Download 9th class bridge course 74-112

CLASS-IX
MPC BRIDGE COURSE
Note: A molecule splits into atoms first
before taking part in a chemical reaction.
CHEMISTRY
 Steps
DAY-1 : SYNOPSIS
 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
1 th
12
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
 75  35   25  37   35.5
100
 Mass
of one atom of an element =
Relative atomic mass  mass of
mass of C- 12
 Molecu le:
1 th
12
the
The term molecule was
introduced by Avogadro. Molecule is the
smallest particle of matter that exists
independently and is formed by the
combination of atoms. Molecule is also
defined as the smallest particle of matter
that can exist and retains all the
properties of that substance.
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2.
the
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  0–27 kg.
 Atomic weight has no units.
 The relative atomic mass of an element
indicates the number of times one atom
of that element is heavier than
1.
3.
4.
to calculate the molecular
weight:
Write the formula of the compound or the
molecule.
Identify the different types of elements
present in it and write their symbols
along with the number of atoms.
Now multiply the number of atoms with
the atomic weights of the respective
elements
Finally add them to get molecular weight.
DAY-1: WORKSHEET
1. 1 amu is equal to the mass of
1)
1
th of C - 12 atom
12
1
th of O-16 atom
14
3) 1g of H2
4) 1.66 × 10–23 kg
2. 1 atomic mass unit =
2)
1 th
mass of a carbon - 12 atom
12
2) 1.66 × 10–24g
3) 6.023 × 10–23g
4) 6.023 × 1023g
3. The ratio of weight of one atom of an
element to its atomic weight is equal to
1) 1 amu
1)
1
th of C – 12 isotopic atom
12
3) 12 amu
4) None
4. The mass of one atom of an element is
40 × 1.66 × 10–24g. The number of protons
in its nucleus is
1) 40
2) 20
3) 10
4) 5
5. The weight of Helium atom in grams is
1) 2
2) 4
–24
3) 6.64 × 10
4) 1.66 × 10–24
6. The symbol of carbon is C. It means that
1) ‘C’ represents one atom of carbon.
2) ‘C’ also represents 1 mole of carbon
atoms.
3) ‘C’ also represents 12g of C.
4) All
2) mass of
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CLASS-IX
7. Atomic weight of an element is x. It
means that weight of one atom of that
element is
1
xg
12
3) 12 × x g
4) 1.66x × 10–24g
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_______.
1) 1
2) 1.66 × 1024
23
3) 6.023 × 10
4) 96500
9. The total mass of 100 atoms of silicon is
1) 2800
2) 2800 amu
3) 28 × 1.66 × 10–22g 4) 280 kg
10. The approximate number of electrons
that are required to make 1 smallest unit
of mass is
1) 6.023 × 1023
2) 1.66 × 1024
3) 1852
4) 2500
1) ‘x’ g
2)
DAY-2 : 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 gram-atomic
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.
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(g)Number of atoms in 1 gram of an
element =
 Gram
6.023×1023
.
Atomic weight
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.
(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.
DAY-2: WORKSHEET
1. Gram atom of any element contains
1) 6.023 × 1023 atoms
2) 3.0115 × 1023 atoms
3) 1.505 × 1023 atoms
4) 12.0 × 1023 atoms
2. Which of the following is correct?
1) Molecular mass of oxygen is 32.
2) Gram molecular mass of sulphur (S8)
is 252 g.
3) The weight of one molecule of O3 is 48
amu.
4) All
3. The ratio of number of molecules present
in 1 gram mole of O2 to one gram mole of
SO2 is
1) 1 : 4
2) 1 : 2
3) 1 : 8
4) 1 : 1
4. The ratio of weights of hydrogen and helium is 1 : 2. Find the ratio of number of
gram atoms.
1) 2 : 1
2) 1 : 1
3) 1 : 4
4) 4 : 1
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CLASS-IX
5. Among all the naturally occuring elements, which one can generate the maximum number of gram atoms from a given
amount?
1) Hydrogen
2) Uranium
3) Calcium
4) Mercury
6. How many gram atoms of the lightest element weigh same as 1 gram atom of
the heaviest element?
1) 1
2) 235
3) 238
4) 100
7. Among all the naturally occuring elements, one gram atom of which element
contains the maximum amount of it?
1) Hydrogen
2) Uranium
3) Calcium
4) Mercury
8. Identify the element whose 2 gram atoms weigh 8g.
1) Hydrogen
2) Helium
3) Oxygen
4) Sulphur
9. Find the number of gram molecules
present in the following:
i) 5g of Neonii) 7 g of nitrogen
(i)
(ii)
1) 0.25
0.25
2) 0.25
0.5
3) 0.5
0.25
4) 1
2
10.1 x = 6.023 × 10 23 molecules = gram
molecular mass of a substance. Then, x
= _____
1) gram atom
2) mole
3) molecular weight 4) gram
DAY-3 : SYNOPSIS
 Important relations related to mole:
a) 1 mole of particles = 6.023  1023 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.
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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.
 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
 Weight of x gram atoms = x  Gram atomic
weight
 Weight of x moles of any compound =
x  Gram molecular weight
DAY-3: WORKSHEET
1. The number of atoms in 8g of Sulphur is
1) 6.02 × 1023
2) 3. 01 × 1023
24
3) 12.04 × 10
4) 1.505 × 1023
2. 12 g of Carbon contains equal number of
atoms as
1) 12 grams of Mg
2)40 grams of Calcium
3) 32 grams of Oxygen
4) 7 grams of nitrogen
3. 6.02 × 1022 particles present in 32 g of
oxygen is
1) 0.1 mole
2) 1 mole
3) 10 moles
4) 100 moles
4. Number of molecules present in 32g of
oxygen is
1) 3.2 × 1010
2) 6.02 × 1023
23
3) 3.2 × 10
4) 6.02 × 1010
5. Which of the following has more number
of molecules?
1) 1 g O2
2) 1 g N2
3) 1 g F2
4) 1 g CO2
6. Which of the following has more number
of atoms?
1) 1 g Ca 2) 1 g C 3) 1 g Cu 4) 1 g Cl2
7. Which of the following pairs of gases
contain equal number of particles?
1) 1 g He, 1 g H2
2) 1 g He, 2 g H2
3) 4 g He, 2 g H2
4) 4 g He, 4 g H2
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CLASS-IX
8. If equal mass of N2 and O2 are taken, the
ratio of number of molecules in these
gases would be
1) 1 : 1
2) 7 : 8
3) 8 : 7
4)28:32
9. Which of the following contains largest
number of atoms?
1) 4 g of H2
2) 1 g of O2
3) 28 g of N2
4) 18 g of H2O
10. Which one of the following pairs of gases
contain the same number of molecules?
1) 16 g of O2 and 14 g of N2
2) 8 g of O2 and 22 g of CO2
3) 28 g of N2 and 22 g of CO2
4) 32 g of O2 and 32 g of N2
DAY-4 : 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.
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.
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.
The mass of proton is 1.67 × 10–24 g. It is
estimated that a proton is 1837 times as
heavy as an electron.
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DAY-4: WORKSHEET
1. The term ‘atom’ was given by
1) Democritus
2) John Dalton
3) William Crookes 4) Maharishi Kanada
2. Which of the following is true according
to Dalton’s atomic theory?
1) Matter consists of small indivisible
particles called atoms.
2) Atoms of same element are alike in
all respects
3) Atoms combine in small whole
numbers to form compound atoms
(molecules)
4) Atom is the smallest unit of matter
which takes part in a chemical reaction.
3. The first atomic theory was given by
1) Democritus
2) John Dalton
c) William Crookes 4) Maharishi Kanada
4. Which is not correct about electrons?
1) Discovered by Chadwick
2) Named by J.L. Stoney
3) Present inside the nucleus
4) It has maximum e/m ratio
5. Which of the following is never true for
cathode rays?
1) They possess kinetic energy
2) They are electromagnetic waves
3) They produce heat
4) They produce mechanical pressure
6. The discharge tube experiment in which
cathode rays are emitted has shown that
1) All nuclei contain positive charge
2) All forms of matter contain electrons
3) Protons are positively charged
4) Mass of proton and that of neutron are
almost equal
7. Assertion : The charge to mass ratio of
the particles in anode rays dependson
nature of the gas taken in the discharge
tube.
Reason: The particles in anode rays
carry positive charge.
1) Assertion is correct and reason is the
correct explanation of assertion.
