Download 8th class Chemistry Bridge program.

CLASS-VIII
MPC BRIDGE COURSE
CHEMISTRY

DAY-1 : SYNOPSIS









Dalton’s atomic theory
Matter consists of small indivisible
particles called atoms.
Atoms of same element are alike in all
respects.
Atoms of different elements are different
in all respects.
Atoms combine in small whole numbers
to form compound atoms (molecules).
Atom is the smallest unit of matter which
takes part in a chemical reaction.
All the points put forward in Dalton’s
atomic theory have been contradicted by
modern research, except that atom is the
smallest unit of matter, which takes part
in a chemical reaction.
Discovery of cathode rays:
The cathode rays were discovered by
William Crookes in 1878.
High electrical voltage was applied
through two electrodes placed in a glass
tube, containing a gas at a pressure of
0.01 mm of mercury.
A zinc sulphide screen placed within the
tube started glowing. It was assumed that
some invisible rays are given out, which
on striking zinc sulphide screen, made
it glow.
Later on, more experiments proved that
cathode does not give out invisible
radiations. Actually tiny particles moved
from cathode towards anode. These
particles were named electrons. The
electrons on striking the screen, made
it glow.
It was found out that electrons have
definite mass and definite electric
charge, which is negative in nature.
Furthermore, the mass and charge of
electrons was same irrespective of the
nature of gas within the discharge tube.
NARAYANA GROUP OF SCHOOLS








Conclusions
Electrons are integral part of all atoms
which are negatively charged.
Since atoms are electrically neutral, an
atom must contain another kind of
particles which are positively charged.
These particles (unknown particles at
that time) were named protons.
Total negative electrical charges on the
electrons of an atom must be equal to
the total positive charges on the protons.
Characteristics of electrons
The absolute charge on an electron is –
1.6 × 10–19 coulombs. Assuming the above
charge equal to unit electric charge, we
can say that relative charge on an
electron is –1.
The absolute mass of an electron is 9.1 x
10 –28 g. However, if the above mass is
compared with 1 atomic mass unit (amu),
then the relative mass of an electron is
1/1837 amu or 1/1837 times the mass
of one atom of hydrogen
Characteristics of Positively Charged
Particles
The electric charge on these particles is
always positive. However, the amount of
positive charge on them varied with the
nature of gas in the discharge tube.
The positive charge on any particle was
found to be a multiple of 1.6 × I0–19 C. It
means the charge could be +1.6 × 10–19C
or + 2 × 1.6 × 10–19 C or + 3 × 1.6×10–19 C,
etc., depending upon the nature of gas.
The mass of these particles was same as
the atomic mass of the gas in a discharge
tube.
These particles were affected by the
electric and magnetic fields but in the
direction, opposite to the cathode rays.
Amongst the positively charged particles
formed by the discharge in gases, it was
found that the particles formed during
the discharge through hydrogen were
lightest. Further on, the magnitude of
charge on these particles was same as
on electron, but positive in nature.
70
CLASS-VIII
The lightest positively charged particle
of hydrogen was named proton.
Characteristics of a Proton
 The electric charge on a proton is +1.6 ×
10–19 C.
–24
 The mass of proton is 1.67 × 10 g. It is
estimated that a proton is 1837 times as
heavy as an electron.
DAY-1: WORKSHEET
1. The term ‘atom’ was given by
a) Democritus
b) John Dalton
c) William Crookes d) Maharishi Kanada
2. Which of the following is true according
to Dalton’s atomic theory?
a) Matter consists of small indivisible
particles called atoms.
b) Atoms of same element are alike in
all respects
c) Atoms combine in small whole numbers
to form compound atoms (molecules)
d) Atom is the smallest unit of matter
which takes part in a chemical reaction.
3. The first atomic theory was given by
a) Democritus
b) John Dalton
c) William Crookes d) Maharishi Kanada
4. Which is not correct about electrons?
a) Discovered by Chadwick
b) Named by J.L. Stoney
c) Present inside the nucleus
d) It has maximum e/m ratio
5. Which of the following is never true for
cathode rays?
a) They possess kinetic energy
b) They are electromagnetic waves
c) They produce heat
d) They produce mechanical pressure
6. The discharge tube experiment in which
cathode rays are emitted has shown that
a) All nuclei contain positive charge
b) All forms of matter contain electrons
c) Protons are positively charged
d) Mass of proton and that of neutron are
almost equal
7. Assertion : The charge to mass ratio of
the particles in anode rays depends on
nature of the gas taken in the discharge
tube.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
Reason : The particles in anode rays
carry positive charge.
a) Assertion is correct and reason is the
correct explanation of assertion.
b) Reason is correct but assertion is
incorrect.
c) Assertion is correct and reason is not
the correct explanation of assertion.
d) Reason and assertion both are
incorrect.
8. Positive rays are
a) Electromagnetic waves
b) Electrons
c) Positively charged gaseous ions
d) Neutrons
9. The ratio of e/m for p+ and  - particle is
a) 1:2
b) 2:1
c) 1:3
d) 1:1
10. Which is not correctly matched:
a) Particle nature of e– was given by 
Bohr.
b) The heaviest sub-atomic particle 
Neutron among e, p, n.
c) Electron is  non fundamental
particle.
d) Outside the nucleus  neutron is
unstable.
11. The mass of the electron is
a) 1.76  10–23 kg
b) 1.67  10–24 kg
c) 9.11  10–28 kg
d) 9.11  10–31 kg
12. The charge to mass ratio of  - particles
is approximately.....the charge to mass
ratio of protons
a) twice
b) half
c) four times
d) six times
13.The charge to mass ratio of proton is
9.55 × 104 C/g and charge on the proton
is + 1.6 × 10–19C. The mass of the proton
would be
a) 1.67  1024 Kg
b) 1.67  10–27 Kg
–24
c) 1.67  10 Kg d) 1.67  10–24 g
14.If S 1 is the specific charge (e/m) of
cathode rays and S2 be that of positive
rays, then which is true?
a) S1 = S2
b) S1 < S2
c) S1 > S2
d) Either of these
71
CLASS-VIII
MPC BRIDGE COURSE
DAY-2 : SYNOPSIS
Thomson’s Atomic model:
• His model of atom has been given
following different names: Water Melon
Model (or) Plum Pudding Model (or) Raisin
Pudding Model.
• He proposed that atom consists of positive
charge in which the negatively charged
electrons are embedded.
• He could not explain how the electrons
are protected from the effect of positive
charge. Hence model is considered to be
a failure.
Rutherford’s Planetary Model:
• This is the first atomic model which has
successfully explained the structure of
atom.
• His explanation is based on the gold foil
experiment.
• His experiment contain (i) source of alpha
particles (ii) lead block (iii) golden foil of
thickness 0.0004mm (iv) ZnS screen.
• He made alpha rays to pass though the
golden foil of very less thickness
(0.0004mm).
Observations
1. Nearly 99% of alpha rays passed
through golden foil without any
deflection.
2. A few alpha particles are deflected
by large angles are deflected back
along that original path.
3. Some of the alpha particles are
deflected with small angles.
Conclusions
Most of the space inside atom is
empty.
There is hard and soft unit at the
centre of the atom called nucleus.
The charge of the nucleus is positive
and is due to the presence of protons.
Main attributes of Rutherford’s Atomic
Model:
(i) Most of the space inside an atom is empty
(ii) Most of the mass and entire positive
charge of the atom is concentrated inside
nucleus.
Hence it is known as Nuclear Model of atom.
Note: Rutherford discovered nucleus for the
first time.
(iii) Size of the nucleus is extremely small
compared to the size of the atom.
(iv) The magnitude of the charge on the
nucleus is different for atoms of different
elements.
(v) The electrons revolve round the nucleus
with very high speed just like the planets
round the sun.
This is the reason why it is called Planetary
Model or Solar Model.
NARAYANA GROUP OF SCHOOLS
(vi) The inward, electrostatic force of
attraction between the positively charged
nucleus and electron is balanced by the
outward centrifugal force. According to
him, these two forces are equal and
opposite and prevent electron from falling
inside the nucleus.
(vii) Total negative charge present outside
the nucleus is equal to the total positive
charge present inside the nucleus.
• Defects of Rutherford’s atomic model:
(i) According to the Classical laws of
mechanics or dynamics of physics, any
charged particle revolving around another
charged particle should lose energy
continuously. Hence electron revolving
round the nucleus should lose energy
and fall inside the nucleus. But nucleus
is found to be stable.
Thus Rutherford’s atomic model does not
explain the stability of an atom.
(ii) It could not explain the distribution of
electrons around the nucleus and does not
tell us anything about their energies.
(iii)
If the electron l oses energy
continuously, then the atomic spectra should
be continuous but it is discontinuous. Hence
It could not explain the line spectrum.
• Quantum theory
(i) Energy is emitted or absorbed not
continuously but discontinuously in the
form of small packets of energy called
quanta.
• The quantum in case of light is called
photon.
(ii)
Each quantum is associated with
definite amount of energy.
(iii) The amount of energy associated with
a quantum of radiation is proportional to
the
frequency
of
radiation.
E    E  h , where h = Planck’s
constant and is equal to 6.625  10–34 Jsec.
72
CLASS-VIII
DAY-2: WORKSHEET
1. Rutherford’s experiment on scattering of
alpha particles showed for the first time
that atom has
a) Nucleus
b) Electrons
c) Protons
d) Neutrons
2. Rutherford’s model is related to explain
a) Discovery of nucleus
b) Spectrum of Hydrogenic species
c) Planetary motion of electrons around
nucleus
d) All of these
3. Rutherford’s  - particle dispersion
experiment concludes?
a) All –Ve ions are deposited at small part.
b) Proton moves around the nucleus.
c) All +Ve ions are deposited at small part.
d) Neutrons are charged particles.
4. When alpha particles are sent through a
thin metal foil, most of them go straight
through the foil because
a) Alpha particles are much heavier than
electrons
b) Alpha particles are positively charged
c) Most part of the atom is empty space
d) Alpha particles move with high velocity
5. Rutherford’s alpha particles scattering
experiment eventually led to the
conclusion that
a) Mass and energy are related
b) Electrons occupy space around the
nucleus
c) Neutrons are deep in the nucleus
d) The point of impact with matter can
be precisely determined
7. Rutherford’s
experiment
which
established the nuclear model of the
atom used a beam of
a)  -particles which impinged on a metal
foil and got absorbed.
b)  -rays which impinged on a metal foil
and ejected electrons.
c) Helium atoms, which impinged on a
metal foil and got scattered.
d) Helium nuclei which impinged on a
metal foil and got scattered.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
8. Assertion A : Size of the nucleus is very
small as compared with size of
the atom.
Reason (R) : Almost all the mass of the
atom is concentrated in the
nucleus.
a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion
b) Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion
c) Assertion is correct and Reason is incorrect
d) Assertion is incorrect and Reason is
correct
9. Very few (1 in 10000 or 20000)
 - particles are deviated through an
angle of 180 oC. This has lead to the
discovery of
a) Proton
b) Neutron
c) Both
d) Nucleus
10. The electrons of Rutherford’s model of
the atom are expected to lose energy
because they
a) are attracted by the nucleus
b) strike each other
c) are accelerated
d) are in motion
11. The value of Planck’s constant (h) is
a) 6.625×10–34 J-sec
b) 6.625×10–28 erg – sec
c) 6.25×10–38 cal – sec
d) 6.25×101–27 erg –
DAY-3 : SYNOPSIS

Atom: The term atom was introduced
by Dalton. Atom is the smallest particle
of matter that takes part in a chemical
reaction. Atom is also defined as the
smallest particle of an element that retains all the properties of an element.

