAL KINETICS







Marks = 5
The branch of chemistry which deals with the study of reaction rates and their mechanisms is
called kinetics.
Rate of a chemical reaction
Change of molar concentration per unit time rate = Change of molar concentration / time
RP
Rate of reaction = -∆[R]/ ∆t = +∆[P]/ ∆t
+ve sign for decreasing in concentration
-ve sign for increasing in concentration
Average rate = decrease in concentration of R / time taken = -[R]/∆t
= increase in concentration of P / time taken = +∆[P]/∆t
Instantaneous rate.
Rate of change of concentration of any one of reactant or product at a particular moment of
time R  P
∆t  0
-∆[R]/∆t = -d[R]/dt
∆t  0
∆[P]/∆t = -d[P]/dt
Calculation of reaction rate
3H2 + N2 2NH3
Rate = - 1/3 d[H2]/dt = -d[N2]/dt = ½ d[NH3]/dt
2HI  H2 + I2
Rate = -1/2 d[HI]/dt = d[I2]/dt
Rate law = it is the mathematical expression in which reaction rate is given in terms of molar
concentration of reactions.


aA + bB  cC + dD
Rate α [A]x[B]Y
Rate = k [A]x[B]Y
Where K = rate constant
X and Y may or may not equal to a and b.
The value of x and y are determined from experiment.
Order of reaction – the sum of powers raise to the concentration terms in the rate law equation
which is determined experimentally.
Rate = k[A]0 zero order
Rate = k[A]1 first order
Rate = k[A]2 second order
Zero order: - The rate of reaction is proportion to the zero power of the concentration of the
reactant.
R→P
Rate law for zero order reaction =
Rate = K [R]0
Derivation for zero order
Rate = K[R]0
Rate = -d[R]0/dt
-d[R]/dt=K[R]0 =K
d[R]=-K dt
On integration,
∫d[R]= -∫ K dt
[R] = -Kt+C ( C = integration constant)
At t=0 [R]=[R]o
[R]o= 0 + C
ie C=[R]0
[R] = - Kt +[R]0
Kt =[R]0 – [R]
K = 1/t x {[R]0 – [R]}
[R]0 = initial concentration
[R] =final concentration
* First Order Reaction – Rate of reaction depend on one conc. term i.e.
Rate of reaction is proportion to first power of concentration of the
reactant.
R →P, its rate law will be →
Rate =K [R]1
Derivation of integrated rate equation for first order reactionRate =K [R]1
[R0] = Initial conc. of reactant [R] = Final conc. of reactant
Rate = -d[R]/dt =K[R]1
d[R]/[R]= -K dt
On integrating both sides
∫ d[R]/[R]= -K∫ dt
Taking log both side
ln [R] = -Kt +C,
When t=0 ,[R] =[R0],
ln[R]=-kt+ ln[R0]
ln[R]/[R0] = -kt
or kt = ln[R0]/[R]
kt =2.303 log [R0]/[R]
So ln [R0] = C
k= 1/t x 2.303 log [R0]/[R]
 Half-life:- It is a time at which the concentration becomes half of its initial concentration.
At t=⅟2
[R]= [R0]/2
For zero order
K=[R0] –{1/2[R0]}/t1/2
K=[R0]/2k
For first order
K= 2.303/t x log[R]0/[R]
K =2.303/t1/2 x log 2[R0]/[R0]
K =2.303/t1/2 x 0.3010
K= 0.693/ t1/2
*Pseudo order of reactionThose reactions which appear to be of high order but actually follow lower Order kinetic
Example: hydrolysis of ethyl acetate, decomposition of ammonia, decomposition of HI
2NH3→ N2+ 3H2; rate law = Rate =K [NH3]0
HI→ H2+ I2; rate law = Rate =K [HI]0
1st order reaction
N2O5→ 2NO2+ 1/2 O2
H2O2→ H2O + ½ O2
2nd Order reaction H2+I2 →2HI
 units of rate constants
Zero order: mol L-1 s-1
First order: s-1
Second order: mol -1 L s1