2) Reason is correct but assertion is
incorrect.
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CLASS-IX
3) Assertion is correct and reason is not
the correct explanation of assertion.
4) Reason and assertion both are
incorrect.
8. Positive rays are
1) Electromagnetic waves 2) Electrons
3) Positively charged gaseous ions
4) Neutrons
9. The ratio of e/m for p+ and  - particle is
1) 1:2
2) 2:1
3) 1:3
4) 1:1
10.Which is not correctly matched:
1) Particle nature of e– was given by 
Bohr.
2) The heaviest sub-atomic particle 
Neutron among e, p, n.
3) Electron is  non fundamental
particle.
4) Outside the nucleus  neutron is
unstable.
DAY-5 : 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).
• Defects of Rutherford’s atomic model:
(i) According to the Classical laws of
mechanics or dynamics of physics, any
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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 loses 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.
• The energy of a photon of light in terms
of wavelength  and velocity of light ‘c’,
is given by
E
hc
.

• A body can emit or absorb energy only in
terms of the integral multiples of
quantum i.e.,E = nh  where n=1,2,3,4 ..
• Main features of Bohr’ atomic model
In order to overcome the drawbacks of
Rutherford’s atomic model, Neil’s Bohr
proposed a new atomic model based upon
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CLASS-IX
quantum theory of radiations.
(i) The electrons in an atom revolve
around the nucleus only in certain
selected circular orbits. These orbits are
associated with definite energies and are
called shells or energy levels.
(ii) These are numbered as 1, 2, 3, 4
.... etc., or designated as K, L, M, N ...etc.
shells
(iii) Only those orbits are permitted in
which the angular momentum of an
electron is a
whole number multiple
of
nh
, where h is Planck’s constant. That
2
is,
Angular momentum of an electron ,
mvr 
nh
,
2
Where n= 1, 2, 3, ......n; m is the mass of
the electron, v is the velocity of the
electron and r is the radius of the orbit.
• n is also known as principal quantum
number.
(iv) As long as the electron revolves in a
particular orbit, its energy remains
constant. Hence the orbits are known as
stationary orbits.
(v) When an electron gains energy, it
jumps to higher energy level by absorbing
a definite amount of energy (equal to the
difference in energy between the two
energy levels). When the electron jumps
back to the lower energy level it radiates
same amount of energy in the form of a
photon.
E  E 2  E1  h ,
where
the
 is
frequency of the radiation emitted or
absorbed when an electron jumps from
one orbit to another orbit.
• Success of Bohr’s atomic model:
(i) Bohr’s model could explain the
stability of atom.
(ii)
Bohr’s theory helped in calculating
the radius and energy of an orbit.
(iii) It could explain the atomic spectrum
of hydrogen.
• Spectrum: The group of wavelengths or
frequencies is known as a spectrum.
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• Electromagnetic spectrum: This is a
spectrum
in
which
all
the
electromagnetic radiations are arranged
in the increasing order of their
wavelengths.
• Drawbacks:
(i) Bohr’s model could not explain the
spectra of atoms containing more than
one electron.
(ii) It was observed that in the presence
of a magnetic field, each spectral line gets
split up into closely spaced lines. This
phenomenon, known as Zeeman effect,
could not be explained by Bohr’s model.
Similarly, the splitting of spectral lines
under the effect of applied electric field
(Stark effect), could not be explained.
(iii)Bohr’s model could not justify the
quantization of angular momentum i.e.,
why the angular momentum of an
electron is mvr 
nh
.
2
DAY-5: WORKSHEET
1. Rutherford’s experiment on scattering of
alpha particles showed for the first time
that atom has
1) Nucleus
2) Electrons
3) Protons
4) Neutrons
2. Rutherford’s model is related to explain
1) Discovery of nucleus
2) Spectrum of Hydrogenic species
3) Planetary motion of electrons around
nucleus
4) All of these
3. Rutherford’s  - particle dispersion
experiment concludes?
1) All –Ve ions are deposited at small
part.
2) Proton moves around the nucleus.
3) All +Ve ions are deposited at small
part.
4) Neutrons are charged particles.
4. When alpha particles are sent through a
thin metal foil, most of them go straight
through the foil because
1) Alpha particles are much heavier than
electrons
2) Alpha particles are positively charged
3) Most part of the atom is empty space
4) Alpha particles move with high velocity
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CLASS-IX
5. Rutherford’s alpha particles scattering
experiment eventually led to the
conclusion that
1) Mass and energy are related
2) Electrons occupy space around the
nucleus
3) Neutrons are deep in the nucleus
4) The point of impact with matter can
be precisely determined.
6. The space between nucleus and
electrons in an atom is
1) full of electromagnetic radiation
2) full of air
3) empty
4) none of the above
7. Rutherford’s
experiment
which
established the nuclear model of the
atom used a beam of
1)  -particles which impinged on a metal
foil and got absorbed.
2)  -rays which impinged on a metal foil
and ejected electrons.
3) Helium atoms, which impinged on a
metal foil and got scattered.
4) Helium nuclei which impinged on a
metal foil and got scattered.
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.
1) Both Assertion and Reason are correct
and Reason is the correct explanation of
Assertion
2) Both Assertion and Reason are correct
but Reason is not the correct explanation
of Assertion
3) Assertion is correct and Reason is
incorrect
4) Assertion is incorrect and Reason is
correct
9. Very few (1 in 10000 or 20000)
 - particles are deviated through an
angle of 180 o C. This has lead to the
discovery of
1) Proton 2) Neutron 3) Both 4) Nucleus
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10. The electrons of Rutherford’s model of
the atom are expected to lose energy
because they
1) are attracted by the nucleus
2) strike each other
3) are accelerated
4) are in motion
DAY-6 : SYNOPSIS
• Sommerfeld’s atomic model:
(i) Electrons revolve in elliptical orbit.
Each ellipse has a major axis and minor
axis.
(ii) He proposed that the angular
momentum of an electron moving in an
elliptical orbit is equal to the integral
multiples of h/2  i.e., mvr 
kh
, where
2
k is called azimuthal quantum number.
(iii) The relation between principal
quantum number (n) and azimuthal
quantum number(k) is as follows:
n length of the major axis
=
k length of the minor axis
(iv) According to him, as the value of k
decreases, the ellipticity of the orbit
increases. Thus, when n=k, the orbit will
be circular.
(v) For a given value of n, the values of k
are 1 to n. In other words for a given
stationary orbit, a set of substationary
orbits are present.
(vi) The substationary orbits are also
known as sublevels or subshells or
subenergy levels. The subshells of a
given shell have slight energy difference.
n=4, k=4
n=4, k=3
n=4, k=2
n=4, k=1
• Success of sommerfeld’s atomic model:
It could explain the reason for further
splitting of spectral lines in electric and
magnetic fields. This is due to the
presence of sub stationary orbits or
subshells.
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CLASS-IX
Quantum numbers
A When each orbital in an atom is specified
by a set of three quantum numbers
(n, l, m) and each electron is designated by a
set of four quantum numbers (n, l, m and s).
(1) Principal quantum number (n)
(i) It was proposed by Bohr and denoted
by ‘n’.
(ii) It determines the average distance
between electron and nucleus, means it
denotes the size of atom.
(iii) It determines the energy of the
electron in an orbit where electron is
present.
(iv) The maximum number of electrons
in an orbit represented by this quantum
number is 2n2. No energy shell in atoms
of known elements possess more than
32 electrons.
(v) It gives the information of orbits like
K, L, M, N----------.
(vi) Angular momentum can also be
calculated by using principal quantum
number.
(2) Azimuthal quantum number (l)
(i) Azimuthal quantum number is also
known as ‘ angular quantum number’.
Proposed by Sommerfeld and denoted by
‘l’.
(ii) It determines the number of subshells or sub-levels to which the electron
belongs.
(iii) It tells about the shape of subshells.
(iv) It also expresses the energies of
subshells s < p < d < f (increasing energy).
(v) The value of l = (n – 1) always. Where
‘n’ is the number of principle shells,
Value of l
Name of
subshell
Shape of
subshell
=
=
0
s
=
Spherical
1
p
2
d
Dumbbell
Double
dumbbell
3……(n-1)
f
Complex
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s – sub-shell  2 electrons, d – sub-shell
 10 electrons
p – sub-shell  6 electrons, f– sub-shell
 14 electrons.
(viii) For a given value of ‘n’ the total
values of ‘l’ is always equal to the value
of ‘n’.
