PUC Chemistry Question Bank
PUC Chemistry Question Bank
PUC Chemistry Question Bank
2. Taxol
2. Taxol
Mixture
The
pure
homogeneous,
substance
irrespective
variable
is 2. In mixture each of its components
of retains its characteristic properties.
its origin
b) 6.002 x 1023
a) 6.005 = Four because the zeroes between the non zero digits
are significant figures
b) 6.022 x 1023 = Four because the exponential term is not
considered.
c) 4.01 x 102 = 3 significant figures
= 1mx1mx1m
= 102cm x 102cm x 102cm = 106 Cm3
= 3.5 x 1000 x 1m3 / 106
= 3.5x103
106
= 3.5 x 10-3 m3
1 minute = 60 seconds
3 day = 3 x 24 Hrs x 60 min x 60 sec
= 259 200 seconds
25. Calculate the molecular mass of the following
a) Ethane (C2H6)
b) Ammonia (NH3)
Ans: a) Ethane (C2H6)
Ethane = C2H6 2 x atomic mass of carbon + 6 x atomic mass of
Hydrogen
= 2 x (12.011u) + 6 x (1.008 U)
= 24.022u + 6.048u
= 30.070u
b) Ammonia (NH3)
Ammonia=(NH3) = 1x atomic mass of Nitrogen + 3 x atomic mass of Hydrogen
= 1x(14.01u) + 3 (1.008u)
= 14.01u + 3.024u = 17.034u
26. Calculate the formula mass of KCl (Potassium chloride)
Ans: formula mass of potassium chloride = Atomic mass of Potassium +
Atomic mass of chlorine
39.10u + 35.5u = 74.60u
27. Calculate the no of molecules present in 2.5 moles of water (H2O)
Ans: 1 mole of water = 6.022 x 1023
Therefore 2.5 moles of water = 2.5x6.022x1023
1
= 15.055 x 1023 molecules
28. Calculate the percent (%) composition of elements in methanol (CH3OH)
Ans: Molecular formula of methanol CH3OH
Molecular mass of methanol = 1x12.01+4x1.008+ 1x16.0
= 32.042 gm
Percent composition of carbon =
= 12.01
32.042
X 100
= 37.48%
Mass of Hydrogen
x 100
Molecular mass of CH3OH
= 49.93%
(g)
(g)
(g)
(g)
(g)
(g)
x 1 = 2 moles
(g)
(g)
= 25 Kg N2 + 5 Kg H2 NH3
1 Kg of N2 = 1000gm N2
25 Kg of N2 = 25 x 1000 = 25000/28 = 892.85 mol
1 Kg of H2 = 1000 gm H2
5 Kg of H2 = 5x 1000 = 5000/2.016 = 2480.15 mol
According to above equation 1 mol of N2(g) requires 3 moles of H2(g)
Hence 892.85 mole of N2 and the mols of H2 required would be
892.85 x 3 mol of H2 / 1 mol N2
= 2678.55 mol = 2.67855 x 103 mol H2
But we have 2480.15 mol of H2. Hence in this process 2678.55
mol of H2 is required.
3 moles of H2(g) 2 moles of NH3
2480.15 mole of H2 1653.43 mole of NH3 gas is formed
2480.15 x 2 = 1653.43 mol of NH3
3
32. A solution is prepared by adding 4.00 gm of a substance A to 18 gm of
water calculate the mass percent of the solute.
Ans: Mass % of solute (A) = Mass of solute (A)
Mass of solution
x 100
= 4 x 100 = 400
= 18.18 %
4+18g H2O 22gm
Mass % of solute = 18.18%
33. Calculate the molarity of sodium Hydroxide (NaOH) in the solution
prepared by dissolving 4 gm in 500ml of the solution.
Ans : Molarity =
No of moles of solute
Volume of solution in litre
4 Marks questions
11
Ans:
Element
At mass
% at mass
Carbon
57.14
12
57.14/12
Nearest
whole no
4.76
Hydrogen
6.16
6.16/1
6.16
6.16/0.68=9.06
Nitrogen
9.52
14
9.52/14
0.68
0.68/0.68=1
Oxygen
27.18
16
27.18/16
1.698
1.698/0.68=2.45
4.76/0.68=7
294.3
147
n=2
Therefore Molecular formula = Empirical formula x n
= (C7H9N1O2)2
= C14H18N2O4
3. Compound contains 4.07% Hydrogen 24.27% Carbon and 71.65%
chlorine. Its molecular mass is 98.96 gm what are its empirical formula
and molecular formula?
Ans:
Element
Symbol %
element
element
element = %
mass
molar
Hydrogen
4.07
4.07/1 = 4.07
4.04/2.018=2.01
Carbon
24.27
12
24.27/12
2.022/2.018
= 2.022
=1.0019
71.65/35.5
2.018/2.018 = 1
Chlorine
Cl
71.65
35.5
= 2.018
12
=2
Carbon
Symbol
% of
element
At mass
of
element
Moles of
the
element =
%
Simplest molar
ratio
40.687
12
40.687/12 3.390/3.389 = 1
Simplest
whole No.
multiplied
by 2
2
= 3.390
Hydrogen
5.085
5.085/1
5.085/3.389=1.5
= 5.085
Oxygen
54.228
16
54.228/16 3.389/3.389=1
= 3.389
13
Step 3
To calculate the molecular mass of the compound
The vapour density of the compound = 59
Molecular mass
= Vapour density x 2
= 59 x 2 = 118
Step 4
To calculate the value of n
n=
molecular mass
= 118
Empirical formula mass
59
=2
Step 5
Molecular formula = Empirical formula x n
= C2H3O2 x 2
= C4H6O4
Therefore Molecular formula is C4H6O4
***********************************************
14
1.
Name the person who first proposed the atomic theory of matter on scientific
basis.
2.
3.
4.
Under what conditions of pressure and voltage, the electrical discharge through the
gases can be observed?
5.
6.
7.
Name the phosphorescent material coated inside the discharge tube behind the
anode.
8.
Give the conditions under which cathode rays travel in straight line.
9.
What is the name given to the particles which constitute the cathode rays?
10.
Does the nature of cathode ray depend on the nature of gas in the discharge tube or
the electrode material?
11.
Name the scientist who was able to determine e/m value of an electron.
12.
13.
14.
15.
Name the fundamental particle of an atom that has highest value for its e/m value.
16.
Does the e/m value of canal rays depend on the nature of gas in the discharge
tube?
17.
Name the gas to be filled in the discharge tube to obtain the smallest and the
lightest positive ion.
18.
Name the smallest and lightest positive ion obtained when hydrogen gas is
subjected to electrical discharge.
19.
20.
21.
22.
Name the electrically neutral particle obtained by bombarding beryllium with particles.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
ZX
6C?
42.
43.
What is the relation between energy (E) and frequency ( )of an electromagnetic
radiation?
44.
What is the name given to the smallest quantity of energy that can be emitted or
absorbed in the form of electromagnetic radiation?
45.
46.
47.
48.
49.
50.
51.
52.
What is spectroscopy?
53.
54.
55.
56.
57.
58.
59.
What is the value of the radius of the first stationary state (Bohr orbit)?
60.
61.
Write the relationship between wave length ( ), velocity (c) and frequency ( )of
a radiation.
62.
63.
64.
65.
66.
67.
68.
69.
70.
71.
72.
73.
What are the possible values of l (azimuthal quantum number) for a given value
of n?
74.
What are the possible values for ml (magnetic quantum number) for a given value
of l?
75.
76.
What is the total value of ml (magnetic quantum number) for a given value of l?
77.
What is the value of l for:- (a) s- sub shell; (b) p- sub shell; (c) d-sub shell; (d)
f-sub shell; (e) g-sub shell; (f) h- sub shell?
78.
79.
Name the quantum number that specifies the shape of an atomic orbital.
80.
Name the quantum number that specifies the size of an atomic orbital.
81.
Name the quantum number that designates the orientation of the atomic orbital.
82.
83.
84.
85.
What is the shape of: (a) s-orbital: (b) p- orbital; (c) d-orbital?
86.
87.
88.
What is the maximum number of electrons that can be accommodated in: (a) sorbital: (b) p- orbitals; (c) d-orbitals; (d) f-orbitals?
89.
90.
Using s, p, d, notations, write the orbitals having following quantum numbers: (a)
n=4, l=0; (b) n=5, l=1; (c) n=3, l= 2.
91.
92.
93.
94.
95.
How many electrons in an atom may have the following quantum numbers: n=3,
l= 0?
96.
97.
Write the electronic configuration of the following elements: (a) Cr (Z=24) (b) Cu
(Z=29) (c) Ca (Z=20).
98.
Write the electronic configuration of the following: (a) Na+ (b) Cl- (c) O-2.
99.
100.
How many unpaired electrons are present in the following: (a) Na (b) P (c) O?
102.
103.
104.
105.
106.
107.
108.
109.
110.
111.
Name two series of hydrogen spectra which fall in infra red region.
112.
113.
114.
115.
116.
117.
Calculate the wave length of the radiation emitted with a frequency of 1,200kHz
(c =3.0x108m/s)
118.
119.
120.
The threshold frequency 0 for a metal is 6.0 X 1013 s-1. Calculate the kinetic
energy of an electron when the radiation of frequency = 1.0 X 1014 s-1 hits the
metal.
121.
What will be the wavelength of a ball of mass 0.2 kg moving with velocity of 10
ms-1?
122.
Calculate the wave number of the spectral line of shortest wavelength appearing in
the Balmer series of hydrogen spectrum (Given RH = 1.09 X 107 m-1)
What are the results drawn from the Cathode ray discharge experiment?
124.
125.
What are the observations made out of Ruther fords -ray scattering experiment?
126.
What are the conclusions drawn regarding the structure of the atom on the basis of
observations in the -ray scattering experiment?
127.
128.
129.
130.
What are the observations made by Hertz after conducting the photo electric effect
experiment?
131.
132.
Answers
Chapter 2: Structure of atom
1.
John Dalton
2.
3.
Like charges repel each other and unlike charges attract each other.
4.
5.
6.
7.
Zinc sulphide
8.
9.
Electrons
10.
11.
J.J. Thomson.
12.
-1.6x10-19C.
13.
9.1094x10-31Kg.
14.
15.
Electron.
16.
Yes.
17.
Hydrogen.
18.
Proton.
19.
20.
1.672x10-27 Kg.
21.
1.675x10-27 Kg.
22.
Neutron.
23.
J. J. Thomson.
24.
25.
26.
Gold.
27.
28.
29.
30.
31.
Atoms of the same element having identical atomic Number but different mass
No.
32.
Atoms of different elements having same mass No. but different atomic No.
33.
34.
A= Mass No,
35.
18.
36.
6.
37.
11.
38.
Electron
39.
Hertz2
40.
3.0x108 m/s.
41.
Number of wave lengths per unit length is called wave number ().
42.
43.
E =h
44.
Proton.
45.
400 nm to 750 nm
46.
6.626x10-34 Js.
47.
The ideal body which emits and absorbs radiations of all frequency.
48.
The ejection of electrons from metal surface when radiation strikes it.
49.
50.
Z= Atomic No.
51.
52.
53.
54.
2.18x10-18 J.
55.
56.
Balmer series
me vr = n h
2
1
n2
1 - 1
n12
n22
57.
En= -RH
58.
= RH
59.
a0 = 52.9 Pm.
60.
rn = n 2 a 0
61.
C=
62.
63.
64.
= h =
mv
cm-1
h
p
65.
66.
67.
68.
H = E
69.
It is the region around the nucleus where electron will most probably be found.
70.
Quantum Numbers specify the energy, size, shape and orientation of an orbital.
71.
n=1,2,3 - - - - - .
73.
l=0,1,2,3 - - - - (n-1).
74.
75.
+ 1 and - 1
2
2
76.
(2l+1) values.
77.
a) l=0,
78.
ml = -2, -1, 0, + 1, + 2.
79.
Azimuthal Q.no(l)
80.
Principal Q. no (n)
81.
82.
83.
84.
No. of nodes = 2
85.
a) Spherical
86.
px , py and pz
87.
88.
a) 2
89.
90.
a) 4s
91.
92.
93.
94.
95.
2 electrons.
96.
b) l=1,
c) l=2,
d) l=3,
b) dumb bell
b) 6
c) 10
b) 5p
c) 3d
e) l=4,
f) l=5
d) 14
97.
99.
3d
100.
a) 1
b) 3
c) 2
102.
103.
It fails to accounts for the finer details of the hydrogen atom spectrum. It could
not explain the ability of atom to form molecules by chemical bonds
104.
Orbit
1 It is a circular path around the
nucleus in which an electron
moves.
Orbital
1 It is the three dimensional region of
space where the probability of
finding the electron is maximum
p Orbital
106.
In the ground state of the atom orbitals are filled in the order of their increasing
energies.
107.
No two electrons in an atom can have the same set of four quantum
numbers.
108.
Pairing of electrons in the orbitals belonging to the same sub shell does not
takes place until each orbital belonging to that sub shell has got one electron
each.
109.
110.
111.
Bracket series
Paschen series
Pfund series ( any two)
112.
Lyman series
Balmer Series
Bracket series
Paschen series
Pfund series
113.
= h
Mv
De Broglie Equation
= RH 1
n12
115.
cm-1
n22
wave number
RH
Rydberg constant
116.