Elementary reaction: The reaction taking place in one step is called elementary reaction.
Complex reaction – when a sequence of elementary reaction gives us the product, the reaction are
called complex reaction.
Molecularity – the number of reacting molecules taking part in an elementary reaction.
Unimolecular –when one molecule involved.
Bimolecular – when two molecule involved.
Ex of Bimolecular: H2+I2 →2HI
Rate determining step: In complex reaction the overall rate of the reaction is controlled by the slowest
step in a reaction called rate determining step.
Rate of reaction become almost doubled for every 10oc rise in temp.
Arrhenius equationK= A e-Ea/RT
Where, A =frequency factor, Ea =activation energy
R =gas constant (8.314 j/K mol)
log K =log A – Ea/2.303RT
logK2/K1 =Ea/2.303R x(1/T1-1/T2)
Activation energy – the excess of energy which must be supplied to the reactant to undergo chemical
reaction. It is the energy required to form intermolecular called activation complex.
Lower the value of activation energy faster will be rate of a reaction.
Collision frequency (Z) : The number of collision that takes place per second per volume of reaction
Effective collision – The collision among the reacting species which result in the product are called
effective collision.
Factors affecting the rate of reaction
1. Concentration- The rate of chemical reaction at a given temperature Depends on concentration
of one or more reactant and products.
Rate =K [A]x [B]y.
2. Temperature – most of reaction are accelerated by increase in temperature For a chemical
reaction with rise in temperature by 10oC, the rate constant is double.
K= A e-Ea/RT
3. Catalyst – A catalyst is a substance which alters the rate of a reaction without itself undergoing
any chemical change.
Catalyst participates in a chemical change by forming temporary bond the reactant and give
complex intermediate.
4. Nature of reaction: if depends on chemical bond.
5. Physical states of reactant: it depends on state. i.e. - solid liquid and gas.
Intext Question
1.
RP . Concentration of reactant from 0.03M to 0.02M in 25 minutes. Calculate the
average rate.
ANS: Average rate = -∆
/∆t= 0.02-0.03/ 25(60) = -0.01M/1500s.
2. 2A Product, concentration of A from 0.5 to 0.4 in 10 minutes . Rate=?
ANS: Average rate = −
=− x(0.4-0.5)/10min = 0.1/20min=0.005M/min.
3. A+B = Products. Rate Law = r k[A]1/2[B]2 , order =?,
ANS: Order=1/2+2=2.5.
4. X→Y. If concentration of X is increase to three times. Rate of formation of Y=?
ANS: r = k[x]2 , r’ =k[3x]2 r’/r =k(3x)2/kx2 = 9. Therefore rate
times.
of conc. of Y increases to 9
5. Rate constant = 1.15 x 10-3/s . How long 5g of this to reduce to 3g?
ANS: t=
log(R0/R) R0 =5g, R =3g ,
t=
log(5/3) =4445s
6. To decompose SO2Cl2 to half is 60 minutes,
given: 1st order reaction. Calculate k.
ANS:
k=
=
=1.925 x 10-4 /sec .
7. Rate of reaction doubles with every 100 rise in temp. . If temp. increases from 295K to
305K . Ea =?
ANS: log(k2/k1)=
(1/295−1/305)
Log2 =
0.301=
(1/T1−1/T2)
x
Ea =51855 J/mol .
8. 2HI→H2 + I2 , Ea =209.5 KJ at 581K .Calculate fraction of molecules of reactant having
energy equal to or greater than Ea .
ANS:
x =n/N = e-Ea/RT
ln x =
, log x =
log x =
=−18.8323
x = Antilog(-18.8323)= 1.471x 10 -19
Examples
1. Rate =k[A]1/2[B]3/2
order =1/2 + 3/2=2
k =2.3 x 10-5Lmol/s
It is 2nd order reaction.
3. 1st order reaction , N2O5→2NO2 + O2 ,[R0]=1.24 X 10-2
After 60 minutes [R] =0.2x 10-2 ,Rate constant =?
2.
log(R1/R2) = k (t2−t1)/2.303
k = 2.303/(t2-t1) X log (R1/R2)
=
=
xlog(1.24 x 10-2/0.2x 10-2)
log( 6.2) /min
k =0.0304 /min
4 1st order, k = 5.5 x 10-14/s,t1/2=?
ANS: t1/2=
t1/2=0.693/5.5 x 10-14 =1.26 x 10-14s
5. Show that in 1st order reaction ,time for 99.9 % is 10 times to half-life of the reaction.
k=
log (R0/R) =
t=6.909/k,
t/t1/2=
log
t1/2=0.693k
x
=10
6. Rate constant at 500K =0.02/s at 700K =0.07 /s.
Calculate Ea and A.
log(k2/k1)=
log(0.07/0.02)=
(T2-T1/T1T2)
(700-500/700x500)
0.544 = Ea x 5.714 x 10-4/19.15
Ea=18230.8J
k=Ae-Ea/RT,
0.02= Ae-18230.8/8.314(500)
A=1.61
7. C2H5I → C2H4 + HI , 1st order reaction, T=600K ,k =1.6 x 10-5
Ea=209KJ/mol, at 700 k=?
logk2 –logk1 =
logk2=logk1 +
=log(1.6x10-5) +
logk2=-2.197
k2=6.36 x 10-3
(1/T2-1/T1)
(1/T2-1/T1)
(1/600-1/700)
Important questions for Practice :
Q1.
Q2.
Q3
Q4
Q5..
Q6.
Q7
Q8.
Q9.
Q10.
Q11.
Q12.
Q13.
Q14
Q15.
Define Pseudo order reaction?
The decomposition reaction of ammonia gas on platinum surface has a rate
constant = 2.3 x 10-5 L mol-1s-1. What is the order of the reaction?
Mention the factors that affect the rate of a chemical reaction.
From the rate expression for the following reactions determine their order
of reaction and dimensions of the rate constants.
a) H2O2 (aq) + 3 I - (aq) + 2H+ 2H2O (l) + 3I-1 Rate = k [H2O] [I-]
b) CH3 CHO (g) CH4(g) + CO(g) Rate = k [CH3 CHO]3/2
A reaction is first order in A and second order in B.
i) Write differential rate equation.
ii) How is the rate affected when concentration of B is tripled?
iii) How is the rate affected when the concentration of both A and B is
doubled?
The decomposition of NH3 on platinum surface is zero order reaction.
What are the rates of production of N2 and H2 if k = 2.5 x 10-4 mol L-1 S-1?
. Derive the Integrated rate equation for first order reaction. Also find half
life period and plot the graph associated to it.
For a first order reaction, show that time required for 99% completion is
twice the time required for the completion of 90% of reaction.
A first order reaction has a rate constant 0.0051min-1 .If we begin with 0.10
M concentration of the reactant, what concentration of the reactant will be
left after 3 hours.
The half-life for radioactive decay of 14C is 5730 years. An archaeological
Artifact containing wood had only 80% of the 14C found in a living tree.
Estimate the age of the sample
What is the effect of temperature on the rate constant of a reaction? How
Can this temperature effect on rate constant be represented quantitatively ?
The rate of a reaction quadruples when the temperature changes from
293K to 313
K. Calculate the energy of activation of the reaction assuming that it does not change with
temperature.
A first order reaction takes 40 min for 30% decomposition. Calculate t1/2..
(a) Distinguish between order of reaction & Molecularity.
(b) For a decomposition reaction the values of rate constant k at two different
temperatures are given below:k1 =2.15 x 10-8 L mol-1s-1 at 650K, k2 =2.39 x 10-7 L mol-1s-1 at
700K Calculate the value of Activation Energy for this reaction.
(i) Write short notes on the following:
(a) Activation energy of a reaction (b) Elementary step in a reaction (c)Rate of a reaction
(ii) The following result has been obtained during the kinetic studies of the
reaction2A + B C+D
Experiment
[A] mol L-1
[B] mol L-1
Intial rate
mol L-1min-1
I
II
III
IV
0.1
0.3
0.3
0.4
0.1
0.2
0.4
0.1
6.0x10-3
7.2X10-2
2.88X10-1
2.40X10-2
Determine the rate law and rate constant for the reaction.
Q16
From the concentrations of C4H9Cl (butyl chloride) at different times given
below, calculate the average rate of the reaction:C4H9Cl + H2O C4H9OH + HCl
during different intervals of time.t/s 0 50 100 150 200 300 400 700 800
[C4H9Cl]/mol L–1 0.100 0.0905 0.0820 0.0741 0.0671 0.0549 0.0439 0.0210 0.017
UNIT - 5
SURFACE CHEMISTRY
1.
Marks = 4
Definition of :
i) Adsorption- A surface phenomenon in which molecules of a substance (adsorbate) are retained on
the surface of another substance (adsorbent). It is an exothermic reaction
ii) desorption- It is an opposite reaction of adsorption. In this process molecules of adsorbate leave the
surface of adsorbent. It is an endothermic reaction
iii) sorption- When adsorption and absorption both take place simultaneously the it is known as sorption.
i) adsorption isotherm- It a graph plotted between extent of adsorption and pressure of the gas at
constant temperature.
ii) homogeneous catalysis- The catalysis process in which all the reactants and catalyst are in same phase.
iii) Heterogeneous catalysis- The catalysis process in which the reactants and catalyst are not in the
same phase.
iv)
Sol- A colloidal solution in which dispersed phase is solid and dispersion medium is liquid.
v)
Emulsion- A colloidal solution in which dispersed phase is liquid and dispersion medium is
also liquid.
Micelle- A colloidal solution which is formed by the association of large no. of species
supplied by dissociation of molecules at certain higher concentration. Soap in water at or above
CMC
CMC- The minimum concentration at or above which formation of micelle takes place.
vi)
vii)
viii) Kraft temperature- The formation of micelle takes place only above a particular temperature
known as Kraft temperature
ix) Tyndall effect-It may be defined as the scattering of light by the colloidal particles in a
colloidal solution kept in a dark room. In this effect the path of the light becomes visible.
x)
Zeta potential- The potential difference between the fixed layer and diffused layer of opposite
charges is called the zeta potential.
xiv) Electrophoresis :
Electrophoresis- It is migration of
colloidal particles towards one of the
electrode under the influence of an
external electric field.
2.
Positive and negatively charged colloidal solutions
Oil in water and water in oil emulsionsOil in water
In this type of emulsion dispersed phase is
oil and medium is water
e.g. milk, vanishing cream
3. Applications of adsorption
Water in oil
In this type of emulsion
dispersed phase is water and
medium is oil
e.g. butter, cream
Applications of colloids-
4. DifferencePhysical and chemical adsorption-
Lyophilic and lyophobic colloids
Lyophilic colloids
In this type of colloid there is greater affinity
between d. phase and d d.mdm
These colloids are reversible
These are formed easily
These are quite stable
Lyophobic colloids
In this type of colloid there is poor affinity between
d. phase and d d.mdm
These colloids are irreversible
These are not formed easily
These are quite unstable
Multimolecular and Macromolecular Colloids Multimolecular
These colloids are formed by combination of
large no. of molecules of a substance.
Exa.- Gold sol, sulphur sol
MacromolecularWhen molecules of dispersed phase are big enough to
form colloidal dimension then they form
macromolecular colloids.
These colloids are quite stable
Exa.- Starch, synthetic rubber
Steps involved in the process of Heterogeneous catalysis-
Examples of Enzymatic reactions-
Examples of Industrial Catalysis -
Formation of Micelle1. In water soap molecules behave like normal electrolyte at lower conc.
2. The dissociated part of soap molecules get attached with dirt particles (Fig. b)
3. All the soap molecules get attached to dirt particles through their tail( hydrophobic end) keeping
head (Hydrophilic end) outward to form micelle
Chemical methods of preparation of colloids-
5. Origin of charge on colloidal particlesPreferential adsorption theory- The colloidal particles acquire +ve or –ve charge by the adsorption
of +ve or –ve ion present in the medium. The colloidal particles adsorbs that ion which is common
with its composition
Some Important Questions :
(Q.) What is meant by critical temperature of gas?
(1 Mark)
(Ans) Critical temperature is the minimum temperature above which a gas cannot be liquefied howsoever
high the pressure may be applied.
(Q.) Give the expression for Fruendlich adsorption isother
(1 Mark)
(Ans) x/m= kp1/n
(Q.) What do x and m represent in the expression x/m=kp1/n
(1 Mark)
(Ans) ‘m’ is the mass of the adsorbent and ‘x’ is the number of moles of the adsorbate when the dynamic
equilibrium has been achieved between the free gas and the adsorbed gas.
(Q.) Why is heterogeneous catalysis also known as surface catalysis
(1 Mark)
(Ans) In heterogeneous catalysis the reaction always starts at the surface of the catalyst. So, it is also known
as surface catalysis.
(Q.) What is a hydrosol?
(1 Mark)
(Ans) A colloid in which the dispersion medium is water is known as hydrosol.
(Q.) Define peptization?
(1 Mark)
(Ans) Peptization is a process of converting a precipitate into colloidal particlesby adding suitable electrolyte.
(Q.) Define Brownian movement?
(1 Mark)
(Ans) Brownian movement can be defined as continuous zig- zag movement of the colloidal particles in a
colloidal sol.
(Q.) Why is Brownian movement important?
(1 Mark)
(Ans) Brownian movement opposes the force of gravity and does not allow the colloidal particles to settle
down, thus making the colloidal solution stable.
(Q.) Differentiate between adsorption and absorption.
(1 Mark)
(Ans) Adsorption
Absorption
a)it occurs only at surface
a) it is a bulk phenomena
b)concentration on the surface b) concentration is same
is more than in the bulk
through out the material
(Q.) What is the effect of temperature on adsorption?
(1 Mark)
(Ans) Adsorption processes, being exothermic, decreases with increase in temperature.
(Q.) When a finely powdered active carbon is stirred into a solution of a dye, the intensity of color in solution
decreases. Why?
(1 Mark)
(Ans) The intensity of color in the solution decreases because of gas adsorbedon the surface of carbon.
(Q.) Why do finely divided substances have larger adsorption power? (1 Mark)
(Ans) Finely divided substances have large surface area for adsorption and hence have larger adsorption
power.
(Q.) What are zeolites?
(1 Mark)
(Ans) Zeolites are aluminosilicates i.e. three dimensional network silicates in which some silicon atoms are
replaced by aluminium atoms.
(Q.) Why are zeolites called shape selective catalysts?
(1 Mark)
(Ans) Zeolites are called shape selective catalysts because their catalytic action depends upon the size and
shape of the reactant and the product molecules as well as on their own pores and cavities.
(Q.) A small amount of silica gel and that of anhydrous CaCl2 are placed separately in two corners of a vessel
containing water vapours. What phenomena will occur in the two corners?
(1 Mark)
(Ans) Adsorption would occur where silica gel is kept in the vessel where as absorption will occur in the corner
where CaCl2 is placed.
(Q.) Name the substance catalysed by Zymase.
(1 Mark)
(Ans) Glucose--Zymase->ethyl alcohol.
(Q.) How can colloidal solution of ferric hydroxide be prepared by peptization?
(1 Mark)
(Ans) A colloidal sol. of ferric hydroxide can be prepared by adding small quantity of ferric chloride solution to
freshly prepared precipitate of ferric hydroxide.
(Q.) What is the cause of Brownian movement?
(1 Mark)
(Ans) Brownian movement is caused by the striking of the colloidal particles with the molecules of dispersion
medium due to their kinetic energy.
(Q.) Define Tyndall effect?
(1 Mark)
(Ans) It is defined as the scattering of light by the colloidal particles present in a colloidal solution.
(Q.) What happens to a gold sol. when gelatin is added to it?
(1 Mark)
(Ans) Gold sol. which is lyophobic starts behaving like lyophilic sol.
(Q.) Write down the relation between pressure of the gas and the amount of it adsorbed?
(1 Mark)
(Ans) x/m = K P1/n
(Q.) Which adsorption may be a multilayered formation phenomenon?
(1 Mark)
(Ans) Physisorption
(Q.) Which is irreversible and why? Physisorption or chemisorption.
(1 Mark)
(Ans) chemisorption. Because of the formation of chemical bond.
(Q.) Name the promoter used in Haber’s process?
(1 Mark)
(Ans) Molybdenum.
(Q.) What is emulsion? What are their different types?
(2 Marks)
(Ans) An emulsion is the colloidal dispersion in which both the dispersed phase and the dispersion mediums
are liquids. They can be of two types:i) Emulsion of oil in water.
ii) Emulsion of water in oil.
(Q.) How are micelles formed in soap solution?
(2 Marks)
(Ans) Soap is sodium salt of fatty acids (RCOONa) which when dissolved in water dissociates to give RCOO- and
Na+. The RCOO- consists of polar group COO- which is hydrophilic and stays at the surface and the non
polar group R which being hydrophobic stays away from the surface. At high concentrations RCOO- ions
are pulled into the solution to form spherical aggregates with R pointing to the centre COO- part
remaining outward. This aggregate is known as ionic micelle.
(Q.)How can lyophobic sols be prepared by mechanical disintegration?
(2 Marks)
(Ans) The coarse suspension of the substance is introduced in the colloid mill that consists of two metal discs
close together rotating at a high speed in the opposite directions. Here the suspension particles are
broken to the colloidal size.
(Q.) Differentiate between chemisorption and physisorption?
(2 Marks)
(Ans) Physisorption:
a)The forces operating are weak vander Waal’s forces
b)The heat of adsorption is low 20-40 KJ Mol-1
c)Does not require any activation energy
d)Forms multimoleculer layer
Chemisorption:
a)Forces acting are similar to those of chemical bonds
b) The heat of adsorption is high 80-240 KJ Mol-1
c) Requires activation energy
d) Forms unimolecular layer
(Q.) Describe the mechanism of peptization?
(2 Marks)
(Ans) When electrolyte is added to the freshly precipitated substance, the particles of the precipitate
preferentially absorb one particular type of ions of the electrolyte and get dispersed due to electrostatic
repulsions giving particles of colloidal size and hence cause peptization.
(Q.) Give any two reasons for the origin of electrical charge on the colloidal particles.
(2 Marks)
(Ans) The two reasons are:
i) Due to electron capture by sol particles during electro dispersion of metals, due to preferential
adsorption of ions from solution
ii) Dissociation of colloidal sols.
(Q.) How is the electrical charge of the colloidal particles responsible for the stability of colloidal sols?
(2 Marks)
(Ans) The electrical charges of the particles prevent them from coming together due to electrostatic repulsion.
All the dispersed particles in a colloidal solution carry the same charge while the dispersion medium has
equal and opposite charge.
(Q.) What is demulsification? Name two demulsifiers.
(2 Marks)
(Ans) The process of separation of the constituent liquids of an emulsion is called demulsification.
Demulsification can be done by centrifugation or boiling.
(Q.) Why lyophilic colloids are called reversible sols while lyophobic sols are called irreversible sols?
(3 Marks)
(Ans) In the lyophilic colloids if the dispersed medium is separated from the dispersion medium the sol can be
made again by simply remixing with the dispersion medium. So they are called reversible sols.
In lyophobic sols if small amount of electrolyte is added, the sols are readily precipitated and do
not give back the colloid by simple addition of the dispersion medium. So they are called
irreversible sols.
(Q.)
Describe
the
preparation
of
the
following
colloidal
solution.
(a) Gold sol
(b) Sulphur sol
(3 Marks)
(Ans) (a) Preparation of Gold sol :- By the reduction of very dilute solution of silver salts with a suitable
reducing agent
2AuCl3 + 3SnCl2 -------------> 2Au + 3SnCl4
Gold sol
(b) Preparation of Sulphur sol :- By the oxidation of H2S in the presence of suitable oxidizing agent like nitric
acid, bromine water, etc.
H2S + Br2 --------------> S + 2HBr
H2S + 2HNO3 ------------> 2H2O + 2NO2 + S
(Q.) What are macromolecular and multimolecular colloids? How are they different from associated colloids?
(3 Marks)
(Ans) Macromolecular colloids:-
i)They are molecules of large size.
ii)They have lyophobic property.
Multimolecular colloids:i) They are formed by the aggregation of large number of atomsor molecules which have diameter
less than 1nm.
ii) They have lyophilic property.
Associated colloids:i) They are formed by the aggregation of large number of ions in concentrated solution
ii) They contain both lyophilic and lyophobic groups
(Q.) What are lyophilic and lyophobic solutions? Give examples for each.
(3 Marks)
(Ans) Lyophilic solutions are those that can be prepared by directly mixing the dispersed phase
with dispersion medium. For example starch dissolved in water.
Lyophobic solutions are those that can not be prepared directly but some special methods are used to
prepare them. For example metal sulphides when mixed with a dispersion medium directly do
not result in any colloid.
(Q.) "Action of soap is due to emulsification and micelle formation". Comment.
(3 Marks)
(Ans) Yes, action of soap is due to emulsification and micelle formation. Soaps are sodium salt of higher fatty
acids like sodium stearate, C17H35COO-Na+
The anionic head of stearate ion (-COO-) is hydrophobic in nature and has great affinity for water, while the
hydrocarbon part (C17H35-) is hydrophilic in nature and great affinity for oil, grease etc. When soap is
used in water, the anions (C17H35COO-) form micelle and due emulsification encapsulate oil or grease
inside. These micelle are removed by rinsing with water; while free dirt (from oil or grease) either settle
down or are washed away by water. Thus the main function of a soap is to entrap oil or grease with the
micelles through emulsification, thereby freeing dirt from grease and oil.
(Q.) Why the sun looks red at the time of setting? Explain on the basis of colloidal properties.
(3 Marks)
(Ans) At the time of setting, the sun is at the horizon. The light emitted by the sun has to travel a longer
distance through the atmosphere. As a result, blue part of the light is scattered away by the dust
particles in the atmosphere. Hence, the red part is visible.
(Q.) Explain the reason for these:
(3 Marks)
(a) Sky looks blue in colour.
(b) Delta is formed at the meeting place of river and sea water.
(c) Blood coagulate on treatment of alum.
(Ans) (a) Sky looks blue in colour because colloidal particles suspended in environment scatter the light and
blue light is scattered maximum.
(b) The charged colloidal particles of river water neutralized by ions present in sea water so coagulation
take place.
(c) The charged colloidal particles present in blood are neutralized by ions of alum.
VALUE BASED QUESTIONS :
Q.1 Use of gas mask is recommended in a crowded place, a place of fire so that poisonous gases are adsorbed.
Does it adsorb gases like N2 and O2 equally? Explain your answer citing a suitable value behind this.
Q.2 Milk and cold dream are examples of colloidal system. Which type of colloid they are? Can they be diluted
by adding water? Based upon your observation/ answer differentiate between them. To protect untimely
curdling of milk how will you utilize the properties of colloidal solution? Extract the value associated with it.
UNIT – 6
GENERAL PRINCIPLE OF EXTRACTION OF ELEMENTS
Marks : 3
1. Differences betweenOre and mineralsOres
The minerals form which elements are
extracted economically and beneficially.
Theses contains higher % of metal compound
Minerals
The combined form of elements found
naturally.
These contains lower % of metal compound
Gangue and slagGangue
The impurity part of an ore.
Slag
A compound form when gangue combines with flux
2. Definition with example –
a. Depressants- The chemical compound which is used to prevent one kind of sulphide ore to go
with froth in the presence of another sulphide ore. Exa.- NaCN
b. Collectors- These substances increase the non-wettability of the ore particles. Exa.- Pine oil, fatty
acids
c. Leaching- When conc. Of ore is done using certain chemicals then this process of concentration is
called leaching. Exa.- Bauxite, Ag and Au ore are concentrated by this method.
d) Pyro-metallurgy- The process of thermal reduction of an oxide by varying the temperature for a
reducing agent is called pyro-metallurgy.
e) Copper matter- It is a combination of Cu2S and FeS obtained on roasting of copper pyrites in
reverberatory furnace.
f) Blister copper- The solidified copper obtained after reduction of copper matte has blistered
appearance. This is called blister copper.
3. Name and formula of some important ores-
4. Thermodynamic principle of metallurgy- Thermodynamic principle of metallurgy is based on the
measurement of ∆G° of formation of oxide of reducing agent form metal oxide. If ∆G° of formation of
oxide of reducing agent is more –ve than ∆G° of formation of metal oxide than the reducing agent can
reduce given metal oxide into metal spontaneously. -ve sign of overall reaction shows its spontaneity.
The possibility of reduction of a metal oxide can also be predicted from Ellingham diagram. A metal
oxide can be reduced by that metal for which ∆G° of formation vs T plot is lower.
Examples1. Mg can reduce Al2O3
2. Al can redue ZnO
3. C can reduce ZnO at higher Temp.
4. CO can reduce FeO at lower temp.
5. At point ‘A’ reaction is at equilibrium
6. Bend in a plot shows change in
Phase
5. Principles of different Refining Processa) Distillation- This purification method is based on the principle that metal to be purified (Zn, Hg)
has lower boiling point that the impurity present in that.
b) Liquation- This purification method is based on the principle that metal to be purified (Sn, Pb) has
lower melting point that the impurity present in that.
c) Electrolytic refining- This purification method is based on the principle that impure metal undergo
oxidation and its ion undergo reduction from solution based on their E° value.
d) Zone refining- This purification method is based on the principle that impurities are more soluble
in molten state than solid state of the metal.
e) Vapur phase refining- This purification method is based on the principle that metal –
a) should form a volatile compound
b) the volatile compound should be easily decomposable
A Few Questions for practice:
1. Give the name and composition of ore chosen for extraction of aluminium.
(1 Mark)
Ans: The ore chosen for the extraction of aluminium is bauxite and its composition is Al2O3.xH2O.
2. What is leaching?
(1 Mark)
Ans: Leaching is the process of extracting a substance from a solid by dissolving it in a liquid. In metallurgy
leaching is used for the ores that are soluble in a suitable solvent.
3. Why cryolite & fluorospar added to alumina during electrolytic reduction?
(1 Mark)
Ans: Cryolite and fluorospar are added to alumina during electrolytic reduction to reduce the melting point of
alumina and to increase its conductivity
4. Reduction with C for Cu2O can be done at lower temp. than ZnO. Why?
(1 Mark)
Ans: In the Ellingham diagram the curve for Cu2O lies higher than ZnO i.e. for the reduction of Cu2O with C the
negative value of Gibbs energy can be reached at a lower temperature than ZnO.
5. Although thermodynamically feasible in practice magnesium metal is not used for the reduction of alumina.
Why?
(1 Mark)
Ans: Magnesium can reduce alumina at the temperature above the intersection point of the curves for Al2O3
and MgO in the Gibbs Energy vs T plot (Ellingham diagram). But the temperature at which this is feasible
is too high to be achieved economically and is also technologically difficult. So this reduction is not done.
6. What is the significance of leaching in extraction of aluminium?
(1 Mark)
Ans: In the extraction of aluminium leaching is used for the concentration of ore by removing the impurities
i.e. silica, iron oxides and titatinium oxides.
7. Define Metallurgy.
(1 Mark)
Ans: Metallurgy is the process of extraction of metals from their ores that includes various steps.
8. Why is hydraulic washing a type of gravity separation?
(1 Mark)
Ans: The process of hydraulic washing is based on the differences in gravity of the ore and the gangue
particles and so is known as gravity separation.
9. What is the use of van Arkel method?
(1 Mark)
Ans: Van Arkel method is used for removal of impurities like oxygen and nitrogen from the metals like
zirconium and titanium.
10. How is distillation used for metal refining?
(1 Mark)
Ans: Distillation is used for the metals with boiling point lower then the impurities. So the metals can be
evaporated and separately obtained as distillate.
11. Why do the anodes used in the electrolytic cell for the reduction of alumina need to be replaced regularly?
(1 Mark)
Ans: The oxygen liberated at the anode during the reduction of alumina, reacts with the carbon of the anode
to form CO and CO burns away the anode and hence the anodes need to be replaced.
12. What is the role of depressant in froth floatation process?
(1 Mark)
Ans: In froth floatation process the depressant selectively prevents one of the ores from coming to the froth
in a mixture of two ores hence enabling the separation of the other one with the froth.
13. State the role of silica in the metallurgy of copper.
(1 Mark)
Ans: Silica in the metallurgy of copper helps in removal of iron oxide as iron silicate (slag).
14. What is the role of graphite rods in the electrometallurgy of aluminium?(1 Mark)
Ans: In the electrometallurgy of aluminium graphite rods act as anodes in the electrolytic cell of reduction
and are the site for release of oxygen
15. Give an example when an element is extracted by oxidation.
(1 Mark)
Ans: Extraction of chlorine from brine is based on oxidation.
16. What will happen if aqueous solution of NaCl is subjected to electrolysis?
(1 Mark)
Ans: If aqueous Solution of NaCl is subjected to electrolysis, Cl2 will be obtained with NaOH and H2 gas as the
side products.
17. What is refining of metals?
(1 Mark)
Ans: Refining of metal is the process of purification of a metal extracted from its ore.
18. What is vapour phase refining?
(1 Mark)
Ans: Vapour phase refining is the method of metal refining by changing the metal into volatile compound that
can be collected separately leaving behind the impurities and can be decomposed to give the pure
metal.
19. Give the principle underlying the process used for refining of gallium.
(1 Mark)
Ans: The process used for the refining of gallium is zone refining and the principle underlying it is that the
impurities are more soluble in the melt than in solid state of the metal.
20. State the principle on which the chromatographic methods of metal refining are based?
(1 Mark)
Ans: Chromatographic methods of metal refining are based on the principle that different components of a
mixture are differently adsorbed on an adsorbent
21. Which is the purest form of iron and what are its uses?
(1 Mark)
Ans: The purest form of iron is wrought iron & is used in making anchors, wires, bolts etc.
22. What are minerals and how are they different from ores?
(2 Marks)
Ans: Minerals are the naturally occurring chemical substances in the earth’s crust obtained by mining. Its
different from ores, as ores are the minerals that are used for the extraction of metals profitably.
23. Name one ore each for iron & copper & give their chemical compositions. (2 Marks)
Ans: The ore of iron is hematite- Fe2O3 and the ore for copper is copper pyrites- CuFeS2.
24. What is the purpose of adding collectors and froth stabilisers during froth floatation? Give an example for
each.
(2 Marks)
Ans: During froth floatation process collectors like pine oil and fatty acids are added to enhance non
wettability of the mineral particles and the froth stabilisers like cresol and aniline stabilise the froth.
25. How can the ores ZnS and PbS be separated from a mixture using froth floatation process?
(2 Marks)
Ans: During the froth floatation process a depressant like NaCN is added to the tank. The depressant
selectively prevents ZnS from coming to the froth but allows PbS to come to the froth and hence helps
the separation of PbS with the froth.
26. Give the equations involved in the concentration of bauxite ore.
(2 Marks)
Ans: i) Al2O3(s) + 2NaOH + 3H2O 2Na[Al(OH)4](aq)
ii) 2Na[Al(OH)4](aq) + CO2(g) Al2O3.xH2O(s) + 2NaHCO3
iii) Al2O3.xH2O(s) Al2O3(s) + xH2O
27. Give one reaction each for roasting and Calcination.
(2 Marks)
Ans: Calcination:
ZnCO3 ZnO(s) + CO2
Roasting:
2ZnS + 3O2 2ZnO + 2SO2
28. Why is coke preferred over CO for reducing FeO?
(2 Marks)
Ans: According to Ellingham diagram the point of intersection of the curves of C, CO and Fe, FeO lies at
temperature lower than that of the point of intersection of CO,CO2 and Fe, FeO curves. This means the
reduction of FeO will occur at much lower temperature with C than with CO. So C is preferred to CO for
reduction.
29. How is cast iron different from pig iron?
(2 Marks)
Ans: Pig iron has 4% carbon and can be easily cast into verity of shapes. Whereas cast iron has lower carbon
content and is extremely hard and brittle.
30. Give the reactions that occur after the copper matte have been fed into silica lined converter.
(2 Marks)
Ans:
2FeS+3O2 2FeO + 3O2
FeO + SiO2 FeSiO3
2Cu2S + 3O2 2Cu2O + 2SO2
2Cu2O + Cu2S 6Cu + SO2
31. Give the reactions taking place at the anode and the cathode during the electrolytic reduction of alumina.
(2 Marks)
Ans: Anode:
C(s) + O2-(melt) CO(g) + 2eC(s) + 2O2-(melt) CO2(g) + 4eCathode:
Al3+(melt) + 3e- Al(l)
32. Explain the process of magnetic separation for concentration of ores.
(3 Marks)
Ans: In magnetic separation ore is carried over a conveyer belt which passes over a magnetic roller. If either
the ore or the gangue is capable of being attracted by the magnetic field then it will collect near the
roller and the particles showing non magnetic behaviour will be collected away from the roller.
33. Differentiate between roasting and Calcination.
(3 Marks)
Ans: Calcination:
i) it involves heating of the ore in the absence of air
ii) it is generally used for carbonate ores
Calcination:
ZnCO3 ZnO(s) + CO2
Roasting:
i) it involves the heating of the ore in the presence of air
ii) it is generally used for sulphide ores
Roasting:
2ZnS + 3O2 2ZnO + 2SO2
34. Give the reactions involved in the reduction of iron oxide to give iron in a blast furnace.
(3 Marks)
Ans: The reactions are as follows:
C + O2 CO2
CaCO3 CaO + CO2
CO2+ C CO
3Fe2O3 + CO Fe3O4 +CO2
Fe3O4 + CO 3FeO + CO2
FeO +CO Fe + CO2
FeO + C Fe + CO
CaO + SiO2 CaSiO3
35. How is copper extracted from low grade ores and scraps?
(3 Marks)
Ans: For extraction of copper from low grade ores and scraps the ore is first leached out using acid or
bacteria. The solution containing Cu2+ is treated with scarp iron or H2 and Cu is obtained.
Cu2+(aq) + H2(g) Cu(s) + 2H+(aq)
Cu2+ + Fe Cu(s) + Fe2+
36. How is gold extracted from its ore?
(3 Marks)
Ans: Extraction of gold involves leaching the metal with CN- giving metal complex.
4Au + 8CN-(aq) + 2H2O + O2(g) 4[Au(CN)2]-(aq) + 4OH-(aq)
the metal is later recovered by displacement method with zinc acting as reducing agent.
2[ Au(CN)2]-(aq) + Zn(s) 2Au(s) + [Zn(CN)4]- (aq)
37. Describe the method used for refining copper metal.
(3 Marks)
Ans: Copper metal is refined by using electrolytic method with impure copper metal as anode and the pure
copper metal strip as cathode. The electrolyte is acidified copper sulphate solution. Copper dissolves
from the anode into the electrolyte and get reduced and deposited on the cathode as pure metal.
Anode: Cu Cu2+ + 2eCathode: Cu2+ + 2e- Cu
Impurities deposit as anode mud.
38. How is nickel refined?
(3 Marks)
Ans: Nickel is refined by Mond’s process which is based upon vapour phase refining.
In this process nickel is heated in stream of carbon monoxide giving a volatile complex, leaving the impurities
behind. The complex is further subjected to higher temperature so that it gets decomposed to giving
pure metal.
330-350K
Ni + 4CO --------------> Ni(CO)4
450-470K
Ni(CO)4 ------------------> Ni + 4CO
39. Describe briefly column chromatography.
(3 Marks)
Ans: Column chromatography is the method of chromatographic refining of metals available in minute
quantities and the impurities are not chemically very much different from the element. In this process
the column of Al2O3 is prepared in glass tube that forms the stationary phase and the solution of the
components to be separated are taken as solution that forms the mobile phase. The components would
separate out based on their different solubilities in the mobile phase and the stationary phase.
40. What criterion is followed for the selection of the stationary phase in chromatography?
(3 Marks)
Ans: Stationary phase is the immobile and immiscible phase in chromatographic method. Stationary phase is
such chosen that the components to be separated present in the mobile phase have different solubilities
in the mobile phase and the stationary phase.
41. How is zinc extracted from zinc blende?
(3 Marks)
Ans: Zinc blende is ZnS. For the extraction of zinc from zinc blende, the ore is first concentrated by the
method of froth floatation. The concentrated ore is then roasted by heating the ore in the presence of
oxygen to give ZnO releasing SO2. The ZnS is further reduced using coke at temperature of 673k giving
zinc metal.
2ZnS + 3O2 2ZnO + 2SO2
ZnO + C Zn + CO
UNIT - 7
p- BLOCK ELEMENTS
Marks= 8
P block elements are placed in groups 13 to 18 of the periodic table. Electronic configuration is ns2
np1-6
Group 15 elements- group 15 includes nitrogen, phosphorous, arsenic, antimony and bismuth .Down
the group metallic character increases nitrogen and phosphorous are non metals arsenic and antimony
are metalloids and bismuth is a typical metal. Electronic configuration is ns2np3 both are half filled
orbitals so they are extra stable. ionization enthalpy decreases down the group due to gradual increase
in atomic size.(Ionization enthalpy is the energy requires to loose electron from its outer most shell to
form positive ion)