(3) Magnetic quantum number (m)
(i) It was proposed by Zeeman and
denoted by ‘m’.
(ii) It gives the number of permitted
orientations of sub-shells.
(iii) The value of ‘m’ varies from –l to +l
through zero.
(iv) It tells about the splitting of spectral
lines in the magnetic field, i.e., this
quantum number proves the Zeeman
effect.
(v) For a given value of ‘n’ the total value
of ‘m’ is equal to n2.
(vi) For a given value of ‘l’ the total value
of ‘m’ is equal to(2l + l).
(vii) Degenerate orbitals :
Orbitals having the same energy are
known as degenerate orbitals. e.g. for ‘p’
subshell px py pz.
(viii) The number of degenerate orbitals
of ‘s’ subshell = 0.
(4) Spin quantum number ( s)
(i) It was proposed by Goldschmidt and
Uhlenback and denoted by the symbol of
‘s’.
(ii) The value of ‘s’ is +½ and –½, which
signifies the spin or rotation or direction
of electron on it’s axis during the
movement.
(iii) The spin may be clockwise or
anticlockwise.
(iv) It represents the value of spin
angular momentum, and is equal
h
s  s  1 .
2
(vi) It represents the orbital angular
momentum, which is equal to
to
h
l l  1 .
2
(v) Maximum spin of an atom = ½ ×
number of unpaired electrons.
(vi) This quantum number is not the
result of solution of Schrodinger equation
as solved for H–atom.
(vii) The maximum number of
in subshell = 2(2l +1).
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electrons
81
CLASS-IX
MPC BRIDGE COURSE
DAY-6: WORKSHEET
3) n = 4, l = 0, m = 0, s = +
1. The ratio of energies of two radiations,
one with a wavelength of 400 nm and
the other with 800 nm.
1) 1 : 4
2) 2 : 1
3) 8 : 1 4) 1 : 6
2. Find the number of photons of light with
wavelength 4000pm that provide 1J of
energy.
1) 4.01×1015
2) 2.01×1016
3) 3.01× 1015
4) 1.01× 1015
3.3 × 10 18 photons of a certain light
radiation are found to produce 1.5 J of
energy. Calculate the wavelength of light
radiations. (h = 6.63 × 10–34 Js).
1) 4527A0 2) 9870A0 3) 3978A0 4) 7956A0
4. If the principal quantum number is 3, the
azimuthal quantum number can have
values
1) 1, 2, 3
2) 3, 2, 1, 0, –1, –2, –3
3) 0, 1, 2
4)
1 1
,
2 2
5. Which one of the following sets of
quantum numbers represent an
impossible arrangement ? n l ml ms
1) 3 2 – 2
1/2 2) 4 0
0
1/2
3) 3 2 – 3
1/2 4) 5 3
0 – 1/2
6. Which of the following represents the
correct set of the four quantum numbers
for 4d-electrons ?
1) 4, 3, 2, + 1/2
2) 4, 2, 1, 0
3) 4, 3, –2, +
1
2
4) 4, 2, 1, –
1
2
7. In an atom, the maximum number of
electrons having the quantum numbers
n = 6, l = 3, s = +
1
are
2
1) 14 2) 7
3) 6
4) 2
8. Which of the following sets of the
quantum numbers is permitted ?
1) n = 4, l = 2, m = + 3, s = +
1
2
2) n = 3, l = 3, m = + 3, s = +
1
2
NARAYANA GROUP OF SCHOOLS
1
2
4) n = 4, l = 3, m = + 1, s = 0.
9. Which of the following sets of quantum
numbers is correct?
1) n = 4, l = 3, m = + 4, s = +
1
2
2) n = 3, l = 2, m = + 3, s = –
1
2
3) n = 2, l = 2, m = + 2, s = +
1
2
4) n = 1, l = 0, m = 0, s = –
1
2
10.For n = 4,
1) The total possible values of ‘l’ are 3.
2) The highest value of ‘l’ is 4.
3) The total number of possible values of
‘m’ is 7.
4) The highest value of ‘m’ is + 3.
DAY-7 : SYNOPSIS
Shapes of orbitals:
(1) Shape of ‘s’ orbital:
(i) For ‘s’ orbital l = 0 & m = 0.So ‘s’ orbital
have only one unidirectional orientation
i.e., the probability of finding the electrons
is same in all directions.
(ii) The size and energy of ‘s’ orbital with
increasing ‘n’ will be
1s < 2s < 3s < 4s.
(iii) s-orbitals are known as radial node
(or) nodal surface. But there is no radial
node for ‘1s’ orbital since it is starting
from the nucleus.
(2) Shape of ‘p’ orbitals:
(i) For ‘p’ orbital, l = 1 and m=+l,0,–1
means there are three ‘p’ orbitals, which
is symbolised as px, py, pz.
(ii) Shape of ‘p’ orbital is dumbbell, in
which the two lobes on opposite side are
separated by the nodal plane.
(iii) p-orbital has directional
properties.
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CLASS-IX
(3) Shape of ‘d’ orbitals:
(i) For the ‘d’ orbital l =2, then the values
of ‘m’ are –2, –1, 0, +1, +2. It shows
that the ‘d’ orbital has five orbitals.
(ii) Each ‘d’ orbital is identical in shape,
size and energy.
(iii) The shape of ‘d’ orbital is double
dumbbell .
(iv) It has directional properties.
(4) Shape of ‘f’ orbital:
(i) For the ‘ f ’ orbital l = 3 then the values
of ‘m’ are –3. –2, –1, 0, + 1, + 2, + 3. It
shows that the ‘ f ’ orbitals have seven
orientations as,
f x x2  y2 , f y x2  y2 , f z x2  y2 , f xyz, f z3 , f 2 , f 2 .

 
 

yz
xz










A
(ii) The ‘ f ’ orbital is complicated in
shape.
Note:
The surface at which the probability of
finding an electron is zero is called a node
or nodal plane.
The spherical ‘s’ orbitals do not have
nodal planes but have nodal regions
equal to n -1 which are present between
spherical ‘s’ orbitals.
p-orbitals have both nodal regions equal
to n - 2 and nodal plane equal to the
value of l ie 1.
The nodal plane for px orbital is yz.
The nodal plane for py orbital is xz.
The nodal plane for pz orbital is xy.
Each d - orbital has nodal regions equal
to n – 3 and nodal planes equal to the
value of l ie 2.
For any orbital, the number of nodal
regions is equal to ‘n – l – 1’ and nodal
planes is equal to ‘l’.
For any orbital, the total number of nodal
regions and nodal planes is equal to n–1
The nodal regions are known as radial
nodes and the nodal planes are known
as angular nodes.
Relative energies of atomic orbitals:
The atomic spectra provide information
regarding the energies of the atomic
orbitals. The order of the energies may
be wirtten as,
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4s < 3d
< 4p < 5s <4d < 5p < 6s < 4f < 5d < 6p <
7s.
A The electrons occupy orbitals with lower
energy. Moeller has given a simple
procedure representing the orbitals of
increasing energies in the above order.
According to this procedure the orbitals
of increasing energies may be shown by
a diagram.
A As per this diagram, the electrons occupy
‘1s’ orbital, next the ‘2s’ orbital will be
occupied. The sequence continues as
shown by the arrows in this diagram. This
procedure is fully valid for lighter
elements. For heavier elements the
order may change. Thus, for example,
the
electronic
configuration
of
Lanthanum, La (Z = 57) should be [Xe]
6s2 4f1.
DAY-7: WORKSHEET
1. When an atom is in a magnetic field, the
possible number of orientations for an
orbital of azimuthal quantum number 3,
is
a) Three b) One
c) Five
d) Seven
2. Which of the d-orbitals lie/s in the xyplane?
a) dxz only
b) dxy only
c) d x 2  y 2 only
d) d xy and d x 2  y 2
3. Which of the following orbitals is
cylindrically symmetrical about z-axis?
a) dxz
b) dyz
c) d z2
d) dxy
4. Which of the following orbitals has/have
two nodal planes?
a) dxy
b) d x2  y 2
c) dyz d) All of these
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CLASS-IX
5. Which of the following orbitals is not
cylindrically symmetrical about z-axis?
a) 3 d2z
b) 4pz
c) 6s
d) 3dyz
6. The number of spherical nodes in ‘6s’
orbital is
a) 6
b) 5
c) 3
d) 0
7. Which of the following orbitals has/have
two lobes along y-axis?
a) dxy
b)dyz
MPC BRIDGE COURSE
A If one electron in an atom has the
quantum numbers n = 1, l = 0 , m = 0
and s = +l/2, no other electron can have
the same four quantum numbers. In
other words, we cannot place two
electrons with the same value of ‘s’ in
‘1s’ orbital.