Atomic mass unit (a.m.u.): It is the
smallest unit of mass and is used to
measure the masses of atoms and
subatomic particles. The mass of one
a.m.u. is equal to the mass of
1
th
12
the
73
CLASS-VIII
MPC BRIDGE COURSE
mass of C-12 atom. The other names of
a.m.u. are Aston, Dalton and Avogram.
Note:1 a.m.u. = 1.66  10–24 g or 1.66  10–27
kg.
 Atomic
combination of atoms. Molecule is also
defined as the smallest particle of matter
that can exist and retains all the
properties of that substance.
Note: A molecule splits into atoms first
before taking part in a chemical reaction.
weight :
 The atomic weight or the relative atomic
mass (RAM) of an element is defined as
the number of times an atom of an
element is heavier than the mass of
1
th of
12
 Molecular
 Relative molecular mass or molecular
weight is defined as the number of times
C-12 isotope’s atom.
 Relative atomic mass of an element
Mass of 1atom of that element
 RAM  1
 (Mass of C 12 atom)
12
indicates the number of times one atom
of that element is heavier than
1 th
12
 75 35   25 37  35.5
100
Mass of one atom of an element =
Relative atomic mass  mass of
mass of C- 12

1 th
12
the
Molecule: The term molecule was
introduced by Avogadro. Molecule is the
smallest particle of matter that exists
independently and is formed by the
NARAYANA GROUP OF SCHOOLS
1 th
12
the mass
of C-12 isotope’s atom.
Average mass of one molecule
Weight of 1 12 th of C  12 atom .
 Relative molecular mass or molecular
weight has no units.
 The molecular weight of an element or
compound indicates the number of times
of
mass of C- 12 isotopes atom. For
example, the atomic weight of calcium is
40. This means that an atom of calcium
is on average is 40 times the mass of 1/
12 the mass of C- 12 isotope’s atom.
 Atomic weights of many elements are not
whole numbers due to the presence of
stable isotopes.
 The number of atoms of a particular
isotope present in 100 atoms of a natural
sample of that element is called its relative
abundance which always remains
constant for a given element.
 Natural chlorine is a mixture of two
isotopes with relative abundances 75%
(Cl-35) and 25% (Cl-37) approximately.
Then, the atomic weight of chlorine is

a molecule is heavier than
 RMM =
 Atomic weight has no units.
 The relative atomic mass of an element
weight:
a molecule is heavier than
1 th
12
the mass
of C-12 isotope’s atom. For example, the
molecular weight of calcium carbonate
is 100, it implies that mass of one
molecule of calcium carbonate is 100
times heavier than
1 th
12
the mass of C-12
isotope’s atom.
 If
the relative molecular mass or
molecular weight of any compound is M,
then its molecular mass is ‘M’ a.m.u.
= Molecular
weight 
 Steps
1 th
12
the mass of C  12 atom .
to calculate the molecular
weight:
Write the formula of the compound or
the molecule.
2. Identify the different types of elements
present in it and write their symbols
along with the number of atoms.
3. Now multiply the number of atoms with
the atomic weights of the respective
elements
4. Finally add them to get molecular weight.
1.
74
CLASS-VIII
DAY-3: WORKSHEET
1. 1 amu is equal to the mass of
a)
1
th of C - 12 atom
12
1
th of O-16 atom
14
c) 1g of H2
d) 1.66 × 10–23 kg
2. 1 atomic mass unit =
b)
1 th
mass of a carbon - 12 atom
12
b) 1.66 × 10–24g
c) 6.023 × 10–23g
d) 6.023 × 1023g
3. The ratio of weight of one atom of an
element to its atomic weight is equal to
a)
a) 1 amu
b) mass of
4.
5.
6.
7.
1
th of C – 12 isotopic atom
12
c) 12 amu
d) None
The mass of one atom of an element is
40 × 1.66 × 10–24g. The number of protons
in its nucleus is
a) 40
b) 20
c) 10
d) 5
The weight of Helium atom in grams is
a) 2
b) 4
c) 6.64 × 10–24
–24
d) 1.66 × 10
The symbol of carbon is C. It means that
a) ‘C’ represents one atom of carbon.
b) ‘C’ also represents 1 mole of carbon
atoms.
c) ‘C’ also represents 12g of C.
d) All
Atomic weight of an element is x. It
means that weight of one atom of that
element is
1
xg
12
d) 1.66x × 10–24g
a) ‘x’ g
b)
c) 12 × x g
MPC BRIDGE COURSE
8. The mass of an atom of an element ‘x’ is
39. The number of atoms of it present in
gram atomic weight of it is_______.
a) 1
b) 1.66 × 1024
c) 6.023 × 1023
d) 96500
9. The total mass of 100 atoms of silicon is
a) 2800
b) 2800 amu
–22
c) 28 × 1.66 × 10 gd) 280 kg
10. The approximate number of electrons
that are required to make 1 smallest unit
of mass is
a) 6.023 × 1023
b) 1.66 × 1024
c) 1852
d) 2500
11.Match the following:
Column - I
Column - II
i) Sodium
p) Monoatomic
ii) Helium
q) Diatomic
iii) Oxygen
r) Triatomic
iv) Ozone
s) Poly atomic
v) Sulphur
a) i, ii - p; iii - q; iv - r; v - s
b) iii, ii - p; i - q; iv - r; v - s
c) i,iii - p; iv - q; ii - r; v - s
d) i,iv - p; iii - q; ii - r; v - s
12.Statement A: A
chemical
formula
represents the composition of a molecule
of the substance in terms of the symbols
of the elements present in a molecule.
Statement B: The chemical formula of
methane is CH 4.
a) ‘A’ is true, ‘B’ is false
b) ‘A’ is false, ‘B’ is true
c) Both ‘A’ and ‘B’ are true
d) Both ‘A’ and ‘B’ are false
13. The units of molecular mass (or)
molecular weight is
a) amu
b) grams
c) Both ‘a’ and ‘b’
d) None
14.
weightof oneatomof anelement
= x.
Itsatomicweight
weightof onemoleculeof acompound
= y.
Itsmolecular weight
Then, x : y is
a) 1 :
NARAYANA GROUP OF SCHOOLS
1
12
b) 2 : 1
c) 1 : 2
d) 1 : 1
75
CLASS-VIII
MPC BRIDGE COURSE
15.Statement A: The number of atoms
present in gram atomic weight of different
elements are equal.
Statement B: The number of molecules
present in gram molecular weight of
different substances is equal.
a) ‘A’ is true, ‘B’ is false
b) ‘A’ is false, ‘B’ is true
c) Both ‘A’ and ‘B’ are true
d) Both ‘A’ and ‘B’ are false
DAY-4 : SYNOPSIS
 Gram
Atomic Weight (GAW):
(a) Atomic weight of an element
expressed in grams is known as its gram
atomic weight. For example, the atomic
weight of hydrogen is 1.008. So, the gramatomic weight of hydrogen is 1.008 g.
(b)Gram atomic weight of any substance
is also called its gram atom. For example,
1 gram atom of carbon weighs 12 gram
and 1 gram atom of nitrogen weighs 14
grams.
(c)Number of
gram atoms
=
Given weight
Gram atomic weight .
For example, the number of gram
atoms in 5 g of hydrogen =5/1 = 5.
(d) Weight of x gram atoms = x  Gram
atomic weight.
(e) 1 gram atom or gram atomic weight
of an element contain = 6.023  10 23
atoms.
(f) Number of atoms in a given substance
( given element) = Number of gram atoms
 6.023  1023.
(g)Number of atoms in 1 gram of an
element =

6.023×1023
.
Atomic weight
Gram Molecular Weight (GMW):
(a) It is the molecular weight of an
element or compound expressed in
grams. For example, the molecular
weight of hydrogen gas is 2. So, the gram
molecular weight of hydrogen is 2 g.
NARAYANA GROUP OF SCHOOLS
(b)Gram molecular weight of a substance
is also called its gram molecule or mole
molecule. For example, the weight of 1
gram molecule or mole molecule of H2O
is 18 grams and the weight of 1 gram
molecule of N2O is 44 grams.
(c)Number of moles =
Given weight
.
Gram Molecular weight
(d) Weight of x moles of any compound
= x  Gram molecular weight.
(e) Number of molecules in a given
substance= Number of gram molecules
 6.023  1023.
(f) Weight of substance in grams =
Number of gram molecules  GMW.

Note:
a) Gram atomic mass of an element and
Molar mass of an element are just the
same.
(b)Gram molecular weight of a substance
and Molar mass of a substance are also
just the same.

Mole: This is unit used to express the
quantity of matter in chemistry.
(a) It is defined as “the amount of a
substance which contains the same number
of chemical units (atoms, molecules or ions)
as there are atoms in exactly 12 grams of pure
carbon”.
(b)12 g of carbon-12 is found to contain
6.023 × 1023 atoms of carbon-12. Thus, a
mole represents a collection of 6.023 ×
1023 chemical units (atoms, molecules or
ions).
(c)The number 6.023 × 1023 is called the
Avogadro’s number. The Avogadro’s
number is denoted by N A or L. Most
commonly the symbol NA is used. Thus, a
mole represents the quantity of material
which contains one Avogadro’s number
(6.023 × 1023) of chemical units (atoms,
molecules, or ions) of any substance.
(d) It is important to note that while
using the unit mole, it is necessary to
specify the chemical unit also. For
example,
76
CLASS-VIII
MPC BRIDGE COURSE
1 mole of hydrogen atoms =
1 mole of hydrogen molecules =
1 mole of carbon dioxide =
1 mole of electrons
=
6.023 × 1023 atoms of hydrogen
6.023 × 1023 molecules of hydrogen
6.023 × 1023 molecules of carbon dioxide
6.023 × 1023 electrons
1 mole of sodium ions (Na+) =
6.023 × 1023 Na+ ions

Symbol of the mole unit. The unit of
mole is given a symbol mol. So, if you
want to express one mole, you may write
it as 1 mol.
DAY-4: WORKSHEET
1. Gram atom of any element contains
a) 6.023 × 1023 atoms
b) 3.0115 × 1023 atoms
c) 1.505 × 1023 atoms
d) 12.0 × 1023 atoms
2. Which of the following is correct?
a) Molecular mass of oxygen is 32.
b) Gram molecular mass of sulphur (S8)
is 252 g.
c) The weight of one molecule of O3 is 48
amu.
d) All
3. The ratio of number of molecules present
in 1 gram mole of O2 to one gram mole of
SO2 is
a) 1 : 4
b) 1 : 2
c) 1 : 8
d) 1 : 1
4. The ratio of weights of hydrogen and helium is 1 : 2. Find the ratio of number of
gram atoms.
a) 2 : 1
b) 1 : 1
c) 1 : 4
d) 4 : 1
5. Among all the naturally occuring elements, which one can generate the maximum number of gram atoms from a given
amount?
a) Hydrogen
b) Uranium
c) Calcium
d) Mercury
6. How many gram atoms of the lightest element weigh same as 1 gram atom of the
heaviest element?
a) 1
b) 235
c) 238
d) 100
7. Among all the naturally occuring elements, one gram atom of which element
contains the maximum amount of it?
a) Hydrogen
b) Uranium
c) Calcium
d) Mercury
8. Identify the element whose 2 gram atoms weigh 8g.
a) Hydrogen b) Helium c) Oxygen
d) Sulphur
NARAYANA GROUP OF SCHOOLS
9. Find the number of gram molecules
present in the following:
i) 5g of Neonii) 7 g of nitrogen
(i)
(ii)
a) 0.25
0.25
b) 0.25
0.5
c) 0.5
0.25
d) 1
2
10. 1 x = 6.023 × 10 23 molecules = gram
molecular mass of a substance. Then, x
= _____
a) gram atom
b) mole
c) molecular weight d) gram
11. Statement A: Avogadro number
1CGS unit of mass
= 1smallest unit of mass
Statement B: The number of atoms
present in 1 gram-atom of hydrogen = the
number of molecules present in 1 gram
mole of hydrogen.
a) ‘A’ is true, ‘B’ is false
b) ‘A’ is false, ‘B’ is true
c) Both ‘A’ and ‘B’ are true
d) Both ‘A’ and ‘B’ are false
12. Which one of the following statements
is incorrect?
a) 1 gram atom of carbon contains
Avogadro number of atoms.
b) 1 mole of oxygen gas contains Avogadro
number of molecule.
c) 1 mole of hydrogen gas contains
Avogadro number of atoms.
d) 1 mole of electrons stands for 6.023 ×
10 23 electrons.
13. The weight of 1/4 mole of calcium atom=
a) 40 grams
b) 20 grams
c) 10 grams
d) 5 grams
14. The weight of 1/4 mole of atom of an
element is 5 grams. Identify the
substance
a) Boron
b) Neon
c) Phosphorus
d) Calcium
15. The weight of 1 molecule of a
substance is 98 amu. The substance is
a) Calcium carbonate b) Sulphuric acid
c) Sodium chloride
d) Phosphoric acid
77
CLASS-VIII
MPC BRIDGE COURSE
 Weight of x gram atoms = x  Gram atomic
DAY-5 : SYNOPSIS