The electronegativity decreases down the group with increasing atomic size. (electronegativity is
the tendency to attract electron towards itself) Metallic character electron towards itself). Metallic
character increases down the group because ionization enthalpy decreases down the group .The
common oxidation states are +3,-3and +5,-5 oxidation states decreases down the group and +3
state increases due to inert pair effect .Inert pair effect is the reluctance of s electron in chemical
bonding.
Disproportionation reaction takes place in nitric acid
3HNO2 → HNO3 + H20 + 2NO
Nitrogen is restricted to a maximum covalent of 4 since only for (one s and three p ) orbitals are
available for bonding
Nitrogen shows anomalous behaviour because of small size, high enthalpy and non availability of d
orbitals . Nitrogen forms pπ-pπ multiple bond. N-N bond is weaker than the single p-p bond because of
high inter electronic repulsions of non bonding electrons and owing to the small bond length .For this
reason also catenation tendency is weaker in nitrogen .Nitrogen cannot form d(∏)-p(∏) bond due to
absence of d orbital .
 The stability of hydrides decreases from NH3 to BIH3 .Also size decrease of hydrides in the group.
Reducing character of hydrides also decreases as size of element increases down the group and
basicity decreases down the group nitrogen does not form pentahalides due to absence of d orbital
 In the laboratory dinitrogen is prepared NH4Cl(aq)+ NaNO2(aq) N2 + 2H2O+NaCl
 Other methods
heat
(NH4)2 Cr2O7
>
N2 +4H2O+Cr2O3
Ammonium dichromate
 pure nitrogen can be obtained by the thermal decomposition of sodium or barium acid
 Ba(N3)2 -> Ba+ 3N2
 Dinitrogen is inert at room temperature because of high bond enthalpy.
 Ammonia is a colourless with pungent odour.
 It is sp3 hybridisation, trigonal pyramid structure .It has sp3 hybridisation, trigonal pyramidal
structure .It has three bond pairs and one lone pair of electrons .NH3 is a Lewis base due to
presence of a lone pair of electrons .It form complexes with metal ions because it donates the
electron pair and forms linkage with metal ions.
 Nitric acid is a colourless liquid .concentrated nitric acid is a strong oxidizing agent and attacks
most metals except noble metals such as gold and platinum. Some metals ex. (r) Al do not dissolve
in conc. Nitric acid because of the formation of a passive film of oxide on the surface .Conc. nitric
acid also oxidises non metals and their compounds .ex. iodine is oxidised to iodic acid .
 Phosphorous have allotropic forms .But important are white red ,black white phosphorous is waxy
solid poisonous ,insoluble in water but soluble in carbon disulphide goes in dark .It is ion stable
,more of tetrahedral by molecules .
 Red phosphorous is iron grey lusture. It is odourless, non-poisonous and insoluble in water and
carbon disulphide. Red phosphorous is much less reactive than white phosphorous. It does not
glow in the dark .Red phosphorous is less reactive than white phosphorous because red
phosphorous has polymeric structure in which p4 units are bonded together with strong single
covalent bonds. In white phosphorous p4 units are held together with vander walls force.
 Phosphine is a colourless gas with rotten fish smell and is highly poisonous, It is slightly soluble in
water .It is weak basic.
 Phosphorous forms two types of halides one is PX3 and other is PX5. Pcl3 is a colourless oily
liquid and hydrolysis in the presence of moisture .It has pyramidal shape ,where phosphorous is sp3
hybridised .PCl5 is a yellowish white powder and in moist air, In gas and liquid phases, it has a
trigonal bipyramidal structure. The three equatorial p-d bonds are equivalent while the two axial
bonds are longer than equatorial bonds. This is due to the axial bond pairs suffer more repulsion as
compared to equatorial bond pairs. In the solid state it exists as an ionic solid [PCl4]+[PCl6]- in
which [PCl4]+ is tetrahedral and anion is [PCl6]- octahedral.
 Phosphorus forms a number of oxoacids. For table 7.5 and figure 7.4 are very important. Ortho
phosphorus acid on heating disproportionates to give ortho phosphoric acid and phosphine
 4H 3PO 3 3H 3PO 4+PH 3
 Hypophosphorus acid is a good reducing agent as it contains two P-H bonds and reduces. 4H 3PO 3
is dibasic as it contains two P-OH bonds. H 3 PO4 is tribasic as it contains 3P-OH bonds.
Group 16 elements















Group 16 elements are oxygen sulphur selenium tellurium and polonium. 16 group also called
chalcogens. The elements of group 16 have 6 electrons in the outer most cell and configuration is
ns2np4.The elements of this group have lower ionization enthalpy compared to 15 group due to
extra stability of half filled of 15 groups element. Oxygen has less negative electron gain enthalpy
than sulphur because of the compact nature of oxygen atom. Metallic character increases from
oxygen to polonium due to electronegativity decreases in a group. The elements shows -2,+2,+4
and+6 oxidation state. The oxygen shows only negative oxidation state due to high
electronegativity. But in OF 2 its oxidation state is +2. In this group +6 oxidation state decreases
down the group and stability of +4 oxidation state due to inert pair effect.
Oxygen shows anomalous behaviour due to small size, high electronegativity and absence of d
orbital. Dimeric halides under go
Disproportionation reaction, example 2Se 2 Cl2  Se Cl 4 + 3Se.
An oxide that combines with water to give an acid is termed as acidic oxide , example SO 2, CO 2.
Non metal oxides are acidic. The oxides which give a base with water are known as basic oxides,
example CaO. Metallic oxides are basic. Some metallic oxides exhibits a dual behaviour. They
show both acidic and basic property. Such oxides are called amphoteric oxides
There are some oxides which are neither acidic nor basic. Such oxides are known as neutral oxides.
Example CO,NO, N2O.
Ozone is an allotropic form of oxygen. Since the formation of ozone from oxygen is an
endothermic process, it is necessary to use a silent electrical discharge in its preparation to prevent
its decomposition. Sulphur form a number of allotropes. Yellow rhombic (alpha sulphur) and
monoclinic (beta sulphur) are important. Sulphur dioxide is a colourless gas with pungent smell
and highly soluble in water. When moist sulphur dioxide behaves as a reducing agent it converts
iron (III) ions to iron (II).
Sulphur forms a number of oxoacids. But sulphuric acid is important. It is manufactured by the
contact process which involves 3 steps1-burning of sulphur in air to form SO 2
S+O2 SO 2.
2- Conversion of SO 2 to SO 3 by the reaction with oxygen in the presence of catalyst (V 20 5)
2SO 2+ O2 V2O5 2SO 3
3-Absorption of SO 3 in H 2SO 4 to give Oleum(H 2S 2O 7)
SO 3+ H 2SO 4 H 2S 20 7
Dilution of Oleium with water gives H 2SO 4 of the desired concentration.
Sulphuric acid is a colourless, dense, oily liquid. The concentrated acid must be added slowly into
water with constant stirring. The larger value of Ka, means that H 2SO 4is highly dissociated into H
+ and HSO4-.Greater the value of dissociation constant, the stronger is the acid. The acid forms 2
series of salt normal sulphate example sodium sulphate and acid sulphate example sodium
hydrogen sulphate.
Group 17 elements (Halogen)
It contains fluorine, chlorine, bromine, Iodine and astatine. These are called halogens means salt
producers. Astatine is a radioactive element. Electronic configuration is ns 2 np 5.High ionization
enthalpy because there is less tendency to loose electon.Ionisation enthalpy decreases down the group
as atomic size decreases.
Halogens have maximum negative electron gain enthalpy in the period due to having 1 electron less
than noble gas configuration. The negative electron gain enthalpy of fluorine is less than that of
chlorine due to small size of fluorine atom and strong inter electronic repulsion 2 p orbitals of
fluorine.
Fluorine and chlorine are gases, bromine is a liquid and Iodine is solid. Halogens are coloured due to
absorption of radiation in visible region, excite the outer electron to higher energy level. By absorbing
different radiations display different colours. F 2 is yellow Cl 2 is greenish yellow, Br 2 is red and I 2 is
violet colour. All the halogens show -1 oxidation state. Fluorine atom has no d orbtals, so it can not
expand and shows only
-1 oxidation state. But Cl,Br and I show +1,+3,+5 and +7 oxidation state.
All the halogens are highly reactive. They react with metals and non metals to form halide. The
reactivity of the halogens decreases down the group. Halogens are strong oxidizing agent because they
accept 1 electron easily to attend noble gas configuration.
F2 is the strongest oxidizing agent.(Oxidising agent is one which oxidises other but itself
reduced.)Fluorine shows anomalous behavior due high ionization enthalpy, highest electronegativity
and absence of d-orbital, small size, low F-F bond dissociation energy. Most of the reactions of
fluorine are exothermic due to small and strong bond formed with other element Hydrogen fluoride is
a liquid with high boiling point due to strong hydrogen bonding. F2O does not exist but OF2 is stable
due to high electronegativity.
 Chlorine is greenish yellow gas with pungent and suffocating odor. It is colorless and pungent
smelling gas. It is easily liquefied to a colorless liquid and freezes to a white crystalline solid. High
value of dissociation constant (Ka) indicates that it is a strong acid in water fluorine forms only one
Oxo acid, due to high electronegativity and small size.
 When two different halogens react with each other, interhalogen compounds are formed, as XX’
,XX3’, XX5’ and XX7 where X is a halogen of larger size and X’ of smaller size and X is more
electropositive than X’. Inter halogen compounds can be prepared by the direct combination or by
the action of halogen on lower interhalogen compounds.
Group 18 elements

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Group 18 consists of six elements helium, neon, argon, krypton, xenon and radon. All these are
gases and chemically unreactive. They form very few compounds because of this they are called
noble gases. General electronic configuration is ns2np6 except helium. Gases exhibit very high
ionization enthalpy due to stable configuration. It decreases down the group with the increase in
atomic size. Noble gases have large positive values of electron gain enthalpy because of stable
configuration and have no tendency to accept the electron. They have very low melting and lowest
boiling point because of interatomic interaction.
Neil Bartlett prepared a red compound formula of O2+PtF-6. Since first ionisation enthalpy of
Oxygen is almost identical with that of xenon. So he prepared red colour compound of Xe 2+PtF-6
by mixing PtF6 and xenon. After his discovery a number of xenon compounds with most
electronegative elements like fluorine and oxygen have been synthesised.
USES
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
Nitrogen: Manufacturing of Ammonia, refrigerant industrial chemicals.
Ammonia: Nitrogenous fertilizer, manufacturing of some inorganic compound. Liquid
ammonia is used as refrigerant.
Nitric Acid: Manufacturing of ammonium nitrate for fertilizer, preparation of nitroglycerin
trinitrotoluene, other uses are pickling of stainless steel, etching of metals and as an oxidise in
rocket fuels.
Phosphine: Home’s signals, smoke screens.
Oxygen:
Oxyacetylene welding, manufacturing of many metals. Oxygen cylinder,
combination of fuel eg hydrazine in liquid oxygen.
Ozone: Used as a germicide, disinfectant for sterilising water. Bleaching oils, flour, ivory,
starch, manufacturing of potassium permanganate.
Sulphur dioxide: Refining petroleum and sugar bleaching wool and silk, anti-chlor,
disinfectant and preservative. Liquid SO2 is used as a solvent to dissolve a number of organic
and inorganic chemicals.
Sulphuric Acid: It is very important industrial chemical. Manufacture of compounds and fertilizer
petroleum refining, manufacture of pigments, paints and dyes, detergent industry, in metallurgy
cleaning fo metals before enameling, electroplating and galvansing, storage batteries, manufacturing of
nitrocellulose product, lab reagent.
Interhalogen Compound are useful fluorinating agents. ClF3 and BrF3 are used for the production of
UF6 in the enrichment of 235U.
(x)
(xi)
(xii)
Helium : It is used in filling balloons due to non-inflammable and light gas, in gas cooled nuclear
reactor. Liquid Helium used as cryogenic agent, produced superconducting magnets for NMR
spectrometer, Magnetic resonance (MRI) for clinical diagnosis, modern diving apparatus.
Neon: Neon is used in discharge table and fluorescent bulbs. Neon bulbs are used in botanical gardens
and in green houses.
Argon: Argon is used in inert atmosphere in high temperature, metallurgical process, filling electric
bulbs, for handling substance that are air sensitive. Xenon and Krypton are used in light bulb designed
for special purpose.
(THE P BLOCK ELEMENTS CHEMICAL REACTION)
NH4 Cl + NaNO3  N2 + 2 H2 O + NaCl
(NH4)2 Cr2O7  N2 + 4H2O + Cr2 O3
Ammonium dichromate
Ba(N3)  Ba + 3 N2
Barium azide
6Li + N2  2Li3N (Lithium nitride)
3 Mg + N2  Mg3N2 (Magnesium nitride)
2NaN3  2Na + 3N2
N2 + 3 H2  2 NH3
2 NH4 Cl + Ca ( OH) 2  2NH3 + 2H2O + CaCl2
(NH4)2 SO4 + 2 Na OH  2 NH3 + 2H2O + Na2 SO4
NH3 + H2O  NH4+ + OHZn SO4 + 2NH4 OH  Zn(OH)2 + (NH4)2SO4
FeCl3 + NH4OH  Fe2O3XH2O + NH4Cl
(BROWN PPT)
Cu+2 + 4NH3aq  [Cu(NH3)4]2+aq
Blue
deep blue solution
Ag+ aq + Cl- 
Ag Cl
WHITE PPT
AgCl +2NH3 aq  [Ag(NH3)2]Cl Colourless
NH4 NO3  N2O + 2H2O
Nitrousoxide
2Pb(NO3)2  4NO2 + 2PbO
NaNO3 + H2SO4  NaHSO4 + HNO3
4NH3 + 5O2  4NO + 6H2O
3NO2 + H2O  2HNO3 + NO (Nitric oxide)
3 Cu + 8 HNO3 (Dil.) 3Cu(NO3)2 + 2NO2 + 2H2O
Cu + 4HNO3 (Conc.)  Cu(NO3) 2 + 2NO2 +2H2O
4Zn + 10HNO3 (Dil.)  4Zn(NO3)2 +5H2O + N2O
Zn + 4HNO3 (Conc.)  Zn(NO3)2 + 2H2O + 2NO2
I2 + 10 HNO3 (Conc.)  2HIO3 + 4H2O + 10NO2
(IODICACID)
C + 4HNO3 (Conc.)  CO2 + 2H2O + 4NO2
S8 + 48HNO3 (Conc.)  8H2SO4 + 16H2O + 48NO2
P4 + 20HNO3 (Conc.)  4H3PO4+ 4H2O + 20NO2
P4 + 3NaOH +3H20 PH3 + 3NaH2PO2 (SODIUM HYPOPHOSPHITE)
P4 + 5O2  P4010
Ca3P2 + 6H2O  3Ca(OH)2 + 2PH3
Ca3P2 + 6HCl 3CACl2 + 2PH3
PH4I + KOH  KI + H2O + PH3
3 CuSO4 + 2PH3  Cu3P2 + 3H2SO4
3HgCl2 + 2PH3  Hg3P2 + 6HCl
MERCURIC PHOSPHIDE
PH3 + HBr  PH4Br
P4 + 6Cl2  4PCl3
P4 + 8SOCl2  4PCl3 +4 SO2 + 2S2Cl2
PCl3+ 3 H2O H3PO3 +3HCl (FUMES )
PCl5 + H2O  POCl3 + 2HCl
POCl3 + 3H2O  H3PO4 + 3HCl
2Ag + PCl5  2AgCl + PCl3
Sn +2 PCl5  SnCl4 + 2PCl3
2KClO3 2 KCl +3 O2
2Ag2O 4Ag(S) +O2(G)
2Pb3O4 (S)  6PbO(S) + O2(G)
2HgO  2Hg (L) + O2(G)
2ZnS +3O2 2ZnO + 2 S02
Al2O3 + 6HCl + 9H2O  2[Al(H2O)6]3+ + 6CAl2O3 + 6NaOH +3H202Na3[Al(OH)6]
PbS+ 4O3  PbSO4 + 4O2
4FeS2 + 11O2  2Fe2O3 +8SO2
SO3 +H2O  H2SO4
CaF2 + H2SO4  CaSO4 + HF
2NaOH + SO2 Na2SO3 + H2O
2Fe3+ + SO2 + 2H2 O  2Fe2+ + SO4-2 + 4H+
NaCl + H2SO4  HCl +Na2SO4
C12H22O11  12 C + 11H2O
Cu + 2H2SO4 (CONC) CuSO4 +SO2+ 2H2O
3S +2H2SO4 (CONC)  3SO2 + 2H2O
C + 2H2SO4 (CONC)  CO2 + 2H2O +2SO2
MnO2 + 4HCl  MnCl2+ Cl 2 + 2H2O
H2 + Cl2  2HCl
4NaCl + MnO2 + 4H2SO4  MnCl2 + 4NaHSO4 +2H2O+Cl2
2kMnO4 + 18 HCl  2KCl + 2MnCl2 + 8H2O + 5Cl2
4HCl +O2  2Cl2 +2H2O
2F2 + 2H2 O  4H+ + 4F- + O2
2NaOH + Cl2  NaCl + NaOCl + H2O
2Cl2 + 2H2 O  HCl + HOCl
2Ca(OH)2 + 2Cl2  Ca(OCl)2 + CaCl2 + 2H2O
2FeSO4 +Cl2 + H2SO4  Fe2(SO4)3 + 2HCl
NaCl +H2SO4  NaHSO4 + HCl
2NaCl +2H2SO4 + MnO2  MnSO4 + Na2SO4 +2H2O + Cl2
NaHSO4 + NaCl  Na2SO4 + HCl
Cl2 + 2NaI  2NaCl + I2
Na2CO3 + 2HCl  2NaCl + H2O + CO2
NaHCO3 + HCl  NaCl + H2O + CO2
Na2SO3 + 2HCl  2NaCl + H2O + SO2
Fe + 2HCl  Fe + H2
U + 3ClF3  UF6 + 3ClF
Xe + F2  XeF2 (XENON IN EXCESS)
Xe + 2F2  XeF4 (1: 5)
Xe + 3F2  XeF6 (1:20)
XeF4 +O2 F2  XeF6 + O2
2XF2 + 2 H2O  2Xe +4HF + O2
4XeF4 +12 H2O  4 Xe + 2XeO3 +24HF +3O2
XeF6 +3H2O  XeO3 +6HF
XeF6 +H2O  XeOF4 + 2HF (PARTIAL HYDROLYSIS)
XeF6 +2H2O  XeO3F2 + 4HF(PARTIAL HYDROLYSIS)
CaCO3 + H2SO4 CaSO4 + H2O + CO2
DISPROPORTIONATION REACTION
Disproportionation reaction, is a specific type of redox reaction, in which a particular element is simultaneously reduced
and oxidized to form two different products.
reaction is comproportionation reaction)
3HNO2  HNO3 + H2O +2NO
4H3PO3  3H3PO4 +PH3
2SeCl2  SeCl4 + 3Se
( it is also known as dismutation Reaction, while the reverse of this
NCERT Intext Questions
1 Why are pentahalides more covalent than trihalides ?
Ans- higher the positive oxidation state more will be polarising power and more covalent character.
2 Why is BiH3 the strongest reducing agent amongst all the hydrides of Group 15 elements ?
Ans-The size of element increases in a group then bond length increases and become weaker.
From N – H bond can break easily and gives hydrogen.
3 Why is N2 less reactive at room temperature?
Ans- Due to presence of a triple bond, bond dissociation energy is higher.
4 Mention the conditions required to maximise the yield of ammonia.
Ans- i. Low temp, but optimum temp is 700K.
ii. high pressure(200 atm)
5 How does ammonia react with a solution of Cu2+?
Ans -Cu+2(aq)+NH4OH(aq)  [Cu(NH3)4]+2 + H2 O
Blue colour complex
6 What is the covalence of nitrogen in N2O5 ?
Ans- Covalancy is 4.
7 Bond angle in PH4+ is higher than that in PH3. Why?
Ans- In PH3 and PH4+ P undergoes sp3 hybridisation. In PH4+ there is no lone pair electron but in PH3
there is one lone pair of electron . due to repulsion , angle decreases in PH3.
8 What is the basicity of H3PO4?
Ans. Basicity is three as it has three P – OH .
9 What happens when H3PO3 is heated?
Ans. H3PO3 undergoes disproportionation reaction H3PO3  PH3 + H3PO4
Here P3+ is oxidised to P5+ and again it reduced to P3-, so it is disproportionation reaction.
10 Why is H2O a liquid and H2S a gas ?
Ans. In H2O, hydrogen bond is formed but in H2S no hydrogen bond exists. Because electronegativity
of S is less than oxygen.
11 Which of the following does not react with oxygen directly?
Zn, Ti, Pt, Fe.
Ans. Platinum (Pt) is a noble element, so it does not react with oxygen.
12 Why does O3 act as a powerful oxidising agent?
Ans. Because ozone on heating gives nascent oxygen.
13 How is the presence of SO2 detected ?
Ans. SO2 is a pungent smell gas. It can be detected by passing through acidified potassium dichromate
solution which turns green due to Cr2+6 changes to Cr2+3.
14 Write the conditions to maximise the yield of H2SO4 by Contact process.
Ans 24. (a) low temp (b) high pressure (c) V2O5 as catalyst
15 Why is Ka2<< Ka1 for H2SO4 in water ?
Ans 25. Ka2 is less than Ka1 because H2SO4-1 has a tendency to donate a proton to water.
16 Sea is the greatest source of some halogens. Comment.
Ans. Sea water contain chloride bromide and iodide of sodium, potasium magnesium and calcium
17 Give the reason for bleaching action of Cl2.
Ans. Chloride liberates nascent oxygen in presence of moisture, which bleaches the coloured
substances present in organic matter.
18 Name two poisonous gases which can be prepared from chlorine gas.
Ans. (a) phosgene Cocl2
(b) mustard gas
19 Why is ICl more reactive than I2?Ans. ICl is more reactive than I2 because I-Cl bond is polar and
weaker than I-I bond which is non-polar and stronger.
NCERT EXAMPLES
Example 1 Though nitrogen exhibits +5 oxidation state, it does not form pentahalide. Give
reason.
Ans - Due to absence of d orbital it does not form pentahalide. But due to involvement of s and p orbit
it exhibits +5 oxidation state.
Example 2 PH3 has lower boiling point than NH3. Why?
Ans- Because of hydrogen bond which is present in NH3.
Example 3 Why does NH3 act as a Lewis base ?
Ans- Nitrogen atom in NH3 has one lone pair of electron for donation (Lewis base is one which donate
one lone pair electron)
Example 4 Why does NO2 dimerise ?
Ans-NO2 contain odd number of valence electron. On dimirisation it gives a stable N2O4 molecule.
Example 5 In what way can it be proved that PH3 is basic in nature?
Ans- PH3 react with acid like HI to give PH4I
Example 6 Why does PCl3 fume in moisture ?
AnsPCl3 hydrogen in the presence
PCl3 + 3H2O H3PO3 + 3HCl
of
moisture
giving
fumes
of
HCl
Example 7 Are all the five bonds in PCl5 molecule equivalent? Justify your answer.
Ans- PCl5 has a trigonal bipyramidal structure and three equatorial P-Cl bonds equivalent, while two
axial bonds are different and longer.
Example 8 How do you account for the reducing behaviour of H3PO2 on the basis of its
structure?
Ans- H3PO2 contains two hydrogen atoms and bonded directly to P atom.
Example 9 Elements of Group 16 generally show lower value of first ionisation enthalpy
compared to the corresponding periods of group 15. Why?
Ans- Due to extra stable of half filled p orbital configuration present in 15 group element.
Example 10 H2S is less acidic than H2Te. Why?
Ans-Bond association decreases down the group as atomic size increases.
increases,
So acidic character
Example 11 Which form of sulphur shows paramagnetic behaviour ?
Ans- Vapour state Sulphur because it has to unpaired electrons.
Example 12 Halogens have maximum negative electron gain enthalpy in the respective periods
of the periodic table. Why?
Ans- Due to size increases in period and high effective nuclear charge. So accept one electron to
acquire noble gas electronic configuration.
Example 13 Although electron gain enthalpy of fluorine is less negative as compared to chlorine,
fluorine is a stronger oxidising agent than chlorine. Why?
AnsI) due to low enthalpy of dissociation of f-f bond.
Ii) high hydration enthalpy of F-
Example 14 Fluorine exhibits only –1 oxidation state whereas other halogens exhibit + 1, + 3, + 5
and + 7 oxidation states also. Explain.
Ans- Due to high electronegative of fluorine it can not exhibit any positive oxidation state. Other
halogens have d orbits so show +1,+3,+5 and +7 oxidation state.
Example 15 When HCl reacts with finely powdered iron, it forms ferrous chloride and not ferric
chloride. Why?
Ans- Libration of hydrogen prevents the formation ferric chloride
Fe + HCl  FeCl2 + H2
Example 16 Noble gases have very low boiling points. Why?
Ans - Noble gasses are mono atomic. They have no inter atomic forces, they have van der Wall’s
forces. So liquefied at very low temp. So they have low boiling points.
Textbook Exercise Questions & Answer
1. Why does the reactivity of nitrogen differ from phosphorus?
Ans- Nitrogen exists as a diatomic molecule by triple bond. It is gas at room temperature .Multiple
bond is not possible in phosphorus due to its large size .It exists as P4 molecule. Nitrogen is less
negative than phosphorus due to high bond dissociation energy.
2. The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be
explained on the basis of sp3 hybridisation in NH3 and only s–p bonding between hydrogen and
other elements of the group].
Ans-The central atom is sp3 hybridised. Electronegativity decreases down the group. The forces of
repulsion decreases down the group so angle decreases.
3. Why does R3P = O exist but R3N = O does not (R = alkyl group)?
Ans- Due to absence of d orbitals in nitrogen. It can not extend its covalency to five but in phosphorus
vacant d-orbits exists. So exists.
4. Explain why NH3 is basic while BiH3 is only feebly basic.
Ans- Nitrogen is smaller than bismuth. So electron density is concentrated on nitrogen is maximum. It
can donate electron easily.
5. Nitrogen exists as diatomic molecule and phosphorus as P4. Why?
Ans-Due to small size, two nitrogen atoms can be joined by three covalent bonds in order to complete
octet. The size of phosphorus is large so it has less tendency to form three bonds. So, p atom complete
its octet by sharing electrons with three other p atoms. So, phosphorus exists as p4 molecule.
6. Write main differences between the properties of white phosphorus and red phosphorus.
AnsWhite phosphorus
Red phosphorus
1. white but turns yellow on exposure
dark red
2. waxy solid
brittle powder
3. less stable at ordinary temp.
more stable at ordinary temp.
4. very active
less reactive
7. Why does nitrogen show catenation properties less than phosphorus?
Ans- Catenation means self linkage property of nitrogen is less than that of phosphorus because the NN bond is weaker than P-P bond.
8. Can PCl5 act as an oxidising as well as a reducing agent? Justify.
Ans-Phosphorus can show maximum oxidation state. Ex PCl5. It can not increases its oxidation state
beyond 5. So, it can not act as reducing agent. But it can act as an oxidizing agent by decreasing its
oxidation state from +5 to +3
Ex .
Ag + PCl5 → 2AgCl + PCl3
Oxidizing agent - is one which oxidize other, but it self is reduced.
Reducing agent – is one which reduce other, but itself is oxidized.
9. Why is dioxygen a gas but sulphur a solid?
Ans-Due to small size and high electronegativity oxygen atom forms Pл-Pл double bond due to weak
van der Walls force, oxygen is gas. But sulphur does not form stable Pл-Pл bond and do not exists as
S2. It is linked by single eight atoms per molecule S8 and strongly held together so it exists as solid.
10. Which aerosols deplete ozone?
Ans -Chloroflurocarbon is oxides of nitrogen and sulphur.
11. How is SO2 an air pollutant?
Ans-SO2 dissolves in rain water and produces acid rain. The acid rain contains sulphuric acid
SO2 + H2O + O → H2SO4
12. Why are halogens strong oxidising agents?
Ans-Halogens have strong tendency to accept electrons due to high electronegativity.
13. Explain why fluorine forms only one oxoacid, HOF.
Ans-Due to small size ,high electronegativity and absence of d-orbital fluorine shows only single
oxidation state (-1).So it forms only one oxoacid HOF.
14. Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding
while chlorine does not.
Ans-Oxygen has smaller size than chlorine ,which favours hydrogen bonding.
15. Write two uses of ClO2.
Ans-(1)It acts as bleaching agent for paper pulp in industry and
textile industry.
(2)It acts as germicide for disinfecting water.
17. What inspired N. Bartlett for carrying out reaction between Xe and PtF6?
Ans- N. Bartlett first prepared an ionic compound O2+[PtF6]- from oxygen and PtF6.Ionisation
enthalpy of xenon gas is close to that oxygen .That is inspired to carry out the reaction between Xe & PtF6.
18. What are the oxidation states of phosphorus in the following:
(i) H3 PO3 (ii) PCl3 (iii) Ca3 P2 (iv) Na3 PO4 (v) POF3?
AnsH3PO3 → +3;
PCl3 → +3;
Ca3P2 → -3
Na3PO4→ +5;
PoF3 → +5
19. With what neutral molecule is ClO- isoelectronic? Is that molecule a Lewis base?
Ans- ClO- is isoelectronic with ClF is Lewis base as Cl atom can accept a pair of electrons. It is due to
presence of vacant d-orbital in chlorine.
20. Arrange the following in the order of property indicated for each set:
(i) F2, Cl2, Br2, I2 - increasing bond dissociation enthalpy.
(ii) HF, HCl, HBr, HI - increasing acid strength.
(iii) NH3, PH3, AsH3, SbH3, BiH3 – increasing base strength
Ans- (i) I2 < F2 < Br2 < Cl2
Increasing bond dissociation enthalpy.
(ii) HF < HCl < HBr < HI
Increasing acid strength
(iii) BiH3 < SbH3 < AsH3 < PH3 < NH3
Increasing base strength
21. Which one of the following does not exist?
(i) XeOF4 (ii) NeF2 (iii) XeF2 (iv) XeF6
Ans- NeF2 does not exist
22. Why do noble gases have comparatively large atomic sizes?
Ans- In case of noble gases the atomic radii corresponds to Vander Waals radii, which are always
large. Vander Waals radii arise simply due to Vander Waal’s force.
UNIT 8
THE d- and f- BLOCK ELEMENTS
Marks =5
Electronic configuration of transition elements is (n-1)1-10 ns1-2. The name of transition metal refers to
d block. The metal of inner transition metal refers to f block. d block contain 3-12 group elements
where d block are progressively filled, where as f block contain elements in which the 4f and 5f
orbitals are progressively filled. Three series of transition metals are present. 3d series9sc-zn) , 4d
series(y – cd), 5d series(La – Hg omitting Ce – Lu)
The name transition is given because of there position between s and p block. the configuration Cr is
3d54s and Cu is 3d104s1. Zn, Cd and Hg are not regarded as transition elements because orbitals are
completely filled.
Properties