A The orbital diagram
c) d x2  y 2 d) All of these
8. When ‘3d’ orbital is complete, the new
electron will enter into the
a) 4p – orbital
b) 4f – orbital
c) 4s – orbital
d) 4d – orbital
9. In a potassium atom, which of the
following order of electronic energy levels
is correct?
a) 4s > 3d b) 4s > 4p c) 4s < 3d d) 4s < 3p
10.Which of the following sets of orbitals
may degenerate?
a) 2s, 2px, 2py
b) 3s, 3px, 3dxy
c) 1s, 2s, 3s
d) 2px, 2py, 2pz
A
A
A
DAY-8 : SYNOPSIS
A The atom is built up by filling electrons
in various orbitals according to the
following rules.
(1) Aufbau’s principle
A This principle states that the electrons
are added one by one to the various
orbitals in order of their increasing
energy starting with the orbital of lowest
energy. The increasing order of energy
of various orbitals is,
Is < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s
< 4d < 5p < 6s < 4f
< 5d < 6p < 7s <5f < 6d < 7p.........
(2) (n+l) Rule
A In neutral isolated atom, the lower the
value of (n + l ) for an orbital, lower is
its energy. However, if the two different
types of orbitals have the same value of
(n + l ), the orbitals with lower value of
‘n’ has lower energy.
(3) Pauli’s exclusion principle
A According to this principle, “no two
electrons in an atom will have same
values of all the four quantum numbers”.
NARAYANA GROUP OF SCHOOLS
A
A
does not
represent a possible arrangement of
electrons . Because there are only two
possible values of ‘s’. An orbital can hold
not more than two electrons.
(4) H und’s Ru le of ma x imum
multiplicity
This rule deals with the filling of
electrons in the orbitals having equal
energy (degenerate orbitals). According
to this rule,
“Electron pairing in p, d and f orbitals
cannot occur until each orbital of a given
subshell contains one electron each or
is singly occupied”.
This is due to the fact that electrons being
identical in charge, repel each other
when present in the same orbital. This
repulsion can however be minimised if
two electrons move as far apart as
possible by occupying different
degenerate orbitals. All the unpaired
electrons in a degenerate set of orbitals
will have same spin.
As we know the Hund’s rule, let us see
how the three electrons are arranged in
‘p’ orbitals.
The important point to be remembered
is that, all the singly occupied orbitals
should have electrons with parallel spins
i.e., in the same direction eitherclockwise or anti-clockwise.
2px 2py 2pz
or
2px 2py 2pz
Electronic configuration of elements:–
A On the basis of the electronic
configuration principles, the electronic
configuration of various elements are
given in the following table :
84
CLASS-IX
MPC BRIDGE COURSE
A The above method of writing the
electronic configurations is quite
cumbersome. Hence, usually the
electronic configuration of the atom of
any element is simply represented by the
notation.
Number of
Electrons
present
x
nl
Number of
Principal
Principle
shell
shells
Symbol of
subshell
Some Unexpected Electronic
Configuration
A Some of the exceptions are important.
Because they occur with common
elements, notably chromium and copper.
A ‘Cu’ has 29 electrons. Its expected
electronic configuration is Is2 2s2 2p6 3s2
3p6 4s2 3d9. But in reality the configuration
is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10 (as this
configuration is more stable). Similarly
‘Cr’ has the configuration of 1s2 2s2 2p 6
3s2 3p5 4s1 3d6 instead of
1s2 2s2 2p6 3s2 3p6 4s2 3d4.
A Factors responsible for the ex tra
stability of half-filled and completely
filled subshells:
(i) Symmetrical distribution: It is well
known fact that symmetry leads to
stability.
Thus
the
electronic
configuration in which all the orbitals of
the same subshell are either completely
filled or exactly half filled are more stable
because of symmetrical distribution of
electrons.
(ii) Exchange energy: The electrons with
parallel spins present in the degenerate
orbitals tend to exchange their position.
The energy released during this
exchange is called exchange energy. The
number of exchanges that can take place
is maximum when the degenerate
orbtials (orbitals of same subshell having
equal energy) are exactly half-filled or
completely filled. As a result, the
exchange energy is maximum and so it
is stable.
NARAYANA GROUP OF SCHOOLS
DAY-8: WORKSHEET
1. No two electrons in an atom can have
a) The same principal quantum numbers.
b) The azimuthal quantum numbers.
c) The same magnetic quantum numbers.
d) An identical set of four quantum
numbers.
2. In the electronic configuration given
below, which rule is violated ?
1s
2s
2P
a) Aufbau rule
b)Pauli’s exclusion principle
c) Hund’s rule
d) The configuration is correct
3. According to Aufbau principle, the 19 th
electron in an atom goes into the
a) 4s-orbital.
b) 3d-orbital.
c) 4p-orbital.
d) 3p-orbital.
4. The orbital diagram, in which both Pauli’s
exclusion principle and Hund’s rule are
violated?
2s
2p
a)
b)
c)
d)
5. Nitrogen has the electronic configuration
1s 2 , 2s 2 , 2p1x , 2p1y , 2p1z .
Which
is
determined by
a) Aufbau’s principle.
b) Pauli’s exclusion principle.
c) Hund’s rule.
d) Uncertainty principle.
6. Pauli’s exclusion principle states that, b
a) Nucleus of an atom contains no
negative charge.
b) Electrons move in circular orbits
around the nucleus.
c) Electrons occupy orbitals of lowest
energy.
d) All the four quantum numbers of two
electrons in an atom cannot be equal.
85
CLASS-IX
7. The statements
(i) In filling a group of orbitals of equal
energy, it is energetically preferable to
assign electrons to empty orbitals rather
than pair them into a particular orbital.
(ii) When two electrons are placed in two
different orbitals, energy is lower if the
spins are parallel are valid for
a) Aufbau principle
b) Hund’s rule
c) Pauli’s exclusion principle
d) Uncertainty principle
8. The orbital diagram in which the Aufbau’s
principle is violated?
2s
2p
a)
2p
2s
2p
b)
2s
c)
2s
2p
d)
9. The electrons would go to lower energy
levels first and then to higher energy
levels, according to which of the
following?
a) Aufbau principle
b) Pauli’s exclusion principle
c) Hund’s rule of maximum multiplicity
d) Heisenberg’s uncertainty principle
10.The number of unpaired electrons in the
ground state of vanadium (Z = 23) is
a) 1
b) 2
c) 5
d) 3
DAY-9 : SYNOPSIS
1. Nece ssity fo r class ificatio n 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 different
elements.
2. Early attempts for classification:
(i) Clas sificat ion into Metals and
Nonmetals: Lavoisier was the first
(a) Such a classification hardly serves
any purpose, as the elements are divided
into two major groups only.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
(b)There is no justification for more
active metals or more active nonmetals.
(c)Certain elements exhibited metallic as
well as nonmetallic character and,
hence, were named as metalloids.
(ii) Classification on the basis of
valency:
(i) Many elements have variable
valencies. This makes the position of
such elements uncertain.
(ii) It has not explained the diverse
nature of elements having the same
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.
• Reason for rejection of Doberiner
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 applied to 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
atomic weights, then the properties of the
elements were repeated at every eighth
element like the eighth note of an octave in
music.
86
CLASS-IX
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).
(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.
(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.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
• The most electronegative elements like
fluorine (F), chlorine (Cl), bromine (Br),
Iodine (I) are placed on the ascending
curve.
Thus, Lothar Meyer observed a periodicity
in the properties of the elements with
atomic mass.
• Main features of Mendeleev’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.
(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.
87
CLASS-IX
• 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.
(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).
4.
5.
6.
DAY-9: 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.
NARAYANA GROUP OF SCHOOLS
7.
8.
MPC BRIDGE COURSE
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.
(i) Elements with both metallic and nonmetallic characters are called_____.
(ii) Arrangement of elements into groups of
three is called ________.
(i)
(ii)
a) Active metals
Octaves
b) Metallic elements Metals
c) Triads
Metals
d) Metalloids
Triads
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
Which of the following is wrong triad?
a) Chlorine, bromine, iodine
b) Lithium, sodium, potassium
c) Carbon, nitrogen, oxygen
d) Calcium, strontium, barium
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.
