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PREPARED BY: PROF. ALTAF GOVT. COLLEGE DAVI GALI
Chapter 1
BASIC CONCEPTS
MULTIPLE CHOICE QUESTIONS
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
Ans:
Isotopes differ in:
(a) properties which depend upon mass
(b)
arrangement of electrons in orbitals
(c) chemical properties
(d) the extent to which they may be affected in electromagnetic field.
Which of the following statements is not true?
(a) Isotopes with even atomic masses are comparatively abundant.
(b) Isotopes with odd atomic masses are comparatively abundant.
(c) Isotopes with even atomic masses and even atomic numbers are comparatively abundant.
(d) Isotopes with even atomic masses and odd atomic numbers are comparatively abundant.
Many elements have fractional atomic masses. This is because:
(a) the mass of the atom is itself fractional.
(b) atomic masses are average masses of isobars.
(c) atomic masses are average masses of isotopes.
(d) atomic masses are average masses of isotopes proportional to their relative abundance.
The mass of one mole of electrons is:
(a) 1.008 mg
(b)
0.55 mg
(c) 0.184 mg
(d)
1.673 mg
27 g of Al will react completely with how much mass of O2, to produce Al2O3:
(a) 8 g of oxygen
(b)
16 g of oxygen
(c) 32 g of oxygen
(d)
24 g of oxygen
The number of moles of CO2 which contain 8.0 g of oxygen:
(a) 0.25
(b)
0.50
(c) 1.0
(d)
1.50
The largest number of molecules are present in:
(a) 3.6 g of H2O
(b)
4.8 g of C2H5OH
(c) 2.8 g of CO
(d)
5.4 g of N2O5
One mole of SO2 contains:
(a) 6.02  1023 atoms of oxygen
(b)
18.1  1023 molecules of SO2
(c) 6.02  1023 atoms of sulphur
(d)
4 g of atoms of SO2
The volume occupied by 1.4 g of N2 at S.T.P is:
(a) 2.24 dm3
(b) 22.4 dm3
(c) 1.12 dm3
(d) 112 cm3
A limiting reactant is the one which:
(a) is taken in lesser quantity in grams as compared to other reactants.
(b) is taken in lesser quantity in volume as compared to the other.
(c) carries the maximum amount of the product which is required.
(d) gives the minimum amount of the product under consideration.
(1)
a
(2)
c
(3)
d
(4)
b
(5)
d
(6)
a
(7)
a
(8)
c
(9)
c
(10)
d
Q.4. (a) What are ions? Under what conditions are they produced? Locate the atoms bearing
3
1
2
negative charges in PO4 . MnO4 . Cr2O7 .
Ans.(a) Ions are those species, which bear positive or negative charges.
Conditions of ion formation:
[1]
Composed by: Malik Khurshid Dhar Bazar Ph: 05824-420161
PREPARED BY: PROF. ALTAF GOVT. COLLEGE DAVI GALI
(i) Ionic compounds when dissolved in suitable solvent i.e. polar solvents produce positive and
negative ions. (ii) Ionic compounds present in the fused state are also free in the form of ions. (iii)
In mass spectrometer, the ions are produced by the bombardment of electrons in the ionization
chamber. These ions are mostly positive. (iv) X-rays when thrown upon gaseous substances,
ionize the molecules of gases.
Position of negative charges in ions:
The positions of negative charges in these ions can be located from the structure of these ions.
The negative charge is present on the electronegative atom of the ion.
O
O
O
O
O
║
—
C
r—
O
—
C
r—
(i)
P
(ii) Mn
(iii)
O
O
 ╱╲  O
–
–
–
O
O
O
O
O
O
O
O
O
Q.5
(b) How does a mass spectrograph show the relative abundance of isotopes of an element.
Ans(b) RELATIVE ABUNDANCE OF ISOTOPES
Definition: The %age of an isotope of an element relative to other isotope of the same element is called
relative abundance of isotope.
Important informations about natural abundance:
1.
All the elements which have isotopes have a definite ratio of natural abundance.
2.
The properties of an element which are described in literature are the properties of the most
abundant isotope of that element.
3.
Relative abundance of isotopes is determined with the help of mass spectrometer.
4.
There are more than 280 different isotopes that occur in nature. This also includes 40
radioactive isotopes.
5.
There are also about 300 unstable radioactive isotopes which have been produced during
artificial disintegration.
6.
The elements which have only one isotope are called mono-isotopic substances.
7.
The elements of odd atomic number mostly do not have more than two stable isotopes.
8.
The elements of even atomic number usually contain large number of isotopes and mass
number multiple of four are particularly quite abundant. For example,
16
24
28
40
56
8O , 12Mg , 14Si , 20Ca and 26Fe form about 50% of earth’s crust.
9.
Out of 280 naturally occurring isotopes, there are 154 isotopes which have even mass
numbers and even atomic numbers.
10. Number of isotopes of some elements are:
Li

Two isotopes
H,C,N,O, Ne 
Three isotopes each
Ni

Five isotopes
Ca, Pd

Six isotopes each
Cd

Nine isotopes
Sn

Eleven isotopes
The natural abundance of some isotopes of common elements are given in table (1.2):
Table (1.2): Natural abundance of some common isotopes
Element
Relative Abundance in %
1
2
Hydrogen
H = 99.985
H = 0.015
12
13
Carbon
C = 98.893
C = 1.107
14
15
Nitrogen
N = 99.634
N = 0.366
16
17
Oxygen
O = 99.759
O = 0.037 18O = 0.204
32
33
Sulphur
S = 95.0
S = 0.76, 34S = 4.22, 36S = 0.014
17
Fluorine
F = 100
35
36
Chlorine
Cl = 75.53
Cl = 24.47
80
81
Bromine
Br = 50.54
Br = 49.49
127
Iodine
I = 100
DETERMINATION OF RELATIVE MASSES OF ISOTOPES BY MASS SPECTROMETRY:
Let us first of all introduce the basic terminology.
[2]
Composed by: Malik Khurshid Dhar Bazar Ph: 05824-420161
PREPARED BY: PROF. ALTAF GOVT. COLLEGE DAVI GALI
Mass spectrometry:
It is a technique in which gas molecules are converted to gaseous ions, which are separated on
the basis of their mass to charge (m/e) ratio.
Mass spectrometer:
The instrument used to separate positively charged particles on the basis of their m/e values and
to get the record on the photographic plate or electrometer is called mass spectrometer.
Mass spectrum:
It is a record of the masses and the relative abundances of molecular ions and positively charged
fragments formed from it by the electron bombardment. The m/e ratios are taken along abscissa
(x-axis), while the relative abundances along the ordinate (y-axis).
Basic functions of mass spectrometer:
These are three basic functions of mass spectrometry:
(i) To vapourize compounds.
(ii) To produce ions from the neutral species in the vapour phase.
(iii) To separate ions according to their m/e ratios, with the help of electric and magnetic fields.
MASS SPECTROMETERS
(a)
Aston’s mass spectrograph:
Aston’s mass spectrograph was first designed to identify the isotopes of an element on the basis
of their atomic masses.
(b)
Dempster’s mass spectrometer:
It has been designed for identification of elements which are found in solid state. Dempster’s
mass spectrometer is superior to that of Aston’s mass spectrometer.
Instrumentation:
The major functioning parts of this instrument are as follows:
(i)
Vapourization chamber:
The substance whose analysis is required for the separation of isotopes is converted into vapours.
The pressure of vapours is reduced to 106—107 torr. These vapours at low pressure are allowed
to enter the ionization chamber.
(ii)
Ionization chamber:
In this chamber fast moving electrons are bombarded which knock out electrons from neutral
atoms. Thus they are converted into ions. These particles may consist of single atoms having
positive charge. They may have different masses depending upon the nature of isotopes of that
element. Different isotopes of same element form ions of different masses and different m/e
values.
(iii)
Electrical field:
A potential difference (E) of 500-2000 volts is applied between the plates to accelerate these
positive ions.
(iv)
Magnetic field:
After acceleration, these ions are sent to the magnetic field of strength (H) through the slit. The
magnetic field is applied in a direction which is perpendicular to the path of positive ions. Due to
this, positively charged ions move in a semi-circular path and then pass through the slit. It may be
noted that a heavy ion will travel slowly in a circular path as compared to the lighter ion. Fig.
(1.2)
(v)
Mathematical explanation:
A mathematical relationship for m/e in terms of radius of curvature of the ions, strength of
electrical field (E) and magnetic field strength (H) is
H2r2
m/e = 2E
m/e is constant quantity for an ion. So if we change E, by keeping H constant, then r will change
to keep m/e constant. When E is increased, r increases and the ion of particular m/e value will fall
on the electrometer at a different place.
[3]
Composed by: Malik Khurshid Dhar Bazar Ph: 05824-420161
PREPARED BY: PROF. ALTAF GOVT. COLLEGE DAVI GALI
(vi)
(vii)
(viii)
If H is increased by keeping E constant, then the radius of curvature r will decrease to keep m/e
value constant in the above equation. So ion will change its position for falling on the
electrometer.
By changing either electric field or magnetic field, the ion of particular m/e value can be made to
fall at the required place i.e. front of the slit and be recorded.
Electrometer or ion collector:
This electrometer records the ions of different m/e values. The electrometer is also called ion
collector.
Current strength and height of peak:
As the number of ions of particular m/e value is greater, greater will be the current generated and
height of the peak.
Comparison with C – 12:
The same experiment is performed with C – 12. The current strengths produced by other particles
are compared with the current strengths of C-12. In this way exact mass number of the isotope is
measured.
Negative
grid
(–)
Analyzer
(Magnet)
Neon gas
inlet
Heated
(–) filament
Electron beam
(+)
To vacuum pump
Detector
20 21 22
Mass
number
Evacuated
glass
envelope
Fig. (1.2) Diagram of a simple mass sepectrometer
(ix)
(x)
Modern spectrograph:
Now-a-days, in modern instruments, each ion strikes a detector, ionic current is amplified and is
fed to the recorder. The recorder makes a graph showing the relative abundance.
Analysis of isotopes of Ne:
When neon gas is analyzed by mass spectrometer, then a computer plotted graph is obtained as
shown in the Fig. (1.3)
100
90.924%
90
80
70
60
50
40
30
20
8.82%
10
0.257%
20
21
22
Fig. (1.3): Computer plotted graph for the isotopes of neon.
Separation of isotopes:
The separation method depends upon the properties of isotopes. Various techniques are as follow:
(i)
Gaseous diffusion
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Composed by: Malik Khurshid Dhar Bazar Ph: 05824-420161
PREPARED BY: PROF. ALTAF GOVT. COLLEGE DAVI GALI
(ii) Thermal diffusion
(iii) Distillation
(iv) Ultracentrifuge
(v) Electromagnetic separation
(vi) Laser separation
FRACTIONAL ATOMIC MASSES:
Most of the elements have their atomic masses in fractions. This is due to the reason that atomic
masses depend upon following two things:
(i)
Number of possible isotopes.
(ii)
Natural abundances of isotopes.
Table (1.3) Atomic masses and abundances of several
naturally occurring isotopes
Isotope
Mass
number
Percent natural
abundance
Relative
atomic mass
Average
atomic mass*
of elements
hydrogen1
hydrogen2
carbon12
carbon13
oxygen16
oxygen17
oxygen18
copper63
copper65
cesium133
uranium235
uranium238
1
2
12
13
16
17
18
63
65
133
235
238
99.985
0.015
98.90
1.10
99.762
0.038
0.200
69.17
30.83
100
0.720
99.280
1.007825
2.0140
12 (by
definition)
13.003355
15.994915
16.999131
17.999160
62.939598
64.927793
132.905429
235.043924
238.050784
1.00794
12.011
15.9994
63.546
132.905
238.024
*(Values, established by international agreement, may differs slightly from calculated averages).
Q.9. Justify the following statements:
(a)
180 grams of glucose and 342 grams of sucrose have same number of molecules, but different
number of atoms present in them. Justify it.
Ans: 180 grams of glucose (C6H12O6), and 342 grams of sucrose (C12H22O11) are one mole of each.
One mole of various substances contain equal number of molecules i.e. 6.02  1023
One molecule of C6H12O6 has 24 atoms. The total number of atoms of glucose in one mole is 24 
6.02  1023. One molecule of C12H22O11 has 45 atoms. The total number of atoms of sucrose in
one mole of sucrose is 45  6.02  1023. It means that one mole of both glucose and sucrose will
have different number of atoms.
(b)
Mg atom is twice heavier than that of carbon, atom.
Ans: The atomic mass of Mg is 24 g mol–1 which is twice in mass as compared to the atomic mass of C
i.e. 12 g mol–1. So Mg atom is twice heavier than that of carbon.
(c)
180 grams of glucose and 342 grams of sucrose have same number of molecules but
different number of atoms present in them.
Ans: 180 grams of glucose (C6H12O6), and 342 grams of sucrose (C12H22O11) are one mole of each.
One mole of various substances contain equal number of molecules i.e. 6.02  1023
One molecule of C6H12O6 has 24 atoms. The total number of atoms of glucose in one mole is 24 
6.02  1023. One molecule of C12H22O11 has 45 atoms. The total number of atoms of sucrose in
one mole of sucrose is 45  6.02  1023. It means that one mole of both glucose and sucrose will
have different number of atoms.
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(d)
Ans:
4.9 g of H2SO4, when completely ionized in water, have equal number of positive and
negative charges, but number of positively charged ions are twice the number of negatively
charged ions.
2–
When one molecule of H2SO4 ionizes, it produces two H and one SO
4 ion. Hydrogen ion
contains +1 charge while sulphate ion has 2 charge. The ions produced by complete ionization
of 4.9 grams of H2SO4 in water will have equal +ve and ve charges but the number of H ions
is twice than number of negatively charged sulphate ions.
(e)
One mg of K2CrO4 has thrice the number of ions than the number of molecules when
ionized in excess of water.
2–
Ans: When K2CrO4 ionizes in water, its one molecule gives three ions i.e. two K and one CrO
4
(chromate) ions. The ratio between number of molecules and number of ions of K 2CrO4 is 1:3. So
one mg of K2CrO4 has thrice the number of ions than the number of molecules when ionized in
water.
(f)
Two grams of H2, 16g of CH4 and 44g of CO2 occupy separately the volumes of 22.414 dm3
at STP although the sizes and masses of molecules of three gases are very different from
each other.
Ans: One mole of an ideal gas at S.T.P occupies a volume of 22.414 dm3. Sizes and masses of
molecules of different gases do not affect the volume. Normally it is known that in the gaseous
state, the distance between the molecules is 300 times greater than their diameter. Therefore two
grams of H2, 16g of CH4 and 44g of CO2 (1 mole of each gas) separately occupy a volume of 22.4
dm3. This is called molar volume (Vm).
Q.10. Calculate each of the following quantities:
(a) Mass in grams of 2.74 moles of KMnO4.
(b) Moles of O atoms in 9.00 g of Mg(NO3)2
(c) Number of O atoms in 10.037 g of CuSO4.5H2O.
(d) Mass in kilograms of 2.6  1020 molecules of SO2.
(e) Moles of Cl atoms in 0.822 g C2H4Cl2.
(f) Mass in grams of 5.136 moles of silver carbonate (Ag2CO3)
(g) Mass in grams of 2.78  1021 molecules of CrO2Cl2
(h) Number of moles and formula units in 100 g KClO3

(i) Number of K ions, ClO 3 ions, Cl atoms and O atoms in (h).
Solution:(a)
Data: Moles of KMnO4 given =
2.74
Molar mass of KMnO4
Formula applied:
Number of moles
Mass of KMnO4
Putting the values
=
=
=
2.74 moles of KMnO4 has mass =
1
39 + 55 + 4  16 = 158 g mol
Mass of KMnO4
Molar mass of KMnO4
Number of moles  molar mass
2.74  158 g = 432.92 g
Ans.
(b) Data:
Mass of Mg(NO3)2
Molar mass of Mg(NO3)2
Number of moles
=
=
9g
24+2(14+316)
=
148 g mol
Mass given of Mg(NO3)2
Molar mass of Mg(NO3)2
=
1
Putting the values
[6]
Composed by: Malik Khurshid Dhar Bazar Ph: 05824-420161
PREPARED BY: PROF. ALTAF GOVT. COLLEGE DAVI GALI
Number of moles
9g
= 0.06
148 g mole1
=
1 mole of Mg(NO3)2 contains
moles of O atoms
=
0.06 moles of Mg(NO3)2 contain
moles of O atoms
=
6
0.06  6
=
(c) Data:
Mass of CuSO4 . 5H2O
Molar mass of CuSO4 . 5H2O
0.36 Ans.
=
=
10.037 g
63.5 + 32 + 4  16 + 5  18
=
249.5 g mol
=
Mass of CuSO4 . 5H2O
Molar mass of CuSO4 . 5H2O
=
10.037 g
= 0.04
249.5 g mole1
1
Formula applied:
Moles of CuSO4 . 5H2O
Substituting the values
Moles of CuSO4 . 5H2O
So,
1 mole of CuSO4.5H2O has
moles of ‘O’
=
0.04 moles of CuSO4.5H2O
have moles of ‘O’
=
Now we calculate number of ‘O’ atoms
Formula applied:
Number of ‘O’ atoms
=
Substituting the values
Number of ‘O’ atoms
=
=
(d) Data:
Number of SO2 molecules
Molar mass of SO2
Mass in kg
Formula applied:
Mass of SO2 molecules
9
9  0.04 = 0.36
Number of moles  NA
0.36  6.02  1023
2.1672  1023 Ans.
=
=
=
2.6  1020
64 g mol–1
?
=
Molar mass
 Number of molecules
NA
Substituting the values
Mass of 2.6 1020 molecules of SO2
=
=
=
64 g mol–1
 2.6  1020
6.02  1023 g mol–1
27.641  103 g
27.641  103
kg
1000
=
2.764  105 kg Ans.
=
=
0.822 g
24 + 4 + 71 = 99 g mol1
(e) Data:
Mass of C2H4Cl2
Molar mass of C2H4Cl2
Formula applied:
[7]
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Number of moles of C2H4Cl2
=
Mass of C2H4Cl2
Molar mass of C2H4Cl2
Putting the values
Number of moles of C2H4Cl2
Since, 1 mole of C2H4Cl2 has Cl
0.0083 moles of C2H4Cl2 has Cl
=
=
=
=
0.822 g
= 0.0083
99 g mol1
2 moles
2  0.0083
0.0166 Ans.
(f) Data:
Moles of Ag2CO3
=
Molar mass of Ag2CO3 =
=
Formula applied:
Mass in grams =
Substituting the values
Mass of 5.136 moles of Ag2CO3 =
=
5.136
2  107.87 + 12 + 3  16
275.74 g mol1
number of moles  molar mass
5.136  275.74
1416.2 g
Ans.
(g) Data:
Molecules of CrO2Cl2 =
Molar mass of CrO2Cl2 =
=
1 mole of CrO2Cl2 contains
molecules
=
23
6.02  10 molecules of
CrO2Cl2 have mass
=
Formula applied:
Mass of molecules of CrO2Cl2 =
2.78  1021
52 + 32 + 71
155 g mol1
6.02  1023
155 g
Molar mass  Number of moelcules
NA
Substituting the values
Mass of 2.781021 molecules of CrO2Cl2 =
=
=
(h) Data:
Mass of KClO3 given
Molar mass of KClO3
Formula applied:
Number of moles of KClO3
155
 2.78  1021
6.02  1023
71.58  102
0.7158 g
Ans.
=
=
100 g
39 + 35.5 + 48 = 122.5 g
=
Mass of KClO3
Molar mass of KClO3
=
100 g
122.5 g mol1
=
0.816 moles
Substituting the values
Number of moles of KClO3
Since 1 mole of KClO3
contains formula units
0.816 moles of KClO3
contains formula units
Ans.
=
6.02  1023
=
6.02  1023  0.816
[8]
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=
(i)(1) 1 mole of KClO3 contains K ions =
0.816 moles of KClO3 contains K ions=
=

(2) 1 mole of KClO3 contains ClO 3 ions =

0.816 moles of KClO3 contains ClO 3 ions =
=
(3)
4.91  1023
Ans.
6.02  1023
6.02  1023  0.816
4.91  1023 Ans.
6.02  1023
6.02  1023  0.816
4.91  1023 Ans.
Similarly
1 mole of KClO3 contains Cl ions
=
4.91  1023 Ans.
1 mole of KClO3 contain ‘O’ atoms
=
3  6.02  1023 Ans.
0.816 moles contains O atoms
=
3  6.02  1023  0.816
=
1.474  1024
Q.11. Aspartame, the artificial sweetner has a molecular formula of C14H18N2O5.
(a) What is the mass of one mole of aspartame.
(b) How many moles are present in 52g of aspartame.
(c) What is the mass in grams of 10.122 moles of aspartame?
(d) How many hydrogen atoms are present in 2.43 g of aspartame.
Solution. (a)
Data:
Molecular formula of aspartame =
C14H18N2O5
Mass of one mole of aspartame =
?
Since, C14H18N2O5 is one mole of aspartame, therefore,
Mass of one mole of C14H18N2O5
=
14C + 18H + 2N + 5O
Mass of one mole of C14H18N2O5
=
14  12 +18  1+2  14+516
=
168 + 18 + 28 + 80
=
294 g mole
So 1 mole of aspartame
=
294 g mol1
Mass of aspartame
=
52 g
Molar mass of aspartame
Formula applied:
=
294 g mol1
Number of moles
=
Mass of aspartame
Molar mass of aspartame
=
52 g
= 0.177
294 g mole1
=
=
=
10.122
?
294 g
(b)
Ans.
Ans.
Data:
Substituting the values
Number of moles
Data:
Number of moles of aspartame
Mass of 10.122 moles of aspartame
Molar mass of aspartame
Formula applied:
Ans.
(c)
[9]
Composed by: Malik Khurshid Dhar Bazar Ph: 05824-420161
PREPARED BY: PROF. ALTAF GOVT. COLLEGE DAVI GALI
Mass of 10.122 moles of aspartame
Putting the values
Mass of 10.122 moles of aspartame
(d)
=
Moles of aspartame  Molar mass
=
10.122 moles  294 g mol1
=
2975.87 g
=
=
2.43 g
294 g mol1
Ans.
Data:
Mass of aspartame
Molar mass of aspartame
Formula applied:
Number of moles of aspartame =
Mass of aspartame
Molar mass of aspartame
Putting the values
Number of moles of aspartame
Moles of H in 1 mole of aspartame
0.00826 moles of aspartame have moles of H
Number of atoms of H
=
=
=
=
=
=
=
2.43 g
= 0.00826
294 g mol1
18 moles
18  0.00826 = 0.1487
Number of moles  NA
0.1487  6.02  1023
0.896  1023
8.96  1022 Ans.
Q.13. In each pair, choose the larger of the indicated quantity, or state if the samples are equal.
(a) Individual particles:0.4 moles of oxygen molecules or 0.4 moles of oxygen atoms.
(b) Mass: 0.4 moles of ozone molecules or 0.4 moles of oxygen atoms.
(c) Mass: 0.6 moles of C2H4 or 0.6 moles of I2.
(d) Individual particles:4.0 g N2O4 or 3.3 g SO2.
(e) Total ions: 2.3 moles of NaClO3 or 2.0 moles of MgCl2.
(f) Molecules: 11.0 g H2O or 11.0 g H2O2.
(g) Na ion: 0.500 moles of NaBr or 0.0145 kg of NaCl.
(h) Mass: 6.02  1023 atoms of 235U or 6.02  1023 atoms 238U
Solution. (a) Both are equal.
Reason:
0.4 moles of O2 and 0.4 moles of ‘O’ atoms, both are equimolar quantities. So they have equal
number of particles.
i.e.
0.4  6.02  1023
=
2.408  1023 particles
(b)
Data:
Mass of 0.4 moles of O3
=
?
Number of moles
=
0.4
Molar mass of O3
=
48 g mol1
Formula applied:
Mass =
Number of moles  molar mass
Putting the values
=
0.4  48 = 19.2 g
Mass of 0.4 moles of ‘O’ atom =
?
Atomic mass of O
=
16 g mol1
Number of moles of O
=
0.4
Mass = Number of moles  atomic mass = 0.4  16 = 6.4 g
0.4 moles of O3 is larger
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(c)
Data:
Number of moles given = 0.6
Formula applied:
Mass = No. of moles  molar mass
Molar mass of C2H4 = 28 g mol1
Putting the values
Mass of C2H4
= 0.6  28
Data:
Number of moles of I2 given = 0.6
Formula applied:
Mass = Number of moles  molar
Molar mass of I2
= 2  126.9
Putting the values
Mass of I2 = 253.8 g mol1 = 0.6253.8
= 16.8 g
= 152.28 g
0.6 moles of I2 is larger quantity
(d) Data:
Mass of N2O4 = 4 g
Individual particles
Molar mass of N2O4
Data:
Mass of SO2 = 3.3 g
=?
Individual particles
=?
= 28 + 64
Molar mass of SO2 = 32 + 32 = 64 g
mol1
= 92 g mol–1
Formula applied:
Formula applied:
Mass of SO2
Number of moles
Number of moles =
Molar
mass of SO2
Mass of N2O4
=
Substituting the values
Molar mass of N2O4
Number of moles of SO2
Substituting the values
3.3 g
4g
=
= 0.055
Number of moles of N2O4=
1
64 g mol1
92 g mol
= 0.043 moles 64 g(1mole) has molecules = 6.02  1023
Now we calculate number of molecules Now we calculate number of molecules
of SO2
of N2O4
Formula applied:
Formula applied:
Number of molecules = moles  NA
Number of molecules = moles  NA
Substituting the values
Substituting the values
23
= 0.0515  6.02  1023
= 0.0436.02  10
= 0.31  1023
= 0.258  1023
= 3.1  1022 molecules
= 2.58  1022 molecules
So, 3.3 g of SO2 contains larger number of individual particles
(e)
(i)
Data:
Moles of NaClO3
= 2.3
Number of ions in 2.3 moles of NaClO3
=?