Transition metals are quite hard, high melting point, high enthalpies of atomization, high ionization
because of strong metallic bond. Covalent bond and unpaired electron. Transition element show
vari9able oxidation states due to the participation of ns and (n-i) d electrons in bonding. Highest
oxidation state shown by transition element is eight.
M = √ n (n+2) M is the magnetic moment n is the number of unpaired electron. Transition
elements and their compounds are paramagnetic in nature because they contain unpaired electrons.
Transition metals from a large number of complex compound due to smaller size of the metal ions ,
high ionic charges and availability of vacant d orbitals.
Transition metal form coloured compounds due to d-d transition.
The transition metals and some of their compounds act as catalyst due to variables oxidation state,
and large surface area.
Interstitial compounds are those which are those which are formed when small atoms like H, C or
N are trapped inside the critical lattice of metals. Transition metal compounds form interstitial
compound because some voids are present in the site. The transition metals form a number of
alloys because of similar atomic size.
Ex Brass (Copper, Zinc) Bronze (Copper- Tin) .
Oxides and Oxo anions
As the oxidation number of a metal increases, ionic character decreases so Mn2O7 is acidic
V2O3 Basic , V2O5 atmospheric , CrO is basic Cr2O3 is atmospheric . MnO2 is atmospheric preparation
K2cr2O7 (Potassium dichromate)
4 FeCr2O4 + 8 Na2CO3 + 7O2
Chromite are
2 Na2CrO4 + 2H+
Na2CrO7 +2 KCl
8Na2CrO4 + 2Fe2O3 + 8CO2
Na2Cr2O7 + 2Na+ + H2O
K2Cr2O7 + 2NaCl
*on increasing pH
Cr2O7-2 + OHCrO4-2 + H2O
Orange
yellow
On decreasing pH
CrO4-2 + H+
Cr2O7-2 + H2O
Yellow
orange

Preparation of Potassium Permanganate (KMnO4)
2MnO2 + 4KOH + O2
2K2MnO4 + 2 H2O
Black
green
3MnO4-2 + 4H+
2MnO4-1 + MnO2 + 2 H2O
Commercial preparation
MnO2
fused with
MnO4-2
Alkali

MnO4-1 permagnet ion
(pink colour)
electrolytic
oxidation
Manganate and permanganate are tetrahedral. Chromate is tetrahedral.
On heating KMnO4
K2MnO4 
MnO2 + O2
Pink
green
USES

KMnO4 - oxidation, bleaching of coal, cotton, silk, analytical chemistry , disinfectant for ater

K2Cr2O7 – oxidant ,chrome tanning in leather industry
f block elements consists of two series1. Lanthanoids – 14 elements Cerium to Lutetium are similar to lanthanium.
2. Actinoids – 14 elements Thorium to lawrencium are similar to actinium





Misch metal 95% lanthanoid metals and 5%iron traces of S, C, Ca, Al.
The different oxidation states exhibited by lanthanoids are +2, +3, +4.
The actinoids exhibits a large number of oxidation states than the corresponding lanthanoid because of
large size.
Electronic configuration of lanthanoid is 4f¹⁻¹⁴ 5dº⁻¹ 6s².
SIMILARITY BETWEEN LANTHANOIDS AND ACTINOIDS