Lower the value of (n+l) for an orbital , the lower is its energy . If two
orbitals have the same value of (n + l) , the orbital with lower value of n will
have the lower energy.
117.
=C
3 X 10 8 ms-1
1200 X 103 s-1
= 0.0025 X 105 m
118.
= 250 m
= 1 = 1/ 580 X 10-10 m
119.
Energy of photon
E = h
h = 6.626 X 10-34 JS
= 4 X 10 12 Hz (s-1)
E = 6.626 X 10-34 Js X 4 X 1012 S-1
= 26.504 X 10 22
= 2.65 X 10 21 J
120.
Kinetic Energy = h ( o)
= (6.626 X 10 -34 Js) (1.0x1014 s-1 6.0 X 1013 s-1 )
= (6.626 X 10-34 Js) (4.0 X 1013 s-1)
= 2.65 X 10-20J
121.
=h
m
= (6.626 X 10 -34 Js)/ (0.2 Kg X 10 ms-1)
= 3.313 X 10 -34 m
122.
= RH (1/n12 1/n22 )
For Balmer series with shortest wavelength
n1 = 2
n2 =
i.
The Cathode rays start form cathode and move towards anode.
ii.
iii.
iv.
v.
124.
i.
Positively charged particles (Canal Rays) depend upon the nature of the gas
present in the cathode-ray tube.
ii.
The charge to mass ratio of the particles depend on the gas from which they
originate.
iii.
iv.
125.
i.
ii.
iii.
A very few -particles bounced back, that is were defected nearly 1800
126.
i.
ii.
The Positive charge of the atom is not spread through out the atom, but
concentrated in a very small volume
iii.
127.
i.
The positive charge and most of the mass of the atom is concentrated in a small
region called nucleus
ii.
iii.
The Electrons move around the nucleus in circular paths called orbits.
iv.
The Electrons and the nucleus are held together by Electrostatic force of
attraction
128.
i.
Electrical and Magnetic waves are perpendicular to each other and both are
perpendicular to the direction of the propagation of the wave.
ii.
Electromagnetic waves do not require medium and they can move in vacuum
iii.
There are many types of electromagnetic radiations. They differ from one
another in wavelength or frequency
iv.
v.
radiation.
129.
i.
The nature of emission of radiation from hot bodies (black body radiation)
ii.
iii.
iv.
130.
i.
The electrons are ejected from the metal surface as soon as the beam of light
strikes the surface
ii.
iii.
iv.
131.
i.
The electron in the hydrogen atom can move around the nucleus in a circular
path of fixed radius and energy and the paths are called orbits
ii.
The energy of an electron in the orbit does not change with time
iii.
E
h
= E2-E1
h
iv.
h
2
i.e., meVr = n h
2
132.
i.
ii.
iii.
Magnetic quantum number gives the information about the spatial orientation
of the orbital.
iv.
UNIT-3
Classification of elements and periodicity in properties
One mark questions:
1. For the triad of elements A, B and C if the atomic weights of A and C are 7 and 39.
Predict the atomic weight of B.
2. Every eighth element has property similar to the 1st element when placed
(arranged) in increasing order of their atomic weight. Name the law for the above
statement.
3. Which property of the element was the basis for the classification of elements by
Mendeleev?
4. State Mendeleevs periodic law.
5. Name the scientist whose experiment on x-ray spectra of elements led to modern
periodic law.
6. State modern periodic law.
7. What is more fundamental property for an atom of an element according to
Moseley?
8. Which quantum number corresponds to the period number in the modern periodic
table?
9. How many elements are there in the 4th period of long form of periodic table?
10. Write the atomic number of the element unniltrium.
11. Give the IUPAC name of the element whose atomic number is 109?
12. Which one of the following subshell is not filled in the 5th period (5s, 5p, 5d, 4d)?
13. In which period does the lanthanoids appear?
14. In which period does the actinide series of elements appear?
15. Name the series of inner transition element found in the 7th period.
16. How many elements are in lanthanide series?
17. Elements of a group have similar chemical properties. Why?
18. The position of helium is in 18th group of p block and not in 2nd group of s block of
long form of periodic table. Justify the statement.
19. To which block of the periodic table do the elements of group-I and II belong?
20. How many groups of elements form p block of the periodic table?
21. What are representative elements?
22. Which group of elements are called chalcogens?
23. Write the general outer electronic configuration of d block elements.
24. Which block of elements are more known for exhibiting paramagnetism and
catalytic properties?
25. A metal X forms coloured ions, is paramagnetic and is used as a catalyst. Predict
the block to which the metal belongs.
52. Give reason for the anamalous properties of 2nd period elements when compared to
the elements in their respective groups.
53. Mention one property common to all actinoid elements.
54. How are H and He related?
55. By what name do we know the 17th group elements?
56. Isoelectronic species do not have the same size. Why?
57. Arrange the following in the increasing order of their metallic character: Cu, K,
Ge, Br.
58. What is the difference between an amphoteric oxide and a neutral oxide?
59. Give an example for a basic oxide.
60. Give an example for a neutral oxide.
61. Name the element that is diagonally related to beryllium.
62. Why do 17th group elements have high negative electron gain enthalpy?
Two Marks questions:
1. Mention one merit and one drawback of Mendeleevs periodic table.
2. What was the name given by Mendeleev to the element if existed and had
properties similar to that of the aluminium? What is the present name of the
element?
3. What observation made by Moseley showed that atomic number and not atomic
mass is more fundamental property of an element?
4. With respect to long form of periodic table what are groups and periods?
5. How many groups and periods are present in the long form of periodic table?
6. How many elements are there in 2nd period? Justify your answer.
7. Which is the i) shortest ii) longest period in the long form of periodic table?
8. What are the subshells filled in i) 2nd period ii) 4th period?
9. Hydrogen is placed separately at the top of the long form periodic table. Justify its
position giving two reasons.
10. The electronic configuration of an element is [Ar] 3d7 4s2. To which block and
period does it belong?
11. Write the electronic configuration of the element with atomic number 118. Predict
the group the element belongs to.
12. Mention any two differences in the properties of metals and non-metals.
13. What are metalloids? Give an example.
14. Determination of size of an atom cannot be precise. Give reason.
15. How does atomic radius vary along a period and down a group in the periodic
table.
16. Explain why atomic size decreases along a period and increases down a group.
3
39. Which one of these is possible: BF4 or BF6 . Justify your answer. (atomic number
of boron = 5)
40. Why is the chemical reactivity of elements at the two extremes (except noble
gases) of the periodic table very high?
41. Why are the elements placed in the extreme left of the periodic table most
metallic?
42. Mention two factors on which ionisation enthalpy depends.
43. Mention one exception (anomaly) each found in the variation of i) ionisation
enthalpy in the 2nd period ii) electron gain enthalpy in the 16th period.
44. Classify these into acidic, basic, amphoteric and neutral oxide: CO, Na2O, Cl2O7,
Al2O3.
Four Marks:
1. a) What is the valence of 2nd group elements?
b) Using the table given below, evolve the formulae of i) aluminium oxide,
ii) aluminium nitride, iii) aluminium fluoride:
Group No. 13 15 16 17
Element Al N O F
2. a) When is a cation and an anion is formed from an atom?
b) Arrange H+, H, H in decreasing order of their size.
c) What does the energy absorbed in the reaction represent?
X (g)
X (+g ) + e
3. What does ionisation enthalpy, electron gain enthalpy and electronegativity
measure for an atom? Which one of these is not a measurable quantity?
4. Given:
Element Electronic configuration
P
1s2 2s2
Q
1s2 2s2 2p6 3s2 3p1
1s2 2s2 2p6 3s1
R
1s2 2s2 2p4
S
i) Arrange P, Q, R, S in decreasing order of their atomic radii.
ii) Which one among these is a chalcogen?
iii) Which one among these has more positive electron gain enthalpy?
iv) Give the formula of the oxide formed from R and S.
5. Mention any two characteristic properties each for s and p block elements.
6. How does metallic and non-metallic character vary along a period and down a
group. Give reasons.
7. Given the outer electronic configurations of A and B as .... 3s2 and .... 3s2 3p5.
a) Locate their position in the periodic table (group and period)
b) Which one of these has i) larger atomic size ii) higher ionisation energy
Value points
1
2
3
23
Law of octaves
Atomic weight or atomic mass
Properties of the elements are periodic functions of their atomic
weight
Moseley or Henry Moseley
Physical and chemical properties or properties of the elements are
periodic functions of their atomic numbers
Atomic number
Principal quantum number
18
103
Unnilennium
5d
6th period or 6
7th period or 7
Actinoids or actinide series
14
Similar outermost electronic configuration or same number and
same distribution of electrons in their outermost orbital
It has completely filled value shell (1s2) and has properties
characteristic of noble gases
s block
6
s and p block
16th group or 16
(n1) d110 ns02
d block
d block
They act as a bridge between chemically most active metals of sblock and less active elements of groups 13 and 14.
(n2) f114 (n1) d01 ns2
Elements appearing after uranium
Metallic character increases
114 pm
It is one half the distance between two atoms bonded by a single
covalent bond
It is half the internuclear distance separating the metal cores in the
metallic crystal
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
Marks
allotted
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
X (g)
+ e
It is the energy required to remove the 2nd most loosely bound
electron
It is more difficult to remove an electron from a positively charged
ion than from a neutral atom
P = Li, Q = Ne, R = Na, S = Ar
It is the enthalpy change that occurs when an electron is added to a
neutral gaseous atom
It is an endothermic process or energy is always absorbed to
remove an electron
Y<X<Z
It is the ability of an atom in a compound to attract the shared
electrons to itself
Electronegativity
4
Electronegativity non-metallic character
Same size, large charge / radius ratio, high electronegativity (any
two)
Radioactivity
Isoelectronic
Halogens
They have different nuclear charge
Br < Ge < Cu < K
Amphoteric oxide has both acidic and basic character. But a
neutral oxide is neither acidic nor basic.
Na2O
CO
Aluminium
By gaining one electron they attain noble gas configuration
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
4
5
6
7
8
10
11
12
13
14
Marks
allotted
i) Position of some elements do not fit in with the scheme of
1
classification or element of lower atomic weight are placed
before the element with high atomic weight.
ii) He correctly predicted the existence of few elements.
1
Eka-aluminium
1
Gallium
1
The regularities in the x-ray spectra of elements could be
1
accounted if a plot of ( is the frequency of x-rays emitted)
versus atomic number and not atomic mass was done.
1
or
1
A graph of versus atomic number and not atomic mass could
1
explain the characteristics of x-ray spectra of elements.
The vertical column of elements are called groups.
1
The horizontal rows of elements are called periods.
1
18 groups and 7 periods
1+1
1
8 elements
nd
1
In the 2 period (n = 2) the sub-shells filled are 2s and 2p only
st
1
1 period
th
1
6 period
i) 2s and 2p
1
ii) 4s, 3d and 4p
1
st
i) It has only one electron in 1 orbital and hence could be
placed in I group
1
ii) It can also gain one electron to achieve a noble gas
configuration like 17 group elements and hence can be placed
1
in 17 group.
Block d
1
1
Period 4
2
2
6
2
6
2
10
6
2
10
6
2
14
10
6
2
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s
1
14
10
6
2
14
10
6
5f 6d 7p or [Rn] 7s 5f 6d 7p
18 group or noble gas group
1
Metals
Non-metals
(any
1. Have high melting point
Have low melting point.
two)
2. Good conductor of heat and Bad conductor of heat and
1+1
electricity
electricity
3. Malleable and ductile
Not malleable and ductile
Elements that show properties similar to both metals and non1
metals
Silicon (or any other)
1
i) Atom is very small
1
ii) Electron cloud of the atom does not have a sharp boundary
1
Value points
15
16
17
18
19
20
21
22
23
24
25
26
27
28
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1+1
Graph-1
Position of elements-1
It increases.
Increase in nuclear charge outweighs (or is more prominent) the
shielding effect
It decreases.
Increase in shielding effect outweighs the increase in nuclear
charge.
In Beryllium electron is removed from 2s electron which is closer
to the nucleus.
In boron the electron is removed from 2p orbital which is far
away from the nucleus and is also shielded by 2s electrons.
In nitrogen the three 2p electrons are in different atomic orbitals.
In oxygen, two of the four 2p electrons occupy the same 2p
orbital resulting in electron-electron repulsion.
In magnesium the electron has to be removed from 3s orbital
which is closer to the nucleus.
In aluminium the electron has to be removed from 3p orbital
which is far from the nucleus and is also well shielded by 3s
electrons.
1
1
1
1
1
1
1
1
1
1
Ionisation enthalpy
Its always positive
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2
2
1
1
1
1
1
1
1
1
1
1
1
1
2
4
5
6
7
Marks
allotted
1
a) 2
1
b) i) Al2O3
ii) AlN
1
iii) AlF3
1
When an atom loses an electron and when an atom gains an
2
electron.
1
H > H > H+
st
1
1 ionisation enthalpy
Ionisation energy measures tendency of an atom to lose an
1
electron.
Electron gain enthalpy measures tendency of an atom to gain an
1
electron.
Electronegativity measures the ability of an atom in a compound
1
to attract shared electrons.
Electronegativity
1
Q>R>P>S
1
S
1
Q
1
R 2S
1
s block : They are metals and form basic oxides.
2
p block : They are non-metals and form acidic oxides.
2
Metallic character increases down a group.