1 mole of NaClO3 contains Na ions
= 6.02  1023
2.3 moles of NaClO3 contain Na ions
= 2.3  6.02  1023
= 13.846  1023

1 mole of NaClO3 cotains ClO 3 ions
= 6.02  1023

2.3 moles of NaClO3 contains ClO 3

Total ions = Na ions + ClO 3 ions
= 13.846  1023
= 13.846  1023 + 13.846  1023
= 27.692  1023
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= 2.7692  1024
Ans.
(ii)
Data:
Moles of MgCl2
Total number of ions in 2 moles of MgCl2
2+
1 mole of MgCl contains Mg ions
=2
=?
2+
2 moles of MgCl2 contain Mg ions
= 2  6.02  1023=12.041023
2
= 6.02  1023
1 mole of MgCl2 contains Cl ions
= 2  6.02  1023
2 moles of MgCl2 contains Cl ions
2+
Total ions = Mg ions + Cl ions
= 2  2  6.02  1023
= 36.12  1023
= 12.04  1023 + 24.08  1023
= 3.612  1024
Ans.
So 2 moles of MgCl2 contain larger number of ions
(f) (i)
Data:
Mass of H2O
Molar mass of H2O
=
11.0 g
=
18 g mol1
Number of molecules in 11.0 g of H2O =
?
Formula applied:
Number of moles of water
=
Mass of water
Molar mass of water
=
11.0 g
18 g mol1
Substituting the values
Number of moles of water
=
0.61 moles
Now we calculate number of molecules in 0.61moles
Formula applied:
Number of molecules
=
Number of moles  NA
=
0.61  6.02  1023
Putting the values
Number of molecules
=
(ii)
Number of moles of H2O2
3.67  1023
=
Amount of H2O2
Molar mass of H2O2
11
= 34
=
Putting the values
0.32 moles
Now we calculate number of molecules of H2O2
Formula applied:
Number of molecules = number of mole  NA
= 0.32  6.02  1023
Putting the values
= 1.926 1023
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So 11.0 g of H2O contains larger number of molecules. Ans.
(g)
Data:
Moles of NaBr
= 0.500
Na ions in 0.500 moles of NaBr
0.500 moles of NaBr contains moles of Na
=?
= 0.5
Formula applied:
Number of ions
=
Number of moles  NA
=
0.5  6.02  1023
=
3.01  1023 ions
Putting the values
Number of ions
Now we calculate Na ions in 0.0145 kg of NaCl
Data:
Mass of NaCl
=
=
0.0145 Kg
14.5 g
Na ions in 0.0145 kg of NaCl = ?
Formula applied:
Number of moles
=
Mass of NaCl
Molar mass of NaCl
=
14.5 g
58.5 g mol1
=
0.2478 mole
Substituting the values
Formula applied:
Number of Na ions =
Putting the values
=
=
Number of moles  NA
0.2478  6.02  1023
1.49  1023
Ans
Number of Na ions in 0.500 M NaBr is larger
(h)
Mass of atoms
=
?
Isotopic mass of 235U
=
235 (1 mole)
Therefore
6.02  1023 atoms of 235U have mass
=
235 g
=
238 g
Similarly
6.02  1023 atoms of 238U have mass
So 6.02  1023 atoms of 238U have larger mass
Q.14. (a) Calculate the percentage of nitrogen in the four important nitrogen fertilizers i.e. NH 3,
NH2CONH2, (NH4)2SO4 and NH4NO3.
(b) Calculate the percentage of nitrogen and phosphorus in each of the following:
(i) NH4H2PO4
ii) (NH4)2 HPO4
(iii) (NH4)3PO4
Solution. (i) NH3 (Ammonia)
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Data:
Molar mass of NH3
Mass of nitrogen in molecule
% age of nitrogen
Formula applied:
% age of nitrogen
=
=
=
14 + 3 = 17 g mol1
14 g mol1
?
=
Mass of nitrogen in molecule
 100
Molar mass
=
14 g mol1
 100
17 g mol1
Putting the values
=
(ii)
NH2CONH2 (Urea)
Data:
Molar mass of urea
82.35 % Ans
=
14 + 2 + 12 + 16 + 14 + 2
=
60 g mol1
28 g mol1
?
Mass of nitrogen in the molecule =
% age of nitrogen
=
Formula applied:
% age of nitrogen
=
Mass of nitrogen in molecule
 100
Molar mass
=
28 g mol1
 100
60 g mol1
=
46.67 % Ans
Putting the values
% age of nitrogen
(iii) (NH4)2SO4 (Ammonium sulphate)
Data:
Molar mass of (NH4)2SO4 =
=
=
(14 + 4)2 + 32 + 64
36 + 32 + 64
132 g mol1
Mass of nitrogen in the compound = 28 g mol1
%age of nitrogen
=
?
Formula applied:
Mass of nitrogen in molecule
% age of nitrogen
=
 100
Molar mass
Substituting the values
% age of nitrogen
=
=
(iv)
28 g mol1
 100
132 g mol1
21.21 %
Ans
NH4NO3 (Ammonium nitrate)
Data:
Molar mass
Mass of nitrogen in 1 mole
=
14 + 4 + 14 + 3  16 = 80 g mol1
=
14 + 14 = 28 g mol1
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%age of nitrogen
=
?
% age of nitrogen
=
Mass of nitrogen in molecule
 100
Molar mass
% age of nitrogen
=
28 g mol1
 100
80 g mol1
Formula applied:
Putting the values
=
(b)
35 % Ans.
(i) NH4H2PO4 (Ammonium dihydrogen phosphate)
Data:
Mass of 1 mole of NH4H2PO4
=
14 + 4 + 2 + 31 + 64
=
115 g mol1
Mass of nitrogen in 1 mole
=
14 g mol1
Mass of phosphorous in 1 mole
%age of phosphorous
=
=
31 g mol1
?
=
Mass of nitrogen in molecule
 100
Molar mass
=
14 g mol1
 100 = 12.17 % Ans
115 g mol1
=
Mass of phosphorus in molecule
 100
Molar mass
Formula applied:
% age of nitrogen
Putting the values
% age of nitrogen
Formula applied:
% age of phosphorus
Putting the values
% age of phosphorous =
31 g mol1
 100
115 g mol1
=
(ii)
26.96 %
(NH4)2HPO4 (Diammonium hydrogen phosphate)
Data:
Molar mass of (NH4)2HPO4
=
2 (14 + 4) + 1 + 31 + 64
=
132 g mol1
Mass of nitrogen in 1 mole
=
28 g mol1
Mass of phosphorous in 1 mole
=
31 g mol1
=
?
=
Mass of nitrogen in molecule
 100
Molar mass
=
28 g mol1
 100 = 21.21% Ans
132 g mol1
%age of phosphorus
Formula applied:
% age of nitrogen
Putting the values
% age of nitrogen
Formula applied:
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% age of phosphorous
=
Mass of phosphorus in molecule
 100
Molar mass
=
31 g mol1
 100 = 23.48 % Ans
132 g mol1
=
=
=
3(14 + 4) + 31 + 4  16
54 + 31 + 64 = 149 g mol1
42 g mol1
Putting the values
% age of phosphorous
(iii)
(NH4)3PO4 (ammonium phosphate)
Data:
Molar mass of (NH4)3PO4
Mass of nitrogen in 1 mole of (NH4)3PO4
Mass of phosphorous in
1 mole of (NH4)3PO4
=
31 g mol1
%age of phosphorous =
?
Formula applied:
Mass of nitrogen in molecule
% age of nitrogen
=
 100
Molar mass
Putting the values
42 g mol1
% of nitrogen =
 100 = 28.18 % Ans.
132 g mol1
Formula applied:
% age of phosphorus =
Mass of phosphorus in molecule
 100
Molar mass
Putting the values
% of phosphorous =
31 g mol1
 100 = 20.81 % Ans
132 g mol1
Q.25. Explain the following with reasons:
(i)
Law of conservation of mass has to be considered during stoichiometric calculations.
Ans:
(ii)
Many chemical reactions taking place in our surrounding involve the limiting reactant.
Ans:(i) The mechanical loss of products during filtration, distillation, washing, drying and crystallization.
(ii) Reversibility of reactions.
(iii) Possibility of side reactions.
(iii)
No individual neon atom in the sample of the element has a mass of 20.18 a.m.u.
Ans. Neon has three isotopes of atomic masses 20, 21 and 22 with relative abundances as 90.92%,
0.26% and 8.82%.
The relative atomic mass of neon, comes out to be 20.18 a.m.u. So 20.18 a.m.u. is the average
atomic mass of all the three isotopes and there is no atom of Ne with this atomic mass.
(iv)
One mole of H2SO4 should completely react with two moles of NaOH. How does Avogadro’s
number help to explain it.
Ans:
The balanced chemical equation between H2SO4 and NaOH.
H2SO4 + 2NaOH  Na2SO4 + 2H2O
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
One mole of H2SO4 releases two moles of H in the solution. It needs two moles of OH ions for
complete neutralization.
(v)
One mole of H2O has two moles of bonds, three moles of atoms, ten moles of electrons and
twenty-eight moles of the total fundamental particles present in it.
Ans:
The molecule of H  O  H has two bonds in it. Therefore, one mole of H2O contains two moles
of bonds and three moles of atoms. Similarly, there are eight electrons in oxygen and one electron
in each of the two H atoms. One molecule of H2O has 10 electrons. So one mole of water
contains 10 moles of electrons. There are 28 moles of all fundamental particles in one mole of
water 10 moles of electrons, 10 moles of protons, 8 moles of neutrons.
(vi)
N2 and CO have same number of electrons, protons and neutrons.
Ans:
No. of electrons in N2 = 7 + 7 = 14, number of protons in N2 = 7 + 7 = 14 and number of neutrons
(14  7) = 7 + 7 = 14.
In CO, number of electrons in C = 6, number of electrons in O = 8,
total number of electrons = 6 + 8 = 14.
Number of protons in C = 6, number of protons in O = 8, total number of protons = 6 + 8 = 14.
Number of neutrons in C = 6, number of neutrons in O = 8, total number of neutrons = 6 + 8 = 14.
[ 17 ]
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Chapter 2
EXPERIMENT TECHNIQUES IN CHEMISTRY
(i)
(ii)
(iii)
(iv)
(v)
Ans:
Q.4.
Ans.
Q.5.
Ans.
Q.6.
Ans.
MULTIPLE CHOICE QUESTIONS
A filtration process could be very time consuming if it were not aided by a gentle suction
which is developed:
(a) If the paper covers the funnel up to its circumference.
(b) If the paper has got small sized pores in it.
(c) If the stem of the funnel is large so that it dips into the filtrate.
(d) If the paper fits tightly.
During the process of crystallization, the hot saturated solution:
(a) is cooled very slowly to get large-sized crystals.
(b) is cooled at a moderated rate to get medium-sized crystals.
(c) is evaporated to get the crystals of the product.
(d) is mixed with an immiscible liquid to get the pure crystals of the product.
Solvent extraction is an equilibrium process and it is controlled by:
(a) Law of mass action
(b)
The amount of solvent used
(c) Distribution law
(d)
The amount of solute
Solvent extraction method is parti-cularly useful technique for separation when the product
to be separated is:
(a) Non-volatile or thermally unstable
(b)
Volatile or thermally stable
(c) Non-volatile or thermally stable
(d)
Volatile or thermally unstable
The comparative rates at which the solutes move in paper chromato-graphy, mainly
depends on:
(a) The size of paper used
(b)
The Rf values of solutes
(c) Temperature of experiment
(d)
Size of chromatographic tank used
(1)
d
(2)
b
(3)
c
(4)
d
(5)
b
Why is there a need to crystallize the crude product?
Need for crystallization:
When a chemical compound is synthesized, it is crude product i.e., it is unrefined. Therefore,
there is a need to purify the compound. This is done by crystallizing the compound using a
suitable solvent.
A water insoluble organic compound aspirin is prepared by the reaction of salicylic acid
with a mixture of acetic acid and acetic anhydride. How will you separate the product from
the reaction mixture?
Separation of aspirin:
During preparation of aspirin, it is separated by filtration. The reaction mixture is added to cold
water. Aspirin crystallizes out. Other products remain in the solution. The aspirin is filtered from
water by sintered glass crucible.
A solid organic compound is soluble in water as well as in chloroform. During its
preparation, it remains in aqueous layer. Describe a method to obtain it from this layer.
Separation of water soluble compound:
The compound can be extracted by solvent extraction technique, as it is mentioned that
compound is soluble in solvents like water and chloroform. If a non polar solvent say CHCl3 is
mixed and the mixture is transferred to a separating funnel, then two layers are formed. By
separating water layer and evaporating it, organic compound is obtained.
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Q.7.
The following figure shows a developed chromatogram on paper with five spots:
(i) unknown mixture X
(ii)
sample A
(iii)
sample B
(iv) sample C
(v)
sample D
Solvent front
Solvent front
y
Spots
X
A
B C
D
Find out the composition of unknown mixture X. Which sample is impure and what is its
composition.
Ans. Unknown mixture X has components B and C. Component D is impure. It contains components
A and C.
Q.8. In solvent extraction technique, why repeated extractions using small portions of solvent are
more efficient than using a single but larger volume of solvent?
Ans. Efficiency of repeated extractions:
It has been experimentally observed that repeated or multiple extractions in which solvent is used
in small portions are more efficient than using a single but large volume of solvent because more
product is extracted with more extractions using smaller portions of solvent.
Q.10. You have been provided with a mixture containing three inks with different colours. Write
down the procedure to separate the mixture with the help of paper chromatography.
Ans. Separation of inks:
Place three spots of three inks on thin pencil line drawn on the Whatman No.1 grade filter paper.
Place the fourth spot of mixture (containing three inks) on the same line. All spots should be
pointed using separate capillary tubes. After drying the spots, hang the filter paper, in the
3
chromatographic tank. As mobile phase rises up to 4 th of the filter paper remove it and point out
the solvent front. Calculate Rf values of all bands. Then by comparing the Rf values of bands of
mixture with the Rf values of the bands of individual inks, components of mixture can be
identified.
[ 19 ]
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Chapter 3
GASES
MULTIPLE CHOICE QUESTIONS
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
Pressure remaining constant at which temperature the volume of a gas will become twice of
what it is at 0C.
(a) 546C
(b)
200C
(c) 546 K
(d)
273 K
3
The number of molecules in one dm of water is close to
6.02
12.04
23
(a) 22.4  1023
(b)
22.4  10
18
(c) 22.4  1023
(d)
55.6  6.02 1023
Which of the following will have the same number of molecules at STP?
(a) 280 cm3 of CO2 and 280 cm3 of N2O
(b)
11.2 dm3 of O2 and 32 g of O2
(c) 44g of CO2 and 11.2 dm3 of CO
(d)
28g of N2 and 5.6 dm3 of oxygen
If absolute temperature of a gas is doubled and the pressure is reduced to one half, the
volume of the gas will
(a) remain unchanged
(b)
increase four times
(c) reduce to ¼
(d)
be doubled
How should the conditions be changed to prevent the volume of a given gas from expanding
when is mass is increased?
(a) Temperature is lowered and pressure is increased.
(b) Temperature is increased and pressure is lowered.
(c) Temperature and pressure both are lowered.
(d) Temperature and pressure both are increased.
The molar volume of CO2 is maximum at
(a) STP
(b)
127C and 1 atm
(c) 0C and 2 atm
(d)
273C and 2 atm
The order of the rate of diffusion of gases NH3, SO2, Cl2 and CO2 is:
(a) NH3 > SO2 > Cl2 > CO2
(b)
NH3 > CO2 > SO2 > Cl2
(c) Cl2 > SO2 > CO2 > NH3
(d)
NH2 > CO2 > Cl2 > SO3
Equal masses of methane and oxygen are mixed in an empty container at 25C. The fraction
of total pressure exerted by oxygen is
1
8
1
16
(a) 3
(b)
(c)
(d)
9
9
17
Gases deviate from ideal behaviour at high pressure. Which of the following is correct for
non-ideality?
(a) At high pressure, the gas molecules move in one direction only.
(b) At high pressure, the collisions between the gas molecules are increased manifold.
(c) At high pressure, the volume of the gas becomes insignificant.
(d) At high pressure, the intermolecular attraction becomes significant.
The deviation of a gas from ideal behaviour is maximum at
(a) 10C and 5.0 atm
(b)
10C and 2.0 atm
(c) 100C and 2 atm
(d)
0C and 2.0 atm
A real gas obeying van der Waal’s equation will resemble ideal gas if the
(a) both a and b are large
(b)
both a and b are small
(c) a is small and b is large
(d) a is large and b is small
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Ans:
(1)
c
(2)
d
(3)
a
(4)
b
(5)
a
(6)
b
(7)
b
(8)
a
(9)
d
(10)
a
(11)
b
Q.4. (a) What is Boyle’s law of gases? Give its experimental verification.
Ans: BOYLE’S LAW
Definition:
The volume of a given amount of a gas is inversely proportional to the pressure of the gas at a
constant temperature.
1
V