Important relations related to mole:
(a) 1 mole of particles = 6.023  10 23
particles ( atoms/ molecules/ions/
electrons/protons/neutrons/nucleons).
(b) The weight of 1 mole atoms of an
element = gram atomic weight of the
element.
(c) The weight of 6.023  1023 atoms of
an element = gram atomic weight of the
element.
(d) The weight of 1 mole molecules of a
compound = gram molecular weight of a
compound.
(e) The weight of 6.023  1023 molecules
of a compound = gram molecular weight
of the compound.
(f) The weight of 1 mole of formula units
of a salt = gram formula weight of the
salt.
UNDERSTANDING OF A MOLE
1 mole of Weighs
Hydrogen
gas
2 grams
Contain 6.023× 1023
Hydrogen
molecules
2× 6.023× 1023
electrons
Contain
2× 6.023× 1023
protons
2× 6.023× 1023
Hydrogen
atoms
0
neutrons
2× 6.023× 1023
nucleons
Contain
Weighs
2× 7 × 6.023×
1023
neutrons
2× 6.023×
nitrogen
atoms
1023
6.023×
oxygen
atoms
2× 14 × 6.023×
1023
nucleons
Contain
1023
6.023  1023
= Atomic weight
 Number of molecule in 1 gram of a
6.023 1023
substance =
Molecular weight
 Weight of an element in grams = Number
of gram atoms  GAW
 Weight of substance in grams = Number
of moles  GMW
 Number of atoms of an element per
molecule can be calculated if MW and
percentage mass of that element are
given by using the formula.
No.of atoms
Contain
2× 7 × 6.023×
1023
protons
6.023× 1023
N2O
molecules
Contain
Contain
2×7× 6.023×
1023
electrons
44 grams
Contain
1 mole of
N2O
weight
 Weight of x moles of any compound =
x  Gram molecular weight
 1 gram atom or gram atomic weight of an
element contains 6.023  1023 atoms.
 1 gram molecule or gram molecular
weight of a substance contains
6.023  1023 molecules.
 Number of atoms in a given substance (
given element) = Number of gram atoms
(ng)  6.023  1023
 Number of molecules in a given substance
( Nm) = Number of moles (n)  6.023  1023
 Number of atoms in 1 gram of an element
8× 6.023× 1023
electrons
8× 6.023× 1023
protons
8× 6.023×
1023
neutrons
16× 6.023× 1023
nucleons
 DAY-6 Some more important relations:
 No of gram atoms or mole atoms =
Given weight
Gram atomic weight .
 Number of moles (n)
Given weight
= Gram Molecular weight
NARAYANA GROUP OF SCHOOLS
MW Percentagemass
At. wt100
[ No te:
Number of atoms is always is a whole
number]
 No. of atoms present in given amount of
substance (Na)
= No. of molecules (N m) × No. of atoms
present in 1 molecule of the substance.
= No. of moles (n) × NA × No. of atoms
present in 1 molecule of the substance.
 No. of subatomic particles (electrons /
protons/ neutrons/ nucleons, etc)
present in given amount of substance (Np)
= No. of molecules (Nm) × No. of subatomic
particles present in 1 molecule of the
substance. = No. of moles (n) × NA × No.
of subatomic particles present in 1
molecule of the substance.
78
CLASS-VIII
DAY-5: WORKSHEET
1. The number of atoms in 8g of Sulphur is
a) 6.02 × 1023
b) 3. 01 × 1023
c) 12.04 × 1024
d) 1.505 × 1023
2. 12 g of Carbon contains equal number of
atoms as
a) 12 grams of Mg
b) 40 grams of Calcium
c) 32 grams of Oxygen
d) 7 grams of nitrogen
3. 6.02 × 1022 particles present in 32 g of
oxygen is
a) 0.1 mole
b) 1 mole
c) 10 moles
d) 100 moles
4. Number of molecules present in 32g of
oxygen is
a) 3.2 × 1010
b) 6.02 × 1023
c) 3.2 × 1023
d) 6.02 × 1010
5. Which of the following has more number
of molecules?
a) 1 g O2 b) 1 g N2 c) 1 g F2 d) 1 g CO2
6. Which of the following has more number
of atoms?
a) 1 g Ca b) 1 g C c) 1 g Cu d) 1 g Cl2
7. Which of the following pairs of gases
contain equal number of particles?
a) 1 g He, 1 g H2
b) 1 g He, 2 g H2
c) 4 g He, 2 g H2
d) 4 g He, 4 g H2
8. If equal mass of N2 and O2 are taken, the
ratio of number of molecules in these
gases would be
a) 1 : 1
b) 7 : 8
c) 8 : 7
d) 28 : 32
9. Which of the following contains largest
number of atoms?
a) 4 g of H2
b) 1 g of O2
c) 28 g of N2
d) 18 g of H2O
10. Which one of the following pairs of
gases contain the same number of
molecules?
a) 16 g of O2 and 14 g of N2
b) 8 g of O2 and 22 g of CO2
c) 28 g of N2 and 22 g of CO2
d) 32 g of O2 and 32 g of N2
11. How much amount of oxygen in (grams)
is present in 32.2 g Na2SO4.10H2O?
a) 20.8
b) 22.4
c) 2.24
d) 2.08
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
12. The number of oxygen atoms in 4.4 g
of CO2 is approximately is
a) 1.2 × 1023
b) 6 × 1022
23
c) 6 × 10
d) 12 × 1023
13. The number of water molecules
present in a drop of water (volume
0.0018ml) at room temperature is
a) 6.023 × 1019
b) 1.084 × 1018
c) 4.84 × 1017
d) 6.023 × 1023
14. 19.7 kg of gold was recovered from a
smuggler. How many atoms of gold were
recovered (At. wt of gold = 197)?
a) 100
b) 6.02 × 1023
c) 6.02 × 1024
d) 6.02 × 1025
DAY-6 : SYNOPSIS
1. Necessity for classification of elements:
As a large number of elements were
discovered, it was realised that it was
not possible to study all the elements and
their compounds by a chemist, unless
they are classified. Following are the
reasons for the classification of elements.
(a) The classification may help to study
them better.
(b)The classification may lead to correlate
the properties of the elements with some
fundamental
property
that
is
characteristic of all the elements.
(c)The classification may further reveal
relationship between the valency.
(iii) Doberiner’s Triads:
(a) This was given by Johann Doberiner,
a German chemist.
(b)He classified elements into sets of
three chemically similar elements, called
triads.
(c)It states that, if the elements of triad
are arranged in the increasing order of
their atomic weights, then the atomic
weight of the middle element is
approximately equal to the average of the
atomic weights of the other two.
This is known as Doberiner’s law of
triads.
79
CLASS-VIII
(d)
MPC BRIDGE COURSE
(b) When the noble gases were
discovered, the idea of octaves could not
be held. For example, with the discovery
of neon (Ne) between F and Na, and argon
(Ar) between Cl and K, it becomes the
ninth element and not the eighth, which
has similar properties.
(v) Lother Meyer’s classification:
(a) This was given by Lothar Meyer, a
German chemist in 1869.
(b)He plotted a graph of atomic volume
(atomic mass/density) versus atomic
mass for various elements. He noticed
that the elements with similar properties
occupied similar positions on the curve.
Examples:
A to m ic
M ea n*
m a ss
Li
6 .9 4
2 3 .0 2
Na
2 2 .9 9
K
3 9 .1 0
Ca
4 0 .0 8
8 7 .2 1
Sr
8 7 .6 2
Ba
1 3 4 .3 4
S
3 2 .0 6
7 9 .8 3
Se
7 8 .9 6
Te
1 2 7 .6 0
Fe
5 5 .8 5
5 7 .2 8
Co
5 8 .9 3
Ni
5 8 .7 1
T A B L for
E 4 .1rejection
. P ro p e rtie s of
o f eDoberiner
le m e n ts in s o m e tr ia d s
• Reason
T r ia d s
Triads:
(a) The
Doberiner’s
method
of
classification could arrange only a limited
number of elements out of those known
at that time in the form of triads.
Therefore, the idea of triads could not be
appliedto all the elements then known.
(b)Elements with dissimilar properties
can also be arranged in the form of triads.
This is against the rule of classification.
(iv) Newland’s law of octaves:
(a) This was given by John Newland’s,
an English chemist and a musician in
1864
(b)It states that, when the elements are
arranged in the order of their increasing
at om ic weigh t s, t h en the properties of the
elements were repeated at every eighth
element like the eighth note of an octave in
music.
Example:
Thus, the properties of sodium (Na) and
potassium (K) are similar to those of
lithium (Li). Similarly, chlorine (Cl)
resembles fluorine (F).
• Reason for rejection of Newland’s law
of octaves:
(a)Newland’s classification failed badly
while dealing with the heavier elements
beyond calcium (Ca).
NARAYANA GROUP OF SCHOOLS
•
•
•
•
(c)For example,
The most electropositive elements like
Lithium (Li), Sodium (Na), potassium (K),
Rubidium (Rb), Cesium (Cs), etc,. occupy
the peak positions.
The moderate electropositive elements
like magnesium (Mg), calcium (Ca),
strontium (Sr) and barium (Ba) are placed
on the descending curve.
The most electronegative elements like
fluorine (F), chlorine (Cl), bromine (Br),
Iodine (I) are placed on the ascending
curve.
Thus, Lothar Meyer obser ved a
periodicity in the properties of the
elements with atomic mass.
(vi)Mendeleev’s periodic classification.
By the year 1869, Dmirti Invanovich
Mendeleev studied the physical and
chemical properties of all 63 elements
known at that time.
Mendeleev’s periodic law: The physical
and chemical properties of the elements
are the periodic functions of their atomic
weights.
80
CLASS-VIII
This means that, if the elements are
arranged in the increasing order of their
atomic weights, then elements with
similar properties get repeated at regular
intervals.
• Mendeleev’s periodic table: It is a
tabular chart, representing systematic
arrangement of elements in vertical
columns, called groups and horizontal
rows, called periods, in the order of their
increasing atomic weights.
• Main featur es of Mendele ev’s
periodic table
(i) In Mendeleev’s table, the elements
were arranged in vertical columns,
called groups.
(ii) There were in all eight groups:
Group I to VIII. The group numbers were
indicated by Roman numerals. i.e., I, II,
III, IV, V, VI, VII & VIII.
(iii) Except VIII, every group is further
divided into subgroups i.e., A and B.
Groups VIII occupy three triads of three
elements each, i.e., in all nine elements
(iv) The properties of the elements in
same group or subgroup are similar.
(v) There is no resemblance in the
elements of subgroups A and B of same
group except valency.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
(vi) The horizontal rows of the periodic
table are known as periods.
(vii) There
were
seven
pe riods,
represented by Arabic numerals 1 to 7.
To accommodate more elements, the
periods 4, 5, 6 and 7 were divided into
two halves. The first half of the elements
are placed in the upper left corner and
second half in the lower right corner.
For example, the elements occupying the
box corresponding to group I and period 4
are potassium (K) and copper (Cu), K is
written in the top left corner, while Cu
is written in the lower right corner.
(viii) A period comprises the entire range
of elements after which the properties
repeat themselves.
(ix) In a period, the properties of the
elements gradually change from metallic
to nonmetallic while moving from left to
right.
(x)There were gaps left in the periodic
table. Mendeleev left these gaps
knowingly, as these elements were not
discovered at that time.
• Merits of the Mendeleev’s Periodic
Table:
Mendeleev’s classification was considered
superior to the others proposed earlier
because of the following reasons:
(i) It is based on the more fundamental
property of atomic weight of an element.
Thus it is better than earlier
classification.
(ii) It helped in systematic study of the
elements. Mendeleev’s classification
condensed the study of about 90
elements (only 65 were known at that
time, but he left a provision for many
more) to the study of only 8 groups of
elements.
(iii) Some gaps were left knowing my
Mendeleev for undiscovered elements.
This accelerated the process of
discovering these elements, as their
properties were predicted by Mendeleev
on the basis of other elements present
in the same group.
81
CLASS-VIII
Property
.
Predicted properties of eka silicon
(1871)
72
5.5
High
MPC BRIDGE COURSE
Winkler's report of
germanium (1886)
72.6
5.36
1231
Atomic mass
Density
Melting
point, K
Action of
Likely to be slightly attacked
Not attacked by HCl,
acid
Reacts with hot HNO 3
action of
Likely not to react
No action with dil. NaOH
alkali
Oxide
MO2 (4.7)
GeO 2 (4.7)
Sulphide
MS2
GeS 2
Chloride
MCl4 (1.9)
GeCl4 (1.88)
Mendeleev's predictions for eka silicon and properties of germanium
(iv) By placing elements strictly
according to the similarity in their
properties, he was also able to correct
certain atomic weights .
• For example, he corrected the atomic
masses of beryllium (Be), gold (Au) and
platinum (Pt).
DAY-6: WORKSHEET
1. G, O, D are the correct symbols of right
elements of the periodic table arranged
in the increasing order of their atomic
weights. The atomic weight of ‘G’ is 40
and that of D is 137. If G, O, D are the
elements of a Dobereiner triad, then find
the atomic weight of ‘O’.
a) 88.5
b) 120.5
c) 99.5
d) 77.5
2. Which of the following is not a Dobereiner
triad?
a) Cl, Br, I
b) Ca, Sr, Ba
c) Li, Na, K
d) Fe, Co, Ni
3. Statement A : Classification of elements
is not useful to reveal the relationship
between the different elements.
Statement B : Classification of elements
may lead to correlate the properties of
the elements with some fundamental
property that is characteristic of all the
elements.
a) Statement ‘A’ is correct but ‘B’ is
incorrect.
b) Statement ‘B’ is correct but ‘A’ is
incorrect.
c) Statements ‘A’ and ‘B’ are incorrect.
d) Both ‘A’ and ‘B’ statements are correct.
4. (i) Elements with both metallic and nonmetallic characters are called_____.
(ii) Arrangement of elements into groups
of three is called ________.
NARAYANA GROUP OF SCHOOLS
(i)
(ii)
a) Active metals
Octaves
b) Metallic elements Metals
c) Triads
Metals
d) Metalloids
Triads
5. Select the following pair of elements in
which their arithmetic mean of atomic
weights is equal to the atomic weight of
strontium.
a) Lithium, Barium b) Sodium, Calcium
c) Calcium, Barium d) Sodium, Barium
6. Which of the following is wrong triad?
a) Chlorine, bromine, iodine
b) Lithium, sodium, potassium
c) Carbon, nitrogen, oxygen
d) Calcium, strontium, barium
7. Which of the following is an achievement
of the triads classification?
a) Relation between all properties of an
element.
b) Relation between only atomic weights
of an element.
c) Relation between the properties of
same elements.
d) Relation between the atomic mass of
all elements.
8. Dobereiner’s law is rejected due to
Statement A : Quite large number of
elements cannot be grouped into tirads.
Statement B : It was possible to group quite
dissimilar elements into triads.
a) Statement ‘A’ is correct but ‘B’ is
incorrect
b) Statement ‘B’ is correct but ‘A’ is
incorrect
c) Statement ‘A’ and ‘B’ are incorrect
d) Both ‘A’ and ‘B’ statements are correct
9. The _______ was the basis of the
classification proposed by Dobereiner,
Newlands and Mendeleev.
a) Atomic number b) Atomic weights
c) Atomic mass
d) None
10. The discovery of which of the following
group of elements gave a death blow to
the Newlands law of octaves.
a) Inert gases
b) Alkaline earths
c) Rare earths
d) Actinides
82
CLASS-VIII
11. The most significant contribution
towards the development of periodic table
was made by
a) Mendeleev
b) Avogadro
c) Dalton
d) Cavendish
12. The number of elements known at that
time when Mendeleev arranged them in
the periodic table was
a) 63
b) 60
c) 71
d) 65
13.The horizontal rows and vertical columns
of a periodic table are called ____and ___
respectively.
a) Groups, periods b) Periods, groups
c) Blocks, partitionsd)Sections,segments
14. Mendeleev’s periodic law is based on
a) Atomic number
b) Atomic weight
c) Equivalent weight d) Valency
15. Which one of the following is incorrect
statement in respect of Mendeleev’s
periodic table?
a) It has made the study of elements
easier and systematic.
b) It has helped in correcting the doubtful
atomic-weights.
c) It has paved the way for the discovery
of new elements.
d) All the above statements are correct.
DAY-7 : SYNOPSIS
Defects (Limitations) of Mendeleev’s
Periodic table
 Anoma lous pai rs : In Mendeleev’s
periodic table, the elements are arranged
on the basis of their atomic weights.
However, there are few such pairs in
which atomic weights of preceeding
elements is higher than that of the
following elements.
Preceeding Elements Following Elements
Argon (40)
Potassium (39)
Cobalt (58.9)
Nickel (58.6)
Tellurium (128.0)
Iodine (127.0)
The above pairs go against Mendeleev’s
periodic law.
 Position of hydrogen: Hydrogen is not
given a definite position. It is placed in
group ‘Ib’ and group ‘VIIb’ of Mendeleev’s
original periodic table.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
 Position of rare earth elements and
actinides : The position of rare earth
elements and actinides cannot be
justified on the basis of atomic weight.
 Positi on of isot opes: Mendeleev’s
periodic table is silent about isotopes. The
position of various isotopes of the
elements cannot be justified on the basis
of atomic weight.
 Posit ion of t ransitio n elemen ts :
Mendeleev’s concept of transition
elements was defective. He regarded
elements of group VIII as transition
elements.
 Over looking chemical similarities : In
Mendeleev’s periodic table, there are
certain relationships which are
excessive.
Examples :
 There was hardly any relationship
between alkali metals and copper, silver
and gold in group I.
 There was hardly any relationship
between fluorine and manganese in
group ‘VIIa’.
 Some obvious similarities between copper
and nickel, platinum and gold were
overlooked.
Modern Periodic Table - Long form of
Periodic Table:
 In 1913, H.G. J. Moseley showed by Xray analysis that the atomic number is more
fundamental property of an element than its
atomic weight. Therefore, he slightly
modified Mendeleev’s periodic law and
replaced the word atomic weight by atomic
number.
 Modern Periodic Law : It states that
physical and chemical properties of all
elements are periodic function of their
atomic numbers.
 On the basis of above law, Moseley
prepared long form of periodic table which
consists of 7 periods and 18 groups.
83
CLASS-VIII