Both shows +3 oxidation states.
f orbitals are progressively filled.
Same number of unpaired electrons.
Electropositive and highly active.
Contraction.
DIFFERENCE BETWEEN LANTHANOIDS AND ACTINOIDS –
LANTHANOIDS
1. Compounds are less basic.
2. Binding energy of 4f is higher.
3. Most ions are colourless.
4. Not radioactive.
5. Tendency to form complexes is less.
ACTINOIDS
1. Compounds are less basic.
2. Binding energy of 5f is lower.
3. Ions are coloured.
4. Radioactive.
5. Tendency is more.
LANTHANOIS CONTRACTION – The steady decrease in atomic size of lanthanoid elements with
increase atomic number is called lanthanoid contraction.
CAUSE – Nuclear charge increases from left to right, to be compensated by shielding effect by 4f
electrons. F electrons orbitals have poor shielding effect. So, there is steady decrease in atomic size.
CONSEQUENCE – Separation of lanthanoids difficult as they have similar properties.
 Similarity in the atomic size of the elements of second and third transition series in the same group. Ex
- Zr - Hf
 Variation in basic strength of hydroxide.
 The basic strength decreases from La (OH)3to Lu(OH)3. Because of size M⁺³ ion decreases.
A few Ionic Euations related to the d- Block Elements
Some questions and Answers for practice :
1. Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it
is a transition element?
ANS: Because silver can exhibit +2 oxidation state, which has incompletely filled d orbital.
2. Why do the transition elements exhibit higher enthalpies of atomisation?
ANS: Because Sc (21) has incompletely filled d orbital so it takes part in the formation of metallic
bond. But zinc has completely filled d orbital so zinc has weakest bond and lowest enthalpy of
atomization.
3. Name a transition element which does not exhibit variable oxidation states.
ANS: Mn exhibits large number of oxidation states because s and d orbitals takes in bond formation.
4. The E◦(M2+/M) value for copper is positive (+0.34V). What is possibly the reason for this?
(Hint: consider its high ΔaH◦ and low ΔhydH◦)
ANS: both fluorine and oxygen have small size ,high electronegativity.
5. Why is the EV value for the Mn3+/Mn2+ couple much more positive than that for Cr3+/Cr2+
or Fe3+/Fe2+? Explain
ANS: Fe⁺² is strongest reducing agent because it will oxidize to Fe⁺³ which is half filled d orbital,
more stable.
6. Calculate the magnetic moment of a divalent ion in aqueous solution. If its atomic number is
25.
ANS: M(27)
M⁺²=[Ar] 3d⁷ 4sº
Magnetic moment, M =√n (n+2)
√3x5 = √15 = 3.8BM
7. What is meant by ‘disproportionation’ of an oxidation state? Give an example.
ANS: Because hydration energy of Cu⁺² ion is high which can overcome 2nd ionization energy. So,
Cu⁺² is more stable.
8. Name a member of the lanthanoids series which is well known to exhibit +4 oxidation state.
ANS: Due to poor shielding by 5f electrons in actinoids as compared to 4f electrons in the
lanthanoids.
NCERT-EXAMPLES
Example 1- Why do the transition elements exhibit higher enthalpies of atomisation?
ANS: due to large number of unpaired electrons, strong bonding atoms results in higher enthalpies of
atomization.
Example 2- Name a transition element which does not exhibit variable oxidation states.
ANS: Scandium does not exhibit variable oxidation state.
Example 3- Why is Cr 2+ reducing and Mn3+ oxidising when both have d 4 configuration.
ANS: Cr⁺²is reducing as it changes from d⁴ to d³ is half filled in t2g Mn⁺² to Mn⁺³ which half filled d⁵
has extra stability.
Example 4- Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic
number is 25.
ANS: With atomic number 25 Mn⁺²has d⁵ cofiguration-50 M = √5(5+2) = 5.92BM
Example 5- Name a member of the lanthanoids series which is well known to exhibit +4
oxidation state.
ANS: Cerium ( Z = 58 )
A few more questions & answers for practice
1. Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state?
Ans – Mn⁺²s 3d⁵, half filled which is stable, while Fe*² is 3d⁶ not stable.
2. To what extent do the electronic configurations decide the stability of oxidation states in the
first series of the transition elements? Illustrate your answer with examples.
Ans – The presence of half filled or completely filled orbitals.
3. Name the oxometal anions of the first series of the transition metals in which the metal
exhibits the oxidation state equal to its group number.
Ans – Mno4⁻¹ oxidation state +7 equal to group number
Cro4⁻² oxidation state +6 equal to group number.
4. In what way is the electronic configuration of the transition elements different from that of the
non transition elements?
Ans – In transition elements d orbitals are progressively filled where as in
Non transition elements outer most s or p orbitals are progressively filled.
5. What are the different oxidation states exhibited by the lanthanoids?
Ans – common stable oxidation state of lanthanoid is +3, others are +2 and +4.
6. Predict which of the following will be coloured in aqueous solution? Ti3+, V3+, Cu+, Sc3+, Mn2+,
Fe3+ and Co2+. Give reasons for each.
Ans - Ti⁺³, V⁺³, Mn⁺², Fe⁺³, Co⁺² as they contain unpaired electrons
7. Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
(i) electronic configuration (iii) oxidation state
(ii) atomic and ionic sizes and (iv) chemical reactivity
Ans – 1) Electronic configuration – In lanthanoids 4f orbitals are progressively filled whereas in
actinoids 5f orbitals are progressively filled
2) Oxidation state – lanthanoids show+3state some shows +2 and +4.
But actinoids show +3, +4, +5, +6, +7 oxidation state. +3 and +4
are Common.
3) Atomic and ionic size decreases from left to right.
atomic and ionic sizes –decreases from left to right in both. But decreases in more in case of
actinoids
4) Chemical reactivity – actinoids are more reactive than lanthanoids.
8. Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
(i) electronic configuration (iii) oxidation state
(ii) atomic and ionic sizes and (iv) chemical reactivity
9. How would you account for the following?
(i) Of the d4 species, Cr2+ is strongly reducing while manganese(III) is strongly oxidising.
(ii) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily
oxidised.
(iii) The d1 configuration is very unstable in ions
Ans- (1). Cr+2 is reducing agent , it change to Cr3+ by loosing electron, Cr3+ is more stable due to half
filled. Mn3+ is oxidising agent, reduced to Mn2+ which is half filled(d5) and very stable.
(2). cobalt(2) is oxidised to Co(3) because Co(3) is more stable than Co(2).
(3). because after loosing electron it become more stable
10. What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in
aqueous solution.
Ans. In disproportionation reaction, an element under goes oxidation and reduction both
eg.
2Cu+ → Cu + Cu2+
11. Which metal in the first series of transition metals exhibits +1 oxidation state most frequently
and why?
Ans. Cu exhibits +1 oxidation state, by loosing one electron, the cation ion acquires a stable
configuration of d orbital.
12. Give examples and suggest reasons for the following features of the transition metal
chemistry:
(i) The lowest oxide of transition metal is basic; the highest is amphoteric/acidic.
(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.
(iii) The highest oxidation state is exhibited in oxo-anions of a metal.
Ans-
(i) In the lowest oxidation state, ionic bond is formed, less number of electrons are involved.
Oxides donate electrons and behave like a base.
In the highest oxidation state, covalent bond is formed. In high oxidation state more electrons
are involved in bonding. Oxide can gain electron, behave like Lewis acid.
(ii) Because oxygen and fluorine are strong oxidising, highly electronegative element.
(iii)Due to high electro negativity of oxygen.
13. What are alloys? Name an important alloy which contains some of the lanthanoid metals.
Mention its uses.
Ans- Alloy are homogeneous mixture of two or more metals. Misch metal is an alloy, which contains
45% lanthanoid metals, iron 5%, traces of S, C, Ca, Al
Use- used to produce bullets, shell and lighter flint.3% misch metal to magnesium used in making jet
engine parts.
14. What are inner transition elements? Decide which of the following atomic numbers are the
atomic numbers of the inner transition elements : 29, 59,74, 95, 102, 104.
Ans- Lanthanoids and Actinoids are called inner transition elements because inner f orbitals are
progressively filled .z=58 to71 are lanthanoids.z=90 to103 are actinoids,so atomic no. 59,95and102
belong to inner transition elements.
Text book ex. Answer 30-Lawrencium (Lr=103)
[Rn] 5f14 6d1 7s2, oxidation state=+3
Text book ex. Answer 36- Te2+→3d2 ,Cr3+→3d3 ,Fe2+→3d6 ,Co2+→3d7 ,Cu2+→3d9 ,V2+→3d8 ,
Mn2+→3d5 ,Fe3+→3d5 ,Ni2+→3d8
15. The chemistry of the actinoid elements is not so smooth as that of thelanthanoids. Justify this
statement by giving some examples from theoxidation state of these elements.
Ans. Lanthanoids exhibits +2 ,+3,+4 oxidation state out of these +3 is most common lanthanoids
element shows +3 oxidation state.
16. Which is the last element in the series of the actinoids? Write the electronic configuration of
this element. Comment on the possible oxidation state of this element.
Ans. Lr, Z=103, is the best actinoids. It electronic configuration is [Rn]865f146d17s2.The possible
oxidation state is +3.
UNIT - 9
CO-ORDINATION COMPOUNDS
Double Salt
It gives test of all the ions present in the
composition of the salt
Marks= 3
Coordination Compound
It does not gives test of all the ions present in the
composition of the salt
IMPORTANT TERMS
a) Coordination entity- A coordination entity is a compound having a central metal
ion/atom bonded with fixed number of ions or molecules and exist as a single entity.
It is written within [metal ion/atom and ligands]
b) Central atom/ion- In a coordination entity the metal ion/atom which is bonded to
a fixed number of ion/molecules (Ligands) is called central metal ion/atom
c) Ligands- The ion or molecules bonded with central ion/atom with coordinate bond
are called ligands. They are of different typesmonodentate or unidentate ligands- These type of ligand are bonded with metal
ion through one site(bond). Examples are H2O, NH3, CNi) Ambidentate ligands- They are unidentate ligands but can donate electron through two different
atoms but at a time only through one atom.
Exa- CN- can donate electron through C as well as Through N atom, NO2- can donate electron
through O as well as Through N atom,
ii) Didentate ligands- These type of ligand are bonded with metal ion through two site/ form two
coordinate bond. Examples are (COO- -)2, Ethylenediamine
iii) Polydentate ligands- These ligands can form more than two coordinate bonds with metal
ion/atom. Example- EDTA
iv) Chelating ligands- Didentate or polydentate ligands which form a closed ring structure around
metal ion are called chelating ligands. Example- oxalate ion. Ethylene diamine etc.
d) Coordination number- Coordination number of a metal ion is equal to the number of coordinate
bonds ligands forms with metal ion. Coordination number of a metal ion is
= the no. of momodentate ligands
= 2 × no. of didentate ligands
= 3 × no. of tridentate ligands
e) Coordination sphere- In a coordination compoud [ ] known as coordination sphere as species
within [ ] are bonded through coordinate bonds
f) Oxidation number of metal ion- It is equal to the charge on metal ion in the complex or
coordination compounds
g) Chelate compound- The coordination formed by chelating ligands with metal ion is called chelate
compound. Chelate compounds are more stable than normal coordination compounds.
Description of a Coordination Compound
q
+x -y
a
+z
Cs
[M (L )d]
Symbol
b
+/-B
-V
AP
Description of symbols
C
Simple cation (may be Na+, K+ etc) ( will give its test)
a
Ionisation sphere (species within this sphere supply ion)
s
Number of simple cation
+z
Charge on simple cation
q
Coordination sphere (Species within [ ] exist as single entity and do not supply ions)
M
Metal ion/atom (Any transition metal with a fixed oxidation state)
+x
Charge on metal ion (It is equal to the oxidation state of TM)
L
Ligand ( Any ionic or molecular species which can donate electron pair to the TM)
-y
Charge on ligand
d
No. of ligand
+ or -B
Charge on complex (It is sum of charge on metal ion and charges on ligands)
b
Ionisation sphere (species within this sphere will supply ion)
A
Simple anion (Will give its test)
p
No. of simple anion
-v
Charge on simple anion
Naming of coordination compounds1.
2.
3.
4.
Name of cation (simple or complex) is written first.
Name of anion (simple or complex) is written at last.
Name of coordination entity is written as single name.
Within coordination sphere ligands name (with suitable prefix and ending) is written first.
When more than one type of ligands are there that their name are written in the order of
alphabates
5. Name of metal ion is follow ligands name with their oxidation no. in Roman.
Examples-
ISOMERISMS IN COORDINATIONCOMPOUNDS
1. GEOMETRICAL ISOMERISM- compounds having same molecular formula, same structural
formula but different spatial arrangement of atoms/ions.
In cis form both the Cl atoms
are in the adjacent position
In trans isomer both the Cl
atoms are in opposite position
Geometrical isomers (cis and trans) of Pt [NH3)2Cl2]
In cis form both the Cl atoms
are in the adjacent position
In trans isomer both the Cl
atoms are in opposite position
In cis form both the Cl atoms
are in the adjacent position
In trans isomer both the Cl
atoms are in opposite position
In cis form both the NH3
atoms are in the adjacent
position
In trans isomer both the NH3
atoms are in opposite position
2. OPTICAL ISOMERISM- compounds having same molecular formula, same structural formula
but different spatial arrangement of atoms/ions and rotate plane of polarized light towards left or right
hand side. It is known as optical activity. Chirality is an essential requirement for showing optical
activity.
Both the structures are
non super imposable
mirror images of each
other
Both the structures are
non super imposable
mirror images of each
other
GEOMETRY OF COORDINATION COMPOUNDS-
BONDING IN COORDINATION COMPOUNDS- (VBT)
In the presence of
strong NH3 ligands
3d electrons pair up.
Low spin compound
is formed.
Magnetic moment
of such compound is
low
They are not able to
show d-d transition
so they are
colourless
In the presence of
weak ‘F’ ligands 3d
electrons do not pair
up.
High spin
compound is
formed.
Magnetic moment
of such compound is
high
They are able to
show d-d transition
so they are coloured
In the presence of
weak ‘Cl’ ligands 3d
electrons do not pair
up.
High spin
compound is
formed.
Magnetic moment
of such compound is
high
They are able to
show d-d transition
so they are coloured
In the presence of
strong CN ligands
3d electrons pair up.
Low spin compound
is formed.
Magnetic moment
of such compound is
low
They are not able to
show d-d transition
so they are
colourless
CRYSTAL FIELD THEORY & CRYSTAL FIELD SPLITING ENERGY-
According to crystal field theory bond formed between ligands and metal ion is ionic. Ligands are
treated as point –ve charged and metal ions are treated as point +ve charged. In order to form ionic bond
ligands approach towards metal ion in this process ligands show repulsive interaction with d orbitals of
transition metal. Five d orbitals of TM split into two sets. eg and tg. The energy difference between two
sets of orbitals is termed as ‘CRYSTAL FIELD SPLITING ENERGY’. Its value depends upon the field
strength of ligands. Stronger ligands like CN-, NH3 CO cause greater splitting and weaker ligands like
F, Cl, H2O cause poor splitting
When splitting is more
electrons can not jump
from lower orbitals to
higher energy orbitals so
they pair up in lower set
of orbitals and show poor
magnetic moment and
colourless. This happens
in the presence of ligands
like CN-, NH3 CO. In
case of weal ligands like
F, Cl, H2O opposite
scene is observed
When splitting is more
electrons can not jump
from lower orbitals to
higher energy orbitals so
they pair up in lower set
of orbitals and show poor
magnetic moment and
colourless. This happens
in the presence of ligands
like CN-, NH3 CO. In
case of weal ligands like
F, Cl, H2O opposite
scene is observed
BONDING IN METAL CARBONYLS
Metal carbonyls are special class of coordination compounds. In metal carbonyls metal is found in
zero oxidation state. There is a special kind of bonding i.e. pπ-dπ bond by back donation of electrons
from metal atom to CO group
In metal carbonyl electron pair
donated from filled orbital of CO to
empty orbitals of transition metal. In
addition to this electron pair is
donated from filled ‘d’ orbital of
transition metal to empty
antibonding orbitals of CO
APPLICATIONS OF COORDINATION COMPOUNDS-
Some questions & Answers for practice :
1. Write the formula for the following: -
(3)
(i) Tetrahydroxozincate (II)ion
(ii) Hexaammineplatinum (IV) ion
(iii) Hexaamminecobalt (III) sulphate
Ans: (i) [Zn(OH)4]2(ii) [Pt(NH3)6]4+
(iii)[Co(NH3)6]2(SO4)3
2. Write the IUPAC name of the following: (i)[Pt(NH3)2Cl(NH2CH3)]Cl
(2)
(ii) [Co(NH3)4Cl(NO2)]Cl
Ans (i) Diamminechloridomethyleamineplatinum (ii) chloride
(ii) Tetraaminechloridonitrito-N-cobalt (iii) chloride
3. What is meant by ambidentate ligands? Give two examples.
Ans : Ligand which can ligate through two different atoms
e.g. CN-, SCN-
(1)
4. Explain on the basis of VBT, the experimental findings that [Ni(CN)4]2- ion with a square-planar structure
is diamagnetic and the [NiCl4]2- ion with tetrahedral geometry is paramagnetic.
(2)
Ans: In [Ni(CN)4]2Ni2+ electronic configuration- 3d8
Hybridization - dSP2
Unpaired electron = 0 , therefore it is diamagnetic
In [NiCl4]2Ni2+ Electronic configuration 3d8
Hybridization SP3
Unpaired electron = 2, therefore it is paramagnetic
5. Aqueous copper sulphate solution (blue in colour) gives( a) a green precipitate with aqueous potassium
fluoride, and(b) a bright green solution with aqueous potassium chloride. Explain. (2)
Ans:
(a)
[Cu(H2O)4]2+ + 4F- → [CuF4]2+ 4H2O
Blue
Green ppt
(b)
[Cu(H2O)4]2+ + 4Cl- → [CuCl4]2+ 4H2O
Blue
Bright green solution
6. Give a chemical test to distinguish between [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br. What kind of
isomerism do they exhibit?
(2)
Ans : (i)
[Co(NH3)5Br]SO4 + Ba2+ →
BaSO4
White ppt
[Co(NH3)5Br]SO4 + Ag+
(ii)
→
No ppt
[Co(NH3)5SO4]Br. + Ba2+ →
No ppt
[Co(NH3)5SO4]Br. + Ag+
→
AgBr
Yellow ppt
7. Name the metal present in (i) Chlorophyll (ii) Haemoglobin (iii) Vitamin B12 (iv) Cis-platin
Ans: (i) Mg
(ii) Fe
(iii) Co (iv) Pt
8. What is the coordination number of central metal ion in (i) [Fe(C2O4)3]3- (ii) [Co(en)2Cl2]+.
Ans: (i) 6
(ii) 6
9. Why are cyclic complexes more stable than open one?
Ans : Cyclic complexes have more stability because of reduced strain in five or six member
rings. e.g. Chlorophyll (cyclic) is more stable than [Co(NH3)5Br]SO4
10. CuSO4 on mixing with NH3 (1:4) does not give test for Cu2+ ions but gives test for SO42ions. Why?
(2)
Ans: It is because when NH3 coordinates to Cu2 ions it forms the complex
[Cu(NH3)4]SO4. Copper ions are present in coordination sphere, therefore, they are
non- ionisable whereas SO42- ions are counter ions which are ionisable.
(2)
(1)
(1)
UNIT-10
HALOALKANES
AND
HALOARENES
Marks = 4
Q :1 Identify A, B in the following:2
─Br
Ans:
A=
Dry ether
+ Mg ─────────→ A
─MgBr
H2 O
───────→ B
B=
Q : 2 Give the IUPAC names of the following:(a) Cl-CH2 C ≡ C-CH2-Br
Ans : A) 1-Bromo,4-Chloro But-2- yne
B) 3-Chloro-4-Methyl-hexane
2
(b) CH3CH2CH(CH3)CH(C2H5)Cl
Q: 3 Explain why Aryl halides are extremely less reactive towards nucleophlic substitution reactions.
Ans : Due to resonance in aryl halide in C-X bond will acquire double bond character so difficult to
substitute halogen.
Q: 4 Haloalkanes are only very slightly soluble in water explain.
1
Ans: Unable to form Hydrogen bond with water.
Q : 5 Write short notes on:- (a) Wurtz reaction (b) Swarts reaction
2
Dry Ether
Ans :A) R-X + 2Na +R-X ------------
R-R + 2Na X
B) R-Br + AgF ----------- RF + AgBr
Q : 6 Arrange the following in increasing order of the property indicated
Bromomethane, Bromoform, Chloromethane, Dibromomethane (Boiling point)
Ans : Chloromethane < Bromomethane < Dibromomethane < Bromoform .
1
Q : 7 What do you mean by asymmetric carbon? Give one example.
1
Ans: If all the four atom or group of atoms attach to a carbon are different then carbon is
called asymmetric carbon. e.g. CH3CHBrCl.
Q : 8 Explain why H2SO4 is not used during the reaction of alcohol with KI.
1
Ans: Because HI produced get oxidize to Iodine in presence of Sulphuric acid.
Q : 9 Explain why Reaction of CH3Br with KCN yields CH3CN while with AgCN yields CH3NC? 1
Ans: Because CN- is an ambidentate nucleophile.
Q : 10 Write short notes on:- (a) Sandmeyer’s reaction (b) Finkelstein reaction
2
CuCl /HCl
Ans : (a) C6H5N2Cl --------------- C6H5Cl
Acetone
(b) C6H5 Br + NaI ---------------- C6H5 I + Na Br
Q11. Convert : -propane-1-ol to 2-iodopropane.
1
Conc. H2SO4
Morkownikove’s Rule
Ans : CH3 CH2 CH2-OH ------------ CH3 CH=CH2 + HI
----------
CH3 CH I CH3
443K
Q12. Arrange the following in increasing order of the property indicated
(a) 1-Bromobutane, 2-Bromobutane, 2-Bromo-2-methylpropane (boiling point)
2
Ans : 2-Bromo-2-methylpropane <2-Bromobutane < 1-Bromobutane,
(b)
─CH2Cl and
─Cl
(reactivity towards SN2 reaction)
Ans :
─CH2Cl
UNIT 10
ALCOHOL, PHENOL & ETHERS
Marks=4
NOMENCLATURE
PREVIOUS YEAR QUESTION ON NOMENCLATURES
1. Draw the structure of Hex-1-en-3ol compounds (Delhi 2012)?
2. Draw the structural formula of 2-Methylpropan-2-ol (delhi2012)
3. Write the name of the following compound:-
(HEX1-EN-3-OL)
2, 5-DIMETHYL HEXANE-1, 3-DI-OL
2,5DIMETHYL PHENOL
2-METHYL-1-METHOXYBUTANE
NAME REACTION
1. KOLBE REACTION- (PHENOL TO SALICYLIC ACID)
2. REIMER –TIMEN REACTION- (PHENOL TO SALICYLALDEHYDE)
3. WILLIAMSON SYNTHESIS- (ALKYL HALIDE TO ETHER)
ALKYL HALIDE SODIUM ALKOXIDE
ETHER
MECHANISM
1. HYDRATION OF ETHENE
H+
H2C=CH2 + H2O
CH3CH2OH
2. DEHYDRATION OF ETHANOLH+
CH3CH2OH
CH2=CH2
3. DEHYDTATION OF ETHANOL ( IN EXCESS OF ALCOHOL)
CH3CH2OH
H+
CH3CH2-OCH2CH3
PREVIOUS YEAR QUESTIONS
1.
Explain the mechanism of acid catalyzed hydration of alkenes to form corresponding
alcohol.
( All India 2012 )
2. Write the mechanism of hydration of ethane to ethanol. (Foreign 2009, 2010,2011 C )
REASON BASED QUESTIONS
Reason based question based on the following facts
ACIDIC CHARACTER
Phenol is more acidic than aliphatic alcohol –
Due to resonance peroxide ion is more stable than phenol
Due to the higher electronegativity of SP2 hybridised carbon of phenol to which –OH is
attached, electron density decreases on oxygen &increases the polarity of O-H bond.
Effect on acidic character of phenol due to presence of EWG and ERG group
The acidic strength of alcohol depends on pKa value: the higher the value of pKa lower is acidic
strength. The pKa is inversely proportional to Ka.
PREVIOUS YEAR QUESTIONS
1.Explain the following behaviorsa. Alcohols are more soluble than water than hydrocarbons of comparable molecular mass
(HINT:- due to hydrogen bonding)?
b.Ortho nitro phenol is more acidic than Orthomethoxy phenol?[HINT:-higher the stability
of phenoxide ion more will be the acidic character,NO2 group (EWG) increases the stability
of phenoxide ion where as methoxy group(ERG) destabilizes the phenoxide ion](ALL INDIA
2012 ).
2.Give reason for the following:A. The boiling point ethanol is higher than that of methanol?
B. Phenol is stronger acid than alcohol
(2009,2011 C)
CHEMICAL TEST FOR DISTINGUISH PAIRS OF COMPOUNDS
1) LUCAS REAGENT TEST-The pri.,sec.,tert. Alcohols can be distinguished by LUCAS
reagent
test (ZnCl2 + conc.HCl)
a) Organic compound +LUCAS reagent --1.If turbidity appears immediately the given
organic
compound is tertiary alcohol (t-alcohol or 3° Alcohol)
b) If turbidity appears after 5 min. the given organic compound is secondary alcohol (sec.
alcohol or 2° Alcohol)
c) If turbidity not appears at room temp.-primary alcohol.
2) FERRIC CHLORIDE TEST (only for phenol)Phenol + neutral Ferric chloride ------- violet color appears.
IMPORTANT REACTION WITH REAGENTS
(BASED ON PREPARATION & PROPERTIES)
Preparation of alcohol
1.
FROM GRIGNARD REAGENT -
2. FROM CARBONYL COMPOUNDS:1. By reduction of aldehyde and ketones -
\
2.By reduction of carboxylic acid and ester :-
Preparation of phenol1.
From chloro benzene
2. From benzene sulphonic acid:-
3. From benzene diazonium chloride:-
4. From cumene
PROPERTIES OF ALCOHOL
1. ESTERIFICATION- alcohols and phenol react with carboxylic acid , acid chloride and acid
anhydride to form esters.
ESTERIFICATION METHOD USED FOR THE CONVERSION OF PHENOL TO ASPIRIN
1.Phenol convt. to salicylic acid .By KOLBE REACTION
2. salicylic acid convt. to aspirin by esterification
2. OXIDATION – Oxidation of alcohol involves formation of carbon oxygen double bond (carbonyl
grp.)
And a mixture of carboxylic acids containing lesser number of carbon atoms is formed.
DEHYDROGENATION
When the vapours of a pri. Or sec. alcohol are passed over a heated copper at 573K, dehydrogenation
takes place and an aldehyde or a ketone is formed while tert. Alcohols undergo dehydration.
REACTION OF PHENOL
Q : 1 Write the names of reagents and equations for the preparation of 2-Methoxy-2-methylpropane. 1
Ans: CH3Br + (CH3)3CO- Na+-----------------------------------> CH3-O-C(CH3)3 + NaBr
Q : 2 Give reasons:
2
a. Ethanol has higher boiling point in comparison to methoxymethane .
Ans : Due to presence of intermolecular hydrogenbond in ethanol.
b. Phenols are more acidic than alcohols.
Ans : Due to –R effect of phenoxide ion produced after the loss of proton from phenol, phenoxide
ion is more stable than alkoxide ion
Q : 3 (a) o-nitrophenol is steam volatile while p-nitrophenol not.
Ans : Due to presence of intramolecular hydrogen bond
(b)Cleavage of phenyl alkyl ether with HI always gives phenol and alkyl iodide
Ans : Due to resonance C-O bond of phenyl will acquire double bond character hence
difficult to break
Q : 4 How are the following conversions carried out: a. Methyl magnesium bromide to 2-methylpropane-2-ol
(2)
CH3COCH3
H+/H2O
Ans : CH3MgBr ---------------------- (CH3)3C-OMgBr -----------------(CH3)3C-OH
Dry Ether
b. Propene to propan-2-ol
HBr,MarkwnkvRule
aq.KOH, ∆
Ans : CH3CH=CH2------------------- CH3CHBrCH3------------------ CH3CHOHCH3
c. Propan-2-one to 2-methyl-2-propanol
H+/H2O
CH3MgBr
Ans : CH3COCH3 --------------------- (CH3)3C-OMgBr -----------------(CH3)3C-OH
Dry Ether
Q : 5 Write IUPAC name-
(2)
i. CH3-O-CH2-CH-CH3
|
CH3
ii.
CH3-CH-CH-CH2-CH2OH
| |
Cl CH3
Ans: i) 2-methoxybutane
ii) 4-Chloro,3-methyl butan-1-ol.
Q : 6 Arrange the following as property indicated: -
(2)
(i) pentan-1-ol, pentanal, ethoxyethane (increasing order of boiling point)
Ans : n-butane <ethoxyethan < pentanal < pentan-1-ol
(ii) pentan-1-ol, phenol, 4-methylphenol, 3-nitrophenol ( increasing order of acid strength)
Ans : pentan-1-ol < 4-methylphenol < phenol < 3-nitrophenol.
Q : 7 Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Phenol and Benzoic acid. (ii) Propan-1-ol and Propan-2-ol.
(2)
Ans:
(i) C6H5OH + NaHCO3 -------------------- No effervescence
C6H5COOH + NaHCO3 ----------------- C6H5COONa + H2O + CO2 (effervescence )
Lucas Reasgent
(ii) CH3-CH2 –CH2OH ------------------- No Turbidity
CH3-CHOH–CH3 ------------------ Turbidity produced with in five minutes
Q:8 Write the mechanism of dehydration of ethanol.
Ans :
Conc.H2SO4
CH3CH2OH-------------------- CH2=CH2 + H2O
2
Mechanism :
STEP1 : Formation of a protonated alcohol
CH3CH2OH + H+-------------- CH3CH2-O+ H
H
STEP: 2 Formation of a carbocation
CH3CH2-O+ H -------------- CH3CH2+ + H2O
H
STEP : 3 Loss of proton :
CH3CH2+ --------------------- CH2=CH2 + H+
Q : 9 A compound ‘A’ with molecular formula C4H10O is a unreactive towards sodium metal . It does not add
Bromine water and does not react with NaHSO3 solution .On refluxing ‘A’ with excess of HI gives ‘B’ which
react with aq. NaOH to form ‘C’. ‘C’can be converted into ‘B’ by reacting with P and I 2 . ‘C’ on heating with
aqueous alkali to form ‘E’ which form ‘F’ on heating with conc. H2SO4. ‘F’ decolourises bromine water .
Identify A to F and write the reactions involved.
Ans : ‘A’is not alcohol therefore it does not react with Sodium metal . ‘A’ is not aldehyde and ketone as it does
not react with NaHSO3 ‘A’ is not unsaturated hydrocarbon as it does not add Br2 (aq) . It is likely to be ether.
Ans : CH3CH2OC2H5 + 2HI-------------------------2C2H5I + H2O
‘A’
excess
‘B’
(C4H10O)
2C2H5I + NaOH (aq) --------------------------------C2H5OH + Na I
‘B’
‘C’
P / I2
C2H5OH
‘C’
------------------------------------- C2H5I
‘B’
Cu
CH3CH2OH --------------------------- CH3CHO
‘C’
573 K
‘D’
-
OH
CH3CHO -----------------------------
‘D’
CH3CHOHCH2CHO
(3-Hydroxybutanal)
3.
UNIT-12
ALDEHYDES, KETONES AND CARBOXYLIC ACIDS
Marks = 6
1. Arrange the following compounds in increasing order of their property as indicated:
(i) HCOOH, CH3COOH, CH2(F)COOH, CH2(Cl)COOH (acid strength)
(ii) Acetaldehyde, Acetone, Formaldehyde, Ethylmethyl ketone (reactivity towards HCN)
Ans: (i) CH3COOH< HCOOH< CH2(Cl)COOH < CH2(F)COOH
(2)
2. Give reasons
(2)
Carboxylic acids are stronger acids than phenols.
Ans : Carboxylic acids are stronger acids than phenols because the carboxylate ion is stabilised more by
the resonace.
3. Give simple chemical test to distinguish between
(2)
(i)
Propanal and propanone
Ans: CH3 CH2CHO + Ag2O ────→ CH3 CH2COOH + Ag (silver mirror)
Propanal
CH3 COCH3 + Ag2O ────→ no reaction
Propanone
(ii)
Phenol and benzoic acid
Ans:
i. Phenol does not give sodium carbonate test whereas benzoic acid give this test.
ii. Phenol give azo dye test , benzoic acid does not.
4. Identify A, B and C
KCN
SnCl2-HCl
Zn-Hg/ HCl
CH3CH2Br ──────→ A ──────→ B ───────→ C
Ans: A= CH3CH2CN ,
B= CH3CH2CHO, C= CH3CH2CH3
5. .Name the following as IUPAC system
(i)
CH3 CH( CH3 )CH2 CH2CHO
Ans:
4- methylpentanal
(ii)
(CH3)3C CH2 COOH
3-methylbutanoic acid
6.
(2)
(2)
Give reasons for the following:
(2)
(a). Ethanal is more reactive towards nucleophilic addition reactions than propanone.
Ans The formula of Ethanal is (CH3CHO) and that of Propanone is CH3 COCH3
It is very clear that, in propanone the presence of two alkyl groups hindere the approach
of nucleophile to carbonyl carbon .
(b).Ethanoic acid is a stronger acid than ethanol.
Ans Due to presence of resonance in ethanoate ion.
7. An
organic compound A (molecular formula C8H16O2) was hydrolysed with dilute sulphuric (5)
acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid
produced (B). (C) on dehydration gives but-1-ene. Identify A, B and C. also write equations for
the reactions involved.
Ans:
A= CH3-CH2-CH2-COO-CH2-CH2-CH2-CH3
B= CH3-CH2-CH2COOH
C= CH3-CH2-CH2CH2OH
Equation
H SO
CH3-CH2-CH2-COO-CH2-CH2-CH2-CH3 ─────→ CH3 CH2 CH2COOH (B)
+ CH3 CH2 CH2CH2OH (C)
2
4
Dehydration
CH3 CH2 CH2CH2OH
─────→
CH3 CH2 CH=CH2
Chromic acid
CH3 CH2 CH2CH2OH
─────→
CH3 CH2 CH2COOH
8. Complete the following reactions by identifying A, B and C.
(2)
Pd/BaSO4
A + H2(g) ────────→ (CH3)2CH-CHO
(i)
Pd/BaSO4
Ans: (CH3)2CH-CO-Cl + H2 (g) ────────→ (CH3)2CH-CHO
(ii)
CH3 CH2 CH=CH2 + B ────────→ CH3 CH2 CHO
+CH2-O
ZnO/H2O
Ans:
CH3 CH2 CH=CH2 + O3───────→ CH3 CH2 CHO
+CH2-O
9. Suggest a reason for the large difference in the boiling points of butanol and butanal, although they
have the same solubility in water.
(2)
Ans: The b.p. of butanol is higher than butanal because butanol has strong intermolecular hydrogen
bonding while butanal has weak dipole-dipole attraction. However, both of them form Hbonding with water and hance are soluble.
10. Name two methods which are used to convert >C=O group into >CH2 group.
Ans: Clemmensen reduction and Wolff-Kishner reduction
(2)
UNIT – 13
NITROGEN CONTAINING ORGANIC COMPOUNDS
Nomeclature of these Compound :
Marks = 4
5.COUPLING REACTION (the reaction between benzenediazoniamchloride with phenol or
aniline to form dyes (coloured) compound)
N2Cl + HBDC
-OH
PHENOL
→
-N=N
OH
dyes (coloured) compound
PREVIOUS YEARS QUESTION
1.
Describe the following giving the relevant chemical equation in each case.
[All India 2012; Delhi 2012]
(i) Carbylamine reaction
(ii) Hofmann’s bromamide reaction
Hints: See the name reaction.
2.