(when ‘n’ and ‘T’ are constant)
P
1
V
=
k.P
PV
=
k
----------- (1)
‘k’ is the proportionality constant and ‘n’ is the number of moles of gas. The value of ‘k’ is
different for the different amounts of the same gas. It also depends upon the pressure and quantity
of the gas. According to equation (1), Boyle’s law has got another definition.
Definition:
The product of pressure and volume of a fixed amount of a gas is a constant quantity at constant
temperature. Let the initial pressure and initial volume are P1 and V1. Final pressure and final
volume are P2 and V2.
Since,
P1V1 =
k
and P2V2 = k
so,
P1V1 =
P2V2
---------- (2)
EXPERIMENTAL VERIFICATION OF BOYLE’S LAW
Boyle’s law can be verified by working on the apparatus shown in Fig (3.1)
V=1
1
o
25 C
dm
3
V=1/2
2
o
25 C
2 atm = P1
dm
3
V=1/3
2
o
25 C
4 atm = P2
dm
3
6 atm = P3
Fig (3.1): Verification of Boyle's law. The temperature and quantity of gas are constant.
Take a gas in a cylinder with movable piston. This is also attached with a manometer to record
the pressure.
The volume ‘V1’ of the given quantity of a gas at pressure P1 = 1 atm. is reduced in proportion to
the increase in pressure, when the temperature is kept constant. When the piston is pressed twice
the pressure becomes two atmospheres. When this piston is pressed three times, then the pressure
becomes three atmospheres. If the initial volume at two atmospheres is 1dm3, then it becomes 0.5
dm3 and then 0.33 dm3, respectively with increasing pressures. The value of the product of
pressure and volume remains the same at same temperature.
Hence,
P1V1
=
2 atm  1 dm3
=
2 dm3 atm.
=k
3
3
P2V2
=
4 atm  0.5 dm
=
2 dm atm.
=k
3
3
P3V3
=
6 atm  0.33 dm
=
2 dm atm.
=k
3
The product of pressure and volume has the units of dm atmosphere.
[ 21 ]
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(b)
Ans.
(c)
Ans.
(d)
Ans.
What are isotherms? What happens to the positions of isotherms when they are plotted at
high temperature for a particular gas.
Isotherms are the graphs between pressure and volume, when temperature is constant. These
graphs are plotted keeping in view the Boyle’s Law.
When the isotherms are plotted at higher temperatures, then they go away from the axis. The
reason is that, the volumes of the gases increase at high temperatures.
Why do we get a straight line when pressures are plotted against inverse of volumes. This
straight line changes its positions in the graph by varying the temperature. Justify it.
1
When the pressure of a gas is plotted against V , we get a straight line at constant temperature.
1
The reason is that P and V are directly proportional to each other, with power unity on both
1
variables. When the temp. changes then value of p changes for same V value.
How do you explain that the value of the constant k in (PV = k) in Boyle’s law depend upon
(i) The temperature of a gas (ii) the quantity of a gas.
(i) In Boyle’s law, temperature is kept constant for fixed number of moles of an ideal gas. So
the product PV remains constant. Anyhow, if the temperature of the gas is increased, then
the volume increases for the same number of moles. So the product PV increases.
Anyhow, if the increase of temperature is carried out by keeping volume constant, then the
pressure increases for the same amount of gas. So the product PV increases.
Similarly by decreasing the temperature the product PV decreases.
(ii) When the quantity of gas is increased at same temperature, then the volume increases and
so the product PV increases.
Q.5(c) Do you think that the volume of any quantity of a gas becomes zero at 273C. Is it not
against the law of conservation of mass? How do you deduce the idea of absolute zero from
this information?
Ans. The volume of gas does not become zero at 273C. All the ideal and real gases are converted
into the liquid state before reaching 273C. It is against the law of conservation of mass that
volume of gas becomes zero. That theoretical temperature at which the volume of gas becomes
zero is absolute zero.
Q.6(b) Throw some light on the factor 1/273 in Charles’s law.
1
Ans. Factor of 273 is very important in Charles’s law. The volume of given mass of the gas increases
1
or decreases by 273 of its volume at 0C. Following equation helps us to calculate the volume of
gas at any temperature.
T 
VT = V0 1 +
 273 
Q.7.(c) How do you justify from general gas equation that increase in temperature or decrease of
pressure decreases the density of the gas?
Ans.
PM
(c)
The formula for the density of an ideal gas is d = RT . According to this equation density is
directly proportional to the pressure of the gas and inversely proportional to the temperature. So
greater the temperature of the gas lesser the density. The increase of temperature increases the
volume and so the density falls down.
When we decrease the pressure, the molecules go away from each other, volumes increase and
density is decreased.
[ 22 ]
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Why do we feel comfortable in expressing the densities of gases in the units of g dm 3 rather
than g cm3, a unit which is used to express the densities of liquids and solids.
Ans. The densities of gases are very low as compared to liquids and solids. So if densities are
expressed in g cm–3, then the answer is very small and may go to fourth place of decimal for some
gases.
When we express the densities in g dm–3, then the values of the densities become reasonable to be
expressed. e.g., the density of CH4 at 0C and 1 atm is 0.71 g dm–3, but if it is expressed in gcm–3,
then it is 0.00071 = 7.1  104.
So, we have to use the exponentials. This creates uncomfortability.
Q.9.(b) Do you think that 1 mole of H2 and 1 mole of NH3 at 0C and one atm-pressure will have
Avogadro’s number of particles?
Ans. One mole of H2 and one mole of NH3 at 0C and 1 atm should occupy 22.4 dm3 of volume and
each gas should have Avogadro’s number of molecules. Actually, NH3 is non-ideal gas so its
volume will be little bit less than 22.4 dm3 because it deviates from Avogadro’s law. Since there
is one mole of each, so the number of molecules will be same.
(c)
Justify that 1 cm3 of H2 and 1 cm3 of CH4 at STP will have same number of molecules.
When one molecule of CH4 is 8 times heavier that of hydrogen.
Ans. According to Avogadro’s law, equal volumes of all the ideal gases at same temperature and
pressure have equal number of molecules. So 1cm3 of H2 and 1 cm3 of CH4 at STP will have an
equal number of molecules.
No doubt, the molecule of methane is eight times heavier than H2, but the sizes of the gas
molecules and their masses don’t disturb the volumes. The reason is that at STP, one molecule of
the gas is at a distance of three hundred times than its diameter.
Q.10. (a)
Dalton’s law of partial pressures is only obeyed by those gases which don’t have
attractive forces among themselves. Explain it.
Ans. DALTON’S LAW OF PARTIAL PRESSURES
Definition:
The total pressure exerted by the mixture of gases is equal to sum of individual partial pressures
at a given temperature.
Explanation:
Whenever ideal gases are mixed, they collide on the walls of the container independently so the
total pressure is contributed by all the gases.
Mathematically,
P
=
pA + pB + pC
----------- (1)
Where ‘P’ is the total pressure of the mixture while pA, pB and pC are the individual partial
pressures.
Definition of partial pressure:
The pressure exerted by an individual gas in a gaseous mixture is called the partial pressure of
the gas.
Example: Air is a mixture of several gases, like
N2
=
78.08%
O2
=
20.95%
Ar
=
0.93%
CO2
=
0.03%
and many other gases, called trace gases.
The total pressure of the atmosphere is equal to the sum of individual partial pressures.
Patm
=
PN2 + PO2 + PAr + PCO2 + Ptrace gases ------(2)
General gas equation and partial pressures:
In the case of mixture of ideal gases, molecules of each gas move independently. So we can apply
the general gas equation on each gas.
(d)
[ 23 ]
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General equation:
PV
=
nRT
----- (3)
Let us have mixture of three gases i.e., H2, O2 and CH4. The partial pressures and number of
moles are present in the general gas equation and have different values. The temperatures and
volume of three gases in the mixture is same.
RT
pH2V = nH2RT,
pH2
=
nH2
,
pH2  nH2
----- (4)
V
RT
pO2V = nO2RT,
pO2
=
nO2
,
pO2  nO2
----- (5)
V
RT
pCH4V = nCH4RT,
pCH4
=
nCH4 V ,
pCH4  nCH4
----- (6)
Adding all the partial pressures to get the total pressure, we add equation (4), (5) and (6).
P = pH2 + pO2 + pCH4
RT
RT
RT
+ nO2
+ nCH4
V
V
V
RT
P = nt V
where nt = nH2 + nO2 + nCH4
According to equation (4), (5) and (6), the partial pressure of any gas is directly proportional to
the number of moles of that gas.
Q.13.(c) Hydrogen and helium are ideal at room temperature, but SO 2 and Cl2 are non-deal. How
do you explain it?
Ans. H2 and He have very low boiling points. So at room temperature, they are far away from their
boiling points. At room temperature, the attractive forces are absent. So they behave ideally.
SO2 and Cl2 have boiling points close to room temperature, but are below 0C. At room
temperature, they are not far away from their boiling points. Sufficient attractive forces are
present at room temperature. So they are non-ideal.
Q.15. Explain the following facts.
(i)
The plot of PV versus P is a straight line at constant temperature and with a fixed number
of moles of an ideal gas.
Ans: The product PV is a constant quantity when temperature and the number of moles are constant.
So, when we plot a graph between pressure on x-axis and PV on y-axis, then a straight line
parallel to the pressure axis is obtained. This is verification of Boyle’s Law.
(ii)
The straight line in (a) is parallel to x-axis and goes away from the pressure axis at higher
pressures.
Ans: When the pressure on the gases is increased beyond certain limits, then PV verses P is no more a
straight line parallel to pressure axis. The straight line becomes a curve depending upon the
nature of the gas. At this situation, gases become non-ideal.
(iii)
The van der Wall’s constant ‘b’ of a gas is four times the molar volume of gas.
Ans. When n moles of a gas are enclosed in a container of volume “V”, the space in which the
molecules are free to move is equal to “V” if the volume occupies by the molecules is negligible.
Since, the molecules have a finite sizes, so, their volume in which they move is less than “V”. If
“b” is the effective volume then, it must be substracted from the total volume. Therefore,
Videal =
Vb
For n moles of a gas
Videal =
(V  nb)
b is also called excluded volume.
Excluded but four times of actual
This volume is not equal to the actual volume
volume
volume of gas. It can be calculated by
assuming bimolecular
collisions between two impenetrable spheres
of gas molecules.
Suppose ‘r’ is the radius and  = 2r, is the
collision diameter of
gas molecules. The space indicating by the
dotted sphere having a
P=
nH2
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(iv)
Ans:
(v)
Ans:
(vi)
Ans:
radius ‘’ will not be available to the pair of the colliding molecules. Thus, this space is the
excluded volume for the pair of molecules and is given below:
As, we know that volume of one gas molecule
4 3
Vm
=
------ (1)
3 r
4
Vpair
=
3
3
Since, 
=
2r
4
So,
Vpair
=
 (2r)3
3
4
=
 (8r3)
3
4
Vpair
=
8  3 r3
------ (2)


This excluded volume is for a pair of molecules. Now, the excluded volume per molecule is:
1
4
V
=
 8  r3
2
3 
4
V
=
4  r3
------- (3)
3 
4 3
Since,
Vm
3 r =
Putting this value in equation (3) for (1).
V
=
4 (Vm)
Pressure of NH3 gas at given conditions (say 20 atm pressure and room temperature) is less
as calculated by van der Waal’s equation than that calculated by general gas equation.
NH3 is a polar gas and has forces of attractions in its molecules when the pressure is high. It
shows non-ideal behaviour at high pressures. The factors ‘a’ and ‘b’ become dominant at this
pressure and van der Waal’s equation gives less pressure than that calculated by general gas
equation.
Water vapours do not behave ideally at 273K.
When water vapours are present at 273 K (0 C), there are sufficient forces of attractions among
its molecules at 0C (freezing point of water). Due to this reason water vapours behave nonideally at 273 K.
SO2 is comparatively non-ideal at 273K, but behaves ideally at 327K.
SO2 gas is close to its boiling point at 273 K. So, at 273 K, attractive forces are dominating and
make the gas non-ideal. But when the temperature of the gas is 327 K, then forces of attractions
are less dominant and gas behaves ideally.
[ 25 ]
Composed by: Malik Khurshid Dhar Bazar Ph: 05824-420161
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Chapter 4
LIQUIDS AND SOLIDS
(1)
(2)
(3)
(4)
(5)
Ans:
MULTIPLE CHOICE QUESTIONS
London dispersion forces are the only forces present among the:
(a) molecules of water in liquid state
(b) atoms of helium in gaseous state at high temperature
(c) molecules of solid iodine
(d) molecules of hydrogen chloride gas
Acetone and chloroform are soluble in each other due to:
(a) intermolecular hydrogen bonding
(b)
dipole-dipole interaction
(c) instantaneous dipoles
(d)
all of the above
NH3 shows a maximum boiling point among the hydrides of V-A group elements due to:
(a) very small size of nitrogen
(b) lone pair of electrons present on nitrogen
(c) enhanced electronegative character of nitrogen
(d) pyramidal structure of NH3
When water freezes at 0C, its density decreases due to:
(a) cubic structure of ice
(b)
empty spaces present in the structure of ice
(c) change of bond lengths
(d)
change of bond angles
In order to mention the B.P. of water at 110C, the external pressure should be
(a) between 760 torr and 1200 torr (b)
between 200 torr and 760 torr
(c) 765 torr
(d)
any value of pressure.
(1)
d
(2)
d
(3)
c
(4)
b
(5)
c
Q.7(c) What are molecular crystals? Give their properties. Justify that molecular crystals are
softer than ionic crystals.
Ans: MOLECULAR SOLIDS
Definition:
The crystalline solids in which the polar or non-polar molecules or atoms are held together by
dipole-dipole or van der Waal’s forces are called molecular solids.
Strength of intermolecular forces:
The intermolecular forces found in molecular crystals are much weaker as compared to the
attractive forces between cation and anion in ionic crystal. These forces are also weaker than
those between the atom of covalent crystals and between lattice points in metallic crystals.
Strength of these crystals directly depends upon the strength of dipole-dipole interactions and
Van der Waal’s forces.
Examples:
(i)
Ice and sugar: They have polar molecules. Their melting and boiling points are high.
(ii)
Iodine, sulphur, phosphorus and carbon dioxide:
They have non-polar molecules. Their melting and boiling points are low.
Properties of the molecular solids:
1.
X-ray analysis: X-ray analysis indicates regular arrangement of atoms in the molecules of these
crystals. We get the exact positions of all the atoms.
2.
Softness: These crystals are soft and easily compressible, because forces of attraction which hold
these molecules are very weak.
3.
Volatility: They are volatile in nature.
4.
M.P and B.P: They have low melting and boiling points.
5.
Electrical conductivity: They are bad conductors of electricity.
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6.
7.
8.
Density: They have lower density as compared to that of other types of crystals.
Solubility: Polar molecular crystals are soluble in polar solvents, whereas non-polar molecular
crystals are mostly soluble in non-polar solvents.
Transparent for light: Sometimes they are transparent to light.
Structure of iodine:
Crystal of iodine has face centred cubic
structure. The lattice, points are occupied by I2
molecules. The lattice points are held together
by van der Waal’s forces. Each unit cell has
fourteen I2 molecules.
In the solid state the I2 molecules are arranged in
such a way that they form layer lattice as shown
in Fig.(4.19). The bond distance of I2 in solid
state is 271.5 pm which is greater than the bond
distance of I2 in gaseous state i.e., 266.6 pm.
Iodine is poor conductor of electricity.
Q.9
Ans:
Fig. (4.19) Structure of iodine molecules
Crystals of salts fracture easily, but metals are deformed under stress without fracturing.
Explain the difference.
METALLIC SOLIDS
Definition:
The crystalline solids in which metal atoms are held together by metallic bonds are known as
metallic solids.
Metallic bond:
The force of attraction that binds positive metal ion to the number of electron within its sphere of
influence is called metallic bond.
Theories of metallic bonding:
Various theories have been put forward to explain the properties of these type of crystals:
(a)
Electron pool or electron gas theory:
This was proposed by Drude and extended by Loren in 1923. According to this theory, each atom
present in a metallic crystal loses all its valence electrons. In this way a pool of electrons is
formed. Fig. (4.20) It is believed that positively charged metal ions are held together by pool of
electrons. These positive ions are located at definite positions. They are situated at measurable
distances from each other in the metallic crystal. The valence electrons are not associated with
any individual ion. They move freely from one part to the other in the crystal lattice.
Following figure indicates positive ions surrounded by free electrons.
e–
e–
e–
e–
e–
e–
e–
e–
e–
e–
e–
Mobile
electrons
e–
e–
e–
Positive
ions
Fig. (4.20) Positive ions surrounded by mobile electrons.
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(b)
Valence bond theory:
This theory was given by L. Pauling. According to this theory, metallic bond behaves just like
covalent bond. Covalent bonds are not localized. These are highly delocalized in metallic crystals.
The delocalized electrons can explain the electrical and thermal conductivities.
(c)
Molecular orbital theory:
This theory has recently been applied to explain the properties of metallic crystals. According to
this theory, the electrons which are in the completely filled orbitals are localized. The atomic
orbitals having valence electrons overlap to form a set of delocalized orbitals known as molecular
orbitals. These are extended all over the crystal lattice. Due to such type of combination of atomic
orbitals a large number of closely spaced states are formed called bands of energy. So this is also
called as band theory. The properties of metallic solids depend upon the gap between two bands.
Q.10 What is the coordination number of an ion? What is the coordination number of the cation
in (a) NaCl structure, and (b) CsCl structure? Explain the reason for this difference.
Ans: IONIC SOLIDS
Definition:
The crystalline solids in which positively and negatively charged ions are held together through
ionic bond or electrostatic forces of attraction is called ionic crystals or ionic solids.
Properties:
1.
Physical state:
They never exist in the form of liquids or gases under ordinary conditions of temperature and
pressure.
2.
Geometrical shape:
Positively and negatively charged ions are arranged in a well-defined geometry. Their geometric
shapes depend upon the nature of ions.
3.
Stability:
Ionic crystalline solids are very stable. Therefore, they need very high energy to separate positive
and negative ions.
4.
Hardness:
Ionic crystals are very hard.
5.
Volatility:
They have low volatility.
6.
M.P and B.P:
Their melting and boiling points are very high.
7.
Non-independent molecules:
Ionic solids cannot exist in the form of individual neutral independent molecules.
8.
Non-directional bonds:
The forces present between oppositely charged ions are non-directional.
9.
Density:
When cation and anions are closely packed, then the density becomes high and a crystal lattice is
formed.
10.
Structure and radius ratio:
The structure of ionic crystalline solid is based upon the radius ratio of positive and negative ions.
NaCl and CsF have same geometry due to same radius ratio in both the cases.
11.
Formula mass:
In case of crystalline solids the term formula mass is used instead of molecular mass. The reason
is that they do not exist in the form of molecules.
12.
Electrical conductivity:
Ionic crystals conduct electricity only in molten state or solution form and not in the form of solid
crystal. Ions have fixed positions, due to strong electrostatic force of attraction.
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13.
14.
Brittleness:
Ionic solids are highly brittle because they consist of parallel layers of cations and anions in an
alternate way over one another. When an external force is applied, one layer of ions slides a bit
over the other and so the like ions come in front of each other. Due to repulsion between like
ions, two layers become brittle. Fig. (4.16)
Reaction in polar solvents:
Ionic solids give ionic reactions in polar solvents, therefore, their reactions are very fast.
One layer of
ions
Force
Repulsion
Other layer
of ions
15.
Q.12
(i)
Ans:
(ii)
Ans:
(iii)
(iv)
Ans:
(v)
Ans:
(vi)
Ans:
(vii)
Ans:
(viii)
Plane along which a
layer of ions slides
Crystal
breaks
here
Fig. (4.16) Explanation of high brittleness of ionic crystals
Isomorphism and polymorphism:
Due to close packing of particles, they have high density.
Isomorphism and polymorphism are associated with crystalline solids.
Explain the following with reasons:
Sodium is softer than copper but both are very good electrical conductors.
Sodium is an alkali metal. It has one electron in the outermost orbital. Copper is a transition
element having ten electrons in outermost 3d orbitals. So, there are greater chances for
overlapping of orbitals in copper and hardness is developed.
Well, the availability of electrons to the applied potential difference is almost equal. So, they are
equally good electrical conductors.
Diamond is hard and an electrical insulator.
There is sp3  sp3 effective overlapping of the carbon atoms in diamond. Tetrahedral angles are
produced around each carbon atom. The bond length of 1.54 A is most suitable for effective
packing. That is why diamond has a hard structure. These are no free electrons so it is insulator.
Sodium chloride and caesium chloride have different structures.
Iodine dissolves readily in tetrachloromethane.
The basic principle is that “like dissolves like.” Since iodine is a non-polar substance, hence it
becomes soluble in non-polar solvent CCl4. I2 is 85 times more soluble in CCl4 than H2O.
The vapour pressures of solids are far less than those of liquids.
Particles of solids are tightly packed. There are no translational motions and no collisions among
the molecules of solid. The molecules of the liquid collide among themselves due to translational
motions. So the molecules present on the surface of liquid, get favourable collisions and can leave
the surface.
Amorphous solid like glass is also called super cooled liquid.
Amorphous solids like glass have random structures and their particles are disarranged just like
liquids. So the amorphous solids are no doubt hard and rigid but look like liquids. That is why
glass is called a super cooled liquid.
Cleavage of the crystals is itself anisotropic behaviour.
Anisotropy is the property of a crystal to obey a certain property better in one direction than the
other. Cleavage is the breakage of a crystal along definite planes. Since cleavage of the crystals
can take place only in particular directions, so it is anisotropic behaviour.
The crystals showing isomorphism mostly have the same atomic ratios.
[ 29 ]
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Ans:
Isomorphism
The phenomenon in which two different substances exist in the same crystalline form is known as
isomorphism.
Different substances having same crystalline shape are called isomorphs of one another. A
crystalline shape depends upon number of atoms and their way of combination only. It does not
depend on the chemical nature of atoms.
Isomorphism found in various compounds is mostly due to same atomic ratio. The crystals of
isomorphic substances grow together in all proportions in the homogeneous mixtures.
Nature of compound, their crystalline forms and ratio of atoms are given in following table:
Isomorphs
NaNO3 , KNO3
K2SO4 , K2CrO4
ZnSO4 , NiSO4
NaF, MgO
(ix)
Ans:
(x)
Ans:
(xi)
Ans:
(xii)
Ans:
(xiii)
Ans:
(xiv)
Ans:
(xv)
Ans:
Crystalline form
Rhombohedral
Orthorhombic
Orthorhombic
Cubic
Atomic ratio
1:1:3
2:1:4
1:1:4
1:1
2–

Negatively charged ions like CO 
3 and NO 3 having same atomic ratio have same triangular
2–
2–
planar structures. Similarly, SO 
4 and CrO 4 are tetrahedral, and have same atomic ratio.
The transition temperature is given by elements having allotropic forms and by compounds
showing polymorphism.
The elements having two or more than two crystalline forms have allotropes. Polymorphism is a
property of a compound having two or more than two crystalline forms.
At transition temperature one crystalline form changes to other. So transition temperature is for
all those elements which show allotropy and those compounds which show polymorphism.
One of the unit cell angles of hexagonal crystal is 120.
In hexagonal crystal one of the angles which is inside the hexagon is 120. The other two angles
are of 90.
The electrical conductivity of the metals decrease by increasing temperature.
This is due to the availability of free electron. The particles at the lattice points have to and fro
motions. With the increase of temperature the extensions and compressions are enhanced. So, the
probability of electrons to more freely becomes less.
In the closest packing of atoms of metals, only 74% space is occupied.
The coordination number of each atom in closely packed metal structures is 12. This is the
maximum coordination number. But due to the circularity of atoms, certain vacant spaces are left
behind which are called interstices. These vacant spaces constitutes 26% of the crystal structure
and 74% is occupied by the atoms.
Ionic crystals don’t conduct electricity in the solid state.
In ionic crystals, ions are tightly packed in a three-dimensional way. They don’t have translatory
motion. So they don’t become responsible for carrying of current.
Ionic crystals are highly brittle.
In ionic crystals, the boundaries of similarly charged ions touch each other in the crystal lattice.
So when a crystal is broken under stress, they become loosely held then similar ions repel each
other. This repulsion causes brittleness.
The number of positive ions surrounding the negative ion in the ionic crystals lattice
depends upon the sizes of the two ions.
In ionic crystals, each ion is surrounded by oppositely charged ions. If the size of the central ion
is greater then the number of opposite ions surrounding it is also greater. It increases coordination
number of central ion. The structure depends upon the coordination number.
[ 30 ]
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Chapter 5
ATOMIC STRUCTURE
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
Ans:
Q.4
(a)
Ans.
MULTIPLE CHOICE QUESTIONS
The nature of positive rays depends on:
(a) the nature of electrode
(b)
the nature of discharge tube
(c) the nature of residual gas
(d)
all of the above
The velocity of photon is:
(a) independent of its wavelength
(b)
depends on its wavelength
(c) equal to square of its amplitude (d)
depends on its source
The wave number of the light emitted by a certain source is 2  106 m. The wavelength of this
light is:
(a) 500 nm
(b)
500 m
(c) 200 nm
(d)
5  107 m
Rutherford’s model of atom failed because:
(a) the atom did not have a nucleus and electrons
(b) it did not account for the attraction between protons and neutrons
(c) it did not account for the stability of the atom
(d) there is actually no space between the nucleus and the electrons
Bohr’s model of atom is contradicted by:
(a) Planck quantum theory
(b)
Quartization of energy of electrons
(c) Heisenberg’s uncertainty principle (d)
Quartization of angular members
Splitting of spectral lines when atoms are subjected to strong electric field is called:
(a) Zeeman effect
(b)
Stark effect
(c) Photoelectric effect
(d)
Compton effect
In the ground state of an atom, the electron is present:
(a) in the nucleus
(b)
in the second shell
(c) nearest to the nucleus
(d)
farthest from the nucleus
Quantum number value for 2p subshell are:
(a) n = 2, l = 1
(b)
n = 1, l = 2
(c) n = 1, l = 0
(d)
n = 2, l = 0
Orbitals having same energy are called:
(a) hybrid orbitals
(b)
valence orbitals
(c) degenerate orbitals
(d)
d-orbitals
When 6d orbital is complete, the entering electron goes into:
(a) 7f
(b)
7s
(c) 7p
(d)
7d
(1)
c
(2)
a
(3)
a
(4)
c
(5)
c
(6)
b
(7)
c
(8)
a
(9)
c
(10)
c
Keeping in mind the discharge tube experiment, answer the following questions:
Why is it necessary to decrease the pressure in the discharge tube to get the cathode rays?
The pressure in discharged tube is decreased to allow the cathode rays and anode rays to move
freely from one electrode to the other. In this way the possibility of collisions between rays and
the gas molecules are minimized.
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(b)
Whichever gas is used in the discharge tube, the nature of the cathode rays remains the
same. Why?
Ans. All the gases are consisted of atoms or molecules. They have electrons in outermost orbitals.
These electrons are detached by the high voltage and due to collisions, these electrons become
free. They are repelled by the cathode and attracted towards the anode. That is why, they are
called cathode rays. They are always electrons and nothing else.
(c)
Why e/m value of the cathode rays is just equal to that of electron?
Ans. Since cathode rays are electrons, so their e/m values are just equal to those of electrons.
(d)
How the bending of the cathode rays in the electric and magnetic fields shows that they are
negatively charged? Give reasons.
Ans. Cathode rays are deflected towards the positive plate when electric field is applied. It shows that
cathode rays are negatively charged. When cathode rays are passed through the magnetic field,
they bend perpendicular to the joining line of two poles. This is due to the negative charge.
Anyhow, positively charged particles will bend in opposite direction to that of electrons.
(e)
Why the positive rays are also called canal rays? Give its reason.
Ans. The positive rays has the ability to pass through holes., Due to this the positive rays are also
called canal rays.
(f)
The e/m values of positive rays for different gases are different, but those for cathode rays
the e/m values are same. Justify it.
Ans. The nature of particles of positive rays in a discharge tube depend upon the nature of the gas
because the nucleus of every gas has its own number of protons and neutrons. Greater the number
of protons and neutrons in the nucleus of an atom or nuclei of the molecule, smaller the e/m
values.
(g)
The e/m value of the positive rays obtained from the hydrogen gas is 1836 times less than
that of cathode rays. Justify it.
Ans. When we use hydrogen gas in the discharge tube, the positive rays are consisted of single protons.
The proton is 1836 times heavier than that of electron. So its e/m value is 1836 times smaller.
Q.7. (a) Give the postulates of Bohr’s atomic model. Which postulates tells us that orbits are
stationary and energy is quantized?
Ans: BOHR’S ATOMIC MODEL
Introduction:
In 1887, the spectra of hydrogen was recorded by Lyman and Balmer. The lines of the spectra
could not be explained. Moreover, the defective model of Rutherford also needed correction. In
the light of all these shortcomings, Bohr gave his own model of atom in 1913.
Postulates of Bohr atomic model:
There are four postulates of Bohr’s model of atom:
1.
The electrons revolve around the nucleus in a circular manner.
2.
As long as an electron remains in the particular orbit, it neither loses the energy nor gains the
energy. Thus in a particular orbit the energy of the revolving electron remains constant. Such
energy levels are called stationery energy levels.
3.
When the electron jumps from a lower energy level to a higher energy level it absorbs the energy
in the form of photon. When the electron jumps from the higher energy level to a lower one, it
loses the energy in the form of photon.
So
E
=
hv
where, E is the energy difference between the levels
h is Planck’s constant and its value is 6.02  1034 Js
v is the frequency of the photon emitted or absorbed.
4.
The electron can move only in that orbit in which the angular momentum of the electron (mvr) is
h
integral multiple of
2
nh
so,
mvr
=
2
where
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(b)
Ans:
‘n’ is a simple integer and is the number of orbit. In other words the angular momentum is
quantised.
The values of angular momentum for first four orbits are
h 2h 3h 4h
,
,
,
2 2 2 2
Derive the equation for the radius of nth orbit of hydrogen atom using Bohr’s model.
MATHEMATICAL DERIVATION OF BOHR’S RADIUS
Consider an atom having one electron. The charge of the nucleus is Ze. The electron of mass
‘m’ is moving in a circular manner with radius ‘r’ with velocity ‘v’, Fig (5.11) The number of
protons in the nucleus is Z and e+ is charge of an proton.
V
e
m
r
+ Ze+
Orbit of
electron
Nucleus
Fig. (5.11) Picture of Bohr’s model of atom
The force with which the electron wants to fly away is the centripetal force.
mv2
F
=
----- (1)
r
This force is being balanced by electrostatic force of attraction of electron (e ) and the nucleus
(Ze). According to coulombs law, the force of attraction is
Ze.e
Ze2
F=
=
----- (2)
2
4or
4or2
Here 0 (epsilon) is a proportionality constant. It is known as permitivity of the vacuum. Its
numerical value is
0
=
8.854  1012 C2J1m1
The electron can only revolve in the orbit, if the centripetal force and the coulombic forces are
equal.
mv2
Ze2
=
----- (3)
r
4or2
Ze2
mv2
=
----- (4)
4or
Ze2
r
=
----- (5)
4omv2
According to equation (5), the radius of the atom is inversely proportional to the velocity of the
electron. It means that when the electron goes to the higher orbits its velocity decreases.
But the determination of the velocity of the revolving electron in a very small sized atom is
impossible. So we eliminate the factor of velocity from equation (5). For this purpose, we take the
help of fourth postulate of Bohr’s model.
nh
mvr
=
2
nh
v
=
2mr
Taking square,
n2h2
v2
=
----- (6)
42m2r2
Putting equation (6) into equation (5)
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r
=
Ze
2
n2h2
42m2r2
Simplifying this equation, taking r on L.H.S. and other factors in R.H.S.
n2h2o
r
=
----- (7)
mZe2
h2o n2
r
=
----- (8)
me2  Z 
Actually, we have re-arranged the equation (7) and the constant factors have been taken outside
the brackets. If we do the calculations of these factors, by putting their values as,
h = 6.625  1034 Js , o = 8.854  1012 C J1 m1
 = 3.1416, m = 9.106  1031 kg,
e = 1.602  1019 C
h2o
Then,
= 0.529  1010 m
me2
‘m’ is the unit of length
since, 1010 m = 1A so equation (8) becomes
n2
r
=
0.529  Z  A ----- (9)
 