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
88
CLASS-IX
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

DAY-10 : 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.
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.
Positi on of trans ition elem ents :
Mendeleev’s concept of transition
elements was defective. He regarded
elements of group VIII as transition
elements.
NARAYANA GROUP OF SCHOOLS




MPC BRIDGE COURSE
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.
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 el ectronic
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.
89
CLASS-IX
transition
elements
are
 The
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.
 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.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
 The elements in group IIA are called
alkaline earth metals.
 The elements in group VII A are called
halogens.
 The elements in groups IIIB, IVB, VB, VIB,
VIIB, VIII, IB and IIB are called transition
elements. In transition elements the outermost
shell (valence shell) and penultimate shell (the
shell before outermost shell) are incomplete.
Common names of the elements:
Merits of long form of periodic table:
 The classification of elements is based
on fundamental property of elements, i.e.,
atomic number.
 It relates position of an element to its
electronic configuration in the valence
shell and hence, it posses similar
chemical properties.
 It explains variations and similarities in
the properties of elements in terms of
electronic configuration.
 Inert gases in which valence shells are
completely filled have been placed at the
end of each period. Such a location of
inert gases shows logical completion of
each period.
 The elements of a subgroup in a given
group have been placed separately. Thus,
dissimilar elements do not fall together.
 It provides clear demarcation of different
kinds of elements such as :
(i) active metals (ii) non-metals
(iii) metalloids
(iv) transition elements
(v) inert gases
(vi) lanthanides
(vii) actinides.
Defects (De-merits) of long form of
periodic table:
 Position of hydrogen is unresolved.
 It fails to accommodate lanthanides and
actinides in the main body of the table.
 The arrangement is unable to reflect
electron configuration of many elements
in transition group, lanthanides and
actinides.
90
CLASS-IX
DAY-10: WORKSHEET
1. Which of the following elements position
cannot be justified on the basis of atomic
weight?
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
number in which the element with
atomic number 24 is present.
a) 2, VB b) 4, VIB c) 5, VIIB d) 3, VB
5. Outer electronic configuration of the
atoms of the four elements are given
below:
(i) 3d0 4s1
(ii) 3s2 3p5
(iii) 4s2 4p6
(iv) 3d8 4s2
In which group and period of the periodic
table are they situated respectively?
(i)
(ii)
(iii)
(iv)
a) 4 and IA 3 and VIIA 4 and 0 4 and VIII
b) 4 and 0 3 and VII A 4 and IA 4 and VIII
c) 2 and IIA 4 and II A
4 and VIIA 4and 0
d) 3 and VIIA 4 and IA
4 and 0 4 and I A
6. Electronic configuration of three
elements are given below:
(i) 1s2 2s1
(ii) 1s2 2s2 2p5
2
2
(iii) 1s 2s 2p6 3s2 3p6 4s2
Choose the right option:
a)
b)
c)
d)
( i)
A lk a line
ea rt h m e ta l
H a lo g en
A lk a li
m e ta l
A lk a li
m e ta l
(ii )
A lk a li
m e ta l
A lk a line
ea rth m e ta l
H alo g en
A lk a line
ea rth m e ta l
NARAYANA GROUP OF SCHOOLS
(i ii)
H alog e n
A lk ali m e tal
A lk alin e
e a rth m et al
H alog e n
MPC BRIDGE COURSE
7. 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
8. 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
9. 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
10.Which of the following pair is against to
Mendeleev’s periodic law?
a) Chromium, Manganese
b) Sodium, Magnesium
c) Copper, Zinc
d) Tellurium, Iodine
DAY-11 : SYNOPSIS
1 Classification of elements into blocks
(i) s - block:
(a)The elements in which the last electron
enters the s-subshell of their outermost
energy level are called s- block elements.
(b)This block is situated at the extreme left
of the periodic table.
(c)It contains elements of groups IA and IIA.
(d)The general electronic configuration of
these elements is ns 1- 2 , where ‘n’
represents the outermost shell.
(e)The elements IA group elements are
called Alkali metals and that of IIA group
elements are known as Alkaline earth
metals.
 General characteristics of s-block
elements are:
• They are soft metals and have low
melting points.
91
CLASS-IX
• They are highly electropositive and have
low ionisation energies .
• They are highly reactive and form ionic
compounds.
• They show oxidation states of +1 and +2
and are good reducing agents.
(ii) p - block:
(a) The elements in which the last
electron enters the p-subshell of their
outermost energy level are called p- block
elements.
(b)This block is situated at the extreme
right side of the periodic table.
(c)It contains elements of groups IIIA, IVA,
VA, VIA, VIIA and VIIIA (exceptionhelium).
(d) The general electronic configuration
of these elements is ns2 np1-6, where ‘n’
represents the outermost shell.
(e) It includes metals, non metals ,
metalloids and inert gases.
 General characteristics of p-block
elements are:
• They form ionic as well as covalent
compounds.
• They have relatively high values of
ionisation energies.
• Most of them are non metals and are
highly electronegative.
• They show variable oxidation states and
form acidic oxides.
(iii) d - block:
(a) The elements in which the last
electron enters the d-subshell of the
penultimate energy level are called dblock elements.
(b)This block is situated in between ‘s’
and ‘p’ blocks of the periodic table.
(c)It contains elements of groups IB, IIB,
IIIB, IVB, VB, VIB, VIIB and VIII
groups.
(d) The general electronic configuration
of these elements is (n-1) d 1-10 ns 1- 2 ,
where (n-1) represents the penultimate
shell and ‘n’ represents the outermost
shell.
(e) It includes weak metals.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
 General characteristics of d-block
elements are:
• They are hard, high melting metals
showing variable oxidation states.
• They form coloured complexes and form
ionic as well as covalent
compounds.
• Most of them exhibit paramagnetism and
possess catalytic properties.
• They form alloys and are good conductors
of heat and electricity.
(iv) f - block:
(a) The elements in which the last
electron enters the f-subshell of the antipenultimate (third to the outer most shell
) shell are called f- block elements.
(b)This block is placed separately at the
bottom of the main periodic table.
(c)It consists of two series of elements
placed at the bottom of the periodic table.
The elements of first series follow
lanthanum and are called Lanthanides
and the elements of second series follow
actinium and are called actinides.
(d) The general electronic configuration
of these elements is (n-2)f1-14 (n-1) d 0-1
ns2, where ‘n’ represents the outermost
shell, (n-1) represents the penultimate
shell and (n - 2) represents the anti
penultimate shell.
 General characteristics of f-block
elements are:
• They are hard, high melting metals
showing variable oxidation states.
• They form coloured complexes and have
high densities.
• Most of the elements of actinide series
are radioactive.
(v) Inert gases:
(a) The elements of VIII A group, the
last column of p-block are known as inert
gases or noble gases or aerogens.
(b)The outermost shell of these elements
is completely filled and are hence nonreactive.
(vi) Representative elements:
(a) The elements of ‘s’ and ‘p’ block
elements, except inert gases are known
as representative elements.
92
CLASS-IX
(b)The outermost shell of these elements
are incompletely filled and hence are
highly reactive.
(c)These are also known as main group
elements or normal elements.
(vii) Transition elements:
(a) The elements of d- block in which the
outermost shell and penultimate shell
are incompletely filled are known as
transition elements.
(b)The name is derived from the fact that
they represent transition (change) in
character from reactive metals (elements
from IA and IIA) on one side and less
active metals of groups IB and IIB on the
other side.
(c)These elements are less reactive than
representative elements.
(viii) Inner transition elements:
(a) The elements in which the last three
principal shells are incomplete are
known as inner transition elements.
(b)The lanthanides and actinides are
included in this category.
(c)These are less reactive than transition
elements.