Description of long form of (Extended
form) of Periodic Table
In long form of the periodic table, the
elements are arranged in the order of
increasing atomic numbers in horizontal
rows called periods, such that all
elements having same number of valence
electrons come under the same vertical
column called group. This not only ensures
periodicity in electronic configuration, but
periodicity in chemical properties also.
Characteristics of long form of periodic
table :
The subgroups ‘A’ and ‘B’ are separated in
this table.
In a group elements, the electrons in
outermost shell participates in chemical
reactions, whereas in B group elements,
the electrons from outermost and inner
shells participates.
The
transition
elements
are
accommodated in the middle of the table
in three series.
The strongly metallic elements (alkali metals
and alkaline earth metals) occupy groups
IA and ‘IIA’ respectively on the left hand of
transition elements.
The non-metallic elements are placed on the
right hand of transition elements.
The rare gases (noble gases) are placed
in zero group at the end (last column) of
periodic table.
The elements occupying left and right
wing vertical columns (groups) are called
normal representative elements.
The rare earths (Lanthanides) and
Actinides are called inner transition
elements. They are kept outside the periodic
table to mark their peculiar properties.
The horizontal rows in the periodic table
are called periods. There are seven
periods in all, such that each period has
consecutive (or continuous) atomic
number.
The number of elements in a period
corresponds to maximum number of
electrons which can be accommodated in
its one shell.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
 The number of period to which an element
belongs is given by its quantum number(n),
i.e., the number of outermost shell as
counted from nucleus.
 The vertical columns in periodic table are
called groups.
 There are 18 groups in long form of
periodic table. The elements in a group
do not have consecutive atomic numbers.
However, each element in a group has
same number of electrons in its outermost
shell and hence, all elements of a group
have same chemical properties.
 The elements in zero group are called
rare gases or noble gases. All the
elements in this group (with exception of
Helium which has 2 electrons) have eight
electrons in their outermost shell.
 The elements in the group IA are called
alkali metals.
 The elements in group IIA are called
alkaline earth metals.
DAY-7: WORKSHEET
1. The elements in group VII A are
a) Rare earth elements b) Alkali metals
c) Transition elements d) Actinides
2. What are the indefinite positions of
hydrogen given in Mendeleev’s periodic
table?
a) 1 b, III b
b) I a, II b
c) I b, VII b
d) VII a, III b
3. An element ‘E’ has atomic number 14. To
which period this element belongs? How
many maximum number of elements are
present in the period to which element
‘E’ belongs?
a) 1st period and 6 elements.
b) 3rd period and 8 elements.
c) 4th period and 8 elements.
d) 5th period and 13 elements.
4. Find the period number and the group
num ber in which the element with
atomic number 24 is present.
a) 2, VB
b) 4, VIB
c) 5, VIIB
d) 3, VB
84
CLASS-VIII
5. Three elements A, B and C have atomic
number Z, Z + 2 and Z + 3 respectively.
Among these, ‘C’ is an alkali metal. To
which groups of the periodic table do the
elements ‘A’ and ‘B’ belong respectively?
a) 16, 18
b) 14, 16
c) 15, 17
d) 12, 14
6. Match the following:
Column - I
Column - II
1) Pnicogens
p) Fluorine family
2) Chalcogens
q) Nitrogen family
3) Halogens
r) Helium family
4) Aerogens
s) Oxygen family
a) 1  p, 2  q, 3  r, 4  s
b) 1  q, 2  r, 3  p, 4  s
c) 1  q, 2  s, 3  p, 4  r
d) 1  r, 2  p, 3  q, 4  s
7. How many elements are present in, (i)
Fifth period (ii) Sixth period
a) (i) 16 (ii) 32
b) (i) 18 (ii) 32
c) (i) 14 (ii) 28
d) (i) 14 (ii) 32
8. Which of the following pair is against to
Mendeleev’s periodic law?
a) Chromium, Manganese
b) Sodium, Magnesium
c) Copper, Zinc
d) Tellurium, Iodine
9. Which of the following statements is
incorrect?
a) The elements of subgroups ‘A’ are all
normal elements.
b) The elements of subgroups ‘B’ are all
transition elements.
c) The elements on the left of the periodic
table are metals, on the right are
non-metals and in the middle are
metalloids.
d) d-block elements are also called
transition elements.
10. The 3 rd period of the periodic table
contains:
a) 8 elements
b) 32 elements
c) 18 elements
d) 19 elements
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
DAY-8 : SYNOPSIS
 Periodic Properties: Properties which are
directly or indirectly related to the electronic
configuration of the elements and show a
regular gradation when we move from left to
right in a period or from top to bottom in a
group are called periodic properties. Some
important periodic properties are atomic
size, ionization energy, electron affinity,
electronegativity, valency, density, atomic
volume, melting and boiling points etc.
ATOMIC SIZE
 Atomic size: It refers to the distance
between the centre of the nucleus of the
atom to the outermost shell containing
electrons.
 Since, absolute value of the atomic size
cannot be determined, it is usually
expressed in terms of the following
operational definitions.
Units: Atomic size is expressed in terms
of angstrom (1A° = 10–10 m).
In a period, on moving from left to right
in a period, the size of the elements
decreases due to a gradual increase in
the nuclear pull.
In a group atomic size increases from top
to bottom due to increase in the number
of shells.
 Ionic size: An atom can be changed to a
cation by loss of electrons and to an anion
by gain of electrons. A cation is always
smaller than the parent atom because, during
its formation effective nuclear charge increases
and sometimes a shell may also decrease. On
the other hand, the size of an anion is
always larger than the parent atom because,
during its formation effective nuclear charge
decreases.
 Isoelectronic ions or species are the
neutral atoms, cations or anions of
different elements which have the same
number of electrons but different nuclear
charge.
 The size of the isoelectronic species
depends upon their nuclear charge.
Greater the nuclear charge, smaller the size.
85
CLASS-VIII
IONISATION ENERGY (IE)
 The energy required to remove the
outermost electron from an isolated
gaseous atom of the element is called
Ionisation energy.
A  g   IE  A +  g  + e  .
 Ionisation energy is expressed in eV/
atom or kJ/mole
1 eV/atom = 96.45 kJ/mole.
 Ionisation energy depends on the
following factors :
(i) Size of the atom: Greater the size of
atom, smaller is the IE.
(ii) Nuclear charge: Greater the
nuclear charge, greater is the IE.
(iii) Screening effect: Greater the
screening effect of the inner electrons,
smaller is the IE.
(iv) Penetration effect: Greater the
penetrating effect of the outermost
occupied orbital, greater is the IE. For a
particular energy level, the penetration
effect is in the order : s > p > d > f.
(v) Electronic co nfiguration: If the
electronic configuration of the atom is
stable, it would have relatively higher IE.
 Ionisation energy in general increases
on moving along the period and decreases
on going down the group.
 Be, Mg, N, P and noble gases have
relatively higher values of IE due to their
stable electronic configuration.
 Alkali metals have the least and noble
gases have the highest ionisation
energies in their respective periods.
 Helium (He) has the highest IE among
all the elements.
 Cesium (Cs) has the least IE among all
the elements (except Fr which is
radioactive).
 The amount of energy required to remove
the outermost electron from an isolated
gaseous atom of the element is called
first ionisation energy (IE1).
 The amount of energy required to remove
the outermost electron from an uni
positive ion of the element is called
second ionisation energy (IE2).
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
 Successive ionisation energies are always
greater than the first ionisation energy.
IE3 > IE2 > IE1
 As we move from top to bottom in the
periodic table, the atomic size increases
and the nuclear pull decreases. The
weaker nuclear pull results in the outer
electron being held loosely, thereby
requiring less energy to remove the outer
electron and hence, ionisation energy
decreases.Thus, down the group the
ionisation energy decreases due to
increase in size.
 While comparing IE 2 , consider the
electronic configuration of A+ ion, keeping
in mind the stability of electronic
configuration with half filled and fully
filled subshells.
 Electronegativity can be defined as the
tendency of an atom in a molecule to attract
the shared pair of electrons towards itself.
 As we move from top to bottom in a
periodic table, the atomic size increases
and the nuclear attraction decrease. This
decreases the electronegativity.
 As we move from left to right in a period
the atomic size decreases and the
nuclear attraction increases. This
increases the electronegativity.
 Fluorine is the most electronegative
element.
 In a period, the highest electronegativity
is of halogens and the lowest is of alkali
metals.
DAY-8: WORKSHEET
1. The atomic radii in case of inert gases
is:
a) Ionic radii
b) Covalent radii
c) vander Waals’ radii
d) None
2. The covalent and van der Waal’s radii of
hydrogen respectively are:
o
o
a) 0.37 A , 1.2 A
o
o
o
o
b) 0.37 A , 0.37 A
o
o
c) 1.2 A , 1.2 A
d) 1.2 A , 0.37 A
3. The size of the species, Pb, Pb 2+, Pb 4+
decreases as:
a) Pb4+ > Pb2+ > Pb b) Pb > Pb2+ > Pb4+
c) Pb > Pb4+ > Pb2+ d) Pb4+ > Pb > Pb2+
86
CLASS-VIII
MPC BRIDGE COURSE
4. Which of the following has largest radius?
b) Small atomic size.
a) 1s2, 2s2, 2p6, 3s2
c) Metallic properties.
b) 1s2, 2s2, 2p6, 3s2 3p1
d) Strongly bound valence electrons.
c) 1s2, 2s2, 2p6, 3s2 3p3
d) 1s2, 2s2, 2p6, 3s2 3p5
5. Which of the following combinations
contains only isoelectronic ions?
a) N3–, O2–, Cl–, Ne
b) F–, Ar, S2–, Cl–
c) P3–, S2–, Cr, Ar
d) N3–, F–, O2–, Ar
6. Arrange the elements S, P, As in the
order of their increasing ionization
energies.
a) S < P < As
b) P < S < As
c) As < S < P
d) As < P < S
7. The factor that is not affecting the
ionisation energy is
10. Arrange the following atoms in order
of increasing first ionisation energies.
K, Cs, Rb, Ca
a) Cs, Rb, K, Ca
b) Rb, Cs, K, Ca
c) Cs, Rb, Ca, K
d) Rb, Cs, Ca, K
11.Choose the correct answer.
a)
lonisation
energy
and
electronnegativity increases along a
period.
b) lonisation energy increases but
electronnegativity decreases along a
period.
a) Size of atom
c) Ionisation energy decreases but
electronnegativity increases.
b) Charge in the nucleus
d) Both decrease along a period.
c) Type of bonding in the crystalline
lattice
d) Type of electron involved
8. The graph of first ionisation enthalpy
versus atomic number is as follows:
12.Which of the following element has the
highest electronegativity ?
a) As
b) Sb
c) P
d) S
13.The chemical elements are arranged in
the order of increasing electro
negativities in the sequence:
a) Si, P, Se, Br, Cl, O
FIRST I.P.
b) Si, P, Br, Se, Cl, O
c) P, Si, Br, Se, Cl, O
d) Se, Si, P, Br, Cl, O
ATOMIC NUMBER
Which of the following statements is
correct ?
a) Alkali metals are at the maxima and
noble gases at the minima.
b) Noble gases are at the maxima and
alkali metals at the minima.
c) Transition elements are at the
maxima.
d) Minima and maxima do not show any
regular behaviour.
9. Atoms which have high first ionisation
energy have
14. The electronegativities of the following
elements: H, O, F, S and Cl increase in
the order:
a) H < O < F < S < Cl
b) Cl< H < O < F < S
c) H < S < O < Cl < F
d) H < S < Cl< O < F
15. Electronegativity values
elements help in predicting
for the
a) Polarity of bonds.
b) Dipole moments.
c) Valency of elements.
d) Position in the electrochemical series.
a) High nuclear charge.
NARAYANA GROUP OF SCHOOLS
87
CLASS-VIII
DAY-9 : SYNOPSIS
ELECTRO POSITIVITY, METALLIC/NONMETALLIC CHARACTER
 Metals have the tendency to form cations
by loss of electrons and this property
makes the elements as electropositive
elements or metals.
M  g   M  g   e
 The tendency of an element to lose
electrons is closely connected to the (IE)
of the element. The smaller the
Ionisation energy (IE) of an element,
the greater will be its tendency to lose
electrons and thus greater will be its
metallic character.
 Tendency to oxidise itself provides
reducing property to the elements thus,
smaller the (IE), greater the metallic
character hence, greater the reducing
nature : (IE) increases on moving along
a period left to right and decreases down
the group, hence metallic and reducing
nature decreases along the period and
increases down the group.
 The most reactive metals are on the left
of the periodic table, whereas the least
reactive metals are in the transition
metal groups closer to the right side of
the table.
 Variation of metallic character and nonmetallic character
In a group, as we move from top of bottom,
the size of atoms increases, resulting in
an increase in electropositive character.
 Thus, the metallic character increases
down the group.
 As we move from top to bottom, the size
of atoms increases resulting in the
decrease in ionisation energy. Thus, the
non-metallic character decreases down
the group.
 So, as we move down the group, the
metallic character increases and the
non-metallic character decreases.
 This is understood from the following
example.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
 In a period as we move from left to right,
the size of atom decreases, resulting in
a decrease in electropositivity.
 Thus, metallic character decreases as we
move from left to right in a period.
 As we move from left to right, the size of
atoms increases, resulting in an increase
in ionisation energy or electronegativity.
 Thus non-metallic character increases,
as we move from left to right in a period.
 Thus metallic character decreases and
non-metallic character increases from
left to right in a period.
DAY-11
ELECTRON AFFINITY (EA]
 The energy released, when an isolated
atom of the element in gaseous state
accepts an electron to form univalent
negative ion is called electron affinity.
 It is measured in eV/atom or kJ/mole.
X  g   e   X   g  + EA