This equation (9) is for any atomic system having one electron, and revolving around Z-protons.
Let us apply this equation on hydrogen atom for which
Z = 1 (H atom has one proton in the nucleus)
r
=
0.529 (n2)A ----- (10)
With the help equation (10) let us calculate the radii of first five orbits of hydrogen atom.
r1 = 0.529A (12)
=
0.529A
2
r2 = 0.529A (2 )
=
2.41A
2
r3 = 0.529A(3 )
=
4.8A
2
r4 = 0.529A (4 )
=
8.4A
2
r5 = 0.529A (5 )
=
13.2A
These calculations show that distances between adjacent orbits go on increasing from the lower to
the higher ones. Fig (5.11A)
4om 
o
5 = 13.22 A
o
r4 = 8.45A
n=1
n=2
+
r3 = 4.75Ao
n=3
n=4
r1 = 0.529A
o
o
r2 = 2.11A
n=5
Fig. (5.11A):
Radii of the first five orbits of hydrogen atom (The figure has not been drawn to scale)
In other words, r2  r1 < r3  r2 < r4  r3 and so on.
Important information
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(c)
Ans.
(i)
(ii)
(d)
Ans.
(e)
Ans:
These calculations tell us that second orbit is four times away than the first one from the nucleus.
The third orbit is nine times away. In other words, the atom becomes more and more thin as one
goes away from the nucleus.
How does the above equation tell you that:
(i) radius is directly proportional to the square of number of orbit.
(ii) radius is inversely proportional to the number of protons in the nucleus.
When we derive the equation for radius of an atom according to Bohr’s model, we get the
following equation
n2
r = 0.529  Z  A
 
According to this equation, radius is directly proportional to square of number of orbit. So when
the number of orbit increases the radius increases sharply. Second Bohr’s orbit is four times away
from the nucleus than the 1st orbit. Third orbit is nine times away.
According to the above equation, radius is inversely proportional to the number of protons in the
nucleus. Greater the number of protons, smaller the radius. i.e. He has two protons in the
nucleus, so its 1st orbit is at half a distance than hydrogen atom. Hence He ion is smaller than
that of hydrogen atom.
How do you come to know that the velocities of electrons in higher orbits are less than those
in lower orbits of hydrogen atom?
According to Bohr’s proposals, the centrifugal force of the electron is equal to the force of
attraction between nucleus and electron.
mv2
Ze2
=
r
4or2
Ze2
r=
4omv2
According to this equation, radius and velocities are inverse to each other. Greater the velocity of
the moving electron, smaller the radius.
Justify that the distance gaps between different orbits go on increasing from the lower to
the higher orbits.
The equation for the radius of the H-atom, after putting the values of different parameters for Hatom is as follows
r
Q.8.
(a)
Ans:
=
1A = 1010 m
0.529 (n2) A
If we put the value of “n” = 1, 2, 3, , we can get the radii which show that the distances
between adjacent orbits go on increasing.
Derive the formula for calculating the energy of an electron in nth orbit using Bohr’s
model. Keeping in view this formula explain the following:
The potential energy of the bounded electron is negative.
The expression for the potential energy of electron is
 Ze2
P.E.
=
40r2
Ze2 
The value of the kinetic energy is calculated from the equalization of force of attraction 
40r2
2
mv
and centrifugal force,  r 
 
1
Ze2
So, 2 mv2 = K.E. =
80r
When we add K.E. and P.E. the total energy is negative.
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2
ETotal =
(b)
Ans:
2
2
Ze
Ze
Ze

= 
80r 40r
80r
Total energy of the bounded electron is also negative.
The expression for the potential energy of electron is
P.E.
 Ze2
40r2
=
The value of the kinetic energy is calculated from the equalization of force of attraction 
Ze2 
40r2
mv2
and centrifugal force,  r 
 
1
Ze2
So, 2 mv2 = K.E. =
80r
When we add K.E. and P.E. the total energy is negative.
ETotal =
(c)
Ans.
Ze2
Ze2
Ze2

= 
80r 40r
80r
Energy of an electron is inversely proportional to ‘n2’ but energy of higher orbits are always
greater than those of the lower orbits.
The formula for the energy of an electron revolving in any orbit is given by the equation
1
E =  2.18  1018 J  2 
n 
Greater the value of ‘n’ greater the value of energy because energy is negative inverse of n.
It becomes more and more, ‘less’ negative. The value of energy approaches zero, when n = 
(d)
Ans.
The energy difference between adjacent levels goes on decreasing sharply.
If we put the value of n as 1,2,3,4 we get the energies of various orbits of hydrogen atom. These
values are as follows:
E1 = 2.18  1018 J
E2 =  0.54  1018 J
E3 =  0.24  1018 J
E4 =  0.14  1018 J
As is clear from these values that energy differences between adjacent levels go on decreasing
from lower to the higher level.
Q.11. (a) Hydrogen atom and He are monoelectronic system, but the size of He is much
smaller than H+. Why?
Ans.
Hydrogen atom and He ion have single electron in the valance shell but He has two protons in
the nucleus, so it will have greater force of attraction. The size of He is smaller than H atom.
rHe
(b)
Ans.
12
= 0.529  2  A = 0.2645 A
 
Do you think that the size of Li+2 is even smaller than He+. Justify with calculations.
+2
+2
The size of Li
is even smaller than He , because Li
has three protons in the nucleus. It has
only one electron. So,
n=1
and
Z=3
1
0.529
2+
rLi = 0.529  3  A = 3 A = 0.176A
 
Q.12.(b) How does the Bohr’s model justify the Moseley’s equation?
Ans: MOSELEY’S LAW
2
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Definition:
The square root of a frequency of a spectral line on the characteristic x-ray spectrum of an
element varies directly as the atomic number of the element producing the spectrum.
The mathematical relationship is as follows:
v  (Z  b)
v = a(Z  b)
---------- (1)
‘v’ is the frequency of the characteristic x-rays,
Z is the atomic number of the element
‘b’ is screening constant
‘a’ is proportionality constant
Equation (1) is called Moseley’s law.
If we plot a graph between Z on x-axis and v on y-axis a straight line is obtained for various
metals. The straight lines shown in the following diagram are for K, L and M lines of
spectra.
K
Line
24
22
Rh45
20
Zr40
L Line
18
Br35
16
Th90
Zn30
14
Tb65
12
Mn25
10
Cs55
At85
8
P15
Nd60
Sn50
CO20
At85
Hg80
Re75
Yb70
Zn40
Rh45
Re75
6
6
15
25
35
45
55
65
75
Th90
M Line
Hg80
85
95
Fig. Plot of v  10 against of Z of the element producing x-ray spectrum v is the
frequency of the line and z is the atomic number.
4
Q.14.(a) Briefly discuss the wave mechanical model of atom. How does it give the idea of orbital?
Compare orbit and orbital.
Ans: Definition:
It is difficult to determine the position as well as the momentum of the electron simultaneously.
According to Bohr’s theory, electron is considered as a material particle. The position and
momentum of an electron can be determined with great accuracy. According to de Broglie,
electron has the wave associated with it. So the measurement of both parameters simultaneously
is difficult.
In order to understand this principle, let us imagine that we throw a photon of x-rays on the
moving electron. These photons have very short wavelengths. The possibility for these photons to
hit the electron is very bright. So one can assess the position of electron when the photon of xrays reflects back. But this photon of very high energy of x-rays has disturbed the velocity of
electron and its momentum has changed. In other words the uncertainty in the momentum has
been created.
In order to avoid the uncertainty in the momentum we throw a photon of visible light on the
electron. This photon is weak as compared to the photon of x-rays. There are less chances for the
uncertainty in the momentum. But due to greater wavelength of this photon, the chance of hitting
on the electron, becomes very very small, so position becomes uncertain.
Mathematical expression for Heisenberg’s uncertainty principle :
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If the x represents the uncertainty of position and p represents the uncertainty of momentum
for a subatomic particle like electron, then
h
x  p
4
( indicates equal to or greater than)
This relation is called uncertainty relation. According to this relation if x is small, p will be
large and vice versa. Hence if one quantity is measured accurately the other quantity is measured
less accurately. This also means that the certainty of determination of one quantity introduces
uncertainty for the determination of other quantity.
Above equation can be written as
h
p

2  x
If x = 0 (we are sure about the position of particle)
h
Then p


2  0
It means that the error in the determination of the momentum is infinite. In other words the
momentum of the particle cannot be measured, when we are sure about the position of electron.
WAVE MECHANICAL MODEL OF ATOM
After the failure of Bohr’s atomic model the scientists were in search of new model. This model
was presented by Schrodinger. He has given an equation called Schrodinger wave equation. This
is called quantum mechanical model. It includes the following ideas :
(i)
Electrons has waves according to de Broglie’s concept.
(ii) Heisenberg’s uncertainty principle is true.
(iii) The formation of the stationary energy levels for the electrons around the nucleus is also
true.
Applications:
This wave mechanical model describes the electron as a three dimensional wave in the electronic
field of a positively charged nucleus. Schrodinger wave equation which is based on the concept of
the probability has following applications:
(1) The equation is used to calculate the energy and the wave function of the particle moving in
a single dimension.
(2) One can calculate the energy and wave function of particle in three dimensional box.
(3) The equation is used to derive an expression for the energy of an electron in hydrogen
atom.
(4) The equation gives the concept of atomic orbital from wave mechanical model.
Q.15. (b) What is (n + 1)rule? Arrange the orbitals according to this rule. Do you think that this
rule is applicable to degenerate orbitals?
Ans:` ELECTRONIC DISTRIBUTION
The maximum number of electrons that can be accommodated in any shell is given by the
formula ‘2n2’.
n = 1 (K)
2(1)2 = 2
n = 2 (L)
2(2)2 = 8
n = 3 (M)
2(3)2 = 18
n = 4 (N)
2(4)2 = 32
n = 5 (O)
2(5)2 = 50
A shell is further divided into subshells
n=1
has only one subshell 1s
n=2
has two subshells 2s, 2p
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n=3
has three subshells 3s, 3p, 3d
n=4
has four subshells 4s, 4p, 4d, 4f
n=5
has again four subshells i.e., 5s, 5p, 5d, 5f
Actually there should have been five sub-shells in n = 5. The number of elements in the periodic
table, which have been discovered up to this time are around 110. So we don’t need the fifth subshell for filling the electrons in any of the atoms.
n = 6 has again four sub-shell i.e. 6s, 6p, 6d, 6f
Further division of subshells:
Subshells are further subdivided into orbitals. These orbitals are named as s, p, d and f.
s subshell has only one orbital.
p subshell has three orbitals. i.e. px, py, and pz.
d subshell has five orbitals i.e. dxy, dyz, dxz, dz2, dx2y2
f subshell has seven orbitals.
Three orbitals of p subshells are triply degenerate.
Five orbitals of d subshells are five times degenerate.
Seven orbitals of f subshells are seven times degenerate.
The maximum number of electrons in ‘p’ subshells are six.
The maximum number of electrons in ‘d’ subshells are ten.
The maximum number of electrons in ‘f’ subshells are fourteen.
The arrangement of the orbitals keeping in view the energy values and principal quantum number
is as follows:
4f
4d
4p
4s
3d
3p
E
3s
2p
2s
1s
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Chapter 6
CHEMICAL BONDING
(1)
(2)
(3)
(4)
(5)
(6)
MULTIPLE CHOICE QUESTIONS
An ionic compound A+B is most likely to be formed when:
(a) the ionization energy of A is high and electron affinity of B is low
(b) the ionization energy of A is low and electron affinity of B is high
(c) both the ionization energy of A and electron affinity of B are high
(d) both the ionization energy of A and electron affinity of B are low
The number of bonds in nitrogen molecule is:
(a) one  and one 
(b)
one  and two 
(c) three sigma only
(d)
two  and two 
Which of the following statements is not correct regarding bonding molecular orbitals?
(a) Bonding molecular orbitals possess less energy than atomic orbitals from which they are formed
(b) Bonding molecular orbitals have low electron density between the two nuclei
(c) Every electron in the bonding molecular orbitals contributes to the attraction between
atoms
(d) Bonding molecular orbitals are formed when the electron waves undergo constructive
interference
Which of the following molecules has zero dipole moment?
(a) NH3
(b)
CHCl3
(c) H2O
(d)
BF3
Which of the hydrogen halides has the highest percentage of ionic character?
(a) HF
(b)
HBr
(c) HCl
(d)
HI
Which of the following species has unpaired electrons in antibonding molecular orbitals?
(a) O2+2
(b)
N22
(c) B2
(d)
F2
(1)
(6)
Ans:
Q.6.
b
b
(2)
b
(3)
b
(4)
d
(5)
c
Write the Lewis structures for the following compounds:
(i) HCN
(ii)
CCl4 (iii)
CS2
(iv)
H3NAlF3
(v) NH4OH (vi)
H2SO4 (vii)
H3PO4 (viii) K2Cr2O7
(ix) N2O5
(x)
Ag(NH3)2NO3
Ans:
(ii)
(
i
)H
C
N
H
(
iv
) HN
H
H
(v
) HN
H OH
H
F
O
(vii) H O P
O
O
(viii)
O
H
H
C
r OC
r O
OO
K
K
O
O O
S
O
O
O
(vi)
(
i
i
i
)S
C
S
H
H
F
A
lF
Cl
Cl C Cl
Cl
[ 40 ]
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O
O
N O N
(ix)
O
H
(x)
O
H
H N
H
Ag
O
N H
H
N
O
O
Q.7. (a) Explain qualitatively the valence bond theory. How does it differ from molecular orbital
theory?
(b)
How the bonding in the following molecules can be explained with respect to valence bond
theory?
Cl2, O2, N2, HF, H2S,
Ans: (a) DIFFERENCE BETWEEN VBT AND MOT
Molecular orbital theory
Valence bond theory
(b)
(i)
In valence bond theory, two atomic (i)
orbitals give an inter-atomic orbital
obtained by space filling of two
unpaired electrons, one being in
each of the two atomic orbitals.
In molecular orbital theory,
molecular orbitals are formed by
linear combination of atomic
orbitals (LCAO) method.
(ii)
The resulting molecule, consists of (ii)
atoms and retain their individual
character.
Atomic orbitals of the resulting
molecule lose their individual
identities.
(iii)
Atomic orbitals are mono-centric.
Molecular orbitals are poly-centric.
(iii)
Bonding in Cl2:
In case of Cl2, any one of 3p-orbitals of one Cl atom (say 3px), undergoes head on overlap with
each other to form Cl—Cl sigma bond. Other two p orbitals are completely filled and they do not
do the overlapping.
Bonding in O2:
In case of O2, half-filled 2px orbital of one oxygen has linear overlap with 2px orbital of other
oxygen atom to form -bond. Two 2py orbitals of both oxygen atoms undergo parallel overlap to
form -bond.
Bonding in N2:
In case of N2, p orbitals are half-filled. One half-filled 2px orbital has linear overlap with other
half-filled 2px orbital to form -bond. There are also two half-filled 2py and 2pz orbitals which
undergo sideways overlap to form two -bonds.
[ 41 ]
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Bonding in HF:
In case of HF, 2px orbital of fluorine atom is half-filled. It undergoes overlap 1s orbital of
hydrogen to form H  F sigma bond.
Bonding in H2S:
In case of H2S, sulphur has two half-filled p orbitals which overlap with two half-filled 1s orbitals
of two hydrogen atoms to form two sigma bonds.
Anyhow, the bond angle is greater than 90 due to repulsion of H-atoms.
Q.8. Explain VSERP theory. Discuss the structures of CH4, NH3, H2O, BeCl2, BF3, SO2, SO3 with
reference to this theory.
Ans: VALENCE SHELL ELECTRON PAIR REPULSION THEORY
Introduction:
This theory was put forwarded by Sidgwick and Powell in 1940. They said that the shapes of the
molecules could be interpreted in terms of electron pairs in the outer orbit of central atom.
Recently, Nylholm and Gillespie developed VSEPR theory, which can explain the shape of
molecules for non-transition elements.
Basic assumptions of VSEPR Theory:
The fundamental idea in this theory is that, the valence electron pair around the central atom
arranges themselves to give a geometry to the molecule. These valence electrons may be lone
pairs or bond pairs. These electrons pairs remain at maximum distance apart to keep the
repulsions at a minimum.
Postulates of VSEPR Theory:
(i)
The lone pairs and the bond pairs both participate in determining the geometry of
molecules.
(ii) The electron pairs arrange around the central polyvalent atom and remain at maximum
distance apart to avoid repulsion.
(iii) The electron pairs in the form of lone pairs occupy more space than the bond pairs.
(iv) The magnitude of the repulsion between the electron pairs in a given molecule decrease in
the following order.
Lone pair– lone pair > lone pair–bond pair>bond pair–bond pair.
These repulsions are called van der Waal’s repulsions.
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(v)
The two electron pairs of a double bond and three electron pair of a triple bond contain
higher electronic charge density. For this reason, they occupy more space than one electron
pair of a single bond.
(vi) While determining the geometry of molecule, the two electron pairs of a double bond and
three electron pairs of a triple bond, behave like a single bond. The reason is that, they tend
to occupy the same region between the two nuclei like a single bond.
Reason for more repulsion of lone pair than bond pair:
A bonding electron pair is attracted by both nuclei of the atoms. A non-bonding electron pair is
attracted by only one nucleus. Lone pair experiences less nuclear attraction, so its electronic
charge is spread out more in space than that of a bonding pair. So the non-bonding electron pair
exerts greater repulsive forces on bonding electron pairs. In this way, they tend to compress the
bond pair.
Types of molecules in VSERP theory:
In order to illustrate the VSERP theory, let us consider that the central atom is ‘A’ and this atom
is polyvalent. More than one ‘B’ type atoms are linked with ‘A’ to give AB 2, AB3, AB4, AB5,
AB6 and AB7 type molecules. It depends upon the valency of the atom ‘A’ that how many ‘B’
type atoms are attached with that. Following table (6-4) gives the shapes of different types of
molecules.
Table (6.4) Shapes of molecules according to VSPER Theory
Type
AB2
AB3
Electron pairs
Arrangement
of pairs
Molecular
geometry
Shape
Lone
2
0
Linear
Linear
B–A–B
3
0
Total Bonding
2
Trigonal
planar
3
2
Trigonal
planar
Bent (or
angular
1
B
Examples
BeCl2
HgCl2
120°
A—B
BH3, BF3
AlCl3
B
A
B
SnCl2, SO2
B
less than 120°
B
4
AB4
Q.9.
Tetrahedral
0
4
3
1
2
2
Tetrahedral
Trigonal
pyramidal
Bent (or
angular
A
CH4,SiCl4,
CCl4, BF 4– ,
NH4+,SO4–2
109.5°
B
B B
A
B
B
B
less than 109.5°
NH3,NF3,
PH3
A
B
H2O, H2S
B
less than 109.5°
The molecules NF3, BF3 and ClF3 all have molecular formula of the type XF3. But they have
different structural formulas. Keeping in view VSEPR theory sketch the shape of each
molecule and explain the origin of differing in shapes.
Ans:
F
Shape of BF3:
Boron has three elements in its valence shell. It
shares its electrons with three fluorine atoms to
form three covalent bonds. Hence, triangular
planar structure is formed with angle of 120º.
There is no lone pair on boron.
120O
F
[ 43 ]
Triangular
planar
structure
F
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Shape of NF3:
Nitrogen has five electrons in its outermost
shell, one lone pair and three unpaired electrons.
It undergoes sp3-hybridization. So it forms three
covalent bonds with three fluorine atoms.
N
There is one lone pair on nitrogen which repels other bond pairs. In this way F—N—F
bond angle reduces. Thus NF3 shows trigonal pyramidal structure like NH3.
Shape of ClF3:
Cl has seven electrons in outermost shell. In excited state, an electron jumps from 2py to 3d
orbital.
Ground state
2s
2p
Excited state
3d
3d
2s
3
2p
dsp -hybridization
In this way, there are three unpaired electrons. It
F
undergoes dsp3 hybridization. Thus fived dsp3
hybrid orbitals are formed. Cl shares three
unpaired electrons with three fluorine atoms to
Cl
F
form three covalent bonds. There are also two
lone pairs on Cl which are repelled by each other
and they also repel other bond pairs. Thus T-
T-shaped
structure
F
shaped structure is formed.