1
Ionisation energy decreases.
1
Non-metallic character increases along a period.
1
Electron gain enthalpy increases.
1
nd
rd
th
rd
2
A: 2 group, 3 period ; B: 17 group, 3 period
i) A ii) B
2
Value points
Chemical bonding
1 Mark Questions
1) Who were the first to propose a theory on chemical bonding using electrons?
2) What is a chemical bond?
3) Write Lewis dot symbols for atoms of the following elements
Mg, Na, B, O, N, and Br.
4) Draw the Lewis structures for the following molecules and ions.
H2, O2, CO2, C2H4, C2H2, HNO3, CO
5) In the periodic table, the group of highly electronegative elements is _____.
6) In the periodic table, the group of highly electropositive elements is _____.
7) Write the general electronic configuration of noble gases.
8) What type of bond is present in NaCl?
9) Which force holds oppositely charged ions together in an ionic bond?
10) Name a cation that contains cation having two non metallic elements in an ionic
compound.
11) How does resonance stabilize a molecule?
12) Give the mathematical expression for dipole moment.
13) Expand VSEPR
14) What is the basis of VSEPR theory?
15) Arrange the repulsive interaction between electron pairs (lp-bp, bp-bp, and lp-lp)
in increasing order.
16) What is lone pair of electrons?
17) What is bonded pair of electrons?
18) Among bonded pair of electron and lone pair of electron, which occupy more
space in a molecule?
19) How many lone pairs of electrons are present in a molecule of ammonia?
20) How many lone pairs of electrons are present in a molecule of water?
21) How many lone pairs of electrons are present in a molecule of ClF3?
22) How many lone pairs of electrons are present in a molecule of SF4?
23) What is hybridization?
24) Write the shape and bond angle of sp hybrid orbitals?
25) Give an example of a molecule having sp hybridization?
26) What is the percentage of s character in sp hybridization?
27) What is the shape and bond angle of sp2 hybrid orbitals?
28) What is the percentage of s character in sp3 hybridizations?
29) Give an example for sp2 and sp3 hybrid molecules
30) Write the shape and bond angle of sp3 hybridized orbitals.
31) Write the shape, hybridization and bond angle of a) NH3 and, b) H2O
32) How many and bonds are there in a) ethylene b) ethyne?
33) What is the type of hybridization found in PCl5?
34) What is the shape and hybridization of SF6?
35) Write the number of axial and equatorial bonds in PCl5
36) Give example of a molecule showing dsp2 hybridization.
37) Write the shape of Br F5 molecule.
38) Arrange the following orbitals in the increasing order of s- character.sp, sp2, sp3
39) Define hydrogen bond.
40) How is the magnitude of hydrogen bonding in different states of matter?
41) Is the force between two nuclei of hydrogen atoms attractive or repulsive?
42) Which of the force (i.e. attractive or repulsive) is stronger?
43) What is bond enthalpy?
44) The electrons present in valence shells should have ___spins in order to be
paired up.
45) Will a bond be stronger when there is more overlap or less overlap of atomic
orbitals?
46) Hydrogen molecule is formed due to overlap of ____ orbitals.
47) What is tetrahedral bond angle?
48) What are the types of overlap depending on the sign (phase) of orbital wave
function?
49) If the wave functions of participating atomic orbitals are same phase, the overlap
is called _________.
50) The energy required for changing the electronic configuration from ground to
exited state is provided from where?
51) What is the angle between any two p orbitals?
52) What are the two types of covalent bonds?
53) What is a pi bond?
54) What is the shape of the electron cloud in a pi bond?
55) Can a pi bond exist without a sigma bond?
56) Is there greater overlap in sigma or pi bonds?
57) Define hydrogen bond.
2 Marks
1) Which type of elements is likely to form anions and give the reason for the same?
2) Which factor favours formation of cation? Explain.
3) Among KCl and NaCl, which is more stable? Give reason.
4) Under what conditions the concept of resonance is applied?
5) The dipole moment in BF3 is zero. Explain
6) The net dipole of NH3 is greater than that of NF3. Why?
7) Write the resonance structures of CO32- and CO2.
8) List the outcome of repulsive interaction between electron pairs in a molecule.
9) Account for the shape of the following molecules based on VSEPR theory
i) Water molecule.
ii) SO2 molecule
iii) Ammonia molecule
iv) SF4 molecule.
10) What are the causes of formation of hydrogen bond?
11) What are the conditions for hydrogen bonding?
12) Draw the shapes of following hybrid orbitals.
sp, sp2,sp3
13) Considering X- axis as the internuclear axis, which out of the following will not
form sigma bond? Why?
1s and 2s
b) 1s and 2px
d) 1s and 2s
3 Mark questions
1) Explain the conditions for the combination of atomic orbitals to form molecular orbitals.
2) Write the assumptions of the molecular orbital theory.
3) Explain the diamagnetic behaviour of Hydrogen molecule on the basis of molecular
orbital theory.
4) Show the non-existence of helium molecule based on molecular orbital theory.
5) Explain the formation of Lithium molecule on the basis of molecular orbital.
1mark Answers:
1). Ans: Kossel and Lewis
2). Ans: The attractive force which holds various constituents (atoms or ions etc) together
in different chemical species is called a chemical bond
3). Ans:
2 2
6 2
12Mg=1s 2s 2p 3s
2 2 6 1
11Na= 1s 2s 2p 3s
2 2 1
5B=1s 2s 2p
2 2 6
8O = 1s 2s 2p
2 2 3
7N =1s 2s 2p
2 2 6 2 6 2 10 5
35Br =1s 2s 2p 3s 3p 4s 3d 4p
4). Ans: Refer to pages 99 and 100
5) Ans: Halogens
6) Ans: Alkali metals
7) Ans: The general electronic configuration of noble gases is ns2np6
8) Ans: Electrovalent bond
9) Ans: Electrostatic force or coulombic force.
10) Ans: NH4+ in NH4Cl
11) Ans: Resonance stabilizes the molecule as the energy of the resonance hybrid is
less than the energy of any single canonical structure.
12) Ans: =Q
Dipole moment=chargedistance
13) Ans: Valence shell electron pair repulsion
14) Ans: Based on repulsive interactions between valence electron pairs in a molecule.
15) Ans: Bond pair-bond pair < bond pair-lone pair < lone pair-lone pair.
16) Ans: The localized pair of valence electrons over the central atom of a molecule
which do not take part in covalent bond formation
. 17) Ans: The pair of electron shared between two atoms in a molecule.
18) Ans: Lone pair.
19) Ans: One lone pair.
20) Ans: Two lone pair.
21) Ans: Two lone pair.
22) Ans: One lone pair
23) Ans: The process of intermixing of atomic orbitals of different energies to get
same number of new orbitals of equivalent energies is called hybridization
24) Ans: Linear shape, bond angle-1800
25) Ans: BeCl2
26) Ans: 50%
27) Ans: Shape-trigonal planar, bond angle-1200
28) Ans: 25 %
29) Ans: BCl3 - sp2
CH4 sp3
30) Ans: Shape: regular tetrahedron
Bond angle: 1090 28
2 Marks answers
1) Ans: Non-metallic elements have high electron gain enthalpy to form anions which
is produced in the process
Cl (g) + e-1 Cl-(g)
H = 348.7kJmol-1
2) Ans: Metallic elements have low ionization energy which facilitates the easy
release of electrons from the metal and formation of cation is
Na Na+ + e
3) Ans: NaCl is more stable than KCl.
Lattice energy of NaCl (788KJ/mol) greater than lattice energy of KCl
(718KJ/mol) because smaller ionic radius of Na+ (95pm) when compared to
K+ (133pm).
4)Ans: According to the concept of resonance, whenever a single Lewis structure
cannot describe a molecule accurately, a number of structures with similar
energy, positions of nuclei, bonding and nonbonding pairs of electrons are
taken as the canonical structures and the hybrid describes the molecule
accurately.
5) Ans: In BF3 = 0, although the B-F bonds are oriented at an angle of 1200to one
another. This is because the bond moments give a net sum of zero as the
resultant of any two is equal and opposite of third.
6)Ans: In case of NH3 the 0rbital dipole due to lone pair of electron on nitrogen atom ,
is in the same direction as the resultant dipole moment of N-H bonds, where
as in NF3, it is in the direction opposite to resultant dipole moment of 3 N-F
bonds. The orbital dipole decreases the effect of the resultant N-F bond
moments which reduces dipole moment of NF3.
7) Ans: For the resonance structures of CO32- and CO2 refer to page no 106( Part-1)
8) Ans: a) Deviation in the shape of the molecule
b) Alterations in the bond angle in the molecule.
9) Ans: i) Water molecule belongs to type of AB2E2
The shape of water molecule should have been tetrahedral if there were all
bond pair. But because of the presence of two lone pair, the shape is distorted
tetrahedral or angular. The reason is lp-lp repulsion is more than bp-bp
repulsion. Thus the angle is reduced to 104.50 from 109.50.
Refer to page no 112 for structure.
ii) SO2 molecule
SO2 molecule belongs to AB2E
The shape of SO2 molecule should have been triangular planar but it is found
to be bent or V-shaped .This is due to the fact that lp-bp repulsion is much
more than bp-bp repulsion. So the angle is reduced to 119.50 from 1200.
iii) Ammonia molecule
Ammonia molecule belong to AB3E type
The shape of NH3 molecule should have been tetrahedral if all the electrons
were bond pair. But because of the presence of one lp,ther is lp-bp repulsion
which is more than bp-bp repulsion and the angle is reduced from 1070 to 104.50.
iv) SF4 molecule.
SF4 molecule belongs to AB4E type.
The shape of SF4 molecule is distorted tetrahedron or a folded square or a
see- saw in which the lp is in an equatorial position where there are only two
lp-bp repulsions. This arrangement is more stable than the arrangement
where the lp is present at axial position where there are three lp-bp
repulsions at 900.
10) Ans: When there is formation of covalent bond between hydrogen and
electronegative elements, the electrons of the covalent bond are shifted
towards more electronegative atom as of which the positively charged
hydrogen forms a bond with the other electronegative atom .
11) Ans: a) Hydrogen atom should be bonded to highly electronegative atoms such as
fluorine, oxygen or nitrogen.
b) Size of the electronegative atoms should be small.
12) Ans: sp Refer to page 117,fig 4.10
sp2 Refer to page 117, fig 4.11
sp3 Refer to page 118, fig 4.12
13) Ans: 2py and 2py will not form a sigma bond because taking x- axis as the
internuclear axis, there will be lateral overlapping between 2py orbitals forming a
pi bond.
14) Ans: For H2O refer page 118, fig 4.14 and for C2H4 refer page 119, Fig 4.15
15) Ans: Electronic configuration of Al = 1s22s22p63s23p2
Exited state configuration is 1s22s22p63s13px13py1
Hence hybridization is sp2
In AlCl4- empty 3pz orbital is also involved so the hybridization is sp3 and
shape is tetrahedral.
16) Ans: In BF3, B atom is sp2 hybridized. In NH3, N atom is sp3 hybridized. After
the reaction hybridization changes from sp2 to sp3.
17) Ans: a) Attractive force between nucleus and its own electron.
b) Attractive force between nucleus and electron of other atom
c) Repulsive force between the two nuclei
d) Repulsive force between two electrons.
18) Ans: Ground state [He] 2s22p2
Excited state [He] 2s12px12py12pz1
19) Ans: This type of covalent bond is formed by the head on overlap of atomic
orbitals along the internuclear axis. The sigma bonds result from the
following types of overlap
a) s-s b) s-p c) p-p
20) Ans: The molecular orbitals are formed by the linear combination of wave functions of
the participating atomic orbitals. They may combine either by addition or by
subtraction. Let A and B represent the wave functions of the two combining
atomic orbitals A and B taking part in chemical combination.
21) Ans: Atomic orbitals
1. Atomic orbitals are monocentric.
2. Atomic orbitals have simple shapes like spherical or dumb-bell.
Molecular orbitals
1. Molecular orbitals are polycentric.
2. Molecular orbitals have complex shapes.
22) Ans: Bonding molecular orbitals
1. Formed by symmetric combination of atomic orbitals.
2. Has more electron density between the nuclei.
Antibonding molecular orbitals.
1. Formed by asymmetric combination of atomic orbitals.
2. Has less electron density between the nuclei.
23) Ans: It is half of the difference between the number of electrons present in
bonding molecular orbitals and the number of electrons present in the
antibonding molecular orbitals.
Bond order=Number of electrons in B.M.O-Number of electrons in A.B.M.O/2
24) Ans: When there is formation of covalent bond between hydrogen and
electronegative elements, the electrons of the covalent bond are shifted
towards more electronegative atom which makes the hydrogen atom
positive and it forms the hydrogen bond with other electronegative atom.
25) Ans: The conditions are
a) Hydrogen atom should be bonded to a highly electronegative atom such as F, O or N
b) Size of the electronegative atoms should be small.
3 Mark answers
1) Ans: The conditions for the combination of atomic orbitals to form molecular orbitals are
1. The combining atomic orbitals must have same or nearly the same energy.
2. The combining atomic orbitals must have the same symmetry about the molecular
axis.
3. The combining atomic orbitals must overlap to the maximum extent.
2) Ans: The assumptions of the molecular orbital theory are
1. The molecular orbital is the region in space comprising the nuclei of the combining
atoms around which there is maximum probability of finding electron density.