Illustrate the following with an example of reaction in each case.
(i) Sandmeyer’s reaction
(ii) Coupling reaction
Hints: (i) application of BDC (ii) See the name reaction.
[Delhi 2011C]
3.
Giving an example for each describe the following reactions. [Delhi 2009, 2008C; Foreign 2008]
(i) Hofmann’s bromamide reaction
(ii) Coupling reaction
Hints: See the name reaction.
REASON BASED QUESTIONS
Reason based question based on the following facts:
(i) Gabriel phthalimide synthesis is not suitable for preparation of aromatic primary amine because aryl
halides do not under go nucliophilic substitution with the anion formed by phthalimide.
(ii) Basic character of amines:
(a) Ammonia is lewis base due to presence of lone pair of electron. Amines are derivatives of ammonia.
(b) Basic character depends on pKa value. Larger the value of pKa or smaller the value of Ka weaker is
base and vice-versa.
(c) Basic character of Primary secondary amine-Condition No. 1: 3oamine > 2o amine > 1o amine (In vapour phase).
This is due to +I effect of alkyl group
Condition No. 2: In aqueous solution = 1oamine > 2o amine > 3o amine
This is due to solvation effect. The greater the size of ion lesser will be solvation
and the less stabilized is the ion.
Condition No. 3: In general the basic characters of amines in aqueous medium i.e. inductive effect,
solvation effect and steric hindrance the sequence are as follows-
Condition No. 4: In case aryl amine (aniline). Aniline are less basic than ammonia because the –
NH2 group is attached directly to the benzene ring, hence the unshared electron
pair on nitrogen is less available for protonation due to the resonance.
Condition No. 5: Effect of the substituent on basic strength of aniline.
(a) The presence of ERG group increases the basic strength of aniline whereas
(b) The presence of EWG group decreases the basic strength of aniline
(iii) Nitration of aniline gives ortho para and meta derivatives-- In strong acidic medium aniline is
protonated to form anilinium ion which is meta directing, hence besides ortho and para deravatives
significant amount of metaderivative also formed.
(iv) Aniline does not undergo friedel craft reaction-- Due to salt formation with aluminium chloride
( lewis acid) hence nitrogen of aniline acquire positive charge act as a strong deactivating group for
further reaction.
PREVIOUS YEARS QUESTION
1.
2.
Assign reason for:
[All India 2009C]
(i) Amines are less acidic than alcohols of comparable molecular masses.
(ii) Aliphatic amines are stronger bases than aromatic amines.
Write chemical equation for the following conversion:
[Delhi 2012]
(i) Nitrobenzene to benzoic acid
(ii) Benzyl chloride to 2-phenylethanamine
(iii) Aniline to benzyl alcohol
CHEMICAL TEST FOR DISTINGUISH PAIRS OF COMPOUNDS
1.
Carbyl amine reaction- To distinguish Pri. , Sec., Aliphatic or Aromatic amines with others amines
2.
HINSBERG’S REAGENT TEST (BENZENE SULPHONYL CHLORIDE): Hinsberg’s
reagent test distinguished primary, secondary and tertiary amine.
Primary amine
Secondary amine
Tertiary amine
Reaction with Hinsberg’s
Reaction with Hinsberg’s
Reaction with Hinsberg’s
reagent
reagent
reagent
Product soluble in alkali
Product insoluble in alkali
No reaction
2. Aniline can be distinguished from other primary amines by coupling reaction.
IMPORTANT REACTION WITH REAGENTS
(BASED ON PREPARATION & PROPERTIES)
AMMONOLYSIS
PREVIOUS YEARS QUESTION
1.
2.
3.
Convert:
(i) Nitrobenzene to phenol
(ii) Aniline to chlorobenzene
How will you convert?
(i) Aniline to benzonitrile?
(ii) ethanamine to ethanoic acid?
How will you convert?
(i) Nitrobenzene to aniline?
(ii) Aniline to iodobenzene?
[Delhi 2011C]
[Delhi 2011C]
[Delhi 2011]
UNIT-14
BIOMOLECULES
Marks =4
KEY POINTS
EXPLANATIONS
Monosaccharides Cannot be hydrolyzed further .eg- glucose, fructose, ribose
Disaccharides
Sucrose (α-D- glucose + β-D-fructose) , Maltose(α-D- glucose + α-D- glucose)
Lactose(β-D-galactose + β-D-glucose )
Polysaccharides
Starch (two components—Amylose and Amylopectin) polymer of α-D- glucose
Amylose
Water soluble, 15-20% of starch.,
unbranched chain ,
C1– C4
glycosidic linkage.
Amylopectin
Water insoluble , 80-85% of starch., branched chain polymer, C1–C4 & C1–
C6 glycosidic linkage
Cellulose
Straight chain polysaccharide of β -D-glucose units/ joined by C1-C4glycosidic
linkage (β-link), not digestible by human / constituent of cell wall of plant cells
Glycogen
Highly branched polymer of α-D- glucose .found in liver, muscles and brain.
reducing sugars
Aldehydic/ ketonic groups free so reduce Fehling’s/ Tollens solution and. Egmaltose and lactose
Non reducing
Aldehydic/ ketonic groups are bonded so cannot reduce Fehling’s solution and
sugars
Tollens’ reagent. Eg- Sucrose
Anomers.
The two cyclic hemiacetal forms of glucose differ only in the configuration of the
hydroxyl group at C-1, called anomeric carbon Such isomers, i.e., α –form and β form, are called anomers.
Invert sugar
Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and
laevorotatory fructose. Since the laevorotation of fructose (–92.4°) is more than
dextrorotation of glucose (+ 52.5°), the mixture is laevorotatory. Thus, hydrolysis
of sucrose brings about a change in the sign of rotation, from dextro (+) to laevo (–
) and the product is named as invert sugar
Glycosidic linkage Linkage between two mono saccharide
Importance of
-Major portion of our food. / used as storage molecules as starch in plants and
Carbohydrates
glycogen in animals/.
-Cell wall of bacteria and plants is made up of cellulose./wood and cloth are
cellulose /
-Provide raw materials for many important industries like textiles, paper, lacquers
and breweries.
essential amino
-Which cannot be synthesised in the body and must be obtained through diet, egacids
Valine, Leucine
Nonessential
-Which can be synthesised in the body,
eg - Glycine, Alanine
amino acids
zwitter ion.
In aqueous solution, amino acids exist as a dipolar ion known as zwitter ion.
peptide linkage
10- structure of
proteins:
20- structure of
proteins:
Peptide linkage is an amide formed between –COOH group and –NH2 group of
two successive amino acids in peptide chain.
sequence of amino acids that is said to be the primary structure of protein
Secondary structure of protein refers to the shape in which a long polypeptide
chain can exist.
They are found to exist in two types of structures viz. α -helix and β -pleated sheet
structure.
3◦ structure of
Further folding of the secondary structure. It gives rise to two major molecular
proteins:
shapes viz. fibrous and globular.
Fibrous proteins
Polypeptide chains run parallel, held together by hydrogen and disulphide bonds,
fibre– like structure. Water insoluble .Eg- are keratin (in hair, wool, silk) and
myosin (present in muscles).
Globular proteins Chains of polypeptides coil around to give a spherical shape. Water soluble. EgInsulin and albumins
Forces which
Hydrogen bonds, disulphide linkages, van der Waals and electrostatic forces of
stabilise 2°& 3°
attraction.
Denaturation of
When a protein is subjected to physical change like change in temperature or
Proteins
chemical change like change in pH, the hydrogen bonds are disturbed. Due to this,
globules unfold and helix gets uncoiled and protein loses its biological activity.
This is called denaturation of protein.
(During denaturation 2° and 3° structures are destroyed but 1º structure remains
intact.)
eg- The coagulation of egg white on boiling, curdling of milk
Fat soluble
These are vitamins A, D, E and K. They are stored in liver and adipose (fat storing)
vitamin
tissues
Water soluble
B & C . these vitamins must be supplied regularly in diet because they are readily
vitamin
excreted in urine
Vitamins –
Vit- A (Fish liver oil, carrots)- Night blindness
/
Vitamin B1
sources(Yeast, milk,)- Beriberi
Deficiency
Vit-B2 (Milk, eggwhite)- Cheilosis / Vit- B6 (Yeast, milk,)- Convulsions / Vitdiseases
B12 (Meat, fish,)- anaemia
Vit C(Citrus fruits)- Scurvy,
/
Vit D(Exposure to sunlight, fish and egg
yolk)- Rickets, osteomalacia
Vit E(wheat oil, sunflower oil)- fragility of RBCs
/ Vit K(leafy vegetables)Increased blood clotting time
DNA
pentose sugar (D-2-deoxyribose) + phosphoric acid + nitrogenious bases
( A , G , C, T )
RNA
pentose sugar (ribose) +
phosphoric acid + nitrogenious bases
(A , G , C, U )
Nucleoside /
Nucleoside  sugar + base
Nucleotides  sugar + base + phosphate
tides
Phosphodiester
Linkage between two nucleotides in polynucleotides
link
Functions of
DNA reserve genetic information, maintain the identity of different species e is
Nucleic Acids
capable of self duplication during cell division, synthesizes protein in the cell.
A FEW QUESTIONS for Practice
1 Mark Questions
(Q.) Define the term bio molecules?
(Ans) Bio molecules may be defined as the complex lifeless chemical substances which form the basis
of life, i.e., they not only build up living systems (creatures) but are also responsible for their growth,
maintenance and their ability to reproduce.
(Q.) Define the term photosynthesis? Give its general chemical equation?
(Ans) Photosynthesis may be defined as a chemical process through which plants make their own
food by the reaction of carbon dioxide and water in the presence of sunlight with the help of plant
chlorophyll.
x CO2 + y H2O ----->Cx(H2O)y + x O2
(Q.) Define Monosaccharides.
(Ans) These are the simplest carbohydrates which cannot be hydrolysed to smaller molecules. Their
general formula is (CH2O)n, where n = 3-7.
(Q.) Define the term Oligosaccharides?
(Ans) Those carbohydrates which give 2 to 10 molecules of monosaccharides in hydrolysis.
(Q.) Define Disaccharides.
(Ans) Carbohydrates which on hydrolysis give two molecules of the same or different
monosaccharides are called disaccharides. e.g.,
C12H22O11 + H2O
C6H12O6 + C6H12O6
sucrose
glucose Fructoes
(Q.) What is difference between Reducing and non-reducing sugars or carbohydrates?
(Ans) All those carbohydrates which contain aldehydic and ketonic group in the hemiacetal or
hemiketal form and reduce Tollen’s reagent or Fehling’s solution are called reducing carbohydrates
while others which do not reduce these reagents are called non-reducing reagents.
(Q.) Explain the term mutarotation?
(Ans) Mutarotation is the change in the specific rotation of an optically active compound with time, to
an equilibrium mixture.
(Q.) Define glycosidic linkage?
(Ans) The two monosaccharide units are joined together through an ethereal or oxide linkage formed
by the loss of a molecule of H2O. Such a linkage between two monosaccharide units through oxygen
atoms is called glycosidic linkage.
(Q.) Give a chemical equation for obtaining Maltose?
(Ans) Maltose is obtained by partial hydrolysis of starch by the enzyme diastase present in malt i.e.,
sprouted barley seeds.
2(C6H10O5)n + n H2O
n C6H12O6
(Q.) What are the main sources of vitamins?
(Ans) The main sources of vitamins are milk, butter, cheese, fruits, green vegetables, meat, fish, eggs,
etc.
(Q.) Give two methods for the preparation of glucose?
(Ans) The methods for the preparation of glucose are:
(i) From sucrose (Cane Sugar).
When sucrose is hydrolysed by boiling with dil. HCl or H2SO4 in alcoholic solution, an equimolar
mixture of glucose or fructose is obtained.
C12H22O11 + H2O
C6H12O6 + C6H12O6
(ii) From starch.
Commercially glucose is obtained by hydrolysis of starch by boiling it with dil. H2SO4 at 393 K under
pressure.
(C6H10O5)n + n H2O
n C6H12O6
(Q.) Define Carbohydrates? Give their basic classification depending upon their behaviour
towards hydrolysis.
(Ans) Carbohydrates are defined as optically active polyhydroxy aldehydes or polyhydroxy ketone
substances which give these on hydrolysis.
These are broadly classified as:
(i) Monosaccharides.
(ii) Oligosaccharides.
(iii) Polysaccharides.
(Q.) What is Milk sugar? Give its characteristics.
(Ans) Lactose occurs in milk so, it is called milk sugar. Lactose on hydrolysis with dilute acids yields
an equimolar mixture of D-glucose and D-galactose. It is a reducing sugar since it forms an osazone. It
undergoes mutarotation and also reduces Tollen’s or Fehling’s solution.
(Q.) Define the term vitamins? State its importance.
(Ans) Vitamins may be defined as group of bio-molecules (other than fats, carbohydrates and
proteins) which are required in small amounts for normal metabolic processes and for the life, growth
and health of human beings and animal organisms.Vitamins neither supply energy nor help in building
tissues of the cells. They play an important role in keeping good health of human beings and animals.
Their deficiency causes serious disturbances and diseases in the body.
(Q.) What do you understand by denaturation of proteins?
(Ans) When a protein in its native form, is subjected to physical change like in temperature or
chemical change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and
helix get uncoiled and protein loses its biological activity. This is called denaturation of protein.
(Q.) Give the D and L configurations of Glyceraldehyde?
(Ans)
(Q.) Give the chemical structure of sucrose & explain why sucrose is non reducing sugar.
(Ans)
The two monosaccharide are held together by a glycosidic linkage between C1 of -glucose and C2
of -fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond
formation, sucrose is a non-reducing sugar.
(Q.) Give a broad classification of vitamins?
(Ans) Vitamins are complex organic molecules.They can be broadly classified as:
(i) Water soluble vitamins: These include vitamin B-complex and vitamin C.
(ii) Fat soluble vitamins: These are oily substances that are not readily soluble in water. However,
they are soluble in fat. These include vitamins A,D,E and K. Nucleic acids are bipolar (i.e. polymers
present in the living system). They are also called polynucleotides since the repeating structural unit of
nucleic acids is a nucleotide.
General structure of a Nucleotide can be given as:
(Q.) Write a short note on cellulose and give its chemical structure.
(Ans)
Cellulose occurs exclusively in plants and it is the most abundant organic substance in plant
kingdom. It is a predominant constituent of cell wall of plant cells. Cellulose is a straight chain
polysaccharide composed only of -D-glucose units which are joined by glycosidic linkage between
C1 of one glucose unit and C4 of t he next glucose unit.
(Q.) Give a short note on Zwitter ion?
(Ans) Amino acids are usually colourless, crystalline solids. These are water soluble , high melting
solids and behave like salts rather than simple amines or carboxylic acids. This behaviour is due to the
presence of both acidic (carboxylic group) and basic (amino group) groups in the same molecule. In
aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving
rise to a dipolar ion known as zwitter ion.
(Q.) How are peptides formed? Show the formation of peptide bond with diagram.
(Ans) Peptides are amides formed by the condensation of amino group of one -amino acid with the
carboxyl group of another molecule of the same or different -amino acid with the elimination of
awater molecule. They are classified as di-, tri-, tetra-, etc.
eg.
UNIT 15
POLYMERS
Marks= 3
(D) Based on
Molecular
Forces
(C) Based On Mode of
Polymerization:
(B) Based on
Structure
(A) Based on
Source
CLASSIFICATION OF POLYMERS(i) Natural
Found in plants and animals, e.g. Proteins, cellulose, natural
rubber, silk, wool.
(ii) Synthetic
Man-made e.g. Nylon, polyester, neoprene, Bakelite, Teflon, PVC,
polystyrene
(i) Linear Polymer
This consist of long and straight chain repeating units e.g.
Polythene (HDPE), PVC, nylon, polyester.
(ii) Branched
Polymers
(iii) Cross Linked
Polymers
This contain linear chains having some branches e.g. amylopectin,
glycogen etc.
Strong covalent bonds are present between various linear polymer
chains. E.g. Bakelite, urea-formaldehyde polymer, melamine,
polymer etc.
(i) Addition
Polymers
These are formed by the repeated addition of monomer molecules
possessing multiple bonds, e.g., polythene, polypropene,
polystyrene, PMMA (polymethylmethacrylate)
(ii) Condensation
Polymers
These are formed by the repeated condensation reaction of
different bifunctional or trifunctional monomers, with the
elimination of small molecules like water, HCL, NH3, alcohol etc.
eg. Bakelite, nylon, polyester, urea-formaldehyde resin.
(i) Elastomers
Forces of interaction between polymer chains are weakest, e.g.
natural rubber, neoprene, and vulcanized rubber.
(ii) Fibers
Strong hydrogen bonds are present between the polymer chains.
They have high tensile strength e.g., nylon, polyester, silk, wool,
orlon, rayon etc.
(E) Based On
Growth Of
Polymerization
(iii) Thermoplastics
They are linear / slightly branched chains molecules capable of
repeated softening on heating and hardening on cooling e.g. ,
polythene, PVC, polystrene, polypropene.
(i) Addition
Polymers or Chain
Growth Polymers
They follow mostly free radical mechanism.
(ii) Condensation
Polymers or Step
Growth Polymers
Because they are formed in gradual steps.
PREVIOUS YEARS QUESTIONS
1.
2.
3.
Define the terms:
(i) Homopolymerization
(ii) Copolymerization
(iii) Addition polymer
(iv) Condensation polymer
[2012]
[2010]
[2007, 09,11C]
[2007, 09,11C]
Write the monomers of:
(i) Neoprene
(ii) PMMA
(iii) Buna-N
(iv) Nylon-6, polypropene
[2011]
[2006C, 2010]
[2006]
[2009C, 2006, 2012]
Write the difference between:
(i) Elastomer & fibre
(ii) Homopolymer & copolymer
[2008C]
[2008C,2010]
A Few question on Polymers
(Q.) Give the structure of natural rubber?
(Ans)
(1 Mark)
(Q.) Define the term polymerization?
(1 Mark)
(Ans) The repeating structural units formed from some simple and reactive molecules
(monomers) are linked together by covalent bonds. This process of formation of polymers from
their respective monomers is called polymerization.
(Q.) Define Elastomers?
(1 Mark)
(Ans) Elastomers are rubber-like solids with elastic properties. In elastomers, the polymer chain
is heldtogether by weak intermolecular forces which allow the polymer to be stretched.
e.g. buna-S, buna-N neoprene.
(Q.) What are fibers?
(1 Mark)
(Ans) Fibers are the thread forming solids which possess high tensile strength and high modulus.
These polymers possess strong intermolecular forces.Thus leading to close packing of chains and
imparting crystalline nature. e.g., polyamides( nylon 6,6) and polyesters( terylene), etc.
(Q.) Define Thermoplastics polymers?
(1 Mark)
(Ans) Thermoplastic polymers are the linear or slightly branched chain molecules capable of
repeatedly softening on heating and hardening on cooling. Some examples of thermoplastics are
polythene, polystyrene, polyvinyl, etc
(Q.) Define Thermosetting plastics?
(1 Mark)
(Ans) Thermosetting polymers are cross linked or heavily branched molecules which on heating
undergo extensive cross linking in moulds and again become infusible. Some examples of
thermosetting plastics are bakelite, urea-formaldehyde resins, etc.
(Q.) Give the method of preparation of polyacrylonitrile?
(2 Marks)
(Ans) The addition polymerization of acrylonitrile in presence of a peroxide catalyst leads to the
formation of polyacrylonitrile. It is used as a substitute for wool in making fibers as orlon or
acrilan.
(Q.) Define copolymerization? Give chemical reaction showing formation of copolymer.(2 Marks)
(Ans) Copolymerisation is a polymerization reaction in which a mixture of more than one
monomeric species is allowed to polymerise and form a copolymer. For example, a mixture of a
1,3 – butadiene and styrene form a copolymer.
(Q.) Describe the method for the preparation of neoprene?
(2 Marks)
(Ans) Neoprene or poly chloroprene is formed by the free radical polymerization of chloroprene.
(Q.) Differentiate between Addition and Condensation polymers?
(Ans)
(2 Marks)
S.No
1.
Addition Polymers
Condensation polymers
They are formed by the repeated addition
They are formed by repeated condensation reaction between
of molecules possessing double or triple
two different bi-functional and tri-functional monomeric units.
bonds.
2.
E.g. polythene
E.g. nylon 66
(Q.) Differentiate between Homopolymers and Copolymers with example.
(2 Marks)
(Ans) Homopolymers: The addition polymers formed by the repeated addition of monomer
molecules possessing double or triple bonds, are known as homopolymers.
E.g., nCH2 = CH2----->---(CH2—CH2)--Ethene polythene
Copolymers: The polymers formed by addition polymerization of two different monomers are termed as copolymers.
E.g., nCH2 = CH—CH = CH2 + nC6H5CH=CH2---->---[CH2—CH = CH—CH2—CH2—CH(C6H5)]n--(Q.) Give the differences between homopolymers and copolymers?
(2 Marks)
(Ans)
S.No.
1
Homopolymers
The addition polymers formed by the polymerisation of
a single monomeric unit are called Homopolymers.
2
E.g. Polythene.
Copolymers
The polymers formed by the addition
polymerisation of two different monomers are
termed as copolymers.
E.g. Buna-S
(Q.) What is synthetic rubber?
(2 Marks)
(Ans) Synthetic rubber is any vulcanisable rubber like polymer, which is capable of getting
stretched twice its length. However, it returns to its original shape and size as soon as the
external stretching force is released.
(Q.) Give the method of preparation of Teflon and its uses.
(3 Marks)
(Ans) Polytetrafluoroethene (Teflon) is manufactured by heating tetrafluroethene with a free radical or
persulphate catalyst at high temperature. It is chemically inert in nature. It is used for making oil seals and
gaskets and also used for non-stick surface coated utensils.
Tetraflouroethane
Teflon
(Q.) What are the different types of structural polymers? Give examples?
(Ans) There are three different types of structural polymers:
(3 Marks)
(1) Linear polymers: These polymers consist of long and straight chains.
e.g. polythene, polyvinyl chloride, etc.
(2) Branched chain polymers: These polymers contain linear chains having some branches.
e.g. low density polythene.
(3) Cross linked polymers (network polymers): These are usually formed from bi-functional and
tri-functional monomers and contain strong covalent bonds between various linear polymer
chains.
e.g. bakelite, melamine etc.
(Q.) Give the method for preparing Bakelite?
(3 Marks)
(Ans) Bakelite is manufactured from Phenol-formaldehyde polymers. It is obtained by the
condensation reaction of phenol with formaldehyde in thepresence of an acid or base catalyst.
The initial product formed is a linear chain called – Novolac used in paints. Novolac on heating
with formaldehyde undergoes cross linking to form infusible solid mass called bakelite.
(Q.) Give the method of preparation and uses of nylon 6?
(Ans) Nylon 6 is obtained by heating caprolactum with water at high temperature.
(3 Marks)
Nylon 6 is used for the manufacture of tyre cords, and fabrics and ropes.
(Q.) How are condensation polymers formed? Explain giving one example.
(3 Marks)
(Ans) The condensation polymers are formed by repeated condensation reaction between two
different bi-functional or tri-functional monomeric units. During this process, a small molecule of
water, alcohol, hydrogen chloride, etc is eliminated. E.g. terylene, nylon 6,6, nylon 6, etc.
Preparation of nylon 6,6:
nH2N(CH2)6NH2 + nHOOC(CH2)4COOH ----->---(NH(CH2)6NHCO(CH2)4CO)n--- + n H2O
(Q.) What are natural and synthetic polymers? Give 2 examples of each.
(3 Marks)
(Ans) Natural polymers: Polymers which are found in plants and animals are called natural
polymers. e.g. proteins, cellulose starch etc.
Synthetic polymers: Man made polymers are called synthetic polymers. They consists of a number of smaller molecules
to form large molecules. e.g. nylon 6,6 and Buna-S.
(Q.) Define the term polyesters? How is it manufactured?
(3 Marks)
(Ans) Polyesters are the condensation products of dicarboxylic acids and diols. e.g. Dacron or
Terylene.
It is manufactured by heating a mixture of ethylene glycol and terephthalic acid at 420 to 460 K
in the presence of zinc acetate-antimony trioxide catalyst.
(Q.) Explainvulcanisation of rubber?
(3 Marks)
(Ans) Natural rubber becomes soft at high temperatures and brittle at low temperatures. It has a
high water absorption capacity. It is soluble in non-polar solvents and is non-resistant to attack
by oxidizing agents.
To improve upon these physical properties, the process of vulcanization is carried out. It consists
of heating a mixture of raw rubber with sulphur with a temperature range between 373 K to 415
K. On vulcanisation, sulphur forms cross links at reactive sites of double bonds and thus the
rubber gets stiffened.
(Q.) Give three examples of biodegradable polymers?
(3 Marks)
(Ans) (1) phbv
(2) Nylon-2
(3) nylon-6
UNIT 16
CHEMISTRY IN EVERY DAY LIFE
Marks=3
1
DRUGS
2
CHEMOTHERAPY
3
4
Drugs are chemical of low molecular masses, which interact with macro
molecular targets and produce a biological response.
The use of chemicals to cure illness/ailments is called chemotherapy.
Drugs
for
a
particular type of
1. On the basis of pharmacological effects
problem.
e.g.
analgesics for pain
relieving.
Action on drug on a
2. On the basis of Drug Action
particular biological
CLASSIFICATIONS
process
OF DRUGS
Drugs
having
3. On the Basis Chemical Action
similar
structure.
E.g. sulpha drugs
Drugs
interacting
with biomolecules
4. On the Basis of Molecular targets
as
lipids
and
protiens
(a) Enzymes have
active sites which
hold the substrate
(i) Catalytic Action of Enzymes
molecule. It can be
attracted by reacting
molecules.
(b) Substrate is
bonded to active
sites
through
hydrogen
bonds,
ionic bonds, vander
waal or dipole dipole interactions
(ii)
Drug
Enzyme
Interactions
(a) Drug complete
ENZYMES AS
with
natural
DRUG TARGETS
substrate for their
attachments on the
active
sites
of
enzymes. They are
called competitive
inhibitors.
(b) Some drugs
binds to a different
site of the enzyme
called
allosteric
sites which changes
the shape of active
sites.
RECEPTORS AS
DRUG TARGETS
5
ANTAGONISTS
6
AGONISTS
7
ANTACIDS
8
ANTI
HISTAMINES
9
TRANQULIZERS
10
ANALGESICS
11
ANTIMICROBIALS
Receptors are proteins that are crucial to
body’s communication process. In the body
message between two neurons and that
between
neurons
to
muscles
is
communicated through certain chemicals.
These chemicals known as chemical
messengers are received at the binding sites
of receptor proteins. To accommodate a
messenger’s shapes of receptors sites are
changed.
The drugs that bind to the receptor site and
inhibit its natural functions.
Drugs mimic the natural messenger by
switching on the receptor.
These are compounds which neutralize
excess acid of stomach
The drugs which interfere with the natural
action of histamines and prevent the allergic
reaction.
The class of chemical compounds used for
treatment of stress, mild or even severe
mental diseases.
They reduce pain without causing
impairment of consciousness, mental
confusion or some other disturbance of the
nervous system.
There are a large number of
receptors in the body that
interact with different
chemical messengers hence
one medicine is not suitable
for different type of diseases.
e.g. Aluminium hydroxide,
Magnesium hydroxide.
e.g. ranitidine, tegament, avil.
e.g. luminal, seconal, equanil,
iproniagid.
e.g. aspirin: Use antipyretic
as well as analgesic. It
prevents platelets coagulation,
because of its anti-blood
clotting action aspirin finds
use in prevention of heart
attacks.
saridon, phenacetin.
They tend to prevent / destroy or inhibit the pathogenic actions of microbes
as bacteria, virus, fungi etc. They are classified as
(i) Antibiotics: Those are chemical substances which are produced by microorganism and used to kill the pathogenic micro organism. e.g. penicillin,
offloxacin
(a) Narrow spectrum Antibiotics
(b) Broad spectrum Antibiotics
These are affective mainly
against gram positive or gram
negative
bacteria.
e.g.
penicillin, streptomycin
They kill or inhibit a wide
range of micro-organism.
E.g.
chloramphenicol,
tetracycline.
12
13
14
15
16
17
18
(ii) Antiseptics or Disinfectant: These
are which either kill / inhibit the growth of
micro - organism antiseptic applied to the
living tissues such as wounds , cuts, ulcers
etc.
e.g. furacine, chloroxylenol &
terpinol (Dettol). Disinfectant is
applied to inanimate objects such
as floors, drainage system. e.g.
0.2% solution of phenol is an
antiseptic while 1% solution is
disinfectant.
These are the chemical substances used to
control pregnancy. They are also called
contraceptives or birth control pills.
These are the chemical compounds which
ARTIFICIAL
give sweetning effects to the foods
SWEETNING
without adding calorie. They are good for
AGENTS
the diabatic patients.
They prevent spoilage of food to
FOOD
PRESERVATIVES microbial growth.
1. SOAPS: They are sodium or potassium
CLEANSING
salts of long chain fatty acids. They are
AGENTS
obtained by the saponification reaction,
when fatty acids are heated with aqueous
sodium hydroxide. They do not work well
in hard water.
They are cleaning agents having
SYNTHETIC
properties of soaps, but actually contain
DETERGENTS
no soap. They can used in both soft and
hard water. They are:
e.g. Mifepristone, Norethindrone.
ANTIFERTILITY
DRUGS
BIO
DEGREDABLE
DETERGENTS
NON BIO
DEGREDABLE
DETERGENTS:
The detergents which are linear or less
branched and can be attacked by microorganism are bio degradable.
The detergents which are highly branched
and can not be decomposed by microorganism are called non-bio degradable
detergents. It creates water pollution.
e.g. aspartame, saccharin, alitame,
sucrolose.
e.g. salt, sugar and sodium
benzoate, BHT, BHA
They are biodegradable.
(i) Anionic Detergents: They are
sodium salts of sulphonated long
chain alcohols or hydro carbons.
E.g. Sodium lauryl sulphonate. Its
anionic part is responsible for
cleansing action
(ii) Cationic Detergents: They are
quarternary ammonium salts of
amines with acetates, chlorides or
bromides. They are expensive and
used to limited extent. E.g.
cetyltrimethylammoniumbromide.
Its cationic part is responsible for
cleansing action
(iii) Non-ionic Detergents: They
do not contain any ions. Some
liquid dish washing detergents
which are of non-ionic type
E.g. Sodium-4 - (1-dodecyl)
benzene / sulphonate.
N.B.: SOLVE ALL THE EXERCISE AS WELL AS INTEXT QUESTIONS.
IMPORTANT TOOLS / NOTES FOR ORGANIC
CHEMISTRY (CLASS XII)
KEY FOR CONVERSIONS IN ORGANIC CHEMISTRY
S.No
1
Reagent
KMnO4 / H+
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
Group Out
-CH2OH
Group In
-COOH
LiAlH4
-COOH
-CH2OH
Cu / 573 K
or
CrO3
PCl5 or SOCl2
Cl2 / Δ
or Cl2 /
UV
Aq NaOH / KOH
KCN
AgCN
Alcoholic KOH
Mg / dry ether
HBr
H2 / Pd-BaSO4
Zn-Hg / HCl
NH3 / Δ
Br2 / NaOH or
NaOBr
HNO2 or
NaNO2/HCl
CHCl3 / alc KOH
P2O5
H3O+
OHLiAlH4
Red P / Cl2
-CH2OH
-CHO
-OH
-H
-Cl
-Cl
-X
-X
-X
-HX
>=<
-COCl
>C=O
-COOH
-CONH2
-OH
-CN
-NC
=
Mg
H, Br
-CHO
-CH2-CONH2
-NH2
-NH2
-OH
-NH2
-CONH2
-CN
-CN
-CN
α-H of acid
-NC
-CN
-COOH
-CONH2
-CH2NH2
-Cl
Remark
Strong Oxidation (20 alcohol
gives ketone)
Strong Reduction (ketone
gives 20 alcohol)
Dehydrogenation
Free radical substitution
Nucleophilic substitution
Steping Up of a compound
Dehydrohalogenation (Stzf)
R-X  R-MgX
Markovnikov
Rosenmund Reduction
Clemmension Reduction
-COOH + NH3  -COONH4
Step Down ( Hoffmann)
HONO
Carbyl amine
Dehydration
Hydrolysis
Reduction
HVZ Reaction
KEY FOR CONVERSIONS IN BENZENE RING :
23
24
25
26
27
28
29
30
31
32
33
34
35
Fe / X2 /dark
CH3Cl / AlCl3(anhyd)
CH3COCl /
AlCl3(anhyd)
Conc.HNO3/con.H2SO4
Conc H2SO4
KMnO4 / H+
CrO2Cl2 / H+
Sn / HCl or Fe/HCl
NaOH / 623K / 300
atm
Zn dust / Δ
NaNO2 / dil HCl / 273278 K
CuCl / HCl or
Cu/HCl
CuBr / HBr or
-H
-H
-H
-X
-CH3
-COCH3
Halogination
Friedel Craft alkylation
Friedel Craft acylation
-H
-H
-R
-CH3
-NO2
-SO3H
-COOH
-CHO
-NO2
-Cl
-NH2
-OH
Nitration
Sulphonation
Oxidation
Mild oxidation(Etard
Reaction)
Reduction
-OH
-NH2
-H
-N2+Cl-
-N2+Cl-
-Cl
Sanmeyer or Gattermann
-N2+Cl-
-Br
Sanmeyer or Gattermann
Diazo reaction
36
37
38
39
40
41
42
43
Cu/HBr
CuCN / KCN
KI
HBF4 / Δ
H3PO2 or
CH3CH2OH
H2O / 283 K
HBF4/ NaNO2, Cu / Δ
C6H5-OH
C6H5-NH2
-N2+Cl-N2+Cl-N2+Cl-N2+Cl-
-CN
-I
-F
-H
-N2+Cl-N2+Cl-N2+Cl-N2+Cl-
-OH
-NO2
-N=N-C6H5-OH
-N=N-C6H5-NH2
Sanmeyer
Coupling ( p-hydroxy)
Coupling ( p-amino)
Reactions of Grignard Reagent
Grignard reagent +
Any one below + H2O
H2O or ROH or RNH2
H-CHO
R-CHO
R-CO-R
CO2
R-CN
HCOOR
RCOOR
R-MgX