Q.10. These species NH 2 , NH3, NH
4 have bond angles of 105, 107.5 and 109.5 respectively.
Justify these values by drawing their structures.
Ans:

Justification of angles of NH
4 , NH3 and NH 2 :
 
N H4 :
H
Nitrogen forms three covalent and one coordinate
covalent bond. After formation of four bonds, there
remains no lone pair of electron. So it has perfect
tetrahedral structure with the angle of 109.5. All
the bonds have equal states.
Normal and
equal repulsion.
Angles 109.5O
+
N
Tetrahedral
structure

N H 3:
Greater
repuslion
Nitrogen forms three covalent bonds with
hydrogen. There is one lone pair on nitrogen which
repels bond pairs. Thus angle between bond pairs
reduces from 109.5 to 107.5. It has trigonal
pyramidal structure instead of tetrahedral.
N
Less
repuslion
 
N H 2 :
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In
 
N H 2 , nitrogen forms two covalent bonds
with two hydrogen atoms. Nitrogen has two lone
pairs of electrons on it. According to valence shell
N
electron pair repulsion theory, lone-pair lone-pair
repulsion is the highest. So in NH
2 , there is
greater repulsion than in NH3. These lone pairs are
repelled by each other. They also repel bond pairs.
Hence, angle is further reduced to 105.
Q.11. (a) Explain atomic orbital hybridization with reference to sp3, sp2 and sp modes of
hybridizations for PH3, C2H4 and C2H2. Discuss geometries of CCl4, PCl3 and H2S by
hybridization of central atoms.
Ans:
Atomic orbital hybridization for PH3:
In 
P H3, phosphorus has five electrons in its valence shell. There are three unpaired and one
lone pair of electrons. It undergoes sp3 hybridization to form four sp3 hybrid orbitals. Three 1s
orbitals of hydrogen atoms overlap with three sp3 hybrid orbitals. Fourth sp3 hybrid orbital
contains one lone pair of electron. This lone pair repels bond pairs and angle reduces to 107.5
and trigonal pyramidal structure is formed. This structure is quite similar to that of 
N H.
3
ATOMIC ORBITAL HYBRIDIZATION AND SHAPES OF MOLECULES
Need for hybridization:
We have studied overlapping between unmodified (unhybridized) atomic orbitals in valence bond
theory. There are some problems in explaining the formation of bonds and geometry of


molecules. For example, the bond angles of CH4, N H3 and H2 O can be explained on the

basis of concept of hybridizations.
Definition:
The process in which atomic orbitals of different energies and different shapes intermix to
form an equal number of orbitals, which are identical in all respects like shape, angle, length
and energy except orientation in space, is called hybridization.
Explanation:
The nature and shapes of the orbitals in the outermost shell of an atom mix up with each other. In
this way, they get the stability by rearranging the geometries of the orbitals. For this purpose they
are mixed up in definite ratio to give new hybrid orbitals. The bond so produced are
comparatively stable.
Change of valency and hybridization:
The hybridization explains valency of elements. In hybridization, usually the first step is the
formation of an excited state. It involves unpairing of electrons followed by promotion of an
electron to higher energy level. As a result there is an increase in the number of unpaired
electrons. So the valency of the element changes as explained in the following Table (6.8):
Table (6.8): Change in valency after promotion of electrons from ground to excited state
Valency
Element
Electronic configuration
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Before excitation
1s
Be(4)
1s
B(5)
1s
C(6)
2s
2p° y
2p° x
2p° z
2s
2p° y
2px
2p° z
2s
2px
2py
2p° z
after excitation
after excitation
1s

1s
1s
2s
2p° y
2px
2p° z
2s
2px
2py
2p° z
2s
2px
2py
2pz
2
3
4
Energy for excitation:
The energy required for this excitation comes from the energy which is released during the
process of hybridization and bond formation with other atoms. Remember that excitation of
electron and hybridization are simultaneous processes.
Significance of hybridization:
Hybridization gives entirely new shape and orientation of valence orbitals of an atom. It is quite
significant in the determination of the geometry of the molecules.
TYPES OF HYBRIDIZATION:
There are various types of hybridizations but we are going to discuss the types of hybridization
due to the mixing of s and p orbitals of small sized atoms.
Hybridizaiton of s and p orbitals has following types:
(a) sp3-hybridization
(b)
sp2-hybridization
(c) sp-hybridization
3
sp -Hybridization:
The type of hybridization in which one ‘s’ and three ‘p’ orbitals mix together to form four sp3hybridized orbitals is called sp3-hybridization.
(i)
Formation of CH4
Electronic configuration of valence shell of C.
C (ground state) = 1s 2 s 2px
2py 2p° z
sp3 sp3 sp3 sp3
C (excited state) = 1s 2s 2px 2py 2pz = 1s
In the ground state, C atom has two partially filled orbitals i.e., 2px and 2py. When one electron
from 2s orbital is promoted to 2pz orbital, carbon changes from divalent to tetravalent. At the
same time hybridization takes place and four sp3-hybrid orbitals are formed. A tetrahedral
geometry is developed. Carbon is at the centre and four equivalent hybrid orbitals are directed
towards the four corners of a regular tetrahedron.
z
z
+
y
y
x
z
+
x
z
y
x
+
y
x
py
px
s
3
Atomic orbitals hybridize to form four sp hybrid orbitals.
[ 46 ]
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sp3
+
+
sp3
+
sp3
sp3
Shown together (large lobes only)
sp3
sp3
3
sp
sp3
3
Fig.(6.7a) Formation of four sp hybrid orbitals
from a set of one s and three p orbitals.
The hybridized orbitals are oriented in space in such a way that angle between them is 109.5.
Fig. (6.7a) In methane molecule, four sp3 hybrid orbitals overlap with four 1s orbitals of four
hydrogen atoms at four corners of tetrahedron forming four sigma bonds. Fig. (6.7b)
H
H
H
C 109
.5 o
C
H
H
H
H
H
Fig. (6.7b) Four sp3-s overlaps in tetrahedral structure of CH4 molecule.
(ii)
Ammonia, NH3:
Electronic configuration of N7 is
1s 2s 2px 2py 2pz
N7 =
hybridized orbitals
= 1s
sp3 sp3 sp3 sp3
In electronic configuration of nitrogen, one 2s and three 2p orbitals undergo hybridization to form
four sp3 hybridized orbitals. These hybridized orbitals are directed towards four corners of
tetrahedron. Out of four hybrid orbitals, one is completely filled due to lone pair of electrons
while remaining three are half-filled. Nitrogen forms three bonds with three H atoms due to sp3-s
overlapping. These three hydrogen atoms are located at three corners of tetrahedron while fourth
sp3 hybrid orbital having lone pair of electrons is located at fourth corner. In this way, a
pyramidal molecule is obtained in which three H atoms form a base while lone pair forms the
apex.
The experimental value of angle in NH3 is 107.5. The decrease in the angle from 109.5 to
107.5 is due to the repulsion between lone pair and bond pairs of electrons. Fig. (6.8)
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lone pair
lone pair
N
N
o
H
H
107.5
H
H
H
H
3
Fig. (6.8) Three sp -s overlaps in NH3 molecule to form a pyramidal structure.
(iii)
Formation of water molecule:
Electronic configuration of valence shell of oxygen is
1s 2s 2px 2py 2pz
hybridized orbitals
O8 =
=
1s
sp3 sp3 sp3 sp3
In electronic configuration of oxygen, one 2s orbital and three 2p orbitals hybridize to produce
four equivalent sp3-hybrid orbitals. Two of which are completely filled by two lone pairs of
electrons and other two are half-filled. These half-filled orbitals overlap with 1s orbitals of two Hatoms. Hence two sigma bonds are formed. Thus H2O shows tetrahedral geometry in which two
hydrogen atoms are located at two corners and remaining two corners are occupied by two lone
pairs. The HOH bond angle is 104.5 instead of 109.5. It is due to the force of repulsion between
two lone pairs and two bond pairs of electrons.
Fig. (6.9)
lone pair
lone pair
lone pair
O
104.5 O
lone pair
O
.5
104
H
H
Fig. (6.9) sp3-overlaps in H2O to form an angular structure.
sp2-Hybridization:
The process in which one s and two p orbitals intermix to form three sp2-hybrid orbitals is called
sp2-hybridization.
Geometry of CCl4:
Cl

C
Cl
Cl
Cl
In CCl4, carbon atom is sp3 hybridized. Thus four sp3 hybrid orbitals are formed which overlap
with four p orbitals of four Cl atoms. As there is no lone pair on carbon, therefore, perfectly
tetrahedral structure is formed with angle of 109.5.
Geometry of 
P Cl
3
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
P
Cl
Cl
Cl

P Cl3 has structure just like NH3. In PCl3 phosphorus is sp3 hybridized. Thus four sp3 hybrid
orbitals are formed. Three sp3 hybrid orbtials overlap with three p orbitals of three Cl atoms.
There is one lone pair of electron on phosphorus which repels bond pairs and tetrahedral angle is
reduced to 107.5. Thus trigonal pyramidal structure is formed.
Geometry of H 
S :
2

S

H
H
3
In H2S, sulphur is also sp hybridized. Two sp hybrid orbitals overlap with two 1s orbitals of two
hydrogen atoms. Sulphur contains two lone pairs which are repelled by each other. They also
repel bond pairs and angle is reduced to 104.5. Thus angular structure is formed.
(b)
The linear geometry of BeCl2 suggests that central Be atom is sp-hybridized. What type of
hybridization a central atom undergoes when the atoms bonded to it are located at the
corners of (a) an equilateral triangle and (b) a regular tetrahedron (c) triangular
bipyramid.
Ans: (a) When atoms bonded to central atoms are located at the corners of equilateral triangle, the
central atom undergoes sp2 hybridization like BCl3, BF3, BH3, AlCl3 etc.
(b) In case of regular tetrahedral structure, the central atom undergoes sp3 hybridization like
CH4, SiH4, CCl4 etc.
(c) In case of triangular pyramidal structure, the central atom is sp3 hybridized.
Examples, 
N H , 
P H etc.
3
3
3
Q.12. (a)
Give the basis of the molecular orbital theory and discuss the molecular orbital
configurations of the following molecules:
2+
2–


(i)
He2 (ii) N2
(iii) O2
(iv) O2
(v) O2
2+
2–


(b)
How does molecular orbital theory explain the paramagnetic character of O2, O2 and O2
species?
Ans: (a) Molecular Orbital Theory (MOT)
Molecular orbital theory was put forward by Hund and Mulliken. Here are the assumptions of
molecular orbital theory.
Assumptions of MOT:
(1) The atomic orbitals in the valence shells of combining atoms overlap in pairs to form
molecular orbitals.
(2) The overlapping orbitals lose their identity.
(3) The energy and size of overlapping atomic orbitals must be same.
(4) The number of molecular orbitals formed is equal to the number of atomic orbitals
combined.
(5) Half of the molecular orbitals are of lower energy called bonding molecular orbitals and
half of higher energy are called antibonding molecular orbitals.
(6) Bonding molecular orbitals formed by head to head overlapping are named as sigma ()
orbitals and antibonding as sigma star (*).
(7) Bonding molecular orbitals formed by side ways overlapping of atomic orbitals are named
as pi () orbital and antibonding as pi-star (*).
(8) The electrons which occupy the valence shells of combining atoms are redistributed into
molecular orbitals according to the rules. (Aufau’s principle, Hund’s rule and Pauli
exclusion principle).
(9) The increasing order of energy of molecular orbitals is :
*  2px  2py = 2pz *2py = *2pz  *2px
1s  *1s  2s 2s
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(10) A bond between two atoms is possible only when the number of electrons in bonding
molecular orbitals is greater than that in antibonding molecular orbitals.
Let us discuss basic terminologies, which will be used in next topics.
Atomic orbital:
The region around the nucleus of an atom in which probability of finding an electron is maximum
is called atomic orbital.
Molecular orbital:
The orbital which is under the influence of two or more nuclei is called molecular orbital.
Bonding molecular orbital (BMO):
Molecular orbital which is of lower energy than parent atomic orbitals is called bonding
molecular orbital.
Bond axis:
The imaginary line between two bonded nuclei is called bond axis.
Paramagnetic substance:
A substance that is attracted towards magnet due to presence of unpaired electrons is called
paramagnetic. Liquid oxygen is paramagnetic in nature due to unpaired electrons in the
antibonding molecular orbitals.
Diamagnetic substance:
A substance which is not attracted towards magnet due to absence of unpaired electrons is called
diamagnetic.
Formation of molecular orbitals:
Two types of molecular orbitals i.e., bonding, of lower energy and antibonding, of higher energy
are formed due to s-s and p-p overlap.
(i)
s-s overlap:
Due to s-s overlap, two types of molecular orbitals are formed. (i) Bonding molecular orbital is
called  (s) and ii) Antibonding molecular orbital designated as *(s). Fig. (6.19a) and (6.19b)
In the formation of H2, 1s1 orbital of one hydrogen atom overlaps with 1s1 orbital of other
hydrogen atom to form two types of molecular orbitals. i.e., bonding molecular orbital (1s)
having low energy and antibonding molecular orbital *(1s) of high energy which remains
empty. Fig. (6.19a) and (6.19b)
*
*
1s
(A.O.) 1s = H
H = 1s (A.O.)
1s
(ii)
(a)
A.O.(1s)
H
1s(anti-bonding)
A.O. (1s)
H
1s (bonding)
p-p overlap:
The overlap is of two types:
(a) Head on approach.
(b) Side ways approach.
Head-on approach:(pp)
If two p orbitals of two atoms approach each other along the same axis. Then it is known as
head-on approach. This type of overlap gives (2px) bonding and *(2px) antibonding molecular
orbital when px approaches Px. These are symmetrical about nuclear axis. Fig. (6.20)
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*2p
z
y
z
+
node
y
x+
A,B.M.O
x
2p
Energy
2p
A.O
A.O
B.M.O
2p
Fig. (6.20) Head-on overlap of two p-orbitals
Sideways approach:
When p orbitals of two atoms approach along the axes which are parallel to form -molecular
orbitals, it is called sideways approach.
In this case bonding molecular orbital (2pz) has electron density equal to zero, on the nuclear
axis (nodal plane) but lies above and below the joining line. Antibonding molecular * (2pz has
least electron density in the internuclear region. Fig. (6.20) 2py and 2pz are called degenerate
molecular orbitals which means they are of same energy. Similar is the case of *(2py) and
*(2pz) molecular orbitals. They are also degenerate.
*(2p z)
node
Energy
+
A.B.M.O.
2p z
Nodal plane
2p z
A.O.
(2p z)
A.O.
B.M.O.
Fig. (6.21) Sideways overlap of two p-orbitals
Relative energies of the molecular orbitals:
Spectroscopic methods are used to measure the energies of molecular orbitals. There are two
types of energy order:
(i) Energy order for diatomic molecules like O2, F2 etc. and their positive and negative ions.
Fig. 6.22(a)
 Is*(1s)   (2s)  * (2s) <  (2px)   (2py) =  (2pz)  * (2py) = * (2pz) < *(2px)
(ii) Energy order for diatomic molecules like N2 and lighter elements (Li2, Be2 and B2). Fig.
6.22(b)
 Is < * Is <  (2s)  *(2s)   (2py) =  (2pz)   (2px)  * (2py) = * (2pz)  *(2px)
*
*
(2px)
2(px)
*
(2pz)
*
(2py)
A.B.M.O
*
(2py)
(2py)
2pz 2py 2px
*
(2pz)
(2pz)
A.B.M.O
2px 2py 2pz
A.O
A.O
2(px)
2pz 2py 2px
A.O
B.M.D
(2px)
2pz 2py 2px
A.O
B.M.O
p (2pz)
(2py)
*
2s
2s
A.O
*
B.M.O
2s B.M.O
(a)
2s
2s
A.O
A.B.M.O
2s
2s
A.O
A.O
2s
B.M.O
(b)
Fig. (6.22a) Molecular orbital energy
diagram for O2,F2 and their
Fig.(6.22b) Molecular orbital energy diagram
for Li2, Be2, B2, and N2 positive and negative ions.
[ 51 ]
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Reasons for the relative energies of molecular orbitals:
As the diagram shows, that in the case of B2, C2 and N2, the 2px is higher in energy than bonding molecular orbitals. This has been observed by spectroscopic techniques. Actually, this is
not true in the case of O2, F2 etc.
The reason which has been suggested for this reversal is based upon the fact that 2s and 2px
orbitals are mixed up for three molecules of B2, C2 and N2.
In the case of small sized atoms like, B, C and N, the energy difference between 2s and 2px
atomic orbitals is small. There is a possibility of mixing of these orbitals. In other words, there
happens a sort of hybridization of 2s and 2px orbitals. As a result of which 2s and *2s
molecular orbitals do not retain their pure s-character. Similarly 2px and *2px molecular orbital
do not have pure p-character. In this way, all the four molecular orbitals i.e. 2s, *2s, 2px and
2px acquire sp-character. Due to this mixing their energy change happens in such a way that their
molecular orbitals 2s and *2s, become more stable and are lowered in energy. The molecular
orbitals like 2px and *2px become less stable and are raised in energy. Since, p orbitals are
not involved in mixing, so the energies of -bonding molecular orbitals remain unchanged.
Under these circumstances, we can believe that 2px is raised to such an extent that it becomes
higher in energies than -bonding molecular orbitals.
The molecules of O2 and F2 have their usual order of molecular orbitals. The reason is high
energy difference of 2s and 2p-orbitals. The energy difference for oxygen and fluorine molecules
are 1595 kJ.mol–1 and 2078 kJ.mol–1 between 2s and 2p-orbitlas respectively. These values for
boron, carbon and nitrogen are 554 kJ.mol–1, 846 kJ/mol–1 and 1195 kJ.mol–1 respectively. These
energy differences are small and they cause the repulsions of molecular orbitals. These energy
differences have been calculated by spectroscopic techniques.
Bond order:
It can be defined in two ways:
The number of bonds formed between two atoms after overlapping of atomic orbitals is called
bond order. OR
Half of the difference between the number of bonding electrons and antibonding electrons is
called bond order.
Calculation of bond order of H2. Fig. 17(a), 6.17(b).
Number of electrons in bonding orbitals (nb) = 2
(iv)
Ans.
Number of electrons in antibonding orbitals (na) = 0
nb – na
2–0
Bond order = 2
= 2 =1
2+

Molecular orbital picture of O2
2+

In molecular orbital diagram of O2 , each oxygen contributes five electrons. Thus eight electrons
go into bonding molecular orbitals and two into antibonding molecular orbital.
82
Bond order = 2 = 3
[KK(2s)2 < (*2s)2 < (2px)2 < (2py)2 = (2pz)2
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2+

(v) Molecular orbital picture of O2
2+

In molecular orbital diagram of O2 , each oxygen contributes seven electrons. Thus eight
electrons go into a bonding molecular orbitals and six electrons in the antibonding molecular
orbital.
86
= 1
2
[KK(2s)2 < (*2s)2 < (2px)2 < (2py)2
= (2pz)2 < (*2py)2 = (*2pz)2
2+
2–


Paramagnetic character of O2, O2 and O2 :
Bond order =
(b)
2+
2–


Consider the molecular orbital diagrams of O2 (given in the text), O2 and O2 in part (a) of this
question.
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* (2p ) and 
*(2p )
In the molecular orbital diagram of O2, there are two unpaired electrons in 
y
z
2+ *

*
orbitals. So O is paramagnetic. In the molecular orbital diagram of O , (2p ) and (2p )
2
2
y
z
orbitals are vacant. So paramagnetic behaviour vanishes and now it is diamagnetic. In case of
2–
*(2p ) and 
*(p ) will contain four electrons.
O, two electrons are absorbed into O . Therefore 
2
2
y
z
There is no unpaired electron. So it is not paramagnetic.
Q,14 (b) How do you compare the bond strengths of
(i)
Polar and non-polar molecules
(ii) - and -bonds.
(c)
Calculate the bond energy of HBr. The bond energy of H  H is 436 kJ mol1 and that of
Br  Br is 193 kJ mol1.
Solution:
(b)(i) Comparison of strengths of polar and non-polar bonds:
In polar bond, bonded atoms have partial positive and partial negative charges. In non-polar bond,
bonded atoms either have no partial positive and partial negative charges or they have partial
positive and negative charge but they cancel the effect of each other. Thus molecular as a whole
is non-polar.
Polar bond has dipole-dipole forces which are stronger than the forces present in non-polar bond. So
polar bond is stronger than non-polar bond.
(ii)
Comparison of strengths of  and  bonds:
-bond is formed by the overlap of head to head overlap of half-filled atomic orbitals while 
bond is formed due to parallel overlap of two half-filled orbitals.
In -bond, the electronic cloud is symmetrical about the line joining the two nuclei. The
overlapping of orbitals along the same axes is greater.
In -bond, the electronic cloud is above and below the central line joining the two nuclei. The
extent of overlapping is less.
So -bond is stronger than -bond.
(c)
Data:
Bond energy of H  B
=
?
Bond energy of H  Hr
=
436 kJ mol1
Bond energy of Br  Br
=
193 kJ mol1
First of all we calculate bond energy for 1 atom of H
Formula applied:
Bond energy of 1 mole of H2
Bond energy of 1 atom of H
=
NA  2
Putting the values
436
Bond energy of 1 atom of H
=
= 3.621  1022 kJ/atom
6.0210232
Now we calculate bond energy for 1 atom of Br
Formula applied:
Bond energy of 1 mole of Br2
Bond energy of 1 atom for Br =
NA  2
Putting the values
193
=
= 1.603  1022 kJ/atom
6.0210232
Adding bond energies of H and
Br atoms
=
3.621 1022 +1.603  1022
=
5.224  1022 kJ/molecule
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The amount of energy for
1 mole of HBr
5.224  1022  6.02  1023
=
314.48 kJ mol1
=
Ans.
Q.16. PF3 is a polar molecule with dipole moment 1.02 D and thus the P-F bond is polar. Si, is in
the proximity of P in the periodic table. It is expected that Si-F bond would also be polar,
but SiF4 has no dipole moment. Explain it.
+s
Ans. PF3 is a pyramidal molecule like NH3.
P
Fluorine is sufficiently electronegative,
so all the bonds are polar. There develops
a net dipole moment as shown below:
s
s
F
F
s
F
In the case of SiF4, the molecule is
perfectly tetrahedral. All the four bonds
cancel the effect of each other and net
dipole moment is zero.
F
s
Si
F
+s
s
F
s
s
F
Q.17. Which of the following molecules will be polar or non-polar. Sketch the structures and
justify your answer.)
(i) CCl4
(ii)
SO3
(iii)
SF4
(iv)
NF3
(v) PF5
(vi)
SO2
(vii)
SF6
(viii) IF7
Ans.
C
l
(i)
Structure of CCl4:
In CCl4, the C-Cl bonds are sufficiently polar
(N
on-polar)
C
but all the four bonds cancel the effect of each
other. The resultant dipole moment is zero. So
C
l
C
l
it is non-polar.
C
l
(ii)
O
Structure of SO3:
It is AB3 type molecule. It has trigonal planar
S +
structure and there is no lone pair of electrons.
It has zero dipole moment. So it non-polar.