It is also called electron gain enthalpy.
 The energy change in the process of
addition of an electron to monovalent
negative ion of the element in gaseous
state is called second electron affinity
(EA2).
X  g   e  X2  g  + EA2 .
 ‘EA 1 ’ is generally positive (energy is
released), whereas, ‘EA 2 ’ is always
negative
(energy is absorbed). In
other words, energy is generally released
when the electron is added to atom of
the element in gaseous state, whereas
energy is always required when an
electron is added to monovalent negative
ion of the element in gaseous state.
 Electron affinity is numerically equal to
ionisation energy but are opposite to each
other
 Electron affinity depends on the following
factors :
(i) Atomic size: Smaller the size of the
atom, greater is the EA.
(ii) Nuclear charge: Greater the
nuclear charge, greater is the EA.
88
CLASS-VIII
MPC BRIDGE COURSE
(iii) Elect ronic co nfigurat ion: If
electronic configuration of the element
is stable, its EA would be exceptionally
low.
 ‘EA’ (electron affinity) in general,
increases on moving across the period
and decreases on going down the group.
 Be, Mg, N and P have exceptionally low
values of ‘EA’ due to their stable electronic
configurations.
 Noble gases (He, Ne, Ar, Kr, Xe, Rn) have
negative values of ‘EA’ due to their stable
electronic configuration.
 Halogens have the highest ‘EA’ in their
respective periods.
 Chlorine has the highest ‘EA’ among all
the elements.
Some trends in the values of electron
affinities :
EA1 : Cl > F > Br > I
EA1 :
S > O > Se > Te
EA1 : C > B > Li > Be
EA1 :
Si > Al > Na > Mg
EA1 : F > O > N > Ne.
REDUCING, OXIDISING CHARACTERS
AND NATURE OF OXIDES
 Redox reactions are common for almost
every element in the periodic table
except for the noble gas elements of group
VIIIA. In general, metals act as reducing
agents, and reactive non-metals, such as
O and the halogens act as oxidising
agents,
Reducing Agents:
Electropositive elements can
lose electrons easily and hence,
can act as good reducing
agents.
The elements of IA and IIA
groups are electropositive and
are good reducing agents.
Down the group, the
electropositivity increases and
also the reducing character.
The Best Reducing AgentAn exception:
Lithium is the strongest
reducing agent due to high
hydration energy of Li+ ion.
Oxidising Agents:
Elements which can gain electrons
easily act as good oxidising agents.
The elements of VI A and VII A
groups gain electrons easily and
are good oxidising agents.
From left to right in a period, the
electron affinity and
electronegativity increases, and
also the tendency to gain electrons.
So the oxidising character
increases from left to right in a
table.
The Best Oxidising Agent:
Fluorine is the strongest
oxidising agent due to its highest
electronegativity.
Variation of acidic and basic character
Metals are characterised by basic
character and non-metal characterised
by acidic character. Down the group, the
NARAYANA GROUP OF SCHOOLS
metallic character increases and the
non-metallic character decreases. Thus,
the basic character increases and acidic
character decreases down the group.
From left to right in a period, the metallic
character decreases and the nonmetallic character increases. Thus, the
basic character decreases from left to
right, whereas, the acidic character
increases.
DAY-9: WORKSHEET
1. The electronic configurations of the
elements X, Y, Z and J are given below.
Which element has the highest metallic
character ?
a) X = 2, 8, 4
b) Y = 2, 8, 8
c) Z = 2, 8, 8, 1
d) J = 2, 8, 8,7
2. Which of the following is arranged in the
order of decreasing electropositive
character ?
a) Fe, Mg, Cu
b) Mg, Cu, Fe
c) Mg, Fe, Cu
d) Cu, Fe, Mg.
3. Which of the following sets of elements
has the strongest tendency to form
positive ions in gaseous state?
a) Li, Na, K
b) Be, Mg, Ca
c) F, Cl, Br
d) O, S, Se
4. Which of the following elements is most
electropositive?
a) C
b) N
c) Be
d) O
5. Out of C, Si, Ge, Sn, Pb metallic nature
is in
a) Ge, Sn, Pb
b) Sn, Pb
c) Ge, Pb
d) Ge, Sn
6. Which of the following elements is most
metallic?
a) Al
b) Mg
c) P
d) S
7. A metal has an atomic number of :
a) 9
b) 18
c) 35
d) 37
–
8. Ionisation energy of F is 320 kJ mol –1.
The electron affinity of fluorine would be:
a) ––320 kJ mol–1
b) ––160 kJ mol–1
c) 320 kJ mol–1
d) 160 kJ mol–1
9. The element having very high ionisation
energy but zero electron affinity is:
a) H
b) F
c) He
d) Be
89
CLASS-VIII
10.The electron affinities of B, C, N and O
are in the order:
a) B < C < N < O
b) B < C < O > N
c) B < C > O > N
d) B > C < O < N
11.The correct order for electron affinity is:
a) S < Se < O
b) Se < S < O
c) Se < O < S
d) O < S < Se
12. Which of the following elements is
expected to have highest
a) Chlorine
b) Carbon
c) Nitrogen
d) Flourine
13. The element with highest electron
affinity belongs to
a) Period 2, group 17
b) Period 3, group 17
c) Period 2, group 18
d) Period 2, group 1
14. Of the following elements of III period,
the strongest reducing agent is:
a) Na
b) Mg
c) P
d) Cl
15. Amongst the following oxides which is
least acidic?
a) Al2O3 b) B2O3
c) CO3
d) NO2
16. The order in which the following oxides
are arranged according to decreasing
order of basic nature is:
a) Na2O, MgO, Al2O3, CuO
b) CuO, Al2O3, MgO, Na2O
c) Al2O3, CuO, MgO, Na2O
d) CuO, MgO, Na2O, Al2O3
DAY-10 : SYNOPSIS
 Why do atoms combine?
Atoms combine to attain stability (i) by
attaining octet configuration. (ii) by
reducing the energy.
 How do atoms acquire stable octet
configuration?
Atoms can complete the valence shell by
acquiring octet configuration in two ways.
1. By transfer of one or more electrons,
from one atom to another. Generally,
electropositive elements lose electrons
and electronegative elements gain
electrons.
2. By sharing one or more electrons between
two or more atoms.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
 Thus, we can conclude that, atoms tend
to acquire 8 electrons in their outermost
shell (except hydrogen, lithium and
beryllium which tend to acquire 2
electrons), in order to attain stable state.
This is called ‘octet rule’.
 What is a Chemical Bond ?
During a chemical reaction, atoms come
closer and are held together by a force of
attraction to form molecules. This force
of attraction is called a chemical bond.
Chemical bonds are responsible for the
existence of molecules
Ionic bond and its formation
The bond that is formed by transfer of
electrons from one atom to another atom
is called an ionic or electrovalent bond.
The cations and anions formed as a result
of electron transfer are drawn towards
each other due to the electrostatic force
(coulomb force) of attraction. They form
an ionic bond or an electrovalent bond.
Note: The bond between two elements is
ionic if the EN difference between them
is greater than 1.7
The number of electrons transferred
during an ionic bond formation is known
as an electrovalency.
 Compounds containing ionic bonds are
called ionic compounds. Examples of ionic
compounds are NaCl(Na+Cl–), CaO(Ca2+O2–
), MgO(Mg2+O2–) and MgCl2 (Cl –Mg++ Cl–).
 Features of donor atoms:
Donor atoms lose electrons with greater
ease if the following conditions are
satisfied:
1) Less ionisation potential (I.P.)
2) More size
3) Less cation charge
Features of acceptor atoms
Acceptor atoms gain electrons more
easily in the following conditions:
1) High
electron
affinity
and
electronegativity
2) Less size
3) Less anion charge
90
CLASS-VIII
MPC BRIDGE COURSE
a) For donor atoms to lose electrons easily,
they should possess less I.P., big size and
less charge on cations. IA group has the
above mentioned characters and therefore
is the best donor group.
b) For acceptor atoms to gain electrons
easily, they should possess more EA, less
size and less negative charge on anions.
Group VII A has these features and so is
regarded as the best acceptor group.
Therefore, IA and VII A get along well
with each other, forming a strong ionic
bond.
Properties of ionic compounds
1. Physical state:
Generally, ionic solids are relatively hard.
It is because of the close packing due to
strong inter-ionic force of attraction
present between oppositely charged ions.
Brittleness of ionic compounds
Even though ionic compounds like rock salt are hard solids, they
break quite easily when dropped on floor.
 ...... + ......  ...... + ......  ...... +
Reason: The behaviour of ionic
compounds is much like a
 ......  ......  ......  ......  ...... 
glass, which breaks into many
 ...... + ......  ...... + ......  ...... +
pieces on falling. This
brittleness is because of shift in
alignment of its ions.
Normally, the allignment is such
that the oppositely charged ions
are next to each other as shown
Normal alignment
+
+