2. The number of molecular orbitals formed is equal to the number of atomic orbitals
taking part in the combination.
3. When two atomic orbitals combine according to L.C.A.O. principle, they form two
molecular orbitals i.e. bonding and antibonding.
4. The bonding molecular orbital has lower energy (more stability) as compared to
antibonding molecular orbital.
3) Ans: It is formed by the combination of the two hydrogen atoms.
Each hydrogen atom has one electron in 1s orbital.
Hence there are two electrons in Hydrogen molecule.
The molecular orbital configuration is 1s2
The bond order is calculated as follows
Bond order = Nb Na/2
= 2-0/2
=1
Hydrogen molecule is diamagnetic since there is no unpaired electron.
4) Ans: It is formed by the combination of the two helium atoms.
Each helium atom has two electrons in 1s orbital.
Hence there are four electrons in Helium molecule.
The molecular orbital configuration is 1s2, *1s2
The bond order is calculated as follows
Bond order = Nb Na/2
=2-2/2
=0
Helium molecule is unstable and does not exist.
5) Ans: It is formed by the combination of the two Lithium atoms.
Each Lithium atom has three electrons and its electronic configuration is 1s2 2s1
Hence there are two electrons in Hydrogen molecule.
The molecular orbital configuration is 1s2 *1s2 2s2
The bond order is calculated as follows
Bond order = Nb Na/2
= 4-2/2
=1
Lithium molecule is diamagnetic since there is no unpaired electron.
6) Ans:
7) Ans:
Bond order
bond length
N2 NN
110nm
O2 O=O
121nm
Greater the bond order, bond enthalpy (energy required to break the bond) increases,
bond length decreases. Hence stability of the molecule increases. Thus N2 is more
stable than O2.
12) Ans: In case of NH3 the orbital dipole due to lone pair of electrons on N atom , is in the
same direction as the resultant dipole moment of N-H bonds, whereas in NF3 it is in
the direction opposite to resultant dipole moment of three N-F bonds. The orbital
dipole decreases the effect of the resultant N-F bond moments which reduces of
NF3.
13) Ans: a) The shape of the molecule depends upon the number of valence shell electron
pairs (bonded or non bonded) around the central atom.
b) Pairs of electrons in the valence shell repel one another since their electron clouds
are negatively charged.
c) These pairs of electrons tend to occupy such positions in space that minimize
repulsion and thus maximize distance between them.
d) The valence shell is taken as a sphere with the electron pairs localizing on the
spherical surface at maximum distance from one another.
e) A multiple bond is treated as if it is a single electron pair and the two or three
electron pairs of a multiple bond are treated as a single super pair.
f) Where two or more resonance structure can represent a molecule, the VSEPR
model is applicable to any such structure.
14) Ans: a) AB2
Linear geometry, Ex: BeCl2
b) AB3
Trigonal planar geometry, Ex: BF3
c) AB4
Tetrahedral geometry, Ex: CH4
d) AB5
orbitals of hydrogen atoms to form 3 N-H sigma bonds. Force of repulsion between a
lone pair and a bond pair is more than the force of repulsion between two bond pairs
of electrons. The molecules thus get distorted and the bond angle is reduced to 1070
from 109.50. The geometry of such a molecule will be pyramidal and can be
explained with sp3 hybridization. Ref Pg 118, Fig 4.13.
22).Ans: In SF6 the central sulphur atom has the ground state outer electronic configuration of
3s23p4. In the exited state the available 6 orbitals i.e., 1s, 3p and 2d are singly
occupied by electrons. These orbitals hybridise to form 6 new sp3d2 hybrid orbitals,
which are projected towards the 6 corners of a regular octahedron in SF6. These 6
sp3d2 hybrid orbitals overlap with singly occupied orbitals of Fluorine atoms to form
6 S-F sigma bonds. Thus it has regular octahedron shape. Ref pg121, Fig 4.18
UNIT 5
States of matter
I. Questions carrying one mark
5.1 What are van der Waals forces?
5.2 What type of van der Waals force exists between HCl molecules?
5.3 Between which type of molecules does dipole induced dipole forces exist?
5.4 Even though HF has lower molecular mass compared to HCl, HF is in the liquid
state and HCl is in the gaseous state at room temperature .why?
5.5 Why are liquids & solids are hard to compress?
5.6 State Boyles law
5.7 What happens to the pressure when the volume of a gas is doubled at constant
temperature?
5.8 Give the relationship between pressure and density of a gas
5.9
Mountaineers carry oxygen cylinders along with them . Which gas law will account
to the above statement?
5.12 Each line of volume vs pressure graph at constant temperature is known as what?
5.13 Define absolute zero temperature
5.14 Represent Charles law graphically
5.15 What is the volume occupied by the gas at -273.15C?
5.16 State Gay Lussacs law
5.17 Equal volumes of all gases under the same condition of temperature and pressure
contain equal number of molecules. Name the gas law for the above statement
5.18 Which law relates temperature and pressure ?
5.19 If the numbers of moles of a gas are doubled by keeping the temperature and
pressure constant, what would be the new volume of the gas?
5.20 Relate density and molar mass of a gas molecule
5.21 Give the value of R in SI unit.
5.22 Write the value of molar volume of an ideal gas at STP, when pressure is 1bar
5.23 State Daltons law of partial pressure
5.24 Write a postulate of kinetic molecular theory of gases that explains the great
compressibility of gases
5.25 Define aqueous tension
5.26 Why do the gases expand &occupy the entire space available to them?
5.27 How does a gas exert pressure?
5.28 Collision of the gas molecules does not change average kinetic energy.why?
5.29 Write van der Waals equation for n moles of a gas
5.30Write the expression for compressibility factor Z for one mole of a gas.
5.31Define Boyle temperature
5.32 The value of Van der walls constant a for a gas is zero, what does it signify?
5.33 Define boiling point temperature at a given pressure for a liquid
5.34 Give the value for standard boiling point of water
5.35 Define surface energy
5.36 What would be SI unit of a quantity PV2 T2/n ?
5.37 The magnitude of coefficient of viscosity of liquid at 25C ,45C, 19C and 57C
are p, q, r, s respectively. Arrange them in the increasing value of co-efficient of
viscosity
5.38 What is meant by most probable speed of gas molecule ?
5.39 Water has higher vapour pressure than mercury. Justify
5.40 Name the property of the liquid that measures the resistance to the flow of liquid
due to internal friction.
5.41 How does the volume of a gas under given temperature and pressure vary with
molar mass?
5.42. The size of the weather balloon becomes larger and larger as it ascends up into
higher attitude. Which gas law explains the above phenomenon?
5.43 Arrange Urms,Ump,Uav in the increasing value for a gas at a given temperature
5.19 What happens to the compressibility factor for gases like CO2 at
i) Very high pressure & ordinary temperature
ii) Low pressure & ordinary temperature
5.20 Can Daltons law of partial pressure be used to calculate pressure of mixture of NH3
& HCl? Justify the answer .
5.21 Two gases A and B have critical temperature of 250K & 125K respectively. Which
one of these can be liquefied easily and why?
5.22 Define surface tension. Write the SI unit of surface tension.
5.23 A drop of liquid assumes spherical shape. why?
5.24 Name the two factors on which the magnitude of surface tension depends?
5.25 A balloon has a volume of 175dm3 when filled with hydrogen gas at a pressure of
1.0 bar. Calculate the volume of the balloon when the pressure is 0.8bar. Assume
that the temperature remains constant.
5.26 A gas cylinder containing cooking gas can withstand a pressure of 14.9bar. The
pressure gauge of the cylinder indicates 12bar at 27C. Due to sudden fire in the
building, the temperature starts rising. At what temperature will the cylinder
explode?
5.27 A weather balloon filled with hydrogen at 1 atm & 27C has volume equal to
1200dm3. On ascending it reaches a place where the temperature is -23C and
pressure is 0.5 atm. What is the volume of the balloon at this temperature?
5.28 Write two factors on which molecular speed of a gas depends?
5.29. The density of a gas is found to be 1.56g /dm3 at 0.98bar pressure & 65C.
Calculate the molar mass of the gas (R=0.083 bar dm3/K/mol).
Answers
I. Answers for Questions carrying 1 mark
Q. NO
5.1
Value point
Attractive intermolecular forces
Marks
1
5.2
5.3
5.4
molecules in HF.
5.5
5.8
Pressure density
OR
PM=dRT
5.9
Boyles Law
5.10
T= 273+t
-273.15C
5.12
Isotherms
5.13
5.15
Zero or 0
5.16
Avogadros law
5.18
5.19
Doubles
5.20
5.21
8.341J/K/mol
5.22
22.7 L/ mol
5.23
5.26
5.29
5.30
Z= PV/RT
5.31
5.33
99.6C
5.35
by one unit
5.36
Nm4K2/mol
5.37
s<q< p < r
5.38
5.39
5.40
Viscosity
5.41
V1/M or V= mRT/MP
5.42
Boyles law
5.43
Value point
i.
ii
5.2
Thermal energy
Inter molecular forces
Marks
1
1
1
Or
P1
V
5.4
Vt = V0 (273.15 + t)
273.15
Substitute t = -273.150 C Vt = 0
Or
volume of a gas at - 273.150 C is zero indicates that
substances fails to exist as gas below this temperature.
5.5
1
Isobar
5.6
It states that equal volumes of all gases under same
conditions of temperature and pressure contain equal
number of molecules
1
V n
5.7
PV=nRT
P=m RT
MXV
PM = d R T since m = d
V
5.8
1
where M is the molar mass
&d is the density
R = P V / n T = 105 x 22.7x10 -3
1 x 273
= 8.314 Pa m3/K/mol
= 8.314 J/K/mol.
5.9
The total pressure of exerted by the mixture of non-reactive
gases is equal to the sum of the partial pressures of
individual gases.
Partial pressure = mole fraction x total pressure.
or
PA = XA X PTotal
5.10
It means that total energy of molecules before and after the
collision remains same.
If there were loss of kinetic energy, the motion of the
molecules would stop & gases will settle down.But this does
not happen shows that collisions are elastic.
5.11
The two gases will have equal number of molecules
5.12
1
1
1
1
5.13
5.14
5.15
1
1
5.16
2
Z=1
5.17
5.18
5.19
5.20
5.21
No
It is applicable to a mixture of non reacting molecules.
Gas A
1
1
1
1
1
1
5.22
5.23
force of attraction
5.24
5.25
ii) Temperature.
By Boyles law
P1 V1 = P2 V2
V2 = 1x175 = 228.75 d m3
0.8
5.26
Gay-Lussacs law
P1 =
P2
T1
T2
P1V1 =
T1
1x1200
P2 V2
T2
=
0.5xV2
300
250
V2 = 2000 dm3
5.28
5.29
i)Temperature
PM = dRT or M = dRT/P
= 44.66g/mol
Value point
Gay-Lussacss law
At constant volume, pressure of a fixed amount of a gas
Marks
1
1
at constant volume
5.2
V T_ Charles Law
V n T/p
PV= nRT where R is constant called as gas
constant.
5.3
tiny
particles, of
as atoms or
molecules
2) There is no force of attraction between these particles
straight line
5) Collisions of gas molecules are perfectly elastic
6) Different particles in the gas have different speeds & hence
different kinetic energies
7) The average kinetic energy of gas molecules is directly
proportional to the absolute temperature.
Note: Write any three- each carry 1 mark
5.4
vapour
iii) Critical volume - Volume of one mole of the gas at a
highest temperature where liquefaction of gas first occurs
or
volume of one mole of a gas at critical temperature
5.5
1
1
1
temperature .
5.6
n = 1x 0.5/ 0.083x300.
= 2x 10-2 mol.
n = m/M , M = m/n = 4x 10-2/2x 10-2
= 2g/mol
5.7
PV=n1 R T1=n2RT2
constant
n1T1 = n2T2
molecular mass
M2 =M1m2 T2/m1T1
Substituting
1
n=m/M
No of moles of oxygen =8/32= 0.25mol
No of moles of Hydrogen=4/2=2mol
PH2=2x0.083x300/1=49.8bar
Total pressure=PO2+PH2=6.225+49.8=56.025bar
5.9
8.8
0.2 mol
44
1
1
V= nRT/P
= 0.2x 0.083 x304.1
1
= 5.048L
I PUC CHEMISTRY
CHAPTER - 06
Thermodynamics
One mark questions
1. Define System.
2. Define surroundings.
3. What is an open system? Give one example.
4. What is closed system? Give one example.
5. What is an isolated system? Give one example.
6. What is an extensive property? Give example.
7. What is an intensive property? Give example.
8. What is isothermal process?
9. What is an adiabatic process?
10. Write an expression for work done during an isothermal reversible expansion of an
ideal gas.
11. State I law of Thermodynamics.
12. Write mathematical expression to represent I law.
13. Define Enthalpy.
14. Write mathematical statement of enthalpy.
15. What is an exothermic reaction?
16. Give one example for exothermic reaction.
17. What is an endothermic reaction?
18. Give one example for endothermic reaction.
o
-1
16. Given : N2(g) + 3H2(g) 2NH3(g) ; rH = -92.4 kJ mol
What is the standard enthalpy of formation of NH3 gas.