DAY-11: WORKSHEET
1. Match the correct pairs:
Electronic configurations Type of element
i) n s 2np 5
1) Alkali metal
1
ii) ns
2) Alkaline earth metal
2
6
iii) ns np
3) Transition Metal
iv) ns 2
4) Inert gas
1 – 10
v) (n – 1) d
ns1 – 2 5) Non metal
a) i  3, ii  1, iii  4, iv  5, v  2
b) i  1, ii  5, iii  4, iv  3, v  2
c) i  5, ii  1, iii  4, iv  2, v  3
d) i  5, ii  4, iii  2, iv  1, v  3
2. The electronic configuration of halogen is:
a) ns2np 6
b) ns2np3 c) ns2np 5 d) ns2
3. Hydrogen by containing one electron
forms H +. In this property, it resembles
with:
a) Transitional metals
b) Alkali earth metals
c) Alkali metals
d) Halogens
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
4. The electronic structure (n – 1) d 1 – 10 ns0
–2
is characteristic of:
a) Transition elements b) Lanthanides
c) Actinides
d) Rare – earths
5. The elements having the electronic
configuration, [Kr] 4d 10 f 14, 5s 2 p 6d 2 , 6s 2
belongs to:
a) s-block
b) p-block
c) d-block
d) f-block
6. An element has the electronic
configuration 1s2, 2s22p 6, 3s23p 63d 5, 4s1.
It is:
a) Li
b) Cr
c) B
d) Be
7. An element has electronic configuration
1s 2 2s 2 2p 63s 2 3p 4 . Predict their period,
group and block.
a) Period = 3rd, block = p, group = 16
b) Period = 5th, block = s, group = 1
c) Period = 3rd, block = p, group = 10
d) Period = 4th, block = d, group = 12
8. In the long form of the periodic table, all
the non-metals are placed under
a) s-block b) p-block c) d-block d) f-block
9. In the modern periodic table, the place of
the element with atomic number 31 is in:
a) s-block b) d-block c) p-block d) f-block
10. Elements of ‘d’ group are called:
a) Transition elements
b) Transuranic elements
c) Metals
d) Metalloids
DAY-12 : 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.
93
CLASS-IX
 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.
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.
NARAYANA GROUP OF SCHOOLS

MPC BRIDGE COURSE
(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 configuration: 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).
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.
As we move from left to right in the
periodic table, the atomic size decreases
and the nuclear pull increases. This
results in the outer electron being held
more tightly, increasing the ionisation
energy. Thus, as we move from left to
right, the ionisation energy increases due
to the decrease in size.
94
CLASS-IX
DAY-12: 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
b) 0.37 A , 0.37 A
o
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+
4. Which of the following has largest radius?
a) 1s2, 2s2, 2p6, 3s2
b) 1s2, 2s2, 2p6, 3s2 3p1
NARAYANA GROUP OF SCHOOLS
5.
6.
7.
8.
2
6
FIRST I.P.
 Some trends among the elements of
second and third periods are given below:
IE1 : Li < B < Be < C < O < N < F < Ne.
IE1 : Na < Al < Mg < Si < S < P < Cl < Ar.
IE2 : Li > C > B > Be.
IE2 : O > F > N > C.
IE2 : Na > Al > Mg.
IE3 : Mg > Na > Al.
 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.
MPC BRIDGE COURSE
c) 1s , 2s , 2p , 3s2 3p3
d) 1s2, 2s2, 2p6, 3s2 3p5
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
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
The factor that is not affecting the
ionisation energy is
a) Size of atom
b) Charge in the nucleus
c) Type of bonding in the crystalline
lattice
d) Type of electron involved
The graph of first ionisation enthalpy
versus atomic number is as follows:
2
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
a) High nuclear charge.
b) Small atomic size.
c) Metallic properties.
d) Strongly bound valence electrons.
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
95
CLASS-IX
DAY-13 : 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
non-metallic 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.
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.
96
CLASS-IX
MPC BRIDGE COURSE
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.






(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.
Among halogens, fluorine has lower ‘EA’
than
chlorine
due
to
greater
interelectronic repulsions in small sized
‘2p’ subshell.
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.
NARAYANA GROUP OF SCHOOLS
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.
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,
Variation of acidic and basic character
Metals are characterised by basic
character and non-metal characterised
by acidic character. Down the group, the
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.
97
CLASS-IX
MPC BRIDGE COURSE
DAY-13: WORKSHEET
DAY-14 : SYNOPSIS
1. The electronic configurations of the
elements X, Y, Z and J are given below.
Which element has the highest metallic
character ?
 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.
 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.
 Electron theory of valency - Kossel and
Lewis’ approach of bonding
A number of attempts were made to
explain formation of chemical bonds in
terms of electrons, but it was only in
1916, Kossel and Lewis’ succeeded
independently in giving a satisfactory
explanation. They proposed a theory,
based on electronic concept of atoms,
known as electron theory of valency.
The main postulates of this theory are:
1. The secret of stability of atoms: Atoms
with eight electrons in the outermost
shell (two in the case of Hydrogen,
Helium, Lithium and Beryllium) are
chemically more stable.
2. Cause of chemical reaction: The cause
for chemical reaction is to attain stability.
An atom achieves this by acquiring the
octet
configuration
(inert
gas
configuration) in its outermost shell.
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
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
NARAYANA GROUP OF SCHOOLS
98
CLASS-IX
3. Type of electrons taking part in a
chem ical rea ction ( or) chem ical
bonding: The electrons present in the
outermost shell of an atom are
responsible for chemical reaction.
The outermost shell is called valence
shell and hence, the electrons present
in it are called valence electrons. The
number of electrons taking part in a
chemical reaction is called valency of
that element.
4. Atta inment o f neare st inert gas
configuration: The atoms of various
elements achieve the nearest inert gas
configuration, either by transfer (losing
or gaining) or by sharing of electrons with
another atom.
This transfer or sharing of electrons
results in the development of an
attractive force between the atoms,
which holds the atoms together by a
bond.
 Electron Dot structure of atoms - Lewis’
Symbols
In the formation of any molecule or
formula unit, only the electrons present
in the outermost shell are shown. The
reason for not showing the inner shell
electrons is that, they are well protected
and do not involve in chemical reaction.
Therefore, valence electrons are
considered for the formation of the
chemical bonds.
G.N. Lewis’ introduced simple symbols
called Lewis’ symbols to denote the
valence electrons in an atom.
 Lewis’ symbols: The symbol of the
element, surrounded by the valence
electrons of its atom, represented in the
form of dots around it, is known as Lewis’
symbol or electron dot symbol.
 Significance of Lewis’ Symbols
(i) The number of dots present around
the symbol, gives the number of electrons
present in the outermost shell i.e.,
number of valence electrons.
(ii) The number of electrons present in
the outermost shell is the common
valency of the element.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
The common valency of the element is
equal to the number of dots around the
symbol (if the dots are < 4, then the
valency is equal to the number of dots
and if the number of dots > 5, then the
valency is 8 – number of dots.)
For example: Li, Be, B and C have
valencies 1, 2, 3 and 4 respectively and
N, O and F have 3, 2, 1 respectively (i.e.
8 – number of dots).
DAY-14: WORKSHEET
1. It was found that atoms having atomic
numbers of 2, 10, 18, 36, 54, 86 are very
stable and do not show any chemical
reactivity, these elements were found to
be gases and are called:
a) Inert gases
b) Diatomic gases
c) Monoatomic gases d) Noble gases
2. Which inert gas has a duplet
configuration in its valence shell?
a) Heliumb) Neon c) Argon d) Krypton
3. Assertion : Duplet configuration implies
that a given atom has 2 electrons in its
valence shell.
Reason :
Elements
with
octet
configuration in their valence shell are
stable.
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.
4. “The duplet and octet configuration of
electrons in the valence shell is most
stable and any atom having this
configuration will be in minimum state
of energy”. This statement was given by
a) Kossel and Lewis
b) Lewis and Debye
c) Kossel and London
d) Lewis and London
99
CLASS-IX
5. Chemical reactivity of an elements
depends on:
a) Outer shell electronic configuration.
b) Reactivity of the nucleus.
c) Core electrons.
d) None.
6. To attain a ______ participating atoms
redistribute their electrons to get a
electronic configuration of the nearest
noble gas either octet or duplet.
a) State of maximum energy
b) State of minimum energy
c) Stability
d) None
7. In each of the following sets, identify the
more stable species.
i) Al or Al3+, ii) N or N3–, iii) H or H2
i
ii
iii
3–
a) Al
N
H
b) Al3+
N 3–
H2
3+
c) Al
N
H
d) Al
N
H
8. Statement A : Pure metal is more stable
than its ore.
Statement B : Ore of metal is more stable
than pure metals.
a) ‘A’ is correct, ‘B’ is incorrect.
b) ‘B’ is correct, ‘A’ is incorrect.
c) ‘A’ and ‘B’, both are correct.
d) ‘A’ and ‘B’, both are incorrect.