+


Line of force
+

+

+

Owing to the impact on falling,
the allignment is disturbed,
such that the ions with similar
charge come next to each other.
Since, like charges repel each
other, the crystal breaks along
Force of
attraction
Allignment on impact
2. Structure of ionic solids:
Unit cell: There is a basic unit in an ionic
crystal, which when repeated threedimensionally, gives complete crystal.
This basic unit is called the unit cell.
3. Melting and boiling points:
Ionic compounds possess high melting
and boiling points.
Reason: Melting and boiling points of ionic
compounds involves breaking of the lattice
structure and setting the ions free. In a
lattice, there are strong electrostatic
forces between oppositely charged ions.
To break these strong electrostatic forces,
considerable amount of energy is
required. Hence, the melting points and
boiling points of ionic compounds are
high.
NARAYANA GROUP OF SCHOOLS
4. Solubility
Ionic compounds are soluble in water.
Reason: Dissolving an ionic solid involves
the setting of opposite ions free from the
lattice into the solvent. This can happen
when the strong electrostatic force of
attraction between the opposite ions is
weakened. Therefore, solvents having
oppositely charged ions, Na+ and Cl– ions.
The mobility of Na + and Cl – results in
conduction.
6. High reactivity
Ionic compounds react instantaneously
in fused state. This is because of easy
formation of free ions, rapid union of
these ions in solutions, form new
compounds.
For example, the reaction between NaCl
and AgNO3 is very rapid in solution state,
resulting in the formation of AgCl and a
precipitate of NaNO3.
7. Directional properties
A given ion in the ionic crystal is
surrounded by a uniform electric field
around it.
Therefore, the electrostatic bonding force
acts equally on the ion in all the
directions. As there is no specific
direction for the electrostatic bonding
force, the ionic bond is a non-directional
bond.
8. Isomorphism
Crystals of different ionic compounds
having similar arrangement of ions as
well as geometry are known as isomorphs,
and the phenomenon is known as
isomorphism.
Eg: ZnSO4. 7H2O & FeSO4. 7H2O
A crystal of an isomorph, if placed in a
saturated solution of other isomorphs,
grows in size.
The valency of elements forming
isomorphous compounds is same.
91
CLASS-VIII
DAY-10: WORKSHEET
1. Number of electrons transferred from one
atom to another during bond formation
in Aluminium Nitride:
a) 1
b) 2
c) 3
d) 4
2. The electrovalency of N in magnesium
nitride:
a) one
b) two
c) three d) four
3. Which one of the following has an
electrovalent linkage?
a) CH4
b) MgCl2 c) SiCl4 d) BF3
4. The electronic structure of four elements
a, b, c and d are:
1) 1s2
2) 1s2 2s2 2p2
3) 1s2 2s2 2p5
4) 1s2 2s2 2p6
The tendency to form electrovalent bond
is greatest in
a) 1
b) 2
c) 3
d) 4
5. Which of the following is easily formed?
a) Calcium chloride b) Calcium bromide
c) Potassium chloride
d) Potassium bromide
6. Which of the following is least ionic?
a) CaF2 b) CaBr2 c) CaCl2 d) CaI2
7. Which of the following is more ionic?
a) Si3N4 b) AlN
c) BN
d) Ca3N2
8. Arrange the bonds in order of increasing
ionic character in the molecules: LiF,
K2O, N2, SO2 and ClF3.
a) N2 < ClF3 < SO2 < LiF < K2O
b) N2 < SO2 < ClF3 < K2O < LiF
c) SO2 < N2 < ClF3 < LiF < K2O
d) ClF3 < N2 < SO2 < K2O < LiF
9. Assertion (A) : Ionic compound tend to be
non-volatile.
Reason (R) : Inter molecular forces in
these compounds are weak.
The correct answer is:
a) Both assertion and reason are correct
and reason is the correct explanation of
assertion.
b) Both assertion and reason are correct
but reason is not the correct explanation
of assertion.
c) Assertion is correct and reason is incorrect.
d) Assertion is incorrect and reason is
correct.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
10. Which of the following true for ionic
compounds?
a) They are hard solids
b) They can be broken down into pieces
very easily
c) They are soluble in non-polar solvents
d) all the above
11. Which of the following boils at higher
temperature?
a) CCl4
b) CO2 c) C6H12O6 d) KCl
12. Which one has greater melting point:
Ag2O or BaO?
a) Ag2O b) BaO
c) both
d) none
13. In which of the following solvents,
should KCl be soluble at 25°C? (D =
Dielectric constant value)
a) C6H6(D = 0)
b) CH3COOCH3 (D = 2)
c) CH3OH (D = 32) d) CCl4 (D = 0)
14. If Na+ ion is larger than Mg2+ ion and
S2– ion is larger than Cl – ion, which of
the following will be least soluble in
water?
a) NaCl b) Na2S c) MgCl2 d) MgS
15. Which one of the following is more
soluble in water?
a) AgF
b) AgI
c) AgCl
d) AgBr
DAY-11 : SYNOPSIS
Covalent bond and its formation
A bond formed by the equal contribution
and equal sharing of electrons between
two atoms or more atoms is known as
covalent
bond
(co-sharing,
valence  valence electron).
Since, the formation of a covalent bond
results in the formation of a molecule, it
is also called molecular bond.
G.N. Lewis did the study of covalent bond.
He explained covalent bond formation by
the electron dot structure called Lewis
Structure.
When is the bond between two atoms
covalent?
When non-metals come together, the
tendency to donate or accept the
electrons is not possible due to the less
electronegativity (EN) difference. Thus,
in order to acquire stable configuration
(an octet or duplet) of a noble gas, sharing
92
CLASS-VIII