17. Explain the spontaineity of Exothermic reactions using Gibbs equation.
18. Explain the spontaineity of endothermic reactions using Gibbs equation.
19. Calculate the entropy change in surroundings when 1.0 mol of H2O( l ) is formed
under standard conditions. Given H = -286 kJ mol-1
20. Under what conditions H < U. Give an example.
21. For an isolated system U = 0; what will be S?
-1
-1
-1
22. For a reaction at 298 K 2A + B C H = 400 kJ mol and S = 0.2 kJ k mol .
At what temperature will the reaction become spontaneous considering H and S to
be constant over the temperature range?
23. Calculate the heat of formation of Carbon-monoxide from the following data.
i) C(s) + O2(g) CO2(g) ; fH = -393.5 kJ
ii) CO(g) + O2(g) CO2(g) ; fH = -282.8 kJ
24. Calculate the work done when a gas expands at a constant temperature from volume
2 x 10-3 m3 to 4 x 10-3 m3 against a constant pressure of 1.2 x 105 Nm-2.
25. Define : specific heat capacity, Molar heat capacity.
-1
26. Standard enthalpy of vapourisation of water at 373 K is 40.66 k J mol . Calculate
internal energy of vapourisation.
3.
4.
5.
-1
iii) H2(g) + O2(g) H2O( l ) ; fH = -286 kJ mol
(iii)
7. Calculate the enthalpy change for the process
CCl4(g) C(g) + 4 Cl(g) and calculate bond enthalpy of C Cl in CCl4(g)
Given : vapH (CCl4) = 30.5 kJ mol-1
fH (CCl4) = -135.5 kJ mol-1
aH ( C) = 715.0 kJ mol-1 where aH is enthalpy of atomisation
aH (Cl2) = 242 kJ mol-1
8. For a reaction ; 2A (g) + B(g) 2D(g)
U298 = -10.5 kJ and S = -44.1 J k-1.
Calculate U298 for the reaction and predict whether the reaction is spontaneous
or not.
9. The equilibrium constant for the reaction is 10. Calculate the value of G ; given
R = 8 J k-1 mol-1 ; T = 300 K
10. Explain the determination of U using bomb caloriemeter.
11. Explain the determination of H using caloriemeter.
*********
Thermodynamics
One mark Answers
1. It is the specific part of the universe in which energy changes are taking place.
2. Rest of the universe which surrounds the system.
3. A system is said to be open if both matter and energy can be exchanged with the
surroundings. Example: Water kept in an open beaker.
4. A system is said to be closed if it exchanges only energy with the surroundings.
Example. Water kept in a closed container.
5. A system is said to be isolated if it neither exchanges matter nor energy. Example:
Coffee taken in a thermos flask.
6. Extensive properties of a system are the properties which depend upon the quantity of
the matter present in the system. Example: Volume, internal energy, Enthalpy, heat
capacity etc.
7. Intensive properties of a system are the properties which do not depend upon the
quantity of the matter present in the system. Example: pressure, temperature, density,
specific heat, surface tension etc.
8. A process is said to be isothermal if the temperature of the system remains constant.
(dT = 0 )
9. A process is said to be adiabatic if no heat exchange between the system and
surroundings takes place. (dq = 0)
P
10. W = -2.303 nRT log 1
P2
V
W = - 2.303 nRT log 2
V1
11. Energy can neither be created nor destroyed, it can be transformed from one form to
another.
12. U = q + w,
U = change in internal energy
q = heat supplied
w = work done on the system
13. It is the sum of the internal energy and pressure volume energy.
14. H = U + PV
H = Enthalpy
U = Internal energy
PV = pressure volume energy
15. A reaction in which heat energy is evolved is an exothermic reaction.
16. C (s) + O2 (g) CO2 (g) ; H = -393.5 k J.
17. A reaction in which heat energy is absorbed is an endothermic reaction.
18. N2(g) + O2 (g) 2 NO (g)
19. The change in heat that takes place in a chemical reaction represented by balanced
chemical equation.
20. The change in enthalpy that takes place when one mole of compound is formed from
its constituent elements, at standard conditions. [298 K, 101.3 k.Pa]
21. The heat change that takes place in a chemical reaction is independent of time taken
and number of intermediate steps involved.
22. A process that can take place on its own with or without initiation is called
spontaneous process.
23. Zn + H2SO4 Zn SO4 + H2
24. Entropy is a measure of randomness or disorder of a system.
25. Joule / Kelvin / mole or JK-1 mol-1
26. All spontaneous processes are thermodynamically irreversible
27. The amount of energy available for doing useful work under conditions of constant
temperature and pressure.
G = free energy
28. G = H TS.
H = Enthalpy
S = Entropy
T = Temperature on Kelvin scale
29. The change in free energy that takes place when the reactants in their standard states
are converted into product in their standard states at 298K and 101.3kPa.
30. Graphite.
31. Increases (or) positive
32. Decreases (or) negative
33. G = H - TS
34. i) Reaction attains equilibrium
ii) Reaction is non spontaneous
iii) Reaction is spontaneous
35. Positive
36. C6H6 ( l ) ) + 15 O2 ( g ) 6 CO2(g) + 3 H2O( l) H = - qkJ
2
37. S > 0 or S system + S surrounding > 0
38. H = U + (g)nRT
39. Expansion of a gas in vacuum is called free expansion.
40. Entropy of perfectly crystalline solid at absolute zero is zero.
41. A process which can be reversed at any instant of time by increasing the opposing
force by an infinitesimal amount.
42. A process which is carried out rapidly so that the system does not get a chance to
attain equilibrium.
43. The energy possessed by the system due to its nature, chemical composition and
thermodynamic state.
44. W = n Cv(T2 T1) for n moles of a gas.
45. The variables like temperature, pressure, volume etc, which define the state of a
system are called state functions.
46. Condition of the system expressed by giving definite values for its properties such as
temperature, pressure, volume.
47. The state of a system at 298K and 101.3 k.Pa is known as standard state of a system.
48. The change in enthalpy that takes place when one mole of a substance is completely
burnt in air or oxygen at a given temperature.
49. Go = - 2.303 RT log KP
50. U = q w
U = 10 15
U = - 5 joules
H2(g) + I2(g)
51. 2 HI(g)
Or any other suitable example.
52. Isolated system.
53. H = U + RT
ii) is the correct answer
54. a)
b) iii) is the correct answer
c) ii) is the correct answer
d) iii) is the correct answer
Explanation: CH4(g) + 202(g) CO2(g) + 2H2O( l )
ng = 1 3 = -2
H = U + ngRT = U - 2RT
H < U
e) i) is the correct answer
Explanation :
According to available data:
i) CH4(g) + 202(g) CO2(g) + 2H2O( l ) ; H = - 890.3 k J mol-1
ii) C(s) + O2(g) CO2(g) ; cH = -393.5 k J mol-1
iii) H2(g) + O2(g) H2O( l ) ; cH = -285.8 k J mol-1
The equation we aim at
C(s) + 2H2(g) CH4(g) ; fH = ?
Equation (ii) + 2 equation (iii) equation (i) and the correct fH
Value is
= (-393.5) + 2 (-285.8) (-890.3) = -74.8 k J mol-1
f) (iv) is the correct answer.
2.
3.
4.
5.
The enthalpy change that accompanies melting of one mole of a solid substance in
standard state is called standard enthalpy of fusion or molar enthalpy of fusion.
fus H
7. Standard enthalpy of sublimation, subH is the change in enthalpy when one mole of
a solid substance sublimes at a constant temperature and under standard pressure.
(1 bar)
8. It is the enthalpy change on breaking one mole of bonds completely to obtain atoms
in the gas phase.
9. It is the change in enthalpy when one mole of covalent bonds of the gaseous covalent
compound is broken to form products in the gaseous phase.
10. G = H T S
At equilibrium G = 0, H = T S
H 400 10 3
T=
=
S
200
T = 2000 K
11. H = + ve , S = + ve
12. It is the change in enthalpy when one mole of a substance is dissolved in a specified
amount of a solvent.
13. G = H - TS
G = -1648 X 103 [(298) (-549.4)]
= -1648000 + 163721.2
= -1484278.8 J/mol = -1484.27 kJ/mol
14. Change in internal energy (U) for an isolated system is zero because it does not
exchange any energy with the surroundings. But entropy tends to increase in case of
spontaneous reaction. Therefore, S > 0 or positive.
-1
16. fH NH3(g) = -(92.4) / 2 = -46.2 kJ mol
17. Gibbs equation is G = H T S (1)
For exothermic reaction H is ve
i) If S is +ve according to equation (1)
G is ve, reaction is spontaneous at all temperature.
ii) If S is ve, according to equation (1)
G is ve at low temperature such that
T S < H. Reaction is spontaneous.
18. Gibbs equation is G = H T S (1)
For endothermic reaction, H is +ve
i) If S is +ve according to equation (1)
G is ve, at high temperature
such that T S > H. Reaction is spontaneous
ii) If S is ve, according to equation (1)
G is always +ve. Reaction is nonspontaneous at all temperature.
6.
19.
S(surroundings) =
10
11
CHAPTER-7
EQUILIBRIUM
ONE MARK QUESTIONS WITH ANSWERS.
CHAPTER WEIGHTAGE: 13
Ans.
373K or 100oC
Ans.
H2O(s) H2O(l)
SO
2
2
______________
P2
SO
3
Ans.
19)
Ans.
21)
Ans.
22)
Explain amphoteric substance with an example?
Ans. Substance which act as acid in presence of a base and base in presence of an acid is
called amphoteric substance.
Example:
CH3COOH + H2O CH3COO- + H3+O
Base
NH3 + H2O NH4+ + OHAcid
23)
What is ionic product of water?
Ans. It is the product of molar concentration of H+ and OH- ions in water or in any aqueous
solution is a constant at constant temperature. i.e., Kw = [H+] [OH-]
24)
Ans.
25)
Explain common ion effect with an example.
Ans. The suppression in degree of dissociation of a weak electrolyte by the addition of
strong electrolyte having common ion is called common ion effect.
Example:
the ionization of acetic acid [weak electrolyte] is suppressed by addition
of sodium acetate [strong electrolyte] containing common acetate ion.
26)
Define solubility product.
Ans. The product of molar concentrations of constituent ions, each raised to the power of its
stoichiometric co-efficient in the equilibrium equation of the electrolyte at given temperature,
is called as solubility product.
27)
What is a buffer solution? Give an Example.
Ans. It is a solution that has ability to resist change in pHupon addition of small amount of
acid or base.
Example:
Mixture of acetic acid and sodium acetate.
28)
Ans.
29)
What is hydrolysis of salt? Give an Example.
Ans. The interaction of cations and anions of salt with water to give acidic or basic or neutral
solution is called hydrolysis.
Example:
when sodium acetate undergoes hydrolysis in water it gives basic
solution due to formation of strong base NaOH and weak acid CH3COOH.
30)
The aqueous solution of K2CO3 is it acidic or basic or neutral? Explain.
Ans. K2CO3 is a salt obtained by neutralization of weak acid H2CO3 and strong base KOH.
Hence the aqueous solution of the salt has more OH- than H+ ions and resultant solution is
basic.
31)
An aqueous solution of NH4Cl is acidic. Why?
Ans. NH4Cl is a salt obtained by neutralization of strong acid HCl and weak base NH4OH. The
aqueous solution of the salt has more H+ ions than OH-- hence the solution is acidic.
32) Write applications of equilibrium constant .
Ans. 1) If the value of Kc is very large [>103] the reaction proceeds nearly to completion.
2) If the value of Kc is small [<10-3] the reaction proceeds rarely.
3) If the value of Kc is in range of 10-3 to 103, the appreciable concentrations of both
reactants and products are present.
33) Explain ionic equilibrium with an example.
Ans. It is the equilibrium established betweens ions and unionized molecules in an aqueous
solution of the weak electrolyte.
Ex. CH3COOH(aq) CH3COO- (aq) + H+(aq)
34) Ka1 of polyprotic acid is higher than Ka2. Why?
Ans. Ka1 of polyprotic acid is higher than Ka2 because it difficult to remove an H+ ion from a
negative ion due to electrostatic forces of attraction.
4 Marks questions with answers
1)
Ans.
Conjugate acid
Species
H3+O
+H+
NH4+
+H+
H2SO4
H2CO3
Conjugate base
-H+
OH-
NH3
-H+
H2N-
+H+
HSO4-
-H+
SO42-
+H+
HCO3-
-H+
CO32-
H2O
2 Marks Problems
1) The Kc for a certain reaction is 4.5 x 107 at 750K, what is Kc for the reverse reaction?
Ans. Kc1
=
1/Kc =
1/4.5x107
=
0.2222x10-7
=
Kc 1
2.222x10-8
2) For a reversible reaction PCl3(g) + Cl2(g) PCl5(g) Kc is 3.0 at 2700 C find the value of Kp?
Ans. Given
t=2700C
For the above reaction
T= t+273
nP=1 , nR=2
=270+273=543K
n = nP nR = 1 2 = -1
R=0.0831barlitre/mol K
W.K.T
KP = Kc(RT)n
Kc=3.0
KP = 3.0x(0.0831x543)-1
KP = 3.0/0.0831x543 = 0.06648
3)
For the synthesis of ammonia N2(g) + 3H2(g) 2NH3(g) the KP value is 41 at 400K.
Calculate KC for the reaction?