9. Valency of the metal atom with respect
to oxygen is maximum in:
a) Mn2O7
b) OsO4 c) MnO2 d) CrO3
10. An element A is tetravalent and
another element B is divalent. The
formula of the compound formed by the
combination of these elements is:
a) A2B
b) AB
c) AB2
d) A2B3
DAY-15 : SYNOPSIS
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.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
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 – ),
2+
2
2+
CaO(Ca O ), MgO(Mg O 2– ) and MgCl 2
(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
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.
2. Structure of ionic solids:
Unit cell: There is a basic unit in an
ionic crystal, which when repeated
three-dimensionally, gives complete
crystal. This basic unit is called the unit
cell.
100
CLASS-IX
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.
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, called polar
solvents should be used. The best polar
solvent is water. Therefore, all ionic
compounds are dissolved in water.
5. Electrical conductivity:
Even though ionic solids consist of
opposite ions, they are bad conductors of
electricity. In ionic solids, a strong
electrostatic force of attraction, making
the ions immobile, holds the oppositely
charged
ions
together.
Hence,
conductivity is not possible.
However, in their fused or aqueous state,
ionic compounds are good conductors of
electricity owing to the presence of
mobile ions.
For instance, NaCl in its fused state or
in its aqueous solution, has free 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.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
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.
DAY-15: 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:
a) 1s2
b) 1s2 2s2 2p2
c) 1s2 2s2 2p5
d) 1s2 2s2 2p6
101
CLASS-IX
MPC BRIDGE COURSE
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
DAY-16 : 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?
a) Both assertion and reason are correct
and reason is the correct explanation of
assertion.
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
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.
b) Both assertion and reason are correct
but reason is not the correct explanation
of assertion.
Representation of covalent bond: The
covalent bond between a pair of two
atoms is represented by a small line[ – ].
c) Assertion is correct and reason is
incorrect.
For example, H 2 can be represented as
H–H.
d) Assertion is incorrect and reason is
correct.
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.
c) SO2 < N2 < ClF3 < LiF < K2O
d) ClF3 < N2 < SO2 < K2O < LiF
9. Assertion (A) :
non-volatile.
Ionic compound tend to be
Reason (R) : Inter molecular forces in
these compounds are weak.
The correct answer is:
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
NARAYANA GROUP OF SCHOOLS
Eg: Covalency of hydrogen molecule is
equal to 1 and that oxygen molecule is
2.
102
CLASS-IX
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:
MPC BRIDGE COURSE
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:
1) Atoms should be of small size.
2) They
should
have
high
electronegativity
and
ionisation
potential.
(or) H – H (hydrogen molecule)
(b)
Formation of chlorine molecule:
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.
(or) Cl – Cl (chlorine molecule)
 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.
Eg: SO2, PCl3, PBr3, SF6, SCl2, IF7, O2,
N2, F2, Cl2, Br2, I2, etc.
(or) O = O (oxygen molecule)
 Note: Number of electron pairs shared
between two atoms of the same element
is equal to the number of electrons short
of octet.
(c)
Formation of 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:
Examples of covalent compounds:
(or)
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.
NARAYANA GROUP OF SCHOOLS
H – Cl (Hydrogen chloride)
After formation of a covalent bond,
hydrogen has stable duplet
103
CLASS-IX
S.No.
1.
2
3.
S.No.
1.
2
3.
MPC BRIDGE COURSE
Type of
Definition
covalent bond
Single
The covalent
covalent bond bond formed
by sharing of 1
pair of
electrons
Double
The covalent
covalent bond bond formed
by sharing of 2
pairs of
electrons
Triple covalent The covalent
bond
bond formed
by sharing of 3
pairs of
electrons
Representation Examples
Type of
Definition
covalent bond
Single
The covalent
covalent bond bond formed
by sharing of 1
pair of
electrons
Double
The covalent
covalent bond bond formed
by sharing of 2
pairs of
electrons
Triple covalent The covalent
bond
bond formed
by sharing of 3
pairs of
electrons
Representation
H × H
H2 , Cl2 , F2
etc
(f) Formation of Carbondioxide molecule
– CO2:
H-H
O2
   × ××
O
O
   × ××
II. Covalent bonds based on the number
of electron pairs shared.
O=O
N2
×
×
×
(or) O = C = O (Carbondioxide)
N N
×
×
 Summary:
DAY-16: WORKSHEET
N N
 H
H ×
Examples
H2 , Cl2, F2
etc
H-H
O2
   × ××
O
O
   × ××
O=O
N2
×
×
×
N N
×
×
N N
configuration, and chlorine has stable octet
configuration.
(b) Formation of water molecule – H2O:
1. When two atoms of chlorine combine to
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
2. 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
3. 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
(c) Formation of Ammonia molecule – NH3:
4. Which of the following is a covalent
compound?
a) H2
b) CaO
c) KCl
d) Na2S
5. Which of the following substance has
covalent bonding?
(d) Formation of methane molecule –
a) Sodium chloride b) Solid neon
c) Copper
d) BeCl2
6. Which is a covalent compound?
:4CH
a) RbF
(e) Formation of carbon tetrachloride –
CCl 4 :
b) MgCl2
c) CaC2
d) NH3
7. 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
(Carbontetrachloride)
NARAYANA GROUP OF SCHOOLS
104
CLASS-IX
MPC BRIDGE COURSE
8. 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
9. 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
10. The molecule, which contains maximum
number of electrons, is:
a) CH4
b) CO2
c) NH3
d) BCl3
DAY-17 : 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.
Exception: Daimond
iii) Solubility: Covalent compounds are
soluble in non-polar solvents.
NARAYANA GROUP OF SCHOOLS
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 Hbonding 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.
105
CLASS-IX
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.
MPC BRIDGE COURSE
 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
vi) Directional property:
Covalent bonds are directional. The
direction is along the inter-nuclear axis
(line joining the nuclei of two reacting
atoms). This directional property leads
to different arrangement of the atoms
along the line of direction. The result
being that different molecules are formed
with the same structure. This
phenomenon is called Isomerism and the
molecules are called Isomers.
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.
NARAYANA GROUP OF SCHOOLS

O

H
 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.
106
CLASS-IX
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-17: 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.
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?
MPC BRIDGE COURSE
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
c) Both i and iii
b) Both i and ii
d) Only iii
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
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.