takes place between them, resulting in
formation of covalent bond. Generally, if
the EN difference between two nonmetals is less than 1.7, a covalent bond
is formed between them due to their
combination.
Representation of covalent bond: The
covalent bond between a pair of two
atoms is represented by a small line[ – ].
For example, H 2 can be represented as
H–H.
Covalency: The number of electron pairs
shared between two atoms of the same
element or different elements during the
formation of a molecule is known as
covalency.
Eg: Covalency of hydrogen molecule is
equal to 1 and that oxygen molecule is 2.
Bond pairs and lone pairs
Bond pair of electrons: The shared pair
of electrons, which result in the
formation of a bond, is called the “bonded
pair”.
Lone pair of electrons: The pair of
electrons, present in the valence shell
but not involved in the bonding is called
the “non-bonded pair” or “lone pair.”
Conditions favourable for covalent bond
formation:
Covalent bonding is all about sharing of
electrons. The ideal conditions necessary
for atoms to share electrons are:
1) Atoms should be of small size.
2) They
should
have
high
electronegativity and ionisation potential.
3) The electronegativity and the
ionisation potential of the combining
atoms should be almost the same.
Favourable conditions for the covalent
bond formation are satisfied by the
elements of VA, VIA, and VIIA groups.
The electrons in the outermost shell
(valence electrons) in the elements of VA,
VIA and VIIA groups are 5, 6 and 7
respectively, and they can have stable
octet configuration by sharing 3, 2 and 1
electron respectively. The molecules
formed among the elements of VA, VIA
and VIIA are mostly covalent molecules.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
Eg: SO2, PCl3, PBr3, SF6, SCl2, IF7, O2, N2,
F2, Cl2, Br2, I2, etc.
 Note: Number of electron pairs shared
between two atoms of the same element
is equal to the number of electrons short
of octet.
Examples of covalent compounds:
F2, Cl2, I2, O2, N2, H2, HCl, H2O, NH3 etc.
Note: The force of attraction present
between the molecules of inert gases is
the Vander Waal’s forces.
DAY-14
Covalent bonds based on the type
of atoms involved in bonding
Based on the types of atoms involved in
bonding, covalent bonds are classified
into homogeneous and heterogeneous
covalent bonds.
1. Homogeneous covalent bond:
It is a covalent bond formed between the
atoms of similar type.
 Examples:
(a) Formation of hydrogen molecule:
(or) H – H (hydrogen molecule)
(b) Formation of chlorine molecule:
(or)
Cl – Cl (chlorine molecule)
(c) Formation of oxygen molecule:
(or) O = O (oxygen molecule)
2. Heterogeneous covalent bond:
It is a covalent bond formed between
the atoms of different types.
 Examples:
(a)Formation of Hydrogen Chloride
HCl:
93
CLASS-VIII
MPC BRIDGE COURSE
DAY-11: WORKSHEET
1. When two atoms of chlorine combine to
(or) H – Cl (Hydrogen chloride)
After formation of a covalent bond,
hydrogen has stable duplet configuration,
and
chlorine
has
stable
octet
configuration.
(b) Formation of water molecule – H2O:
2.
(c) Formation of Ammonia molecule –
NH3:
3.
(d) Formation of methane molecule –
CH4:
4.
5.
(e) Formation of carbon tetrachloride –
CCl 4 :
6.
7.
(Carbontetrachloride)
(f) Formation of Carbondioxide molecule
– CO2:
(or) O = C = O (Carbondioxide)
II. Covalent bonds based on the number
of electron pairs shared.
NARAYANA GROUP OF SCHOOLS
8.
9.
form one molecule of chlorine gas, the
energy of the molecule is:
a) Greater than that of separate atoms
b) Equal to that of separate atoms
c) Lower than that of separate atoms
d) None of these
In the formation of covalent bond,
a) Transfer of electrons take place
b) Electrons are shared by only one atom
c) Sharing of electrons take place
d) None of these
Silicon has 4 electrons in the outermost
orbit. In forming the bonds:
a) It gains electrons b)It loses electrons
c) It shares electrons d) None of the above
Which of the following is a covalent
compound?
a) H2
b) CaO
c) KCl
d) Na2S
Which of the following substance has
covalent bonding?
a) Sodium chloride b) Solid neon
c) Copper
d) BeCl2
Which is a covalent compound?
a) RbF
b) MgCl2 c) CaC2 d) NH3
An element ‘Y’ has the ground state
electronic configuration 2, 8, 8. The type
of bond that exists between the atoms of
‘Y’ is:
a) Ionic
b) Covalent
c) Metallic
d) Vander Waal’s
The bond between two identical nonmetal atoms has a pair of electrons:
a) Unequally shared between the two
b) Transferred fully from one atom to
another
c) With identical spin
d) Equally shared between them
The maximum number of covalent bonds
by which the two atoms can be bonded to
each other is:
a) Four
b) Two
c) Three
d) No fixed number
94
CLASS-VIII
10. The molecule, which contains
maximum number of electrons, is:
a) CH4
b) CO2
c) NH3
d) BCl3
11. The molecule which contains only
bonded pairs or no lone pair of electrons
on the central atom is:
a) H2O
b) NH3
c) BeCl 2 d) BrF5
12. The number of electron pairs involved
in the bond formation in hydrogen
cyanide molecule is:
a) Two
b) Three c) Four
d) Five
13. In which of the following molecule(s),
multiple bond is present?
a) CO2
b) O2
c) N2
d) HCN
14. Match the following :
a) NH3 i) 4 bond pairs and no lone pairs
on the central atom
b) H2O ii) 2 bond pairs and 2 lone pairs
c) O2 iii) 3 bond pairs and 1 lone pair
d) CCl4 iv) 2 bond pairs and 4 lone pairs
a) a - (iv), b - (i), c - (iii), d - (ii)
b) a - (iii), b - (ii), c - (i), d - (iv)
c) a - (iii), b - (ii), c - (iv), d - (i)
d) a - (ii), b - (iii), c - (ii), d - (iv)
DAY-12 : SYNOPSIS
Properties of covalent compounds
i) Physical state.
Generally, covalent compounds are
gases, liquids or soft solids at room
temperature. Reaso n : Covalent
compounds are composed of molecules.
There exists a intermolecular force of
attraction known as Vander waal’s force
of attraction between these molecules’.
These Vander waal’s forces are weak and
hence covalent compounds exist as soft
solids, liquids and gases.
ii) Melting and boiling Points:
Covalent compounds possess low melting
and boiling points.
Reason: Melting and boiling involves the
breaking of intermolecular force of
attraction between the molecules. As the
intermolecular forces are weak between
the molecules, less energy is required
to break them. Hence, melting and boiling
points of covalent compounds are low.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
Exception: Daimond
iii) Solubility: Covalent compounds are
soluble in non-polar solvents.
Reason: Like dissolves like. Covalent
compounds have molecules without the
opposite polarity and hence they are nonpolar compounds. Therefore, covalent
compounds are soluble in non-polar
solvents like kerosene, benzene, ether,
carbon disulphide, carbon tetrachloride,
etc.
Exception:
Covalent compounds, being non-polar, are
insoluble in polar solvents like water. But
some covalent compounds like alcohol,
urea and sugar are soluble in water.
Because, this is due to the attraction
between covalent molecules like alcohol,
urea, sugar and the water molecule. This
attraction is called “Hydrogen bonding”.
The attractive force of Hydrogen bond
pulls out-the urea molecule from its solid
phase into the solvent (water) and the
urea gets dissolved.[Note: We shall learn
about H-bonding in detail, in future
classes].
iv) Conductivity:
Charge carriers are required for
substances to conduct electricity. Ionic
solutions have charge carriers in the
form of mobile ions. Metals have charge
carriers in the form of electrons. But
covalent compounds do not have any
charge carriers and hence they are bad
conductors of electricity.
Exception:
Covalent compounds do not conduct
electricity. But graphite is a good
conductor of electricity even though it is
a covalent compound. Graphite has a
structure comprising layers. Each layer
consists of a network of ‘hexagonal carbon
rings’. Each carbon atom in the ring is
covalently bonded to three adjacent
carbon atoms. Therefore, out of the four
valence electrons of carbon, three get
trapped in the covalent bond, leaving the
fourth electron free. This free electron
in graphite is responsible for the
conductivity of graphite.
95
CLASS-VIII
v) Speed of reaction: Reactions of covalent
compounds are slow.
Reas on: The reactions of covalent
compounds involve, firstly, the breaking
of the existing bonds in the molecules
and then the formation of new bonds.
Breaking of bonds requires extra energy,
and until sufficient energy is available,
the reaction is not initiated. Further, to
initiate the reaction, the reacting
molecules should collide. And among all
the collisions, only a small fraction is
fruitful. Since a covalent reaction
involves a number of operations, it is slow.
DAY-16 Comparison between ionic and
covalent compounds
Polar covalent bond and its formation
Polar covalent bond:
We have seen that covalent bond is all
about sharing of electrons. Consider a
covalent compound AB formed from A and
B. If both A and B exercise the same
amount of influence on the shared
electrons, then A and B are said to share
the electron pair equally. This is seen in
H2, where both atoms A and B belong to
the same element. Such compounds are
called non-polar covalent compounds.
 It is interesting to see what happens
when one of the atoms in the covalent
bond influences the shared electron pair,
more than the other. If this happens,
then the electron pair is said to be shared
unequally between A and B.
 During the unequal sharing, the atom
which exercises more influence on the
electron pair gets the partial negative (  )
charge and the other atom gets the partial
positive (  ) charge.
For example, in H 2 O molecule the
electronegativity of “O” is more, and its
influence is greater on the shared pair.
This results in a partial negative charge
on oxygen and partial positive charge on
hydrogen and electron pair is unequally
shared.

H

O
MPC BRIDGE COURSE
 The covalent bond formed due to unequal
sharing of electrons is called a Polar
covalent bond and the molecule is called
a Polar molecule.
Examples of polar molecules are HF, NH3,
HCl, etc.
 Co-ordinate covalent bond: The bond in
which the shared pair of electrons is
contributed by only one atom but, is
shared by both atoms or molecules is
known as co-ordinate covalent or dative
bond.
Sidgwick and Powell proposed the coordinate covalent bond.
Eg : NH3  BF3, NH4, NH4 , H3O+
H
F
H N
B
H
F
F
H
F
H N
B
H
F
F
 The atom which contributes the shared
pair is called the donor atom, and the
other atom which makes use of it is
known as the acceptor atom. The coordinate covalent bond is represented by
an arrow (  ) directed from the donor to
the acceptor.
According to Lewis, a base is one, which
donates a lone pair of electrons and an
acid is one which accepts the lone pair
of electrons. So, in a co-ordinate covalent
bond donor atom called as Lewis base and
accepted atom called as Lewis acid.
DAY-12: WORKSHEET
1. Substance X has a melting point of
1500°C. It conducts current at room
temperature. Substance Y is gas at room
temperature. It does not conduct current.
Substance Z is soluble in water and the
solution conducts current. However, in
solid state Z does not conducts current
but melts at 1020°C to form a conducting
liquid. Deduce whether X, Y, Z.
i) have giant or molecular structure.
ii) are bonded covalently or ionically.