Ans. Given
For the above reaction
KP = 41
nP=2 , nR=4
T = 400K
n = nP nR = 2 4 = -2
R = 0.0831bar litre / mol K
W.K.T KP = Kc(RT)n
Kc = Kp/(RT)n
=
41/(0.0831x400)-2
KC = 41x(0.0831x400)2 =
45300.8
4)
For the reaction 2NOCl(g) 2NO(g) + Cl2(g), The equilibrium constant is 3.75 X 10-6
at 1069K. What is equilibrium constant for the reaction 2NO(g) + Cl2(g) 2NOCl(g)?
Ans. Given Kc = 3.75x10-6
W.K.T
KC1 =1/Kc = 1/3.75x10-6
= 0.2666x106 = 2.666x10+5
5)
The equilibrium concentrations of ammonia, hydrogen and Nitrogen for synthesis of
ammonia are 1.2x10-2M, 3x10-2 M and 1.5x10-2M respectively, at given temperature
calculate equilibrium constant Kc.
Ans. Given [NH3] = 1.2x10-2M
N2 + 3H2 2NH3(g)
-2
[H2] = 3x10 M
Kc = [NH3]2/[N2][H2]3
[N2] = 1.5x10-2M
= (1.2x10-2)2/(1.5x10-2)x(3x10-2)3
= 0.03556x104+2+6
= 0.03556x10+4 = 3.556x102
6)
The equilibrium concentrations of hydrogen iodide and hydrogen for the reaction
H2(g) + I2(g) 2HI(g) are 5.86x10-2M and 0.86x10-2M respectively. At given temperature
calculate equilibrium constant Kc.
Ans. Given [HI] = 5.86x10-2
Kc = [HI]2/[H2][I2]
=
[H2] = [I2] = 0.86x10-2
(5.86x10-2)2/0.86x10-2 x 0.86x10-2
=
42.43x10-4+2+2
= 42.43
7)
For the reaction 2NO(g) + Br2(g) 2NOBr(g) the equilibrium concentrations of Nitric
oxide, bromine and nitrosyl bromide are 0.0352M, 0.0178M and 0.0518M respectively at
constant temperature, calculate equilibrium constant KC.
Ans. Given [NO] = 0.0352
KC = [NOBr]2/[NO]2[Br2]
[Br2] = 0.0178
KC = (0.0518)2/(0.0352)2(0.0178)
[NOBr] = 0.0518
=
121.66
8) The equillibricum partial pressure of HI , H2 and I2 for the reaction 2HI(g) I2(g) + H2(g)
are 0.04 atm , 0.08atm and 0.08 atm respectively, Calculate equilibrium constant KP for the
reaction.
Ans. Given PHI = 0.04atm
KP = PH2xPI2/P2HI
PI2 = 0.08atm
0.08x0.08/(0.04)2
PH2 = 0.08atm
0.16
9)
For the reaction A+3B 4C,the partial pressure of B and C are 0.8atm and 0.4 atm
respectively. The equilibrium constant KP for the reaction is 24, what is the partial pressure
of A at equilibrium?
KP = Pc4/PA.PB3
Ans. Given KP = 24
PB = 0.8tm
PA = Pc4/KP.PB3
Pc = 0.4atm
= (0.4)4/24x(0.8)3
PA = ?
PA = 0.00208atm
10)
The equilibrium constant KC for a reaction is 1.3x102 at 300K. Calculate standard free
energy change for the reaction. State whether the reaction is spontaneous or nonspontaneous.
Ans. Given R = 8.314J mol-1 K-1
G0 = -2.303RT log KC
T = 300K
= -2.302x8.314JK-1mol-1x300Klog 1.3x102
Kc=1.3 x 102
= -5744.14(2 + log 1.3)
0
G = -12142.54 Jmol-1
Since G0 is negative the reaction is spontaneous
11)
standard free energy change for a reaction is 12 KJ at 300K. Calculate equilibrium
constant for the reaction at same temperature.
G0 =-2.303RT log KC
Ans. Given G0 = 12KJ = 12x103J
T = 300K
log KC =- G0/2.303RT
R = 8.314Jmol-1K-1
log KC =-12000J/2.303x8.314JK-1mol-1x300K
log KC = -2.0891
KC
= antilog[+1-1-2.0891]
__
Kc
= antilog[ 3. 9109]
= 8.145x10-3
12)
Ans.
13)
Ans.
14)
Ans.
15)
Ans.
16)
Ans.
[OH-] = Antilog[12.2]
[OH-] = 1.585 x 10-12
17)
The ionization constant 0.05M propanoic acid is 1.3 x 10-5 . Calculate degree of
ionization.
Ans. Given Ka=1.3 x 10-5
C =0.05M
Given Kb =
1.75 x 10-5
C =0.05M
= 1.87x10-2
The solubility product of silver bromide is 5.0 x 10-13 at 298K. find its solubility.
Ksp
=
5 x 10-13
AgBr
Ag+ + BrKsp
=
[Ag+][Br-]
Ksp
=
s.s
-13
5 x 10 =
s2
s =
= 0.7071 x 10-6mol/L
3 Marks Problems
1)
For the reaction N2(g) + 3H2(g) 2NH3(g), The partial pressures of N2 and H2 are 0.8
and 0.4 atmosphere respectively at equilibrium. The total pressure of the system is 2.80
atmospheres. What is KP for the above reaction?
Ans.
PH2 = 0.4atm
P = 2.8atm
2) The equilibrium constant at 298K for the reaction Cu(s) + 2Ag+(aq) Cu2(aq) + 2Ag(s)
is 2.0 x 1015. The concentration of Cu2+ and Ag+ in solution are 1.8 x 10-2 mol/L and
3.0 x 10-9mol/L respectively. Predict the direction in which reaction proceed.
Ans. Given[Cu2+] = 1.8 x 10-2 mol/L
Qc = [Cu2+(aq)][Ag(s)] / [Cu(s)][Ag+(aq)]2
[Ag+] = 3.0 x 10-9 mol/L
= 1.8 x 10-2/(3.0 x10-9)2
Kp = 2.0 x 1015
= 0.2 x 10-2+18 = 2 x 1015
By convention [Ag(s)] = 1 and [Cu(s)] = 1
Qc = KP = 2 x 1015 , hence reaction is at equilibrium
3)
2 moles of N2O4 taken in a flask of 10L capacity is heated to 350K. At equilibrium 50%
of N2O4 was found to be dissociated to give NO2. Find the equilibrium constant for the
reaction.
Ans. Initial concentration of N2O4 = 2/10 mol/L
50% of 2N2O4 is dissociated implies, x = 50/100 = 0.5
N2O4
2NO2
Initial concentration
2/10
0
Equilibrium concentration
2(1-x) /10
2(2x)/10
2(1-0.5) /10 = 0.1
2(2x0.5)/10 = 0.2
2
Kc = [NO4] /[N2O4]
=(0.2)2/0.1 = 0.4
4)
Calculate the pKa value of 0.1M weak mono basic acid whose degree of ionisation
1.52 x 10-2
Ans. Given = 1.52 x 10-2
Ka = C2
C = 0.1M
=0.1 x (1.52 x 10-2)2= 0.2310 x 10-4
= 2.310 x 10-5
pKa = -log Ka
= - log[2.310 x 10-5] = -[log2.310 (5)log10]
= +5 log2.310 = +5 0.3636 = 4.6364
5)
Calculate the hydrogen ion concentration of 0.1M weak mono basic acid whose
dissociation constant is 4 x 10-10 at 298K.
Ans.
Given C = 0.1M
Ka = 4 x 10-10
Wkt [H+] = C = 0.1 x 6.324 x 10-5 =
= 6.324 x 10-5
0.6324x10-5M
6)
Ans.
4 Marks Problems
1)
The initial molar concentration of reactants A and B are 0.1 M and 0.2M respectively
in the reaction A +B 2C at equilibrium. The concentration of A in the mixture was found to
be 0.06m. Calculate the equilibrium constant.
Ans. Initial concentration of A = 0.1M
Concentration of A at equilibrium = 0.06M
Concentration of A reacted = 0.1 0.06m = 0.04M
A + B 2C
Initial concentration
0.1
0.2
0
Equilibrium concentration 0.1 0.04 0.2 0.04 2 x 0.04 = 0.08
=0.06 =0.16
2
Kc = [C] /[A][B] = (0.08)2/0.06 x 0.16 = 0.6667
2)
2 moles of HI when heated in a closed container, at equilibrium 20% of HI found to be
dissociated. Calculate the equilibrium constant for the reaction 2HI(g) H2(g) + I2(g)
Ans. Initial concentration of HI = 2 moles
20 moles of HI are dissociated out of 100 moles at equilibrium
? moles of HI are dissociated out of 2 moles
Concentration of dissociated HI = 2 x 20 /100 = 4/10 = 0.4 moles
2HI
H2
+ I2
Initial concentration
2
0
0
Equilibrium concentration 2 0.4 =1.6
1.6/2 = 0.8 1.6/2 = 0.8
2
= 0.8 x 0.8/(1.6)2 = 0.25
KC = [H2][I2]/[HI]
3)
What is the equilibrium concentration of each of the substances in equilibrium when
the initial concentration of ICl was 0.78M for the reaction 2ICl I2 + Cl2 , KC = 0.14.
Ans. Let x mol be the equilibrium concentration of I2 and Cl2
2ICl
I2 + Cl2
Initial concentration 0.78
0 0
Equilibrium concentration 0.78 2x
x x
2
KC = [I2][Cl2]/[ICl]
0.14 = X . X/(0.78 2X)2
0.14[0.78 2X]2 = X2
Taking the square root both sides
[0.14]1/2[0.78-2X] = X
0.374 x 0.78 - 0.374 x 2X = X
0.2917 = X + 0.748X
0.2917=1.748X
X = 0.2917/1.748 = 0.1668
[Cl2] = [I2] = X = 0.1668M
[ICl] = 0.78-2X=0.78-2 x 0.1668
[ICl] = 0.78-0.3336 = 0.446M
4)
Calculate the degree of ionisation and Ka of 0.025M ammonia solution, if the
ionisation constant of ammonia is 1.77 x 10-5 at 298K.
Ans.
=
= 2.66 x 10-2
WKT, Ka X Kb = Kw
Ka= Kw/Kb = 10-14/1.77 x 10-5 = 0.5649 x 10-9
C = 0.025
5)
The Ksp values of BaSO4 and PbSO4 are 1.1 x 10-10 and 1.6 x 10-8 respectively. Which
salt is more soluble?
Ans. Let solubility of BaSO4 is s1 and solubility of PbSO4 is s2
BaSO4 Ba2+ + SO42Ksp = [Ba2+] [SO42-]
1.1 x 10-10 = s1.s1
S1 =
= 1.0489 x 10-5mol/L
PbSO4 Pb2++ SO42Ksp = [Pb2+][SO42-]
1.6 x 10-8 = s2 . s2
S2 =
= 1.2649 x 10-4mol/L
S2 > s1 hence PbSO4 is more soluble than BaSO4
SUBJECT : CHEMISTRY
CHAPTER-08 : REDOX REACTIONS
QUESTIONS CARRYING ONE MARK:
....... +2H2O .
H2O + 2O2
3. Define oxidation and reduction in terms of oxygen and hydrogen. Give one example
for each.
4. What is oxidation number? What is the oxidation number(O.N) of Cl in KClO3?
5. Define oxidation and reduction in terms of oxidation number.
6. How are the oxidizing agent and reducing agents defined in terms if oxidation number?
7. Write separate equations for the oxidation and reduction reactions occurring in the
following redox reaction: 2Fe + 2HCl
FeCl2 + H2
2H2O + O2
8. For 2H2O2
(1)
(2)
(3)
i) What is the oxidation number of Oxygen in (2)?
ii) What type of Redox reaction is it?
9. Explain whether the following reaction is a redox reaction or not:
CaCO3(s)
CaO(s) + CO2(g)
(ii) P in H3PO4.
11. What is a redox couple? Identify the redox couples in the reaction:
Zn(s) + 2Ag+(aq)
Zn2+(aq) +2Ag(s)
15. Using Stock notation, represent the following compounds: Fe2O3, CuO, MnO and
MnO2
16. Calculate the oxidation number of phosphorus in the following species:
(a) HPO3 2
and
(b)PO43-
S + H2O
18. Assign oxidation number to the underlined elements in each of the following
species:(a) NaH2PO4 (b) NaHSO4 (c) H4P2O7 (d) K2MnO4
19. Justify that the following reactions are redox reactions:
(a) CuO(s) + H2(g)
(b) Fe2O3(s) + 3CO(g)
Cu(s) + H2O(g)
2 Fe(s) + 3CO2(g)
19. Give an example of a redox combination reaction. Mention the species that undergo
oxidation and reduction.
20. Give an example of a redox decomposition reaction. Mention the species that
undergo oxidation and reduction.
21. Give an example of a redox displacement reaction. Mention the species that undergo
oxidation and reduction.
22. Give an example of a redox disproportionation reaction. Mention the species that
undergo oxidation and reduction.
23. F2 does not undergo disproportionation. Why?
24. What type of redox reactions are the following?
(a) 3Mg(s) + N2(g) Mg3N2(s)
(b) 2KClO3(s) 2KCl(s) + 3O2(g)
(c) Cr2O3 (s) + 2 Al (s) Al2O3 (s) + 2Cr(s)
(d) (d) 2NO2(g) + 2OH(aq) NO2(aq) +NO3 (aq)+H2O(l)
25. Name the redox indicator used in the titration of
(i). KMnO4 v/s FAS.(or H2C2O4).