Product
R-H
R-CH2-OH (10 alc)
R-CH(OH)-R (20 alc)
R2C(OH)-R (30 alc)
R-COOH
R-CO-R
Aldehyde
Ketone
NB: i) During reaction generally changes take place in the functional group only so see the functional
group very carefully.
ii) Remember structural formula of all the common organic compounds ( with their IUPAC and
common names)
iii) Wurtz Reaction and Aldol Condensation are not included in the table although they are
Very important for conversions so study them .
iv) By taking examples practice all the above cases (from 1 to 43 and Grignard)
v) Practice only from NCERT book.
vi) Start practicing NOW !
How to use the table? See below.
Example: See no 7 in the table
Directional Properties of groups in benzene ring for electrophilic substitution
Ortho-para directing group: -R, -OH, -NH2, -X, -OR, -NHR, -NR2, -NHCOCH3, -CH2Cl, -SH, Ph
Meta-directing group:
NH3+
-NO2 , -CHO, -COOH , COOR , -CN, -SO3H , -COCH3, -CCl3 , -
NAME REACTIONS (ORGANIC CHEMISTRY)
1. Finkelstein -
CH3Br + NaI
CH3-I + NaBr
2. Swarts -
CH3Br + AgF
CH3F + AgBr
CH3
3. FriedelCrafts-
+
Anhydrous AlCl3
H3C
Cl
Alkylation
COCH 3
CH3COCl
4. FriedelCrafts-
Anhydrous AlCl3
Acylation
5. Wurtz -
H3C
Cl
+
Cl
Cl
CH3
H3C
CH3
+
Na Cl
Cl
2Na
+
6. Fittig
2Na
+
Dry ether
Na Cl
Cl
+
7. Wurtz-Fittig
2Na
Cl
CH3
OH
8. Kolbe’s
reaction
Na Cl
OH
ONa
Na OH
+
CH3
Dry ether
i) CO2
ii) H+
COOH
OH
OH
ONa
CH3Cl + Na OH
CHO
H+
CHO
9. ReimerTiemann
CH3-Br + CH3-ONa
10. Williamson’s
11. Stephen
H3C
CN
+
CH3-O- CH3 + NaBr
SnCl2 + HCl
H3C
CH
NH
CH3
H3O+
H3C
CHO
CHO
CrO2Cl2
12. Etard
H3O+
CHO
CO / HCl
13. Gatterman –
Koch
Anhydrous AlCl3
O
14. Rosenmund
C
C
reduction
H3C
Cl
H3C
reduction
Conc. HCl
CH3
O
C
16. Wolff-Kishner
H3C
reduction
R-CHO + 2 Cu2+ + 5 OH-
19.Iodoform
Reaction
CH3 ii) KOH / Ethylene glycol / 
I2 / NaOH
C
H3C
CH3
CH2 CH3
H3C
CH2 CH3
R-COO- + 2Ag
NH3
O
H3C
i) NH2-NH2
R-CHO + 2 [Ag(NH3)2]+ + 3 OH-
18. Fehling’s
test
H
Zn - Hg
C
15.Clemmensen
H3C
Pd / BaSO4
O
17. Tollens’ test
O
H2
OR, NaOI
+ 2H2O + 4
Silver mirror
R-COO- + Cu2O
+ 3H2O
CHI3 + CH3COONa
Yellow ppt.
OH
dil NaOH
20. Aldol
Condensation
2 H3C
HCHO
21. Cannizzaro
22. Hell-Volhard-
+
Conc. NaOH
COOH
O
H3C
bromamide
degradation
+
H3C
CHO
OH
i) Cl2 / Red Phosphorus
ii) H2O
23.Hoffmann
H3C CH CH
HCOONa
HCHO
H3C
Zelinsky
(HVZ)