H
NARAYANA GROUP OF SCHOOLS
96
CLASS-VIII
X
Y
Z
a)
Giant structure,
Covalent
Covalent
Ionic
b)
Covalent
Giant structure, Covalent
Ionic
c)
Giant structure, Ionic
Ionic
Covalent
d)
Ionic
Giant structure, Ionic
Covalent
2. CCl4 is insoluble in water because,
a) H2O is non-polar.
b) CCl4 is non-polar.
c) They do not form inter molecular
H–bonding.
d) They do not form intra molecular
H–bonding.
3. Which of the following when dissolved in
water forms a non conducting solution?
a) Green vitriol
b) Chile or Indian salt petre
c) Alcohol
d) Potash alum
4. Which of the following properties would
suggest that a compound under
investigation is covalent?
a) It conducts electricity on melting.
b) It is a non-electrolyte.
c) It has a high melting point.
d) It is a compound of a metal and a nonmetal.
5. Assertion (A) : Graphite is a good
electrical conductor.
Reason (R) : The free electrons in graphite
conducts electricity.
a) Both assertion and reason are correct
and reason is the correct explanation of
assertion.
b) Both assertion and reason are correct
but reason is not the correct explanation
of assertion.
c) Assertion is correct and reason is incorrect.
d) Assertion is incorrect and reason is
correct.
6. Which of the following statements is
incorrect?
i)
Covalent bond is highly directional.
ii) Ionic compounds do not exhibit space
is isomerism.
iii) Molecules react quickly than ions.
a) Both ii and iii
b) Both i and ii
c) Both i and iii
d) Only iii
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
7. The electronegativity values of C, H, O
and N are 2.5, 1, 3.0 and 2.5 respectively.
The most polar bond is
a) S - H b) O - H c) N - H d) C - H
8. Which of the following molecules is non
polar but it contains polar bonds in it?
a) BCl3
b) H2
c) NH3
d) CHCl3
9. The two compounds that are covalent
when taken pure but produce ions when
dissolved in water.
a) NaCl, NaBr
b) H2O, Cl2
c) HCl, H2O
d) HF, NH3
10. Assertion (A): Though flourine is more
electronegative than silicon, SiF4 is nonpolar.
Reason (R) : The four fluorine atoms have
the same force of attraction on silicon.
a) Both assertion and reason are correct
and reason is the correct explanation of
assertion.
b) Both assertion and reason are correct
but reason is not the correct explanation
of assertion.
c) Assertion is correct and reason is incorrect.
d) Assertion is incorrect and reason is
correct.
11. In which of the following molecules,
the shared pair of electrons is contributed
by only one individual atom or molecule?
a) NH3
b) CO
c) H2O
d) CO2
12. NH3 and BF3 form an adduct readily
because, they form
a) A coordinate bond b) A covalent bond
c) An ionic bond d) A hydrogen bond
13. Which of the following contains a coordinate bond?
a) Water
b) Ammonia
c) Ammonium ion d) Ethylene
DAY-13 : SYNOPSIS
 The measurable properties of gases are
volume, temperature, pressure and
amount of gas.
 Volume : Gases always occupy the
complete volume of the container on
account of their high expansibility. Thus,
the volume of a gas is always equal to
the volume of container.
97
CLASS-VIII
 Units of volume :
(a) 1 millilitre (1 ml) = 1 cm3 (1 cc)
(b) 1 litre (1l)= 1 cubic decimetre (dm3)
:
Temperature
is
 Temperatur e
measurement of hotness or coldness of
a body. For gases, temperature is
measured to know how fast or slow the
gas molecules are moving. More fast the
gas molecules move, more is the
temperature.
 Units of temperature :
(a) Celsius temperature is measured in
degrees Celsius = °C
(b) Kelvin temperature is measured in
kelvin = K.
[Details of Kelvin scale will be discussed
in up coming classes]
However, rise or fall in temperature of 1
K = 1°C.
(c) Temperature in kelvin = 273 +
temperature in °C, K = 273 + °C
Note: Temperature in Kelvin scale is also
called absolute temperature.
 Pressure: The force exerted by the gas
molecules per unit area is the pressure
exerted by the gas. Gases exert pressure
in all the directions due to the collisions
of the gas molecules.
 The SI and CGS units of pressure are N/
m2 (Pascal) and dyne/cm 2 respectively.
 The absolute unit of pressure is
atmosphere.
 The other units pressure are bar, torr, cm
of Hg and mm of Hg.
5
5
2
 1 atm = 10 Pascal = 10 N/m
6
2
= 10 dyne / cm = 76 cm of Hg
= 760 mm of Hg = 760 torr.
Measurement of atmospheric pressure:
Atmospheric pressure(P) can be obtained
by a barometer using the formula:
P=h×d×g
where, ‘h’is the height of liquid column,
‘d’ is the density of the barometric liquid
and ‘g’is the acceleration due to gravity.
If ‘h’ is expressed in meters, ‘d’ in Kg/m3
and ‘g” in m/s2, the pressure obtained by
this formula is in terms of Newton/m 2’
or ‘dyne/cm2’. Pressure measured in this
way is called Absolute pressure.
NARAYANA GROUP OF SCHOOLS




MPC BRIDGE COURSE
In the formula ‘P = hdg’, the terms ‘d’
and ‘g’ for a given barometer are constant
and the atmospheric pressure is
proportional to the height of the Mercury
column.
So, in day to day usage, the atmospheric
pressure is expressed in terms of the
height of the Mercury column. The
height of the Mercury level is an index of
the atmospheric pressure. For normal
atmospheric pressure, the height of the
Mercury column is 760 mm or 76 cm.
Amount of gas: The amount of gas is
measured in terms of moles.
I mole of any substances is the weight of
it equal to its gram molecular weight. For
example, 1 mole of hydrogen gas weighs
2 grams.
1 mole of oxygen gas weighs 32 grams
and so on.
The formula to calculate no. of moles is:
Given weight of gas
n = Gram molecular weight of the gas
Note: 1 mole of any gas contain 6.023× 10 23
molecules.
Plotting graph between two terms:
Note: The detailed procedure of plotting
graph, with reasons will dealt in
mathematics in future classes. In this
class, let us have a brief out look on simple
graphs.
Let us assume that we want to plot the
graph between two terms, say ‘a’ and ‘y’.
Follow the simple steps to plot the graph.
i) If a  b or a = kb then the graph is
straight line passing through origin.
a
b
1
k
ii) If a 
or a 
then the graph is a
b
b
hyperbola.
98
CLASS-VIII
MPC BRIDGE COURSE
a) 1000 cc
c)
a
b
iii) If a  b2 or a = kb2 then the graph is
a parabola.
b)
4000
 cc
3
3000
 cc
4
d) 1000 cc
7. The value of atmospheric pressure on the
surface of earth at sea level is nearly
Take density of mercury = 13600 kg/m 3,
density of water = 1000 kg /m 3
a) 105 Pa
b) 103 Pa
c) 101 Pa
d) 104 Pa
8. The liquid used in Barometer is :
a
b
iv) If a  f  b  the graph is straight line
parallel to b-axis.
a
b
DAY-13: WORKSHEET
1. The temperature –2730C on Kelvin scale
is:
a) 273 K b) 546 K c) 0 K
d) 819 K
2. Which of the following is (are) the unit(s)
of volume?
a) litre
b) millilitre
c) cubic meter
d) decimeter cube
o
3. The value of 273 C on Kelvin scale is:
a) 0 K
b) 546 K c) 273 K d) 819 K
4. At what temperature, both centigrade
scale and kelvin scale show the same
reading?
a) –40o
b) –273o c) 0o
d) Never
possible
5. A volume of 1m3 is equal to :
a) 1000cm3
b) 100cm3
3
c) 10dm
d) 106cm3
6. The radius of spherical container is
10cm. What is the volume occupied by
any amount of gas filled into it?
NARAYANA GROUP OF SCHOOLS
a) mercury
b) kerosene
c) water
d) alcohol
9. The column of mercury in a barometer is
76 cm Hg. Calculate the atmospheric
pressure if the density of mercury = 13600
kgm–3. (Take g = 10 ms –b)
a) 1.03 × 105 Pa
b) 1.03 × 103 Pa
c) 1.03 × 104 Pa
d) 1.03 × 106 Pa
10. If the mercury in the barometer is replaced by water, what will be the resulting height of the water column ?
Density of water = 1000 kgm –3 density of
mercury = 13600 kgm–3
a) 0.76m b) 10 m c) 10.3 m d) 11 m
11. A body is moving with uniform speed.
Which of the following is the right graph
for distance travelled with respect to line?
a) s
b)
s
t
t
c)
s
d)
t
s
t
99
CLASS-VIII
12. Which of the following is the right
graph for pressure applied on a gas with
respect volume change at constant
temperature?
MPC BRIDGE COURSE
 The graphs drawn at constant
temperature are known as isotherms.
 Graph between P and 1/V:
P
a) V
b)
V
1/V
P
 Graph between PV and V:
P
c)
d)
V
V
PV
P
P
13. Which of the following contain
maximum no. of moles?
P : 4 g of He
R : 48 g of O3
Q : 16 g of CH4
a) P
b) Q
c) R
d) All
14. Find the no.of atoms present in 4gm of
oxygen.
a) 6.023×1023
b) 3.0115×1023
23
c) 1.206×10
d) 6.023×1026
P
Modified Boyle’s law: We can draw a
relation between pressure and density of
a given mass of gas at constant
temperature as follows:
We know PV = K (constant) as per Boyles'
law.  P 

DAY-14 : SYNOPSIS
 Gas laws: Gas laws give the relation
between the different measurable
properties of gases.
 Boyle’s law: It gives the relation between
pressure and volume of a given mass of
gas at constant temperature.
 It states that “the volume of a given mass
of gas, varies inversely with the pressure,
provided the temperature remains constant”.
1
 Mathematically, V  P or PV  k (T=
constant )  P1V1  P2 V2  P3 V3  Pn Vn
 Graph between P and V: The graph
between P and V is a curve known as
hyperbola as shown in the figure.
P
Hy
pe
rb
ola
m
P K
K  
 Constant
d
d m
P1 P1

d1 d 2 ( or ) P  d .
The above equation is called Modified
Boyle’s law.
 Charles’ law: This law was proposed by
Charles’ and it gives the relation between
the volume and absolute temperature.
 It states that “the volume of a fixed mass of
a gas is directly proportional to its absolute
temperature, provided the pressure remains
constant”.

Mathematically, V  T
or
V
 K (P =
T
V1 V2

T1 T2
 The Charles' law is based on Charles'
observation, according to which the
volume of a given mass of a gas at
constant pressure increases or decreases
constant)

V
NARAYANA GROUP OF SCHOOLS
100
CLASS-VIII
1
by
of its volume at 0 0C for every
273
degree rise or fall of its temperature
respectively.
t 

 Vt  V0 1  273  where, Vt is the volume


of the gas at t0C and V0 is the volume of
the gas 0°C.
 Based on Charles' observation, it was
found that volume at –2730C should be
expected to be zero. This temperature is
called absolute zero.
 All the properties of the gases become
zero at absolute zero.
 Graph between Vt and t
V
MPC BRIDGE COURSE
Standard Temperature and Pressure
As volume depends on temperature and
pressure, they should invariably be
mentioned
during
volume
measurements.
The
standard
temperature and pressure at which we
measure the volume of gas is 0°C and 1
atmosphere
of
pressure.
This
temperature and pressure is called S.T.P
( Standard temperature and pressure) or
N.T.P.(Normal temperature and pressure.
DAY-14: WORKSHEET
1. Which of the following is right graph
between P and V at constant
temperature?
[ ]
a)
b)
Vo
-273
0
-200 -100 tOC
P
100
V
T
 The graphs drawn at constant pressures
are known as isobars.
NARAYANA GROUP OF SCHOOLS
V
1
V
200 300
 Kelvin Scale : This scale of temperature is
given by Kelvin.
 The starting point of Kelvin scale is
absolute zero i.e., –273 0 C which
corresponds to one Kelvin.
 The difference between any two
successive points on the scale is same
as that of centigrade scale.
 The Kelvin scale is also called absolute
scale of temperature.
0
 T ( K ) = t ( C) + 273.
 Graph between V and T: The graph
between V and T is a straight line passing
through origin as shown in the figure.
PV
c)
PV
d)
P
P
V
2. Pressure of a gas is 2 atm at 5l. If its
pressure is increased by three units
then what will be its new volume ?
a) 2 l
b) 3 l
c) 4 l
d) 5 l
3. Pressure of the gas containing in a 5 lit
cylinder is 2 atm if volume of cylinder
increases to 15 litre what will be the
new pressure of the gas?
a) 12 atm
b) 6 atm
c) 1.2 atm
d) 0.6 atm
4. A gas occupied a volume of 250 ml at 700
mm Hg pressure and 250 o C. What
additional pressure is required to reduce
the gas volume to its 4/5th value at the
same temperature?
101