(ii) Na2S2O3 v/s I2.
1. When blue coloured solution of copper sulphate is stirred with a zinc rod, the blue
colour of the solution fades off and the zinc rod is coated with reddish copper metal.
Write the chemical reaction taking place in the above observation and identify the
species undergoing oxidation and reduction.
2. A solution of silver nitrate turns blue slowly on stirring with a copper rod which in turn
gets coated with a white deposit of silver. Write a chemical reaction for this observation
and identify the oxidizing and reducing agents in it.
3. Balance the following equations by the oxidation number method.(3marks each)
(i) Fe2+ + H+ + Cr2O7 2 Cr3+ + Fe3+ + H2O
(ii)MnO4 (aq) + I (aq)
(iii) MnO4 (aq) + SO2 (g) Mn2+ (aq) + HSO4 (aq) (in acidic solution)
(iv) H2O2 (aq) + Fe2+ (aq) Fe3+ (aq) + H2O (l) (in acidic solution)
(v) Cr2O7 2 + SO2(g) Cr3+ (aq) + SO42 (aq) (in acidic solution)
4. Balance the following equations by half reaction method (ion-electron method). (3
marks each)
(a) MnO4 (aq) + SO2 (g) Mn2+ (aq) + HSO4 (aq) (in acidic solution)
(b) MnO4 (aq) + I (aq) MnO2 (s) + I2(s) (in basic medium)
(c) H2O2 (aq) + Fe2+ (aq) Fe3+ (aq) + H2O (l) (in acidic solution)
(d) Cr2O7 2 + SO2(g) Cr3+ (aq) + SO42 (aq) (in acidic solution)
5. In the reactions given below, identify the species undergoing oxidation and reduction:
(i) H2S (g) + Cl2 (g
(ii) 2 Na (s) + H2 (g)
(iii) 2Fe(s) + 2HCl(aq)
\
6. Justify that the reaction: 2Cu2O(s) + Cu2S(s)
6Cu(s) + SO2(g) is a redox
reaction. Identify the species oxidized/reduced, which acts as an oxidant and which acts
as a reductant.
CHAPTER-08 : REDOX REACTIONS
ANSWERS:
QUESTIONS CARRYING ONE MARK:
2Fe3+ + 2H2O .
8. The term Oxidation number denotes the oxidation state of an element in a compound
ascertained according to a set of rules formulated on the basis that electron pair in a
covalent bond belongs entirely to more electronegative element.
9. Oxidation number of oxygen = -2.
Hence, oxidation number of Cr, (x) in Cr2O72- = 2x + 7x(-2) = 0, x = +6
10. Oxidation number of K = +1, oxygen, O = -2.
Hence, oxidation number of Mn, (x) in KMnO4 = (+1) + x + 4(-2) = 0, x = +7
11. Zero.
12. It increases.
13. It decreases
14. In Hydrides, hydrogen has an oxidation state of -1.
15. In peroxides, oxygen has an oxidation state of -1.
16. Zero.
17. A setup consisting of a metal in contact with its salt solution is called an electrode.
18. The potential attained by a metal in contact with a solution containing its own ions is
called electrode potential.
19. The potential attained by a metal in contact with its salt solution of concentration 1
moldm-3 at 298 K.
20. The oxidant is O3.
21. +2
1. A chemical reaction in which both oxidation and reduction are taking place
simultaneously is called a redox reaction.
reduction
Ex: Zn(s) + Cu2+ -- Zn2+ + Cu.
Oxidation
2. :
Fe Fe2++2e-
Reduction reaction:
reaction: 2HCl +2e- H2
8.
2H2O2 (-1)
(1)
9.
+2 + 4 2
+2 2
+4 2
CaCO3 (s) CaO(s) + CO2(g)
It is not a redox reaction because the oxidation number of no element changes.
10.
10 (i) Let the O.N. of S be x
O.N. of H = +1, O = -2 O.N. of S in H2SO4 = 2(+1) + x + 4(-2) = +6.
(ii) Let the O.N. of P be x.
O.N. of H = +1, O = -2
11.
11 A redox couple is defined as having together the oxidized and reduced forms of a
substance taking part in an oxidation or reduction half reaction.
The redox couples in the reaction are, Zn2+/ Zn(s)
Zn(s) and Ag+/Ag.
/Ag.
12.
12 A series of electrode potential values arranged in the increasing or decreasing order
constitute an electrochemical series.
13.
13 An ion which is present in a redox reaction, but does not take part in a reaction during
electron transfer is called a spectator ion.
Ex: SO42- ion in the reaction: Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu.
14. (a) Ni(II)SO4
(c) Tl2(I)SO4
(b) Sn(IV)O2
(d) Fe2(III)(SO4)3
15.
15 Fe2O3 - Fe2(III)O3 ,
MnO - Mn(II)O,
CuO Cu(II)O
MnO2 Mn(IV)O2.
16.
16 (a) Let the O.N of P in HPO3 2 be x.
(+1) + x + 3(-2) = -2
x = +3
(b)Let the O.N of P in PO43- be x.
X + 4(-2) = -3
x = +5
17.
17
4+
20
SO2 + H2S S + H2O.
Step 2: Multiply H2S by 2 to equalize the oxidation numbers on either side of the
equation.
4+ 2 x (2-)
0
SO2 +2 H2S S + H2O.
Step 3: Now, balance S atoms on RHS.
SO2 + 2H2S 3 S + H2O.
x = +5.
x = +6
23.
23 Among halogens, fluorine (F2) is the most electronegative element; it cannot exhibit any
positive oxidation state. Hence it does not show a disproportionation tendency.
24.
24 (a) 3Mg(s) + N2(g) Mg3N2(s) - Redox combination reaction
(b) 2KClO3(s) 2KCl(s) + 3O2(g) - Redox decomposition reaction
(c) Cr2O3 (s) + 2 Al (s) Al2O3 (s) + 2Cr(s) - Redox displacement reaction
Answers:
reduction
1.
Oxidation
Oxidation
In this reaction, Zn loses 2e - to Cu and hence is undergoing oxidation; Cu2+ is
undergoing reduction to Cu.
2.
reduction
Cu(s) + 2Ag+(aq)
Oxidation
Cu2+(aq) +2Ag(s)
oxidation
2+
6+
3+
3+
2
2+
+
3+
Fe + H + Cr2O7 Cr + Fe3+ + H2O
reduction
Step 2: Multiply Cr3+ by 2 and Fe2+ and Fe3+ by 6 to equalize the oxidation numbers on
either side of the equation.
2+
1+
6+
2x3+ 3+
2+
+
2
6Fe + H + Cr2O7 2 Cr3+ +6 Fe3+ + H2O
Step 4: Finally balance H atoms by adding 14H+ on LHS to get a balanced equation as:
6 Fe2+ + 14H+ + Cr2O7 2 2 Cr3+ + 6Fe3+ + 7H2O
.
Step 1: Write skeletal equation with O.N of each element Undergoing change in
oxidation number.
Oxidation
7+
1
MnO4 (aq) + I (aq)
reduction
6+
0
MnO2 (s) + I2(s)
Step 2: Multiply I- by 6 and MnO4- by 2 to equalize the oxidation numbers on either side
of the equation.
2 x (7+)
6 x (1- )
2 x (4+)
0
Step 4: Finally balance H and O atoms by adding 4H2O on LHS to get a balanced
equation as:
2MnO4 (aq) + 6I (aq) + 4H2O
3. (iii) MnO4 (aq) + SO2 (g) Mn2+ (aq) + HSO4 (aq) (in acidic solution)
.
Step 1: Write skeletal equation with O.N of each element undergoing change in
oxidation number.
Oxidation-2e7+
4+
2+
6+
MnO4 (aq) + SO2 (g) Mn2+ (aq) + HSO4 (aq)
reduction-5eStep 2: Multiply SO2 by 5 and MnO4- by 2 to balance +ve charges on both sides.
2 x (7+)
5 x (4+ )
2 x (2+)
5 x (6+)
3. (iv) H2O2 (aq) + Fe2+ (aq) Fe3+ (aq) + H2O (l) (in acidic solution)
.
Step 1: Write skeletal equation with O.N of each element undergoing change in
oxidation number
Oxidation-1e-
12+
3+
22+
3+
H2O2 (aq) + Fe (aq) Fe (aq) + H2O (l))
reduction-2 x 1e-
Step 2: Since the number of charges on both sides are not equal, 2Fe2+ on LHS and
2Fe3+ on RHS
2 x (1-)
2x(2+)
2x (3+)
(2-)
2+
3+
H2O2 (aq) +2 Fe (aq) 2Fe (aq) + H2O (l))
Step 3: Now, put 2H2O to balance O atoms.
H2O2 (aq) + Fe2+ (aq) Fe3+ (aq) +2 H2O (l))
Step 4: Finally add2 H+ on LHS to get a balanced equation as:
H2O2 (aq) +2 Fe2+ (aq) +2H+(aq) 2Fe3+ (aq) +2 H2O (l))
3.(v)
3.(v) Cr2O7 2 + SO2(g) Cr3+ (aq) + SO42 (aq) (in acidic solution)
solution)
Step 1: Write skeletal equation with O.N of each element Undergoing change in
oxidation number.
Oxidation-2e2 x (6+)
4+
3+
6+
2
2
3+
Cr2O7 + SO2(g) Cr (aq) + SO4 (aq)
reduction-3eStep 2: Multiply SO2 by 3 and Cr3+ by 2 on RHS .
Cr2O7 2 +3 SO2(g) 2Cr3+ (aq) + 3 SO42 (aq)
4.. (b)
(b) MnO4 (aq) + I (aq) MnO2 (s) + I2(s) (in basic medium)
Step1: Assign O.N. to the atoms undergoing oxidation / reduction.
4. (c) H2O2 (aq) + Fe2+ (aq) Fe3+ (aq) + H2O (l) (in acidic solution)
Step1: Assign O.N. to the atoms undergoing oxidation / reduction.
oxidation-1eH2O2 (aq) + Fe2+ (aq) Fe3+ (aq) + H2O (l)
reduction-2e-
Step2: Write out oxidation and reduction separately and balance the atoms other than H
and O.
Oxidation half reaction: Fe2+ Fe3+
Reduction half reaction: H2O2 H2O
Step3: Multiply the oxidation reaction with the extent of reduction and reduction reaction
by the extent of oxidation and add.
Oxidation half reaction: [Fe2+ Fe3+ ] x 2
Reduction half reaction: [H2O2 H2O] x 1
2Fe2+ + H2O2 2Fe3+ + H2O
Step4: Add 2H+ on LHS and H2O on RHS to balance H and O atoms in the acid medium to get
a balanced equation.
2Fe2+ + H2O2 + 2H+ 2Fe3+ +2 H2O.
balanced equation.
2Cr2O7 2(aq) + 6 SO2(g) +4H+ 4 Cr3+ (aq) +6 SO42 (aq)+2H2O.
OR,
2 NaH (s)
In this reaction, the species undergoing oxidation is: Na ( the O.N. of Na increases
from 0 to +1).
The species undergoing reduction is: H2 ( the O.N. of H2 decreases from 0 to -1)
FeCl2(aq) + H2(g)
In this reaction, the species undergoing oxidation is: Fe ( the O.N. of Fe increases
from 0 to +2)
The species undergoing reduction is: HCl ( the O.N. of H in HCl decreases from +1 to 0 )
*******************
I P U C QUESTION BANK
SUBJECT:- CHEMISTRY, UNIT 9 : HYDROGEN
Questions carrying one mark:
1. Which is the most abundant element in the universe?
2. Name the isotope of hydrogen that do not contain neutron.
3. What is the composition of water gas?
4. What is the chemical used in clarkes process to remove the temporary hardness of
water?
5. What volume of oxygen is produced by one litre of 10 volume H2O2 at STP?
6. Name the isotope of hydrogen containing two neutrons.
7. What is the role of heavy water in a nuclear reactor?
8. What is a syn gas?
9. Give an example of an ionic hydride.
10. Give an example of a covalent hydride.
11. What is demineralised water ?
12. Arrange LiH, NaH, and CsH in the increasing order of ionic character.
13. Arrange H2, D2, T2 in the increasing order of their boiling points.
14. Which isotope of hydrogen is radioactive?
15. What causes temporary hardness of water?
16. What causes permanent hardness of water?
17. Why H2O has higher boiling point compared to H2S?
18. What are Non-stoichiometric hydrides?
19. Why is H2O2 not stored in glass containers?
20. What is calgon?
21. What is the chemical name of zeolite used in softening of hard water?
22. Out of ice and water, which has low density?
1
Boil
____________ + 2 CO2.
i)
H2(g) + MmOo(s)
ii)
CO(g) + H2(g)
Catalyst
iii)
C3H8(g) + 3 H2O(g)
Catalyst
iv)
Zn(s)
Heat
+ NaOH (aq)
32. Discuss the consequences of high enthalpy of H-H bond in terms of chemical
reactivity of dihydrogen.
33. What do you understand by (i) electron- deficient
(ii) electron-precise
(iii)
34. How do you expect the metallic hydrides to be useful for hydrogen storage?
Explain.
35. How does the atomic hydrogen or oxy hydrogen torch function for cutting and
welding purposes? Explain.
36. Among NH3, H2O, and HF, which has highest magnitude of hydrogen bonding and
why?