O
(iii)
Structure of SF4:
It is AB5 type structure. Due to the presence of
O
F
F
one lone pair sulphur SF4 molecule has a
distorted trigonal bipyramidal shape. All the
dipoles do not cancel the effect of each other.
So it is polar molecule.
F
F
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(iv)
Structure of NF3:
Just like NH3, NF3 gives trigonal pyramidal
geometry. Due to the presence of lone pair on
N
nitrogen the effect of dipoles are not cancelled.
So it is a polar molecular.
(v)
F
Structure of PF5:
PF5 molecule has trigonal bipyramidal shape.
All the P  F bonds are sufficiently polar. All
F
F
dipoles cancel the effects of each other. So it is
F
non-polar.
(vi)
F
Structure of SO2:
It is AB3 type molecule. It has one lone pair on
S+
sulphur. It has V-shaped structure due to lone
pair of electron on sulphur. Its dipole moment

O

O
is 1.6 D. So it is polar.
(vii)
Structure of SF6:
SF6 gives octahedral structure in which all
S  F bonds are equal and are equal to 90.
Individual dipoles cancels the effect of each
other. So it is non-polar.
(viii) Structure of IF7:
IF7 molecule gives pentagonal bipyramidal
structure. All the dipoles cancel the effects of
each other. So it is non-polar.
Q.18. Explain the following with reasons:
(i)
Bond distance is the compromise distance between two atoms.
Ans:
Atoms attract each other and energy of the system is lowered. When they reach at a certain
distance, their forces of attraction are the maximum. If they are brought further closer to this
distance, they start repelling each other and energy of the system increases.
It means they have minimum energy at a certain distance. This is the bond distance at which they
compromise with each other to have minimum energy and is called compromise distance.
(ii)
The distinction between a coordinate covalent bond and a covalent bond vanishes after
bond formation in NH4, H3O and CH3NH3.
Ans:
There are four bonds between nitrogen and hydrogen atoms in NH4 . No doubt one of the bonds
is coordinate covalent and three bonds are covalent, but every bond has 25% coordinate covalent
bond character and 75% covalent bond character.
In H3O , each bond has 33 % coordinate covalent bond character and 66 % covalent bond
character.
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(iii)
Ans:
(iv)
Ans:
(v)
Ans:
(vi)
Ans:
(vii)
Ans:
The bond angles of H2O and NH3 are not 109.5 like that of CH4, although O and N-atoms
are sp3 hybridized.
Like CH4, the molecules of H2O and NH3 are also AB4 type molecules. Carbon, oxygen and
nitrogen atoms undergo sp3-hybridization. CH4 is perfectly tetrahedral with the angle of 109.5.
In case of ammonia, there are three bond pairs and one lone pair. Lone pair-bond pair repulsion is
greater than bond pair-bond pair repulsion. Due to this reason, angle reduces to 107.5.
In case of H2O, there are two lone pairs on oxygen.
Due to this increased repulsion of two lone pairs, the angle further reduces to 104.5.
-bonds are more diffused than -bonds.
The -bond is formed by head to head overlap of two half-filled atomic orbitals. The electronic
cloud density is symmetrical along the bond axis. The electronic cloud density of -bond is not
symmetrical along the bond axis. It consists of two regions, above and below the bond axis. So bond is more diffused.
The abnormality of bond length and bond strength in HI is less prominent than that of HCl.
Chlorine has higher electronegativity than iodine. So, the polarities of HCl and HI bonds are
unequal. Therefore, abnormality of bond length and bond strength of HCl is more prominent than
HI.
Solid sodium chloride does not conduct electricity. But when electric current is passed
through molten sodium chloride or its aqueous solution, electrolysis takes place.
In solid NaCl, the oppositely charged ions are fixed at their positions. So they do not conduct
electricity in the solid state. In the molten state or solution state the ions are free to move towards
the respective electrodes.
The melting points, boiling points, heats of vaporization and heats of sublimation of
electrovalent compounds are higher as compared with those of covalent compounds.
Electrovalent or ionic compounds have high melting and boiling points due to the close packing
of oppositely charged ions. The positively charged ions are surrounded by negatively charged
ions and vice versa. That is why, they have very high melting points, boiling points, heat of
vaporizations and heats of sublimation.
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Chapter 7
THERMOCHEMISTRY
MULTIPLE CHOICE QUESTIONS
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
Ans:
If an endothermic reaction is allowed to take place very rapidly in the air, the temperature of the
surrounding air:
(a) remains constant
(b) increases
(c) decreases
(d) remain unchanged
In endothermic reactions, the heat content of the:
(a) products is more than that of reactants
(b)
reactants is more than that of products
(c) both (a) and (b)
(d)
reactants and products are equal
Calorie is equivalent to:
(a) 0.4184 J
(b)
41.84 J
(c) 4.184 J
(d)
418.4 J
The change in heat energy of a chemical reaction at constant temperature and pressure is called :
(a) enthalpy change
(b)
bond energy
(c) heat of sublimation
(d)
internal energy change
Which of the following statements is contrary to the first law of thermodynamics?
(a) An equivalent amount of heat energy can neither be created nor destroyed
(b) One form of energy can be transferred into an equivalent amount of other kinds of energy
(c) In an adiabatic process, the work done is independent of its path
(d) Continuous production of mechanical work without supplying an equivalent is amount of
heat possible
For a given process, the heat changes at constant pressure (qp) and at constant volume (qv) are
related to each other as:
(a) qp = qv
(b)
qp < qv
qv
(c) qp > qv
(d)
qp = 2
For the reaction
NaOH+HCl
NaCl+H2O,
the change in enthalpy is called:
(a) heat of reaction
(b)
heat of sublimation
(c) heat of neutralization
(d)
heat of combustion
The net heat change in a chemical reaction is same whether it is brought about in two or more
different ways in one or several steps. It is known as:
(a) Henry’s law
(b)
Hess’s law
(c) Joule’s principle
(d)
law of conservation of energy
Enthalpy of neutralization of all the strong acids and strong bases has the same value because:
(a) neutralization leads to the formation of salt and water
(b) strong acids and bases are ionic substances
(c)
acids always give rise to H ions and bases always furnish OH ions
(d)
the net chemical change involves the combination of H
and OH
(1)
c
(2)
a
(3)
c
(4)
a
(6)
c
(7)
c
(8)
b
(9)
d
ions to form water
(5)
d
Q.5.(a) Differentiate between the following :
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(i) Internal energy and enthalpy
Answers : (a) INTERNAL ENERGY OF A SYSTEM
Definition:
The energy stored in a substance in terms of its kinetic and potential energies by virtue of its
constituent particles is called its internal energy.
Explanation:
We know that energy is liberated or absorbed in the chemical reaction. The reason is that the
chemical reactions take place as a result of interaction of various atoms and molecules. It is,
therefore, assumed that the atoms and molecules are associated with the sum of energy of their
own which is called their internal energy. This energy includes energies due to translational,
rotational and vibrational motions of molecules. Fig (7.3) The kinetic and potential energies of
nuclei and electron within the molecules also contribute to the energy of system. It is denoted by
E. It should be noted that the internal energy is different for different substances. It is the
combined kinetic energy and potential energy of all the particles that make up the chemical
system. The internal energy depends upon motion of the particles i.e., molecules, their
arrangement, inter-molecular and intra-molecular forces.
(a) Molecules of He gas are
(b) A diatomic molecule is
(c) A molecule is rotating
having translational motions.
vibrating
on an axis.
Fig (7.3) : Translational, vibrational and rotational movements of molecules.
The absolute internal energy of a system cannot be determined experimentally. However, change
in internal energy E can be measured.
Change of internal energy:
Internal energy E of a system is a state function and depends only on the initial and final states of
a system.
Internal energy E1 at initial state is being contributed by the translational motion, rotational
motion, vibrational motion, orbital motion of electrons, spin motion of electrons and all types of
motions present in the nuclei of the atoms.
Similarly, E2 depends upon all these things. The measurement of all these types of motions is
impossible that is why, we cannot measure E1 and E2, but we only measure E, where
E = E2  E1
Internal energy change:
Internal energy change E is the amount of heat evolved or absorbed by the system at constant
volume.
Effects of increasing the internal energy of a chemical system:
Whenever internal energy of a chemical system is increased, then anyone of the following effects
or some of the effects are noted:
(i)
Increase of temperature.
(ii) Change of phase.
(iii) Happening of a chemical reaction.
Ways of transfer of energy:
There are two fundamental ways of transferring energy:
(i) Heat (ii) Work
Work :
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Work is always done by a system, when it exerts a force to overcome resistance. It is expressed in
terms of force and distance.
Work = force  distance
In the expansion of gases, the work is expressed in terms of pressure and change in volume. So,
W = P  V
The arrangement in Fig. (7.4) shows that CO2 gas produced creates a pressure and pushes the
piston backwards and volume is changed from V1 to V2 i.e. V. The work done is PV and is
negative in this case.
So, W=  PV
----------- (3)
V1
V2
CO2(g)
External pressure
V = V2
- V1
Dilute HCl
Marble chips
Fig 7.4 : Pressure volume work.
CaCO3 + 2HCl  CaCl2 + H2O + CO2(g)
Q.6. (b) Explain that burning of a candle is a spontaneous process.
Answers : SPONTANEOUS REACTIONS
Definition:
The reaction which takes place of its own, without any external assistance and moves from nonequilibrium state towards an equilibrium state is called spontaneous reaction.
Explanation:
A spontaneous process or reaction is unidirectional, irreversible and real. Due to spontaneity, the
reaction attains the equilibrium stage and it will not change until disturbed by some external aid.
Examples of spontaneous processes or reactions:
1.
Water flows from higher to the lower level.
2.
Gas expands spontaneously from a region of high pressure to a region of low pressure.
3.
Heat flows from hot to a cold body.
4.
Neutralization of a strong acid and strong base is a spontaneous reaction.
NaOH(aq) + HCl(aq)
NaCl(aq) + H2O(l)
5.
The burning of CH4 in the air produces heat which is used for cooking and running
industries.
CH4(g) + 2O2(g) —— CO2(g) + 2H2O
6.
When a piece of Zn is added to CuSO4 solution, the blue colour of solution disappears. It is
due to spontaneous redox reaction. (reaction)
Zn + CuSO4 —— ZnSO4 + Cu
Non-spontaneous reactions
The reaction which is reverse of spontaneous reaction and takes place by supplying energy to the
system from external source is called non-spontaneous reaction.
Examples of non-spontaneous processes or reactions :
1.
Flow of heat from a cold end of metal rod to the hot end.
2.
Flow of electrical current from lower potential to higher potential.
3.
The transfer of heat from cold interior part of the refrigerator to hot surroundings.
4.
The combination of N2 and O2 to form NO by electric discharge.
5.
Pumping of water uphill.
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Spontaneity of reaction and its endothermicity:
It is a common experience that a spontaneous process proceeds with a decrease in energy. So, it is
expected that a chemical reaction would proceed spontaneously, if the reaction system decreases
in energy by transferring heat to the surrounding. In other words, we may expect that all
exothermic reactions to be spontaneous. This is not always true. There are many endothermic
processes, which proceed spontaneously.
(i)
Evaporation of water :
H2O(l)
—— H2O(g)
Hv = + 44.0 kJ mol–1
(ii)
Dissolution of NH4Cl(s) in water:
NH4Cl(s) —— NH4(aq) + Cl (aq)
Hs = + 14.0 kJ mol1
Conclusion:
So this energy change alone can not help us to decide about spontaneity of reaction. We should
study the free energy of the system. To understand free energy the concept of entropy is
necessary.
(c)
Is it true that a non-spontaneous process never happens in the universe? Explain it.
Answer :
Non-spontaneous processes happen in the universe, but energy has to be provided to them. The
free energy change of such processes are positive. The amount of temperature has to be increased
to carry out such reactions.
Q.7 (b) How will you differentiate between E and H? Is it true that H and E have the same
values for the reactions taking place in the solution state.
Ans: H is the heat change at constant pressure, while E is the heat change at constant volume. In the
solution state the pressure is constant. So, there is no difference in E and H in such situation.
There is no change in volume when the reaction happens in liquid or solid state.
Hence,
PV = 0
H = E + PV
H = E + P  0
H = E
Q.10.(a) State the laws of thermochemistry and, show how are they based on the first law of
thermodynamics.
Ans: HESS’S LAW OF CONSTANT HEAT SUMMATION
Definition:
The enthalpy change of a system depends upon its initial and final states only. It is independent of
the path followed by the system.
In other words the energy released or absorbed in a chemical reaction will be the same whether it
completes in one step or several steps.
Explanation:
Consider a hypothetical reaction in which a substance A is converted into D by a single step or
through three steps. In a multi-step reaction B and C are intermediates which are stable enough to
be isolated
A
D
C
B
According to Hess’s Law:
H = H1 + H2 + H3
Examples:
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1.
(a)
(b)
Hess’s law can be verified from the formation of sodium carbonate by two ways:
When Na2CO3 is prepared in one step :
2NaOH + CO2  Na2CO3 + H2O
H = 89.16kJ mol1
When Na2CO3 is prepared by two steps:
NaOH + CO2  NaHCO3
H1 =  48.06kJ mol1
NaOH + NaHCO3
 Na2CO3 + H2O
H2 = 41.02 kJ mol1
According to Hess’s Law
H = H1 + H2
89.16kJ mol1 = 48.06  41.02 =  89.08 kJ mol1
2.
From the discussion of these two examples, we can say that the energy change in cyclic process is
always equal to zero.
Carbon dioxide can be produced in two ways, but the enthalpy change is same from both ways.
First way :
C(s) + O2(g)
 CO2(g) H = 393.7 kJmol1
1
Second way: C(s) + 2 O2(g)
 CO(g) H1 = 111kJ mol1
1
CO(g) + 2 O2(g)  CO2(g) H2 = 283 kJ mol1
Therefore,
H = H1 + H2
293.7 kJ mol1 =  111  283
= 394 kJ mol
C+O2
CO2
1
– O2
2
1
1
– O2
2
CO
(b)
Ans:
What is a thermochemical equation? Give three examples. What information do they
convey?
TYPES OF THERMOCHEMICAL REACTIONS
A chemical reaction that occurs with the evolution or absorption of heat is called a
thermochemical reaction.
There are two types of thermochemical reactions :
1.
Exothermic reactions :
Those chemical reactions which release the heat are called exothermic reactions.
In an exothermic reaction, energy is given to the surroundings. Actually the total energy content
of reactants is greater than those of products. This released energy is denoted with negative sign.
Examples: The energy change means enthalpy change.
(i)
(ii)
(iii)
2.
C(s) + O2(g)
2H2(g) + O2(g)
N2(g) + 3H2(g)

——
CO2(g)
2H2O(l)
2NH3(g)
H = 393.7 kJmol1
H = 285.58 kJmol–1
H = 41.6 kJmol–1
Endothermic reactions :
Those chemical reactions which absorb the heat are called endothermic reactions.
In an endothermic reaction, energy is absorbed from the surroundings. Actually the total energy
content of reactants is less than those of products. This absorbed energy is denoted by positive
sign.
Examples:
The energy change means enthalpy change.
(i)
2H2O(l)  2H2(g) + O2
H = + 285.5 kJ mol1
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(ii)
N2(g) + O2(g)
2NO(g)
H = + 180.5 kJ
Following sketch (7.1) clears the idea of an exothermic and an endothermic reaction.
Reactants
Products
H1
H2
Gain in enthalpy :
H is positive
H = H2 H1 = +ve
Loss of enthalpy
H is negative
H = H2 H1 = -ve
-
-
Increasing
enthalpy
Increasing
enthalpy
Products
H2
Reactants
(a) Exothermic reactions
H1
(b) Endothermic reactions
Fig (7.1) : Enthalpy changes in thermochemical reactions.
(c)
Why is it necessary to mention the physical states of reactants and products in a
thermochemical reaction? Apply Hess’s law to justify your answer.
Answer:
The substances exist because they posses energy. Whenever the a substance changes its physical
state or a phase change happens then it changes its energy. So in thermochemical reactions it is
necessary to mention the physical states of reactants and products.
Let us take an example of water synthesis from H2 and O2, when both of the reactants are in the
gaseous state.
If liquid water is produced then 285.8 kJ mol1 of energy is evolved. But when water is produced
in the vapour state 241.5 kJ mol1 of energy is evolved. Actually 44.3 kJ mol1 of energy is used
up to vapourize 1 mole of water at its boiling point. This is called heat of vaporization of water.
Hence, we will write the two reactions as follows :
H2(g) + ½ O2(g)
 H2O(g)
H = 241.5 kJ mol1
H2(g) + ½ O2(g)
 H2O(l)
H = 285.8 kJ mol1
Hess’s law : The Hess’s law can explain the above facts as follows :
H2(g) + ½ O2(g)
 H2O(g) H1 = 241.5 kJ mol1
H2O(g)
 H2O(l) H2 = 44.3 kJ mol1
If this reaction takes place in one step.
H2(g) + ½ O2(g)
 H2O(l) H = 285.8 kJ mol1
According to Hess’s law
H = H1 + H2
285.8 = 241.5 + (44.3)
285.8 kJ = 285.8 kJ
This explanation is just according to Hess’s law.
[ 63 ]
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Chapter 8
CHEMICAL EQUILIBRIUM
(1)
(2)
(3)
(4)
(5)
MULTIPLE CHOICE QUESTIONS
For which system does the equilibrium constant Kc has units of (concentration)1?
(a) N2 + 3H2
2NH3
(b)
H2 + I2
2HI
(c) 2NO2
N2O4
(d)
2HF
H2 + F2
Which statement about the following equilibrium is correct?
2SO2(g) + O2(g)
2SO3(g)
H = 188.3 kJ mol1
(a) The value of Kp falls with a rise in temperature
(b) The value of Kp falls with increasing pressure
(c) Adding V2O5 catalyst increase the equilibrium yield of sulphur trioxide
(d) The value of Kp is equal to Kc
The pH of 103 mol.dm3 of an aqueous solution of H2SO4 is:
(a) 3.0
(b)
2.7
(c) 2.0
(d)
1.5
10
2
6
The solubility product of AgCl is 2.0  10 mol .dm . The maximum concentration of Ag+ ions
in the solution is:
(a) 2.0  1010 mol dm3
(b)
1.41  105 mol dm3
(c) 1.0  1010 mol dm3
(d)
4.0  1020 mol dm3
An excess of aqueous silver nitrate is added to aqueous barium chloride and precipitate is
removed by filtration. What are the main ions in the filtrate?
(a)
Ag
(c)
Ba
(1)
Ans:
and NO3
+2
and NO3
c
only
(b)
only
(2)
(d)
a
(3)
Ag
Ba
b
and Ba
+2
+2
and NO3
(4)
b
and NO3
and Cl
(5)
c
Q.4. (a) Explain the term “reversible reaction” and “state of equilibrium.”
(c)
Write Kc for the following reactions :
(i)
Sn2+ + 2Fe3+
Sn4+ + 2Fe2+
(aq)
(ii)
(iii)
(iv)
(v)
Answers:
(aq)
Ag (aq) + Fe2+ (aq)
N2(g) + O2(g)
4NH3(g) + 5O2(g)
PCl5(g)
(i)
Kc
=
(aq)
(aq)
Fe3+ (aq) + Ag(s)
2NO(g)
4NO(g) + 6H2O(g)
PCl3(g) + Cl2(g)
4
+
2
+
[Sn  (aq)][Fe  (aq)]2
[Sn2+ ][[Fe3+ ]2
(aq)
(aq)
[Fe3+ (aq)][Ag (s)]
(ii)
Kc
=
[Ag (aq)] [Fe2+ (aq)]
[NO(g)]2
(iii)
Kc
=
[N2(g)][O2(g)]
[NO(g)]4[H2O(g)]6
(iv)
Kc
=
[NH3(g)]4[O2(g)]5
[PCl3(g)][Cl2(g)]
(v)
Kc
=
[PCl5(g)]
Q.5. (a) Reversible reactions attain the position of equilibrium which is dynamic in nature and not
static. Explain it.
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(b)
Why do the rates of forward reactions slow down when a reversible reaction approaches the
equilibrium stage?
Ans: STATE OF CHEMICAL EQUILIBRIUM
Definition:
A state of reversible reaction when two opposing reactions occur at the same rate and the
concentrations of reactants and products don’t change with time is called state of chemical
equilibrium.
Explanation :
When a reversible reaction proceeds, then the rate of forward step decreases and that of backward
step increases. A stage reaches when the rates of two opposing reactions become equal and the
system attains a state of equilibrium.
Graphical explanation of reversible reactions:
Y
When we plot a graph between time
on x-axis and the concentrations of
[C]
Concentration
the species on the y-axis then the
following type of graph is obtained.
The stage of equilibrium is that when
the graph becomes parallel to the
D]
=[
[A]
time axis. (Fig. 8.1)
O
= [B
]
X
Time
teg
Fig(8.1) Graph between time and concentration
Chemical equilibrium as ‘dynamic’ equilibrium:
When the equilibrium stage is reached then the two rates become equal. The number of molecule
of A and B which are converted into products are compensated by the backward reaction when C
and D are converted to A and B.
A+B
C+D
In this way all the molecules of A and B find the chance to be converted into C and D and vice versa.
y
[H2]=[I2]
Concentrations
Explanation:
Kinetic molecular model of a reaction
system helps us to understand the
dynamic nature of chemical equilibrium.
The molecules of reactants and products
are busy in ceaseless motion, and so the
collisions happening among A and B
give C and D. The collisions among C
and D can produce A and B.
The graphical explanation of following
reactions is shown in Fig (8.2).
[HI]
O
Time
teg
x
Fig(8.2) State of dynamic equilibrium
H2(g)+I2(g)
450°C
2HI(g)
Q.6.
When a graph is plotted between time on x-axis and the concentrations of reactants and
products on y-axis for a reversible reaction, the curves become parallel to time axis at a
certain stage.
(a) At what stage the curves become parallel?
(b) Before the curves become parallel the steapness of curves fall. Give reasons.
Ans: STATE OF CHEMICAL EQUILIBRIUM
Definition:
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A state of reversible reaction when two opposing reactions occur at the same rate and the
concentrations of reactants and products don’t change with time is called state of chemical
equilibrium.
Explanation :
When a reversible reaction proceeds, then the rate of forward step decreases and that of backward
step increases. A stage reaches when the rates of two opposing reactions become equal and the
system attains a state of equilibrium.
Graphical explanation of reversible reactions:
Y
[C]
Concentration
When we plot a graph between
time on x-axis and the
concentrations of the species on
the y-axis then the following
type of graph is obtained. The
stage of equilibrium is that
when the graph becomes
parallel to the time axis. (Fig.
8.1)
D]
=[
[A]
O
= [B
]
Time
X
teg
Fig(8.1) Graph between time and concentration
Concentrations
Chemical equilibrium as ‘dynamic’ equilibrium:
When the equilibrium stage is reached then the two rates become equal. The number of molecule
of A and B which are converted into products are compensated by the backward reaction when C
and D are converted to A and B.
A+B
C+D
In this way all the molecules of A and B find the chance to be converted into C and D and vice versa.
y
Explanation:
Kinetic molecular model of a reaction
[H2]=[I2]
system helps us to understand the
[HI]
dynamic nature of chemical equilibrium.
The molecules of reactants and products
are busy in ceaseless motion, and so the
collisions happening among A and B
give C and D. The collisions among C
and D can produce A and B.
x
O
teg
Time
The graphical explanation of following
Fig(8.2) State of dynamic equilibrium
reactions is shown in Fig (8.2).
450°C
H2(g)+I2(g)
2HI(g)
(c)
The rate of decrease of concentrations of reactants and rate of increase of concentrations of
products may or may not be equal, for various types of reactions, before the time of
equilibrium. Explain it.
Answer : (c)
When the number of moles of reactants and products are equal, then the rates of increase and
decrease of concentration are equal. But when the number of moles in the balance chemical
equation are different, then the rate of increase of concentration of specie will be different from
the others.
The dissociation of N2O4 to give NO2 is a reversible reaction. One mole of N2O4 breaks and two
moles of NO2 are generated. The speed with which the graph of NO2 rises is greater then the
speed of falling of the graph of N2O4.
N2O4
2NO2
Q.10. Explain:
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(a)
The change of volume disturbs the equilibrium position for some of the gaseous phase
reactions but not the equilibrium constant?
Ans: Those gaseous phase reversible reactions, which happens with changing number of moles are
affected by the change of volume at equilibrium stage. Their equilibrium position is disturbed, but
equilibrium constant is not changed.
(b)
The change of temperature disturbs the equilibrium position and the equilibrium constant
of reaction?
Ans: All the reversible reactions are distributed by changing their equilibrium position and equilibrium
constant by disturbing the temperature. Actually, change of temperature changes the energy
contents of reactants and products.
(c)
The solubility of glucose in water is increased by increasing the temperature.
Ans: The solubility of glucose in water is a endothermic process. Increase of temperature pushes the
system to that side where heat is absorbed, and that is the side where solubility increases.
Q.11. (a) What is ionic product of water? How does this value vary with the change in temperature?
Is it true that this value increase 75 times when the temperature of water increases form
0C to 100C?
(b)
What is the justification for the increase of ionic product with temperature?
(c)
How do you prove that at 25C in 1dm3 of water, there are 107 moles of H3O and 10–7
moles of OH
Ans: IONIC PRODUCT OF WATER
Pure water is a very poor conductor of electricity due to its very lower ionization. This selfionization of water is extremely low but it is measurable. It is a reversible process and
equilibrium constant can be written :
H2O + H2O
H3O + OH
or it can be written as :
HOH
H + OH
The equilibrium constant value of above reaction is 1.810–16 moles dm–3 at 25ºC.
[H+][OH]
Kc
=
= 1.810–16 moles dm–3
[H2O]
Let us suppose that water is 1000g or 55.5 moles. In other words the concentration of water is
55.5 moles dm–3.
So
[H2O] =
55.5
The concentration of water at equilibrium stage is almost equal to the initial concentration.
Thinking this factor to be constant, multiply it on the left-hand side with ‘Kc’
Kc [H2O] = [H] [OH], 1.810–1655.5 = [H] [OH]
[H] [OH]
=
1.0110–14
Kw = [H] [OH]
=
110–14 where Kw = Kc [H2O]
Kw = ionic product of water = dissociation constant of water
Hence, Kw = [H][OH] = 1014 at 25C
The concentration of ions of pure water increase with temperature and the value of ‘K w’
approaches 7.5  1014 at 100C. Table (8.4). The value increases 75 times when temperature fo
pure water increases from 0ºC to 100ºC. The increase in Kw is not regular.
Table (8.4) Values of Kw at various temperatures
Temperature(C)
Kw
0
0.11  10–14
10
0.30  10–14
25
1.0  10–14
40
3.0  10–14
100
7.5  10–14
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Whenever some quantity of an acid or a base is added in water, then ‘K w’ remains the same at a
given temperature but [H] and [OH] are no more equal. In neutral water, the concentration
of [H] and [OH] are equal.
In neutral water [H]
=
[OH]
Therefore
or
or
[H] [H]
[H]2
=
10–14
=
10–14
[H]
[OH]
=
107 moles dm3 in neutral water
Similarly,
=
107 moles dm3 in neutral water.
Q.12. (c) Is it true that the sum of pKa and pKb is always equal to 14 at all temperatures for any
acid? If not why?
Ans: Relationship of pKa, pKb and pKw
Whenever, weak acid or a base is dissolved in water the conjugate acid base pairs are produced.
There is a close relationship between Ka of the acid, Kb of the conjugate base and Kw of water.
Let us discuss the dissociation of HA.
HA(aq)
H (aq) + A (aq)
[H (aq)[A (aq)]
Ka
=
-------- (1)
[HA(aq)]
The strength of the conjugate base A can be checked by its reaction with water.
A (aq) + H2O
Kb
=
HA(aq) + OH (aq)
[OH (aq)][HA (aq)]
-------- (2)
[A (aq)]
Multiply equation (1) and (2)
Ka  Kb
=
[H(aq)] [OH (aq)]
Ka  Kb
=
Kw
-------- (3)
–14
The value of Kw is always 10 at 25C. Hence if we know the Ka of the acid, we can calculate
the Kb of the conjugate base and vice versa.
Kw
1
Ka
=
so Ka 
Kb
Kb
Kw
1
Kb
=
so Kb 
Ka
Ka
This proves that a strong acid having higher Ka value will have a base of low Kb value and vice
versa.
Let us take the log of equation (3)
log Ka  Kb
=
log Kw
log Ka + log Kb =
log Kw
Multiply it with negative sign
log Ka  log Kb
=
 log Kw
pKa + pKb
=
pKw
Since, PKw = 14, so the pKa and pKb of the conjugate acid base pair has a very simple
relationship with each other.
pKa + pKb
=
14 at 25C
Conclusions:
Since pka + pKb = pKw
We can draw two conclusions:
1.
Conjugate base of a very weak acid is relatively very strong.
2.
The conjugate base of a very strong base is relatively very weak acid. Table (B). shows,
that Ka and Kb are related with each other.
TABLE (B) : Some conjugate acid-base pairs
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Ka
(Strong acid)
6.8  104
1.8  105
4.3  107
5.6  1010
5.6  1011
(Negligible acidity)
Acid
HNO3
HF
CH3COOH
H2CO3