NARAYANA GROUP OF SCHOOLS
107
CLASS-IX
MPC BRIDGE COURSE
MATHEMATICS KEY
DAY-5 KEY
DAY-1 KEY
1) 3
2) 2
3) 3
4) 1
5) 2
6) 1
7) 2
8) 1
9) 2
10) 1
11) 2
12) 3
13) 2
14) 1
15) 1
16) 3
17) 1
18) 1
19) 1
DAY-2 KEY
1) 1
2) 4
3) 4
4) 1
5) 4
6) 3
7) 2
8) 2
9) 1
10) 1
11) 3
12) 1
13) 2
14) 1
15) 1
16) 3
17) 3
18) 4
19) 1
20) 1
21) 1
22) 4
1) 2
2) 1
3) 3
4) 1
5) 4
6) 3
7) 1
8) 2
9) 4
10) 3
11) 3
12) 1
13) 4
14) 3
15) 1
16) 4
17) 1,2,3,4
18) 1
19) 3
20) 3
21) 4
22) 1
23) 3
24) 2
25) a-s; b-r; c-q; d-r
26) 0
27) 9/7
28) 3
DAY-6 KEY
DAY-3 KEY
1) 3
2) 4
3) 3
4) 1
5) 2
6) 2
7) 1
8) 3
9) 4
10) 2
11) 3
12) 1
13) 4
14) 1
15) 1
16) 1
17) 1
18) 4
19) 1
20) 3
21) a-s; b-s; c-q; d-p,r
1) 1
2) 3
3) 4
4) 1
5) 2
6) 2
7) 3
8) 3
9) 4
10) 3
11) 1
12) 1
13) 1
14) 4
15) 4
16) 1
17) 2
18) 4
19) 1
DAY-7 KEY
1) 3
2) 1
3) 4
4) 2
5) 2
6) 1
7) 1
8) 1
9) 2
10) 2
11) 3
12) 1
13) 1
14) 1
15) 2
DAY-4 KEY
DAY-8 KEY
1) 3
2) 4
3) 2
4) 3
5) 3
6) 1
7) 1
8) 3
9) 1
10) 4
11) 2
12) 2
13) 1
14) 3
15) 3
16) 2
NARAYANA GROUP OF SCHOOLS
1) 1
2) 2
3) 1
4) 3
5) 2
6) 2
7) 3
8) 1
9) 2
10) 1
11) 2
12) 2,3
13) 1
14) 2
15) 3
108
CLASS-IX
MPC BRIDGE COURSE
DAY-9 KEY
DAY-13 KEY
1) 4
2) 3
3) 2
4) 3
1) 4
5) 2
6) 3
7) 1
8) 3
10) 1,2
9) 4
10) 1
11) 1
12) 2
13) 4
14) 4
15) 1
2) 4
3) to 9) Graphs
DAY-14 KEY
16) 1,2,3 17) 1
18) 3
19) 1
20) a-p,q,r,t; b-q; c-q,r; d-p,q,r,t
21) 1,2
DAY-10 KEY
1) 2
2) 3
3) 3
4) 3
5) 4
6) 1
7) 2
8) 1
9) 3
10) 2
11) 2
12) 1
13) 2
14) 2
15) 3
16) 2
17) 3
21) 2
18) 3
19) 2
20) 1
1) 2
2) 2
3) 3
4) 1
5) 3
6) 4
7) 2
8) 1
9) 2
10) 2
11) 1
12) 2
13) 1
14) 1
15) 1
16) 1
DAY-15 KEY
1) 2
2) 4
3) 4
4) 1
5) 2
6) 1
7) 3
8) 2
9) 3
10) 2
11) 3
12) 2
13) 2
14) 1
15) 4
16) 1
DAY-11 KEY
DAY-16 KEY
1) 1
2) 3
3) 1
4) 1
1) 2
2) 1
3) 4
4) 3
5) 1
6) 2
7) 3
8) 2
5) 1
6) 4
7) 3
8) 3
9) 1
10) 2
11) 2
12) 1
9) 4
10) 2
11) 1
12) 1
13) 4
14) 1
15) 3
16) 1
13) 1
14) 4
15) 1
17) 1
18) 1
19) a-t; b-q; c-s; d-p
DAY-17 KEY
DAY-12 KEY
1) 3
2) 3
3) 2
4) 2
1) 3
2) 1
3) 1
4) 3
5) 4
6) 3
7) 1
8) 1
5) 1
6) 1
7) 3
8) 1
9) 3
10) 2
11) 1
12) 4
9) 1
10) 1
11) 1
12) 2
13) 3
14) a-s; b-r; c-p; d-q
13) 2
14) 3
15) 1
16) 3
17) 1
18) 1
19) 3,4
NARAYANA GROUP OF SCHOOLS
109
CLASS-IX
MPC BRIDGE COURSE
PHYSICS KEY
DAY-7 KEY
DAY-1 KEY
1) 1
5) 3
2) 2
6) 1
3) 4
7) 2
1) 4
2) 2
3) 3
4) 1
5) 3
6) 1
7) 1
8) 4
9) 1
10) 2
11) 1
12) 3
4) 2
8) 1
DAY-8 KEY
DAY-2 KEY
1) 3
2) 1
3) 4
4) 3
1) 2
2) 2
3) 3
4) 3
5) 3
6) 1
7) 1
8) 2
5) 2
6) 1
7) 4
8) 1
9) 3
10) 2
11) 1
12) 3
13) 2
9) 1
DAY-9 KEY
DAY-4 KEY
1) 3
2) 2
3) 2
5) 1
6) 1
7) 3
4) 2
1) 2
2) 4
3) 2
4) 2
5) 3
6) 1
7) 3
8) 3
9) 4
10) 2
11) 2
12) 3
DAY-4 KEY
DAY-10 KEY
1) 3
2) 3
3) 4
4) 3
5) 1
6) 2
7) 2
8) 4
9) 3
10) 2
1) 1
2) 2
3) 1
4) 2
5) 3
6) 4
7) 2
8) 1
9) 1
10) 1
11) 3
12) 1
DAY-5 KEY
DAY-11 KEY
1) 2
2) 1
3) 3
4) 1
1) 2
2) 1
3) 2
4) 3
5) 1
6) 2
7) 3
8) 2
5) 1
6) 1
7) 2
8) 3
9) 2
10) 2
11) 2
13) 4
9) 2
10) 2
11) 3
12) 4
13) 3
14) 4
DAY-12 KEY
DAY-6 KEY
1) 2
2) 3
3) 1
5) 1
6) 4
7) 1
NARAYANA GROUP OF SCHOOLS
4) 2
1) 2
2) 1
3) 1
4) 1
5) 3
6) 2
7) 2
8) 3
9) 1
10) 1
11) 1
110
CLASS-IX
MPC BRIDGE COURSE
DAY-13 KEY
DAY-18 KEY
1) 1
2) 3
3) 2
4) 1
1) 1
2) 1
3) 1
4) 1
5) 4
6) 2
7) 2
8) 2
5) 2
6) 1
7) 1
8) 1
9) 2
10) 2
11) 2
12) 2
9) 2
10) 1
11) 1
12) 4
13) 4
CHEMISTRY KEY
DAY-14 KEY
DAY-1 KEY
1) 1
2) 2
3) 2
4) 3
1) 1
2) 1,2,3,4
5) 3
6) 1
7) 4
8) 3
4) 1
5) 2
6) 2
9) 4
10) 4
8) 3
9) 2
10) 3
DAY-15 KEY
3) 2
7) 3
DAY-2 KEY
1) 1
2) 2
3) 3
4) 4
1) 1
2) 4
3) 3
4) 3
5) 1
6) 2
7) 2
8) 3
5) 2
6) 3
7) 4
8) 2
9) 4
10) 3
11) 1
12) 1
9) 4
10) 4
13) 1
14) 1
DAY-3 KEY
DAY-16 KEY
1) 1
2) 1,2
3) 1,2
4) 2
7) 4
8) 3
1) 2
2) 1
3) 2
4) 3
5) 3
6) 4
5) 3
6) 2
7) 3
8) 1
9) 2,3
10) 3
9) 1
10) 2
11) 1
DAY-4 KEY
DAY-17 KEY
1) 2
2) 2
3) 4
4) 1
5) 1
6) 4
7) 3
8) 3
9) 2
10) 4
11) 2
12) 2
NARAYANA GROUP OF SCHOOLS
1) 1
2) 1,2,3,4
4) 1
8) 3
5) 2
9) 2
3) 2
6) 2
10) 3
7) 3
DAY-5 KEY
1) 1
2) 4
3) 3
4) 3
5) 2
6) 3
7) 4
8) 2
9) 4
10) 4
111
CLASS-IX
MPC BRIDGE COURSE
DAY-6 KEY
DAY-12 KEY
1) 2
2) 2
3) 3
4) 3
1) 3
2) 1
3) 2
4) 1
5) 3
6) 4
7) 2
8) 3
5) 3
6) 3
7) 3
8) 2
9) 4
10) 3,4
9) 1,3,4
10) 1
DAY-7 KEY
DAY-13 KEY
1) 4
2) 4
3) 3
4) 4
1) 3
2) 3
3) 1
4) 3
5) 4
6) 2
7) 3
8) 1
5) 2
6) 2
7) 4
8) 1
9) 3
10) 4
9) 3
10) 2
DAY-8 KEY
DAY-14 KEY
1) 4
2) 3
3) 1
4) 1
1) 1,3,4
2) 1
3) 2
4) 1
5) 3
6) 4
7) 2
8) 2
5) 1
6) 2,3
7) 2
8) 2
9) 1
10) 4
9) 2
10) 3
DAY-9 KEY
DAY-15 KEY
1) 1
2) 4
3) 2
4) 4
1) 3
2) 3
3) 2
4) 3
5) 3
6) 3
7) 1
8) 4
5) 3
6) 4
7) 4
8) 1
9) 2,3
10) 1
9) 3
10) 1,2
DAY-10 KEY
DAY-16 KEY
1) 1,2
2) 3
3) 2
4) 2
1) 3
2) 3
3) 3
4) 1
5) 1
6) 3
7) 1
8) 3
5) 4
6) 4
7) 4
8) 4
9) 2
10) 4
9) 3
10) 4
DAY-11 KEY
DAY-17 KEY
1) 3
2) 3
3) 3
4) 1
1) 1
2) 2
3) 3
4) 2
5) 3
6) 2
7) 1
8) 2
5) 1
6) 4
7) 2
8) 1
9) 3
10) 1
NARAYANA GROUP OF SCHOOLS
9) 4
112