H3C CH CH2 CHO
CHO
H2C
COOH
Cl
Br2
C
NH2
H3C
NaOH
NH2

24. Carbylamine
R-NH2 + CHCl3 + 3 KOH
+
N2 Cl
R-NC + 3 KCl + 3 H2O
-
Cl
CuCl / HCl
+ N2
25. Sandmeyer.
+
N2 Cl
-
Cl
Cu / HCl
+ N2
26. Gatterman
+
27.Coupling
N2 Cl
Reaction
-
+
OHH
OH
N
N
OH
ELECTRON DISPLACEMENT EFFECTS
+ I : O- , COO- , (CH3)3C , (CH3)2CH , CH3CH2 , CH3
(electron donating)
- I : NR3+ , SR2+ , NH3+, NO2 , SO2R , CN , COOH , F , Cl , Br , I , OR , OH, NH2 (e-withdrawing)
+R (+M)
:
OH , NH2 , OR , NHR , X
-R (-M)
: NO2 , CN , CHO , COOH , COCH3
(electron donating)
(electron withdrawing)
DIRECTIVE INFLUENCE OF SUBSTITUENTS IN BENZENE RING
(for electrophilic substitution reactions)
EFFECT OF THE GROUP
DIRECTING
ACTIVATING / DEACTIVATING
+I
Ortho / Para
Activating
+I,+R
Ortho / Para
Activating
-I<+R
Ortho / Para
Activating
-I>+R
Ortho / Para
Deactivating
-I
Meta
Deactivating
-I, -R
Meta
Deactivating
Example:
-I > +R
: - X , - CH=CH2 , -CH=CH-COOH , -CH2-Cl
These groups are deactivating but exceptionally o / p directing due to +E effect by the attacking
reagents electron density increases at -ortho and -para position .
If two groups are present initially
1.When both the groups present in benzene ring are o/p directing than the order of influence :
O- > NH2 > NR2 > OH > OCH3 > NHCOCH3 > CH3 > X
2.When both the groups present in benzene ring are meta directing than the order of influence :
(CH3)3N+ > NO2 > CN > SO3H > CHO > COCH3 > COOH
3.When one group is o/p and another is m – directing than o/p directing group takes priority
Distinction By Single Chemical Test
1. All aldehydes ( R-CHO) give Tollens’ Test and produce silver mirror.
RCHO + 2 [Ag(NH3)2]+ + 3 OH-  RCOO- + 2 Ag
Tollens’ Reagent
+ 2H2O + 4 NH3
silver ppt
Note: HCOOH(methanoic acid ) also gives this test, ketones(RCOR) do not give this test
2. All aldehydes (R-CHO) and ketones(RCOR) give 2,4-DNP test
RCOR + 2,4-DNP  Orange ppt
R-CHO + 2,4-DNP  Orange ppt
3. Aldehydes and ketones having CH3CO- (keto methyl) group give Iodoform Test. Alcohols having
CH3CH- group also give Iodoform Test.
|
OH
CH3CHO + 3I2 + 4 NaOH  CHI3
+ HCOONa + 3 NaI + 3H2O
Yellow ppt
The following compounds give Iodoform Test: ethanol (C2H5OH), propan-2-ol
(CH3CH(OH)CH3),ethanal(CH3CHO), propanone(CH3COCH3),butanone (
CH3COCH2CH3), pentan-2-one (CH3COCH2 CH2CH3) , acetophenone ( PhCOCH3 )
4. All carboxylic acids ( R-COOH) give Bicarbonate Test
RCOOH + NaHCO3  RCOONa + CO2
+ H2O
Brisk effervescence
5. Phenol gives FeCl3 Test
C6H5OH + FeCl3  (C6H5O)3Fe + 3 HCl
(neutral)
(violet color)
6. All primary amines (R/Ar -NH2) give Carbyl Amine Test
R-NH2 + CHCl3 + KOH(alc)  R-NC
+
KCl + H2O
offensive smell
7. Aniline gives Azo Dye Test ( Only for aromatic amines)
C6H5NH2 + NaNO2 + HCl  C6H5N2+Cl- ;
then add β-naphthol  orange dye
8. All alcohols (ROH) give Sodium (Na) metal test
R-OH + Na  R-ONa + H2
(bubbles)
9. For esters (RCOOR) : Hydrolyses first. Then see the products ( acid & alcohol) and give a test to
identify them
10. All alkenes (C=C) and alkynes (C≡C) decolorizes Br2 water from red to colorless
11. Lucas Test to distinguish primary, secondary and tertiary alcohols
Lucas reagent: ZnCl2/HCl
30-alcohol + Lucas reagent  immediate turbidity
20-alcohol + Lucas reagent  turbidity after sometime
10-alcohol + Lucas reagent  no turbidity
SAMPLE QUESTION PAPERS
With BLUE PRINT
& MARKING SCHEME
BLUE PRINT
S.NO
.
NAME OF UNIT
VSA
SA(I)
SA(II)
LONG
TOTAL
1
SOLID STATE
2 X2
4
2
SOLUTION
2
3
5
3
ELECTROCHEMISTRY
2
3
5
4
CHEMICAL KINETICS
5
SURFACE CHEMISTRY
6
5
1
5
3
4
GENERAL PRINCIPLE OF
EXTRACTION
3
3
7
P BLOCK
3
8
THE d-& f-BLOCK
ELEMENTS
9
CO-ORDINATION
COMPOUNDS
10
HALOALKANES& HALO
ARENES
11
ALCOHALS, PHENOLS &
ETHERS.
1
12
ALDEHYDE, KETONES &
CARBOXYLIC ACIDS
1
13
AMINES
1
3
4
14
BIOMOLECULES
1
3
4
15
POLYMERS
1
2
3
16
CHEMISTRY IN
EVERYDAY LIFE
1
2
3
TOTAL
8
10
2
1
5
3
8
5
2
3
2X2
4
3
4
5
9
3
6
70
MODLE QUESTION PAPER
CLASS XII
SUBJECT CHEMISTRY
TIME: 3Hrs
M.M. 70
General Instructions:
(i)
All questions are compulsory.
(ii)
Questions 1 to 8 are short answer questions. Each carries one mark.
(iii) Questions 9 to 18 are short answer questions. Each carries 2 marks.
(iv) Questions 19 to 27 are also short answer questions. Each carries 3 marks
(v) Questions 28 to 30 are long answer questions each carry 5 marks.
(vi) There is no overall choice, however, an internal choice has been provided in one question of
two marks, one question of three marks and all three questions of five marks each. You
have to attempt only one of the given choices in such questions.
(i)
Use log table wherever necessary. Use of calculators is not permitted.
1. Adsorption of a gas on the surface of solid is generally accompanied by decrease in
entropy still it is spontaneous process. Explain.
2. Square planer complexes with co-ordination number of four exhibit geometric al
isomerism whereas tetrahedral complexes do not. Why?
3. How will you prepare methanol from formaldehyde without using a reducing agent?
4. Give the IUPAC name of
CH3 CH3
CH3
CH2 C
Br
C
CH2 CH 2
OH
Br
5.
6.
7.
8.
9.
Write steps involved in the conversion of 1-propaneamine to acetone.
Why can not vitamin C be stored in our body.
Define Elastomers?
Define non-biodegradable detergent with one example.
Analysis shows that a metal oxide has empirical formula of M0.96O . calculate the
percentage of M2+ and M3+ ion in the crystal.
10.Account for the following:
a) some of glass object recovered from ancient monuments look like milky instead of
being transparent
b)Zinc oxide is white but turn yellow on heating. Explain
11. What do you understand by “board spectrum antibiotics?
12 Differentiate between Homopolymers and Copolymers with example.
OR
Describe the method for the preparation of neoprene?
13 The osmotic pressure of human blood is 7.65 atm at 37°C. For injecting
Glucose solution it is necessary the glucose solution has same osmotic pressure as of
human blood. Find the molarity of glucose solution having same osmotic pressure as
of human blood. Describe the method for the preparation of neoprene?
14 . Define conductivity and molar conductivity for the solution of an electrolyte.
Discuss their variation with concentration.
15 What is lanthanoid contraction? Why lanthanides are known as f-block elements?
16) Copper I compounds are white and diamagnetic but copper II compounds are
coloured and paramagnetic. Why?
17) Complete the following reactions:
(a) C6H5ONa + C2H5Cl -------->
(b) CH3CH2CH2OH + SOCl2 ------->
18.) Why is aniline a weaker base than methylamine?
19 a) Define glycosidic linkage
b) What are the main sources of vitamins?
c ) Define the term Oligosaccharides?
OR
a) Explain the term mutarotation?
b)What is difference between Reducing and non-reducing sugars or
carbohydrates?
20 How are the following preparations carried out?
(a) Salicylic acid from phenol.(b) n-propyl alcohol from ethene.
21 a) Explain the coupling reaction with example.
b) Describe the Gabriel-phthalimide reaction.
22 a) Explain ideal and non-ideal solutions with respect to intermolecular interactions in
a binary solution of A and B.
b) Aquatic animals are more comfortable in cold water than in warm water. Explain?
23
24 (a) What is the effect of temperature on adsorption?
b) Why are zeolites called shape selective catalysts?
c) Define tyndall effect
25 (a) What is leaching?
(b)How can the ores ZnS and PbS be separated from a mixture using froth floatation process?
26 (a) Why does NH3 has higher boiling point than PH3?
(b) Why is the ionization energy of group 15 elements higher than that of group 14elements.
(C) Acidic strength increases in this order : HF<HCl<HBr<HI. Give reason.
27 (a) How does +2 oxidation state becomes more and more stable in the first half of
the first row transition elements with increasing atomic number?
(b)Vanadium pentaoxide acts as a good catalyst. Why
28 a) The initial concentration of N2O5 in the following first order reaction
N2O5(g)
2 NO2(g) + 1/2O2 (g) was 1.24 × 10–2 mol L–1 at 318 K. The
concentration of N2O5 after 60 minutes was 0.20 × 10–2 mol L–1.
Calculate the rate constant of the reaction at 318 K.
b) A reaction is second order with respect to a reactant. How is the rate of
reaction affected if the concentration of the reactant is
(i) Doubled (ii) Reduced to half
OR
a)The rate constants of a reaction at 500K and 700K are 0.02s–1 and 0.07s–1 respectively.
Calculate the values of Ea and A.
b)For a reaction, A + B
Product; the rate law is given by, r = k [ A]1/2 [B]2.
What is the order of the reaction?
c) Identify the reaction order from the following rate constants.
(i)
k = 2.3 × 10–5 L mol–1 s–1
29. a)
OR
a)
b) Describe the following(Write reactions only)
(i) Aldol condensation
(ii) Cannizzaro reaction (iii)Friedel-Crafts reaction.
30 a) Why is ammonia a mild reducing agent while BiH3 the strongest reducing agent
among all the hydrides?
b) Explain the brown ring test for nitrates. Give the equations involved for the
reactions occurring during the brown ring test.
OR
a) Give reasons:
(i) Size of oxygen atom exceptionally small
(ii) SF6 exceptionally stable
(iii) PCl3 fumes in atmosphere
b) How does Cl2 react with
(i) cold and dilute NaOH
(ii) hot and concentrated NaOH
Answer Scheme (Sample Question Paper)
1.
Because adsorption is an exothermic process
1mark
2.
Because relative position of ligands attached to central atom are same with
respect to one another
1mark
3.
Give cannizarro reaction.
1mark
4.
3,4-dibromo-3,4-dimethylhexan-1-ol.
1mark
5.
CH3CH2CH2NH2
mark.rule H+/H o
2
6.
HONO
CH3CH2CH2OH Conc. sulphuric acid,443K
CH3CHOHCH3 Cu powder 573k
CH3COCH3
Because it is soluble in water so excreted in urine
CH3CH=CH2
1mark
1mark
7. Elastomers are rubber-like solids with elastic properties. In elastomers, the
polymer chain is held together by weak intermolecular forces which allow the
polymer to be stretched.
e.g. buna-S, buna-N neoprene
1mark
8. Detergent containing branched hydrocarbon chain are not easily degraded by
microorganism and hence are called non-biodegradable detergent e.g. sodium-4(1,3,5,7-tetramethyloctyl) benzenesulphonate.
1mark
9. Ratio of M2+ : M3+ =96:100
Total charge on M2+and M3+
(+2)X +3(96-X) =200
X=88% of M2+ and M3+ =91.7 and 8.3 respectively
1+1mark
10 a) Due to annealing over a number of years glass acquires some crystalline
character
b) because it loose oxygen
ZnO
Zn2+ + 1/2O2 + 2e-
Metal excess defect electron in interstitial voids the colour is yellow as the
remaining colour of white light are absorbed by these electron
1+1mark
11 These are effective against several types of bacteria. For example tetracycline,
chloramphenicol, Ofloxacin which are used as antibiotics.
1+1mark
12) Homopolymers: The addition polymers formed by the repeated addition of
monomer molecules possessing double or triple bonds, are known as homopolymers.
E.g.,
nCH2 = CH2 -----> ---(CH2—CH2)---
Ethene
polythene
Copolymers: The polymers formed by addition polymerization of two different
monomers are termed as copolymers.
E.g., nCH2 = CH—CH = CH2 + nC6H5CH=CH2 ----> ---[CH2—CH = CH—CH2—CH2—
CH(C6H5)]n--1+1marks
OR
Neoprene or poly chloroprene is formed by the free radical polymerization of
chloroprene.
1+1mark
13. (Ans)
=
Or 7.65 =
=
= Molarity = 0.30
1+1mark
14. Conductivity : conductance of a one cm3 of a solution is known as conductivity.
Conductivity decrease with dilution i.e. decrease with decrease of concentration
Molar conductance : Molar conductance is the conductance of a solution having one mole
of electrolyte when whole of the solution is present between the two electrode.
Molar conductance increase with dilution i.e. increase with decrease of
concentration.
1+1mark
15. ) In the lanthanide series as the atomic number increases there is a progressive
decrease in the size of the atoms and trivalent ions which is known as lanthanide
contraction.
The last electron enters in the f-orbital, so lanthanides are knowns as f- block
elements
1+1mark
16. ) In copper I ion all orbitals are completely filled so its compounds are white
and diamagnetic. The electronic configuration of copper II ion is
1s22s22p63s23p63d9. it has one unpaired electron so it is paramagnetic and forms
blue coloured compounds.
1+1mark
17. ) (a) C6H5ONa +
C2H5Cl
Sodium Ethyl chloride
(b) CH3CH2CH2OH + SOCl2
Propanol-1
Thionyl choride
C6H5-O-C2H5 +
NaCl
Phenetolephenoxide
CH3CH2CH2Cl + HCl + SO2
1-chloropropane
1+1marks
18. Aniline and methylamine both have nitrogen with lone pair of electron. In aniline the
phenyl group is electron attracting. It tend to decrease the electron density on nitrogen
atom and hence decreases electron releasing tendency of nitrogen. While in
methylamine CH3- group is electron repelling in nature. It tends to increase the electron
density on the nitrogen atom and helps in electron releasing tendency of nitrogen. So,
the aniline is a weaker base than methylamine.
1+1mark
19. a)The two monosaccharide units are joined together through an ethereal or oxide
linkage formed by the loss of a molecule of H2O. Such a linkage between two
monosaccharide units through oxygen atoms is called glycosidic linkage
b)The main sources of vitamins are milk, butter, cheese, fruits, green vegetables, meat,
fish, eggs, etc.
c) Those carbohydrates which give 2-10 molecules of monosaccharides in
hydrolysis.
1+1+1mark
OR
a)Mutarotation is the change in the specific rotation of an optically active compound with
time, to an equilibrium mixture
b)All those carbohydrates which contain aldehydic and ketonic group in the
solution are called reducing carbohydrates while others which do not reduce
these reagents are called non-reducing reagents
1+1+1marks
20) (a) Salicylic acid from phenol: -
(b) n-propyl alcohol from ethane: - By oxo process
CH2=CH2 + (CO + H2)
CH3-CH2-CHO
CH3-CH2-CH2-OH
3/2+3/2
21.a) Benzene diazonium chloride reacts with phenol in which the phenol molecule at
its para position is coupled with the diazonium salt to form p-hydroxybenzene. This type
of reaction is known as coupling reaction.
b) In this reaction potassium phthalimide is reacted with an alkyl halide to get N-alkyl
phthalimde. N-alkyl phthalimide on hydrolysis with dil. HCl under pressure.
3/2+3/2
22. a) ) For the given binary solution of A and B, it would be ideal if A-B interactions are
equal to A-A and B-B interactions and it would be non-ideal if they are different to each
other.
The deviation from ideal behaviour will be positive if A-B interactions are weaker
as compared to A-A and B-B. The deviation will be negative if A-B interactions
are stronger as compared to A-A and B-B.
b) This is because “Kh” (Henry constant) values for both N2 and O2 increase with
increase in temperature indicating that the solubility of gases increases with decrease in
temperature 2+1 marks
23. a) For correct two reasons
b) For Correct value
(1+1 M)
1M
24. a) Adsorption processes, being exothermic, decreases with increase in temperature.
b) Zeolites are called shape selective catalysts because their catalytic action depends
upon the size and shape of the reactant and the product molecules as well as on their
own pores and cavities.
c) ) Itis defined as the scattering of light by the colloidal particles present in a colloidal
solution.
1+1+1 mark
25.a) Leaching is the process of extracting a substance from a solid by dissolving it in a
liquid. In metallurgy leaching is used for the ores that are soluble in a suitable solvent.
b) During the froth floatation process a depressant like NaCN is added to the tank. The
depressant selectively prevents ZnS from coming to the froth but allows PbS to come to
the froth and hence helps the separation of PbS with the froth.
3/2+3/2 marks
26.a) NH3 has higher boiling point than PH3 because of the presence of inter molecular
hydrogen bonding in NH3, as the electronegativity difference is quite high in case of
Nand H.
b) The ionization energy of group 15 elements is higher than that of group 14 elements
because the elements of group 15 have extra stable half-filled p orbital configuration and
their size is smaller due to the higher nuclear charge
c) Acidic strength increases from HF to HI because, down the group, as the size of the
halogen increases, the Bond dissociation enthalpy decreases and it becomes easier for
that halogen atom to lose its H+ ion
1+1+1 marks
27. a) The sum of first and second ionization enthalpies increases with increasing atomic
number so the standard reduction potentials become less and less negative. Hence the
+2 oxidation state becomes more and more stable.
b) Vanadium shows different-different oxidation states because it has vacant dorbitals.so vanadium pentaoxide acts as a good catalyst.
marks
2+1
28
a)
b) Reaction is of second order w.r.t. reactant
i)therefore when concentration is double rate of reaction become four times
rate =[2R]2 =4[R]2
ii) when concentration is half rate of reaction become ¼ of the initial rate.
rate =[1/2R]2 =1/4[R]2
3+1+1 marks
OR
a)
b)
2+1/2 =2.5
c)
(3+1+1 marks)
29. (a)
1+ 1+1+1+1 mark
OR
(b) give reaction only
(c)give reaction only
3+1+1 marks
30 (a) From N to Bi the size of atom increases and tendency to form covalent bonds
with H decreases. Therefore, the thermal stability also decreases from the hydrides,
down the group. As a result, their tendency to liberate hydrogen increases for hydrides
down the group. Hence, NH3 is a mild reducing agent while BiH3 is a very strong
reducing agent.
(b) The brown ring test is carried out by adding dilute ferrous sulphate solution to
an aqueous solution containing nitrate ions and then carefully adding conc. H2SO4 along
the sides of the test tube. The Fe2+ ions reduce nitrates to nitric oxide, which reacts with
Fe2+ ions to form a brown colored complex that forms a brown ring at the interface
between the solution and sulphuric acid layers. The equations involved in the brown ring
test are:
NO3- + 3Fe2+ + 4H+ = NO + 3Fe3+ +2H2O
[Fe(H2O)6]2+ +NO = [Fe(H2O)5(NO)]2+ + H2O
2+3marks
OR
i) The size of oxygen is exceptionally small because of its high electron density and
hence higher attraction between the electron cloud and the nucleus.
ii) In SF6 the six F atoms protect the central sulphur atom from the reagents making the
compound extra stable.
iii) PCl3 fumes in atmosphere due to its hydrolysis in atmosphere.
PCl3 + 3H2O = H3PO3 + 3HCl
b) ) When Chlorine reacts with cold sodium hydroxide, it forms sodium chloride.
(i) 2NaOH + Cl2 = NaCl + NaOCl + H2O
When Chlorine reacts with hot sodium hydroxide, it forms sodium chloride and other
products.
(ii) 6NaOH + 3Cl2 = 5NaCl + NaClO3 + 3H2O
1+1+1+1+1 mark