NH 4

HCO 3
OH
Kb
(Negligible basicity)
1.5  1011
5.6  1010
2.3  108
1.8  105
1.8  104
(Strong base)
Base

NO 3
F
CH3COO

HCO 3
NH3
2–

CO 3
2–
O
The table (B) shows that the product of Ka and Kb is equal to 1014.
1
So,
Ka 
and pKa
=
pKb
Kb
Q.13.(b) Acetic acid dissolves in water and gives proton to water, but when dissolved in H2SO4, it
accepts protons. Discuss the role of acetic acid in both cases.
Ans: LOWRY-BRONSTED CONCEPT OF ACIDS AND BASES
Definition:
Acids are those species, which release a proton or have a tendency to release a proton.
Bases are those species, which accept a proton or have a tendency to accept proton.
Conjugate acid - base pair:
Suppose an acid HA is dissolved in water
HA + H2O
H3O
+
A
acid
water
conjugate acid
conjugate base
In this reaction water has accepted the proton. So it acts as a acid. A can accept a proton,
so it is a conjugate base of the acid HA.
HCl + H2O
H3O
+
Cl
acid
water
conjugate acid
conjugated base
CH3COOH + H2O
H3O
+ CH3COO
acid
water
conjugate acid
conjugate base

HCl is a stronger acid than acetic acid, Cl is a weaker base than CH3COO ion.
When CH3COOH is dissolved in H2SO4, it excepts proton from the H2SO4, so CH3COOH
acts as a base.
CH COOH + H SO
CH COOH + HSO
3
2
4
3
2
4
Conclusion:
Weaker acids have strong conjugate bases and stronger acids have weak conjugate bases.
Q.14. In the equilibrium
PCl5(g)
PCl3(g) + Cl2(g) H = 90 kJ mol1
What is the effect on
(a) the position of equilibrium
(b)
equilibrium constant?
(i) if temperature is increased
(ii)
volume of the container is decreased
(iii) catalyst is added
(iv)
chlorine is added
Explain your answer.
Answers:
(i) Increase of temperature disturbs the equilibrium position and equilibrium constant of
reaction. The reaction is pushed to the forward direction because it is endothermic.
(ii) The increase of volume of container shifts the equilibrium position to the forward direction.
The decrease of volume brings it back. Actually the reaction is happening with increasing
number of moles and increasing volumes. Equilibrium constant is not affected.
(iii) There is no effect of catalyst on this reaction.
(iv) When chlorine is added from outside at equilibrium stage, the reaction is pushed to the
backward direction according the Le-Chatlier’s principle.
[ 69 ]
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Chapter 9
SOLUTIONS
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
Ans:
MULTIPLE CHOICE QUESTIONS
Molarity of pure water is:
(a) 1
(b)
18
(c) 55.5
(d)
6
18 g of glucose is dissolved in 90 g of water. The relative lowering of vapour pressure is equal to:
1
(a) 5
(b)
5.1
1
(c) 51
(d)
6
A solution of glucose is 10%. The volume in which 1 g mole of it is dissolved will be:
(a) 1 dm3
(b)
1.8 dm3
(c) 200 cm3
(d)
900 cm3
An aqueous solution of ethanol in water has vapour pressure:
(a) equal to that of water
(b)
equal to that of ethanol
(c) more than that of H2O
(d)
less than that of water
Azeotropic mixture of two liquids boils at a lower temperature than either of them, when:
(a) it is saturated
(b)
it shows positive deviation from Raoult’s law
(c) it shows negative deviation from Raoult’s law
(d) it is metastable
In azeotropic mixture showing positive deviation from Raoult’s law, the volume of the mixture is:
(a) slightly more than the total volume of the components
(b) slightly less than the total volume of the components
(c) equal to the total volume of the components
(d) none of these
Which of the following solutions has the highest boiling point?
(a) 5.85% solution of sodium chloride
(b)
18.0% solution of glucose
(c) 6.0% solution of urea
(d)
All have the same boiling point
Two solutions of NaCl and KCl are prepared separately by dissolving same moles of them in the
fixed amount of solvent. Which of the following statements is true for these solutions?
(a) KCl solution will have higher boiling point than NaCl solution
(b) Both the solutions have different boiling point
(c) KCl and NaCl solutions possess same vapour pressure
(d) KCl solution possesses lower freezing point than NaCl solution
The molal boiling point constant is the ratio of the elevation in boiling point to:
(a) molarity
(b)
molality
(c) mole fraction of solvent
(d)
mole fraction of solute
Colligative properties are the properties of:
(a) Dilute solutions which behave as nearly ideal solutions
(b) Concentrated solutions which behave as nearly non-ideal solutions
(c) Both (a) and (b)
(d)
Neither (a) nor (b)
(1)
c
(2)
c
(3)
b
(4)
c
(5)
b
(6)
a
(7)
b
(8)
a
(9)
b
(10)
a
Q.5. (b)One has one molal solution of NaCl and one molal solution of glucose.
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(i) Which solution has greater number of particles.
(ii) Which solution has greater amount of solvent.
(iii) How do you convert these concentrations into weight by weight percentage?
Answer:
(i)
Since both solutions are one molal so the number of molecules of solutes should be same. But
actually NaCl is an electrolyte. Each formula unit of NaCl gives two ions and two particles. If
NaCl is 100% dissociated in the solution then one molal solution of NaCl in 1 kg of water will
give twice the Avogadro’s number of particles of solutes.
Glucose is a non-electrolyte. Its one molal solution in 1 kg of water will give Avogadro’s number
of particles.
(ii)
One mole of NaCl (58.5 g) will have a smaller volume than one mole of glucose (180g). Anyhow,
the amount of solvent in both solutions is 1 kg so the amount of the solvent in both solutions is
equal.
(iii)
Data: Number of moles of NaCl
=1
Mass of NaCl dissolved
= 58.5 g
Mass of solvent water
= 1000 g
Total mass of the solution
= 1000 + 58.5 = 1058.5 g
Percentage of NaCl
=?
Formula applied:
Mass of NaCl
Percentage of NaCl
= Total mass of solution  100
Putting the values
58.5
Percentage of NaCl
=
 100 = 5.52 %
1058.5
Now calculate %age of glucose
Data:
Mass of glucose dissolved
= 180 g
Mass of solvent water
= 1000 g
Percentage of glucose
=?
Total mass of solution
= 1000 + 180 = 1180 g
Mass of glucose dissolved
Percentage of glucose
=
 100
Total mass of solution
Putting the values
180
Percentage of glucose
= 1180  100 = 15.2%
Q.6. Explain the following with reasons.
(i)
The concentration in terms of molality is independent of temperature but molarity depends
upon temperature.
Ans: In molal solutions the mass of the solvent and that of the solute are also fixed. The masses of the
substances are not temperature dependent.
In molar solutions we have the volumes of solutions. Volume of a liquid is a temperature
dependent. So the molality is not influenced by temperature but molarity does change.
(ii)
The sum of mole fractions of all the components is always equal to unity for any solution.
Ans: Mole fraction is the ratio of number of moles of one component to total number of moles. Hence
mole fraction of any substance is always less than unity. The mole fractions of three components
A, B and C, in a solutions are
nA
nB
nC
XA =
, XB =
, XC =
nA + nB + nC
nA + nB + nC
nA + nB + nC
Adding these mole fractions ;
nA
nB
nC
XA + XB + XC =
+
+
= 1
nA + nB + nC nA + nB + nC nA + nB + nC
(iii)
100g of 98% H2SO4 has a volume of 54.34 cm3 of H2SO4 because its density is 1.84 g cm3.
[ 71 ]
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Ans:
98% H2SO4 means that 100 g of H2SO4 solution has 98 g of H2SO4 and only 2 g of water.
mass
Density =
volume
mass
Since, Volume =
density
Putting the values
100 g
Volume = 1.84 g cm3 = 54.34 cm3
It means that the 98% H2SO4 having total mass of 100 g has a volume of 54.34 cm3.
(iv)
Relative lowering of vapour pressure is independent of temperature.
Ans: The relative lowering of vapour pressure and mole fraction of solute are related as:
p
= x2
P
Vapour pressure and lowering of vapour pressure depend upon temperature. So, when the
temperature of a solution is increased both the factors P and P increase in such a way that the
ratio remains the same.
(v)
Colligative properties are obeyed when the solute is non-electrolyte and also when the
solutions are dilute.
Ans: In the case of electrolytes, ions are produced, number of particles of the solutes increase and the
amounts of colligative properties also increase. Colligative properties are obeyed when the
solutions are dilute, so that the solute particles are behaving independently.
(vi)
The total volume of the solution by mixing 100 cm3 of water with 100 cm3 of alcohol may not
be equal to 200 cm3. Justify it.
Ans: Molecules of water and alcohol have bent structures. The forces of attractions among water and
alcohol in pure states are greater than after mixing. So the total volume of the mixture of the two
components is greater than the sum of individual volumes.
(vii)
One molal solution of urea in water is dilute as compared to one molar solution of urea but
the number of particles of the solute is same. Justify it.
Ans: In one molal solution of urea, 60 g of urea is dissolved in 1000 g of water, which is
approximately 1000 cm3 of water. In one molar solution of urea, 60 g of urea is added in water to
make total volume of solution as 1000 cm3. So the volume of water in molar solution is less than
1000 cm3. Hence molar solution is concentrated and molal solution is dilute.
(viii) Non-ideal solutions donot obey the Raoult’s law.
Ans: The molecules of components in non-ideal solutions have forces of attractions for each other. The
values of vapour pressures of individual components are not proportional to their mole fractions
as in Raoult’s law.
Q.12. How do you justify that
(a)
Boiling points of solvents increase due to the presence of solutes.
Ans: The surface of the solution has molecules of solute as well. They do not allow the solvent to leave
the surface as rapidly as in pure solvent. To boil the solutions, we have to increase the
temperature of solutions in comparison to pure solvents.
(b)
Freezing points are depressed due to presence of solutes.
Ans: The lowering of vapour pressure compels the solutions to freeze at those temperatures, which are
below the freezing point of pure solvent. The reason is that the vapour pressure temperature curve
meets the solid phase of pure solvent at lower temperature than the pure solvent.
(d)
Beckmann thermometer is used to note the depression of freezing point.
Ans: Beckmann thermometer can measure up to 1/20th of the degree. The elevation of boiling points
and the depressions of freezing points for dilute solutions are very small quantities. Hence, can
measure these very small increase or decrease of temperatures.
(e)
In summer the anti-freeze solutions protect the radiator from boiling over.
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Water boils at 100C. It is used in the radiators to decrease the temperature of the working
engine. If we add some suitable solutes which increase the boiling point of water, above 100C,
then easy boiling over of water is avoided.
(f)
NaCl and KNO3 are used to lower the melting point of ice.
Ans: NaCl and KNO3 are electrolytes and are sufficiently soluble in water. They double the number of
particles after dissociation in water. In this way they, can manage to decrease the freezing point of
water to a greater extent as compared to a non-electrolyte.
Q.15. Freezing points of solutions are depressed when non-volatile solutes are present in volatile
solvents. Justify it. Plot a graph to elaborate your answer. Also give one method to record
the depression of freezing point of a solution.
Ans: Depression of freezing point (Cryoscopy)
Freezing point:
Definition:
It is that temperature at which the solid and liquid phases of the substance co-exist
At the freezing point the solid and liquid have the same vapour pressures.
We know that vapour pressure of a solvent decreases in the presence of a solute.
So when a graph is plotted between
temperature on the x-axis and vapour
pressure on the y-axis, a curve ABC is
obtained for the pure solvent. The AB
portion of the curve is for pure solvent in
the liquid state. The portion BC of this
curve is for the solid solvent. Because
when the solvent goes to solid state, there
happens a sharp change of vapour pressure.
The point B corresponds to the freezing
point of the pure solvent T1 Fig (9.9).
Vapour Pressure
Ans:
A
B
o
p
Solution
Solid
solvent
p
E
C
O
D
Pure solvent
Freezing
temperature =
depression
Tf
T2
T1
Tf
o
x
Temperature (C )
Fig (9.9) : Depression of freezing point..
When the graph is plotted for the solutions, it lies below the curve of the pure solvent. This curve
DE, touches the curve BC at the point E. This point E corresponds to the freezing point of the
solution T2. So this freezing point is lower than the freezing point of the pure solvent. The vapour
pressure corresponding to the point E is lower than the vapour pressure correspond to point B.
Depression of freezing point = Tf = T1  T2
Depression of freezing point is directly proportional to the molality of the solution.
Tf  m
Tf = Kf  m
------- (1)
and Kf = Tf
The units of Kf are C kg mol1.
Where Kf is the molal freezing point constant of the solvent.
We know that
Mass of solute
1
molality = Molar mass of solute  Mass of solvent in kilograms
W2
1
m=

M2 W1/1000
W2 1000
m=

------- (2)
M2
W1
Putting equation (2) in (1)
W2 1000
Tf = Kf.

------- (3)
M 2 W1
So if we know the mass of the solute, mass of the solvent, molar mass of the solute and molal
freezing point constant then we can calculate the depression of freezing point. This equation can
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be converted into following form to calculate the molar mass of non-volatile, non-electrolyte
solute.
Kf  W2  1000
M2 =
------- (4)
Tf  W1
Measurement of depression of freezing point
Introduction:
There are two important methods for the measurement of depression of freezing point:
1.
Beckmann’s method.
2.
Rast’s camphor method
We will discuss Beckmann’s method which is most important.
Beckmann’s method:
Apparatus:
This is consisted of three major parts:
1.
A freezing tube ‘A’ with a side arm ‘M’ to contain the solvent or the solution. It is fitted with
Beckmann thermometer.
2.
An outer larger tube into which the freezing tube is adjusted.
3.
A large container containing the freezing mixture and having a strirrer.
Procedure:
20 to 25 g of the pure solvent is taken in the freezing point tube. A thermometer is adjusted in
such a way that the bulb of the thermometer is completely immersed in the solvent. First of all
determine the approximate freezing point of the solvent by directly cooling the freezing point
tube in the cooling bath. Fig (9.10)
Beckmann
thermometer
Stirrer
A
M
Air Jacket
Pure
solvent
Cooling
mixture
Fig (9.10) : Beckmann freezing point apparatus
Now melt the solvent and place the freezing point tube again in the freezing bath and allow the
temperature to fall. When this temperature comes down to within about a degree of the
approximate freezing point, then place it cautiously in the air jacket. The temperature is allowed
to fall further to about 0.5 below the freezing point and stir vigorously. In this way, the solid will
separate and the temperature will rise due to latent heat set free. In this way note the highest
temperature reached. Repeat this process to get the concordant value of the freezing point. Let
this freezing point is T1.
The solvent is remelted by removing the tube from the bath.
A weighted quantity of say 0.2 g of the solute is introduced through the side tube. The freezing
point of this solution is determined by the same method, and let it be T2.
Further quantities of the solute may be added, new depressions of freezing point may be noted
and following formula is applied to get the molecular mass of the solute.
Kf  W2  1000
M2 =
Tf  W1
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Chapter 10
ELECTROCHEMISTRY
(1)
(2)
(3)
(4)
(5)
MULTIPLE CHOICE QUESTIONS
The cathodic reaction in the electrolysis of dil. H2SO4 with Pt electrodes is
(a) reduction
(b)
oxidation
(c) both oxidation and reduction
(d)
neither oxidation nor reduction
Which of the following statements is correct about Galvanic cell?
(a) anode is negatively charged
(b)
reduction occurs at anode
(c) cathode is positively charged
(d)
reduction occurs at cathode
Stronger the oxidizing agent, greater is the
(a) oxidation potential
(b)
reduction potential
(c) redox potential
(d)
E.M.F. of cell.
If the salt bridge is not used between two half cells, then the voltage:
(a) decrease rapidly
(b)
decrease slowly
(c) does not change
(d)
drops to zero
If a strip of Cu metal is placed in a solution of FeSO4
(a) Cu will be precipitated down
(b)
Fe is precipitated out
(c) Cu and Fe both dissolve
(d)
No reaction takes place.
(1)
Ans:
Q.4.
(c)
a
(2)
d
(3)
a
(4)
d
(5)
b
Calculate the oxidation number of chromium in the following compounds:
(i) CrCl3
(ii)
Cr2(SO4)3
(iii)
K2CrO4
(iv)
K2Cr2O7
2–
(v) CrO (vi)
Cr O (vii)
Cr O
3
2
3
2
7
Ans: (b)
i)
CrCl3
Let the oxidation number of Cr
Oxidation number of Cl
By applying formula
(O.N. of Cr) + 3 (O.N of Cl)
x + 3 (1)
x3
= x
= 1
=
=
=
0
0
0
x = 3
(ii)
Ans.
Cr2(SO4)3
Let the oxidation number of Cr =
x
oxidation number of S =
+6
oxidation number of O =
2
By applying formula
2(O.N. of Cr) + 3(O.N.S) + 3  4(O.N of O)
2x + 3 (+ 6) + 12  2 =
0
2x + 18  24 =
0
2x
=
6
x
=
(iii) K2CrO4
Let the oxidation number of Cr
Oxidation number of K
Oxidation number of O
=
=
=
[ 75 ]
3
=
0
Ans.
x
+1
2
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By applying formula
2(O.N. of K) + (O.N. of Cr) + 4 (O.N. of O)
2(+1) + (x) + 4(2)
2+x8
x
=
=
=
=
x = 6
K2Cr2O7
Let the oxidation number of Cr =
x
Oxidation number of K
=
+1
Oxidation number of O
=
2
By applying formula
2(O.N. of K) + 2(O.N. of Cr) + 7(O.N. of O)
2(+1) + 2 (x) + 7 (2)
2 + 2x  14
2x
2x
0
0
0
82
Ans.
(iv)
x
CrO3
Let the oxidation number of Cr
Oxidation number of O
By applying formula
(O.N. of Cr) + 3(O.N. of O)
x + 3(2)
x6
=
=
=
=
=
0
0
0
42
12
= 6
Ans.
(v)
=
=
x
2
=
=
=
0
0
0
x = 6
(vi)
Cr2O3
Let the oxidation number of Cr
Oxidation number of O
By applying formula
2(O.N. of Cr) + 3(O.N. of O)
2(x) + 3(2)
2x  6
2x
=
=
x
2
=
=
=
=
0
0
0
6
x = 3
(vii)
Cr2O72
Let the oxidation number of Cr
Oxidation number of O
By applying formula
2(O.N. of Cr) + 7(O.N. of O)
2(x) + 7(2)
2x  14
2x
2x
Ans.
Ans.
=
=
x
2
=
=
=
=
=
 2
2
2
2 + 14
12
x = 6
[ 76 ]
Ans.
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(d)
Calculate the oxidation numbers of the elements underlined in the following compounds:
(i) Ca(ClO3)2
(ii)
Na3PO4(iii)
Cr2(SO4)3
(iv) K2MnO4
(v)
Na2CO3
(vi)
HNO3
(vii) HPO3
Ans:
(i)
Ca(ClO3)2
Let the oxidation number of Cl =
x
Oxidation number of Ca
=
+2
Oxidation number of O
=
2
By applying formula
(O.N. of Ca) + 2(O.N. of Cl)+2  3 (O.N. of O)
(+2) + 2(x) + 6(2)
2 + 2x  12
2x
2x
=
=
=
=
=
0
0
0
12  2
10
x = 5
Ans.
=
=
=
0
0
83
= 5
Ans.
=
=
=
=
0
0
0
18
(ii)
Na3PO4
Let the oxidation number of P
=
x
Oxidation number of Na
=
+1
Oxidation number of O
=
2
By applying formula
3(O.N. of Na) + (O.N. of P) + 4(O.N. of O)
3(+1) + (x) + 4(2)
x
x
(iii) Cr2(SO4)3
Let the oxidation number of S
=
x
Oxidation number of Cr
=
+3
Oxidation number of O
=
2
By applying formula
2(O.N. of Cr) + 3(O.N. of S)+3  4 (O.N. of O)
2(+3) + 3(x) + 12 (2)
6 + 3x  24
3x
x = 6
K2MnO4
Let the oxidation number of Mn
=
Oxidation number of K =
Oxidation number of O =
By applying formula
2(O.N. of K) + (O.N. of Mn) + 4(O.N. of O)
2(+1) + (x) + 4(2)
2+x8
x
Ans.
(iv)
x
+1
2
=
=
=
=
0
0
0
82
x = 6
(v)
Ans.
Na2CO3
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Let the oxidation number of C
=
x
Oxidation number of Na
=
+1
Oxidation number of O
=
2
By applying formula
2(O.N. of Na) + (O.N. of C) + 3(O.N. of O)
2(+1) + (x) + 3(2)
2+x6
x
=
=
=
=
0
0
0
62
x = 4
HNO3
Let the oxidation number of N
=
x
Oxidation number of H
=
+1
Oxidation number of O
=
2
By applying formula
(O.N. of H) + (O.N. of N) + 3(O.N. of O)
(+1) + (x) + 3(2)
1+x6
x
Ans.
(vi)
=
=
=
=
0
0
0
61
= 5
Ans.
=
=
=
=
0
0
0
61
x = 5
Ans.
x
(vii) HPO3
Let the oxidation number of P
=
x
Oxidation number of H
=
+1
Oxidation number of O
=
2
By applying formula
(O.N. of H) + (O.N. of P) + 3(O.N. of O)
(+1) + (x) + 3(2)
1+x6
x
3+
2+
Q.11. Is the reaction Fe  + Ag — Fe  + Ag spontaneous? If not, write spontaneous
reaction involving these species.
Ans:
3+
2+
Fe  + Ag
 Fe  + Ag
cathode anode
3+
In above reaction, Fe is being reduced while Ag is being oxidized. Therefore, Fe  will act
as cathode while Ag as anode.
Ecell = Eoxi + Ered
Ecell =  0.7994 + ( 0.44)
Ecell =  0.7994  0.44
Ecell =  1.2394
Since e.m.f. of the cell is negative, therefore, the cell-reaction is non-spontaneous. If the
electrodes are reversed, the cell-reaction becomes spontaneous i.e.,
2+
3+
Fe  + Ag  Fe  + Ag
Q.12. Explain the difference between:
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(a) Ionization and electrolysis.
(b) Electrolytic and voltaic cell
(c) Conduction through metals and molten electrolytes.
Ans: (a) Difference between ionization and electrolysis
Ionization
Electrolysis
1. The process of splitting up of an ionic
1. The process of decomposing a substance
compound into charged particles in molten
usually in solution or in molten state is called
state or when dissolved in water is called
electrolysis.
ionization.
2. In this process, there is no need of electrodes.
2. In this process, electrodes are required.
3. It occurs in an ordinary container like beaker
3. It occurs in an electrolytic cell.
etc.
4. There is no need of electricity for ionization.
4. There is a need of electricity for electrolysis.
5. Since there is no need of electrodes, therefore,
5. In electrolysis, cations and anions move
cations and anions do not move towards
towards respective electrodes.
respective electrodes.
6. In electrolysis, ions are oxidized or reduced to
6. During ionization, only ions are produced.
neutral atoms or molecules.
7. Example:
7. Example:
H2O
2NaCl(l)  2Na(s) + Cl2(g)
NaCl(s) —— Na(aq) + Cl (aq)
at cathode at anode
(b)
Electrolytic and voltaic cell
(c)
1.
2.
3.
4.
5.
6.
7.
Difference between conduction through metals and conduction through molten electrolyte
Conduction through metals
Conduction through molten electrolyte
In metals, conduction takes place due to free
1. In electrolytes, conduction takes place due to
electrons. So it is also called electronic
free ions.
conduction.
In this case, conduction takes place through
2. In this case, conduction takes place through
external circuit.
aqueous medium.
There is no need to convert metal into molten
3. There is a need to convert solid into molten
state or solution form.
state or solution form.
It takes place when a metal is connected to
4. It takes place when current is passed through
electrodes.
the electrolyte in an electrolytic cell.
It causes no change in the properties of
5. It causes decomposition of the electrolyte as a
conductor.
result of chemical reaction.
There is an increase in the resistance as
6. There is a decrease in the resistance as
temperature rises.
temperature rises.
7. Examples: Salts, acids and base are used as
Examples: All metals are conductors.
electrolytes in molten state or solution form.
Q.15. Will the reaction be spontaneous for the following set of half reactions. What will be the
value of Ecell?
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3+
Cr
Ans:
2+

  Cr(s)
(i)
(ii)
MnO2(s) + 4H + 2e  Mn(aq)
+ 2H2O(l)
(aq) + 3e
(Standard reduction potential for reaction
(i) =  0.74 V and for the reaction (ii) = + 1.28 V).
If emf of the cell is positive, the reaction will be spontaneous. In above reaction (i), chromium is
being reduced while in reaction (ii) manganese is being oxidized. The emf of the cell will be
calculated as:
Ecell = Eoxi + Ered = 1.28 + ( 0.74)
Ecell = 1.28  0.74
Ecell = + 0.54 volts
Q.16
(a)
Ans:
(b)
Ans:
The e.m.f. of cell is positive, so the cell reaction is spontaneous. The value of cell is + 0.54 volts.
Explain the following with reasons:
A porous plate or a salt bridge is not required in lead storage cell.
The oxidizing agents and reducing agents in lead storage cell are in the solid state, and they do
not come in contact with each other in the absence of salt bridge. The internal resistance of the
cell is very low.
The standard oxidation potential of Zn is 0.76 V and its reduction potential is 0.76 V.
The standard reduction potential of Zn for the reaction
Zn
+ 2e
 Zn, is  0.76 V. When this reaction is reversed, then it is an
oxidation half reaction.
2+
Zn  Zn
+ 2e
E = + 0.76 V
The value of oxidation potential will be the same but sign is reversed. i.e., +0.76 V.
Na and K can displace hydrogen from acids but Pt, Pd and Cu can not.
The metals like Pt, Pd and Cu have sufficiently high positive value of reduction potentials,
therefore, they cannot liberate hydrogen from acids. On the other hand, Na and K are close to top
of the electrochemical series and have very low reduction potentials and can liberate hydrogen.
The equilibrium is set up between metal atoms of electrode and ions of metal in a cell.
A piece of metal placed in a solution of its own ions gives electrons to form positively charged ions.
These ions enter into the solution. Meanwhile, positively charged ions which are already present in
solution absorb electrons from piece of metal. They become neutral and get deposited.
In the long run, an equilibrium state is established at which rates of both the processes become
equal. Then, no further change occurs in the potential difference.
A salt bridge maintains the electrical neutrality in the cell.
Salt bridge is a U-shaped glass tube having a saturated solution of some strong electrolyte like
KCl, K2SO4 or KNO3. It prevents the physical contact between the two electrolytic solutions. At the
time of electronic current in the outer circuit negative ions move from cathode compartment to anodic
compartment. In this way, the solutions of both half cells remain neutral.
Lead accumulator is a chargeable battery.
The lead accumulator is a secondary cell. It is charged by passage of a direct current through it.
For this purpose, the anode and cathode of an external source are joined to the anode and the
cathode of the cell, respectively. The redox reactions at respective electrodes are reversed.
Overall reaction during charging is:
2PbSO4(s) + 2H2O(l)  Pb(s) + PbO2(s) + 4H+(aq) + 2SO42(aq)
Hence lead accumulator is a chargeable battery.
Impure Cu can be purified by electrolytic process
The arrangement is done in such a way that the impure copper is made the anode and pure copper
at cathode. These electrodes are dipped in CuSO4 solution. Copper from the anode is removed in
2+
(c)
Ans:
(d)
Ans:
(e)
Ans:
(f)
Ans:
(g)
Ans.
+2
the form of Cu ions and goes to the cathode and deposits there by accepting the electrons. In
this way it is purified.
Chapter 11
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Composed by: Malik Khurshid Dhar Bazar Ph: 05824-420161
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REACTION KINETICS
(1)
(2)
(3)
(4)
(5)
Ans:
MULTIPLE CHOICE QUESTIONS
In zero order reaction, the rate is independent of:
(a) temperature of reaction
(b)
concentration of reactants
(c) concentration of products
(d)
None of these
2
If the rate of reaction for 2A + B  products is k[A] [B] and A is present in large excess, then
order of reaction is:
(a) 1
(b)
2
(c) 3
(d)
none of these
The rate of reaction:
(a) increases as the reaction proceeds
(b)
decreases as the reaction proceeds
(c) remains the same as the reaction proceeds
(d) may decrease or increase as the reaction proceeds
With increase in 10C temperature, the rate of reaction doubles. This increase in rate of reaction
is due to:
(a) decrease in activation energy of reaction
(b) decrease in the number of collisions between reactant molecules
(c) increase in activation energy of reactants
(d) increase in number of effective collisions
The unit of the rate constant is same as that of the rate of reaction in:
(a) first order reaction
(b)
second order reaction
(c) zero order reaction
(d)
third order reaction
(1)
b
(2)
a
(3)
a
(4)
d
(5)
a
Q.4.
What is chemical kinetics? How do you compare chemical kinetics with chemical
equilibrium and thermodynamics?
Answer:
Difference between chemical kinetics and chemical equilibrium
1.
Chemical kinetics
It is the study of rates and mechanism of
reaction
1.
Chemical equilibrium
It is the state at which rate of forward
reaction is equal to the rate of reverse
reaction.
2.
It is related to rate of reaction whether it 2.
is forward or reverse.
It is related to the state of reaction when rates
of forward and reverse reactions become
equal.
3.
In the study of rate of reaction, it is 3.
determined either by the rate of decrease
in the concentration of reactant or rate of
increase in the concentration of product.
It is studied by simultaneously considering
the rate of decrease in the concentration of
reactants and rate of increase in the
concentration of products.
4.
Kinetic study of the reaction is possible 4.
whether it is reversible or irreversible.
The state of chemical equilibrium is possible
only if reaction is reversible.
5.
Chemical kinetics belongs to the path or 5.
mechanism of reaction.
The state of chemical equilibrium belongs to
the rates of forward and reverse reactions.
6.
Chemical kinetics is concerned with rate 6.
of reaction and the factors affecting
reaction rate.
It depends upon the state of equilibrium and
the factors affecting upon it.
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Composed by: Malik Khurshid Dhar Bazar Ph: 05824-420161
PREPARED BY: PROF. ALTAF GOVT. COLLEGE DAVI GALI
Q.5.
Ans:
The rate of chemical reaction with respect to products is written with positive sign, but with
respect to reactants is written with negative sign. Explain it with reference to the following
hypothetical reaction.
aA + bB
 cC + dD
Whenever, we write down the rates of the above reaction, they may be written with respect to any
of the reactants or products. The rate of above mentioned reaction can be expressed with respect
A, B, C and D, as follows:
1 d[A]
1 d[B]
1 d[C]
1 d[D]
Rate = –
=–
=+
=+
a dt
b dt
c dt
d dt
The rate with respect to reactants is negative, because the concentration of reactants decrease with
the passage of time. The rates with respect to products are positive because the concentrations of
products increase with the passage of time.
Q.7. Differentiate between
(i)
Rate and rate constant of a reaction.
Ans: RATE OF REACTION
Definition:
The change in concentration of a reactant or a product per unit time is called the rate of
reaction explanation.
Let the change in concentration be denoted by c and the change in time by t. then
change in concentration
Rate of reaction =
time period of change
c
t
Rate of reaction =
The symbol (delta) means “the change in.”
The change in concentration may be for anyone of the reactants or a product. The following graph
(11.1) shows that concentrations of the reactant or reactants decrease with the passage of time and
the concentrations of product or products increase with the passage of time for the reaction.
y
Concentration
R
ea
ct
ts
d uc
P ro
en
ts
o
x
Time
Fig (11.1) : Change of concentration with time for a chemical reaction.
(ii)
Ans:
Homogeneous and heterogeneous catalysis
Activation of a catalyst
Those substances which promote the activity of a catalyst are called promoters or activators.
Tetraethyl lead Pb(C2H5)4 is added to petrol, because it saves the petrol from pre-ignition. This
compound of lead is used to improve the octane number of petrol.
Examples:
(a) The catalytic activity of Ni in the hydrogenation of vegetable oils is promoted by using
copper and tellurium.
(b) Iron is used as a catalyst in Haber’s process. The small quantities of high melting oxides,
like Al2O3, Cr2O3 or trace earth oxides increase the efficiency of iron.
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(iii)
Fast step and the rate determining step.
Ans:
RATE DETERMINING STEP
Definition:
Rate determining step is that step of a chemical reaction which controls the rate of reaction. It is
mostly the slowest step.
Explanation:
There are only a few reactions in chemistry which complete in a single step. There are many
reactions which happen in more than one elementary steps. All the elementary steps in the
mechanism do not have the same rates. One of the steps is much slower than the other. This step
limits the rate of reaction. That is why, it is called rate limiting or rate determining step.
Rate determining step and the mechanism of reaction
If the reaction happens in a single step, then its rate is being determined by this single step, but if
the reaction is taking place in more than one steps, then we have to write down the series of steps
and we look into the details of reaction. We come to know about the reactants and products along
with the intermediates. This total information about the reaction is called the mechanism of
reaction.
(iv)
Enthalpy change of reaction and energy of activation of reaction.
Ans:
ENERGY OF ACTIVATION
Definition:
It is the minimum amount of energy more than the average energy which is just sufficient to
convert the reactants into products.
Explanation:
The concept of energy of activation is gathered from the collision theory of reaction rates. This
theory is based on two postulates:
(i)
Collision is must for doing a reaction.
(ii)
Proper orientation of reacting molecules is necessary.
In order to understand collision, look at to fig (11.4a). The molecules of A2 and B2 collide
to give AB molecules. The formation of activated complex is clear from the diagram.
A
A
A
B
A
B
B
Covalent bonds between
atoms of A and between
atoms of B
A
A
B
B
B
A-A and B-B bonds are
partially broken, and
A-B bonds are partially
formed (activated complex).
A-A, B-B bonds broken,
A-B bonds formed
in AB molecules.
Figure (11.4a). Reaction between molecules of A 2 and B 2.
There are many collisions which happen among the molecules, but all of them don’t lead
towards the chemical reaction. Those collisions which lead towards the chemical reaction are
called effective collisions. These effective collisions require energy equal to energy of activation
and also demand proper orientation. The idea of proper orientation can be understood from the
reaction between NO2Cl molecules and Cl atoms. The effective collision is that one which
happens between Cl atoms and Cl atom of NO2Cl. Fig (11.4b) and (11.4c).
NO2Cl + Cl
—— NO2 + Cl2
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Cl
Cl
N O
Cl
N O
O
Cl
O
(a) Before collision
N O
O
Cl
collision
+
Cl
After collision no reaction
has occurred
O
N Cl
O
O
N
Cl
Cl
O
(b) Before collision
O
Figure (11.4b) : A collision that cannot produce Cl2 molecule.
Cl
N
+
Cl
Cl
O
collision
After a successful collision
Figure (11.4c) : A collision that can produce NO 2 and Cl 2 molecules.
ACTIVATED COMPLEX
When the colliding molecules come close to each other at the time of collision, they slow down,
stop and fly apart again. Their kinetic energy is converted to potential energy. When a graph is
plotted between potential energy on y-axis and the reaction co-ordinate (the path on which the
reaction travels) on the x-axis, a curve is obtained showing the maximum. The point of maximum
energies is called the position of formation of activated complex. Fig (11.5)
Minimum energy
for reaction
Energy
y
Activation
energy
Energy
content
of reactants
Energy
of reaction
(DH is
negative)
Energy content
or products
O
(Ea1)
Activation energy
of backward step
(Ea2)
x
Reaction Co-ordinate
Figure (11.5a) : Pathway of an exothermic reaction.
Exothermic reactions:
In exothermic reaction the potential energy of the products is lower than the potential energy of
the reactants. This difference is denoted by H. The value of H depends upon the initial state
(reactants) and the final state (products). It does not depend upon the route of reaction. It also
does not depend upon rate of reaction and energy of activation (Ea). So we must put in an amount
of energy (Ea1) equal to the activation energy to get the top of the barrier. Anyhow, we are
compensated more than that when the products are formed. If the reaction is reversible than the
products climb a higher hill to achieve the activated state i.e., energy of activation Ea2.
Ea1 = energy of activation for forward step.
Ea2 = energy of activation for the backward step.
The magnitude of H is also the difference of these two energies of activation. Fig (11.5a)
Ea1  Ea2 = E or H
y
Energy
Minimum energy
for reaction
Ea2
Activation
energy of backward step
Energy
content of
products
Activation
energy
Ea1
Energy
of reaction
(DH is
positive)
Energy content of reactants
x
Reaction Co-ordinate
Figure (11.5b) : Pathway of an endothermic reaction.
O
Q.8.
Justify the following statements:
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(i)
The rate of a chemical reaction is an ever-changing parameter under the given conditions.
Ans:
When the reaction progresses, the reaction is very fast at the beginning, slow somewhere in the
middle and very very slow at the end. The reason is that rate depends upon concentrations
according to law of mass action. The concentrations decrease every moment, so rate decreases
every moment.
(ii)
The reaction rate decreases every moment but the rate constant ‘k’ of the reaction is a
constant quantity, under the given conditions. Justify it.
Ans:
Rate constant is the rate of chemical reaction when the concentrations of reactants are unity. For a
hypothetical reaction,
A + B 
C+D
Rate = k[A][B]
The concentration of A and B decrease with the passage of time, so the rate decreases. But the
rate constant ‘k’ remains the same for the reaction, throughout its progress under the given
conditions.
(iii)
50% of a hypothetical first order reaction completes in one hour. The remaining 50% needs
more than one hour to convert itself into products.
Ans:
When the half life is one hour it means that 50 % of reaction is completed in one hour and 50 %
of rest of 50 % again requires one hour. In other words 75 % is completed in two hours and 87.5
% in 3 hours. So, it will take many hours to get 100 % completed.
(iv)
The radioactive decay is always a first order reaction.
Ans:
Radioactive substances have a single species at a moment, whose nucleus is being broken up
without the help of any external agency. So, only one reactant is present and it follows the first
order mechanism.
(v)
The units of rate constant of second order reaction is dm3.mol1 sec1, but the unit of rate of
reaction is mol.dm3 sec1.
Ans:
Since
Rate
=
 [C]
mole dm3
=
= moles dm3 s1
s
t
In case of second order reaction is
Rate
= k [A] [B]
k
mole dm3 s1
Rate
= [A] [B] = mole dm3 mole dm3
k = dm3 mol1 s1
(vi)
The sum of the co-efficients of a balanced chemical equation is not necessarily important to
give the order of reaction.
Ans:
There are many reactions in which the coefficients of overall balanced equation don’t become the
powers in the rate expression of the reaction. In such reactions there are more than one steps.
Only that step controls the rate which is slowest. The coefficients of that slowest step determines
the order of reaction.
(vii)
The order of a reaction is obtained from the rate expression of a reaction and the rate
expression is obtained from the experiments.
Ans:
There are many reactions which are consisted of more than one steps. One of the steps may be
slowest which controls the overall rate of reaction and the order of reaction. That step is
determined experimentally, so order of reaction is an experimental quantity.
Q.14. The collision frequency and the proper orientation of molecules are necessary conditions for
determining the proper rate of reaction. Justify the statement.
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Ans: ENERGY OF ACTIVATION
Definition:
It is the minimum amount of energy more than the average energy which is just sufficient to
convert the reactants into products.
Explanation:
The concept of energy of activation is gathered from the collision theory of reaction rates. This
theory is based on two postulates:
(i)
Collision is must for doing a reaction.
(ii) Proper orientation of reacting molecules is necessary.
In order to understand collision, look at to fig (11.4a). The molecules of A2 and B2 collide
to give AB molecules. The formation of activated complex is clear from the diagram.
A
A
A
A
B
B
A
B
B
B
B
Covalent bonds between
atoms of A and between
atoms of B
A
A-A and B-B bonds are
partially broken, and
A-B bonds are partially
formed (activated complex).
A-A, B-B bonds broken,
A-B bonds formed
in AB molecules.
Figure (11.4a). Reaction between molecules of A 2 and B 2.
There are many collisions which happen among the molecules, but all of them don’t lead
towards the chemical reaction. Those collisions which lead towards the chemical reaction are
called effective collisions. These effective collisions require energy equal to energy of activation
and also demand proper orientation. The idea of proper orientation can be understood from the
reaction between NO2Cl molecules and Cl atoms. The effective collision is that one which
happens between Cl atoms and Cl atom of NO2Cl. Fig (11.4b) and (11.4c).
NO2Cl + Cl
—— NO2 + Cl2
Cl
Cl
N O
Cl
O
(a) Before collision
Cl
N O
O
Cl
collision
N O
O
+
Cl
After collision no reaction
has occurred
O
N
Cl
Cl
O
(b) Before collision
O
N Cl
O
O
Figure (11.4b) : A collision that cannot produce Cl2 molecule.
Cl
collision
N
+
Cl
Cl
O
After a successful collision
Figure (11.4c) : A collision that can produce NO 2 and Cl 2 molecules.
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