37. What is auto-protolysis of water? Mention its significance.
38. Consider the reaction of water with F2 and suggest, in terms of oxidation and
reduction, which species are oxidized/reduced?
39. Complete the following chemical reactions:
i)
PbS (s) +
H2O2 (aq)
ii)
Mn O4 (aq)
+ H2O(aq)
iii)
iv)
v)
44. How can saline hydrides remove traces of water from organic compounds?
45. What is the difference between the terms hydrolysis and hydration?
46. Mention any two uses of dihydrogen.
47. Calculate the strength of 10 volume solution of hydrogen peroxide.
3
UNIT-9.
HYDROGEN.
Model answers.
i)
i)
Similar to halogens it can gain one electron and form hydride (H--) ion.
ii)
28. Hydrogen molecule has a very high bond dissociation enthalpy . It forms a covalent
bond with another H-atom and exists as a diatomic molecule.
CO(g) +
lectrolysis
Traces of acid/base
2H2(g) + O2(g)electrodes.
Pure water is not an electrolyte. Addition of electrolyte makes the ions available for
electrolysis.
31. I) H2(g) + MmOo(s)
mM(s) + 0 H2O(l).
ii)CO(g)
+ H2(g)
CH3OH(l)
Catalyst
iii)C3H8(g) + 3 H2O(g)
3CO(g) + 7 H2(g)
Catalyst
Heat
NH4+(aq) + OH(aq)
H3O+(aq) + HS(aq)
Na+ ions with Ca2+ and Mg2+ ions present in hard water
+ 2Na+(aq)
44. Saline hydrides are ionic and contain Hion which react with water liberating H2
gas.
For example: NaH + H2O NaOH + H2.
45. Hydrolysis is reaction in which H+ and OHions of water react with a compound to
form products.Hydration is association of one or more molecules of water to form
hydrated compounds.
46.Dihydrogen is used i) in the synthesis of ammonia
22.7 L at STP
+ 2NaCl
*******************
8
Chapter 10
S -BLOCK ELEMENTS
Question and answers carrying 1 mark
1. What are s- block elements?
s-block elements are those in which the last electron enters the
outermost s-orbital.
As the s-orbital can accommodate only two electrons, two groups (1 & 2) belong
to the s-block of the Periodic Table.
2. Name the elements present in the 1st Group of the Periodic Table
lithium, sodium, potassium, rubidium, caesium and francium. They are
collectively known as the alkali metals.
3. Why I group elements are called alkali metals ?
These are called so because they form hydroxides on reaction with water which are
strongly alkaline in nature.
4. Name the elements present in the 2nd Group of the Periodic Table:
beryllium, magnesium, calcium, strontium, barium and radium. These elements
with the exception of beryllium are commonly known as the alkaline earth
metals.
5. Why II group elements are called alkaline earth metals ?
These are called so because their oxides and hydroxides are alkaline in nature and
these metal oxides are found in the earths crust.
6. What is the reason for the diagonal relationship ?
Diagonal relationship is due to the similarity in ionic sizes and /or charge/radius
ratio of the elements.
7. Which is smaller in size between a metal ion and its parent atom?
The monovalent ions (M+) are smaller than the parent atom.
8. Which group elements show very low ionization enthalpy in the periodic
table?
First group elements (alkali metals)
9. How the ionization enthalpy varies in alkali metals
Ionization enthalpy decrease down the group from Li to Cs.
10. Arrange the first group elements in the decreasing order of Hydration
Enthalpy
The hydration enthalpies of alkali metal ions decrease with increase in ionic sizes.
Li+> Na+ > K+ > Rb+ > Cs+
11. why Li salts are hydrated?
Li+ has maximum degree of hydration and for this reason lithium salts are mostly
hydrated, e.g., LiCl 2H2O
12. Write the chemical composition of washing soda.
Na2CO310H2O.
13. Give reason for the higher melting point and boiling point of alkali earth
metals than alkali metals.
The melting and boiling points of these metals are higher than the corresponding
alkali metals due to smaller sizes.
14. Why Be and Mg do not impart colour to the flame ?
The electrons in beryllium and magnesium are too strongly bound to get excited by
flame. Hence, these elements do not impart any colour to the flame.
15. Name the gas liberated when alkali metals react with dil acid?
The alkaline earth metals readily react with acids liberating dihydrogen gas .
M + 2HCl MCl2 + H2
16. Name the alkaline earth metal used in radio therapy.
Radium salts are used in radiotherapy, for example, in the treatment of cancer.
17. Give reason .the compounds of alkaline earth metals are less ionic than
alkali metals
This is due to increased nuclear charge and smaller size.
18. How is Calcium Hydroxide (Slaked lime), Ca(OH)2 Prepared?
Calcium hydroxide is prepared by adding water to quick lime, CaO.
19. How milk of lime reacts with chlorine?
Milk of lime reacts with chlorine to form hypochlorite, a constituent of
bleaching powder.
8. Give reason for the colour imparted to the flame by alkali metals
The alkali metals and their salts impart characteristic colour to an
oxidizing flame. This is because the heat from the flame excites the outermost
orbital electron to a higher energy level. When the excited electron comes back to
the ground state, there is emission of radiation in the visible region.
9. Why are Cs and K used as electrodes in photoelectric cells?
The alkali metal atoms have the largest sizes in a particular period of the
periodic table. With This property makes caesium and potassium useful as
electrodes in photoelectric cells.
10. What happens when alkali metals react with dihydrogen?
The alkali metals react with dihydrogen at about 673K (lithium at
1073K) to form hydrides. All the alkali metal hydrides are ionic solids with
high melting points.
11. Name the most power full reducing agent & give reason for it .
The alkali metals are strong reducing agents, lithium being the
most and sodium the least powerful reducing agent.
Note--- With the small size of its ion, lithium has the highest hydration enthalpy
which accounts for its high
negative E0 value and its high reducing power.
12.Give reason for the low solubility of LiF & CsI in water.
The low solubility of LiF in water is due to its high lattice enthalpy
whereas the low solubility of CsI is due to smaller hydration enthalpy of its two
ions. Other halides of lithium are soluble in ethanol, acetone and ethylacetate; LiCl
is soluble in pyridine also.
ionization enthalpy decreases The first ionisation enthalpies of the alkaline earth
metals are higher than those of the corresponding Group 1 metals. This is due to
their small size as compared to the corresponding alkali metals. It is interesting to
note that the second ionisation enthalpies of the alkaline earth metals are smaller
than those of the corresponding alkali metals.
24. How does the of Hydration Enthalpy of alkaline earth metals vary &
compare it with alkali metals
The hydration enthalpies of alkaline earth metal ions decrease with increase
in ionic size down the group. Be2+> Mg2+ > Ca2+ > Sr2+ > Ba2+ The hydration
enthalpies of alkaline earth metal ions are larger than those of alkali metal ions.
Thus, compounds of alkaline earth metals are more extensively hydrated than those
of alkali metals, e.g., MgCl2 and CaCl2 exist as MgCl2.6H2O and CaCl2 6H2O
while NaCl and KCl do not form such hydrates.
25.What is the colour imparted to the flame by Ca,Sr and Ba?
Calcium, strontium and barium impart characteristic brick red, crimson and apple
green colours respectively to the flame. In flame the electrons are excited to higher
energy levels and when they drop back to the ground state, energy is emitted in the
form of visible light.
28. Account for the reducing nature of Be even though it has less negative
value of reduction potential .
Like alkali metals, the alkaline earth metals are strong reducing agents. This
is indicated by large negative values of their reduction potentials However their
reducing power is less than those of their corresponding alkali metals. Beryllium
has less negative value compared to other alkaline earth metals. However, its
reducing nature is due to large hydration energy associated with the small size of
Be2+ ion and relatively large value of the atomization enthalpy of the metal.
29. write the general equation for the reaction of alkali earth metals with NH3.
Alkaline earth metals dissolve in liquid ammonia to give deep blue black
solutions forming ammoniated ions.
39.Which gas is liberated when Calcium carbonate is reacted with dil acid ?
It reacts with dilute acid to liberate carbon dioxide.
(M = K,
Rb, Cs)
In all these oxides the oxidation state of the alkali metal is +1. Lithium shows
exceptional behaviour in reacting directly with nitrogen of air to form the nitride,
Li3N as well. Because of their high reactivity towards air and water, they are
normally kept in kerosene oil.
3. Explain the reactivity of alkali metals towards water.
The alkali metals react with water to form hydroxide and dihydrogen.
(M = an alkali metal)
It may be noted that although lithium has most negative E0 value, its reaction with
water is less vigorous than that of sodium which has the least negative E0 value
among the alkali metals. This behaviour of lithium is attributed to its small size and
very high hydration energy. Other metals of the group react explosively with
water.
They also react with proton donors such as alcohol, gaseous ammonia and
alkynes.
The blue colour of the solution is due to the ammoniated electron which absorbs
energy in the visible region of light and thus imparts blue colour to the solution.
The solutions are paramagnetic and on standing slowly liberate hydrogen resulting
in the formation of amide.
(where am denotes solution in ammonia.) In concentrated solution, the blue
colour changes to bronze colour and becomes diamagnetic.
6.What are the uses of alkali metals ?
Lithium metal is used to make useful alloys, for example with lead to
make white metal bearings for motor engines, with aluminium to make aircraft
parts, and with magnesium to make armour plates. It is used in thermonuclear
reactions. Lithium is also used to make electrochemical cells. Sodium is used to
make a Na/Pb alloy needed to make PbEt4 and PbMe4. These organolead
compounds were earlier used as anti-knock additives to petrol, but nowadays
vehicles use lead-free petrol. Liquid sodium metal is used as a coolant in fast
breeder nuclear reactors. Potassium has a vital role in biological systems.
Potassium chloride is used as a fertilizer. Potassium hydroxide is used in the
manufacture of soft soap. It is also used as an excellent absorbent of carbon
dioxide. Caesium is used in devising photoelectric cells.
7.What is the reason for the increasing stability of peroxide & superoxide of
alkali metals down the group?
On combustion in excess of air, lithium forms mainly the oxide, Li2O (plus
some peroxide Li2O2), sodium forms the peroxide, Na2O2 (and some superoxide
NaO2) whilst potassium, rubidium and caesium form the superoxides, MO2. Under
appropriate conditions pure compounds M2O, M2O2 and MO2 may be prepared.
The increasing stability of the peroxide or superoxide, as the size of the metal ion
increases, is due to the stabilisation of large anions by larger cations through lattice
energy effects. These oxides are easily hydrolysed by water to form the hydroxides
according to the following reactions :
The oxides and the peroxides are colourless when pure, but the superoxides are
yellow or orange in colour. The superoxides are also paramagnetic. Sodium
peroxide is widely used as an oxidising agent in inorganic chemistry.
(vii) LiF and Li2O are comparatively much less soluble in water than the
corresponding compounds of other
alkali metals.
10. Mention the Points of Similarities between Lithium and Magnesium
The similarity between lithium and magnesium is particularly striking and arises
because of their similar sizes : (atomic radii, Li = 152 pm, Mg = 160 pm; ionic
radii : Li+ = 76 pm, Mg2+= 72 pm. )
The main points of similarity are:
(i) Both lithium and magnesium are harder and lighter than other elements in the
respective groups.
(ii) Lithium and magnesium react slowly with water. Their oxides and hydroxides
are much less soluble and
their hydroxides decompose on heating. Both form a nitride, Li3N and Mg3N2,
by direct combination
with nitrogen.
(iii) The oxides, Li2O and MgO do not combine with excess oxygen to give any
superoxide.
(iv) The carbonates of lithium and magnesium decompose easily on heating to
form the oxides and CO2.
Solid hydrogencarbonates are not formed by lithium and magnesium.
(v) Both LiCl and MgCl2 are soluble in ethanol.
(vi) Both LiCl and MgCl2 are deliquescent and crystallise from aqueous solution as
hydrates, LiCl2H2O
and MgCl28H2O.
11. How is Sodium Carbonate (Washing Soda), Na2CO310H2O manufactured
by Solvay Process
Principle-- In this process, advantage is taken of the low solubility of sodium
hydrogencarbonate whereby it gets precipitated in the reaction of sodium chloride
with ammonium hydrogencarbonate. The latter is prepared by passing CO2 to a
concentrated solution of sodium chloride saturated with ammonia, where
ammonium carbonate followed by ammonium hydrogencarbonate are formed.
The equations for the complete process may be written as :
The amalgam is treated with water to give sodium hydroxide and hydrogen gas.
carbon. Calcium and barium metals, owing to their reactivity with oxygen and
nitrogen at elevated temperatures, have often been used to remove air from vacuum
tubes. Radium salts are used in radiotherapy, for example, in the treatment of
cancer.
In the vapour phase BeCl2 tends to form a chloro-bridged dimer which dissociates
into the linear monomer at high temperatures of the order of 1200 K.
The ionic radius of Be2+ is estimated to be 31 pm; the charge/radius ratio is nearly
the same as that of the Al3+ ion. Hence beryllium resembles aluminium in some
ways. Some of the similarities are:
(i) Like aluminium, beryllium is not readily attacked by acids because of the
presence of an oxide film on
the surface of the metal.
(ii) Beryllium hydroxide dissolves in excess of alkali to give a beryllate ion,
[Be(OH)4]2 just as aluminium
hydroxide gives aluminate ion, [Al(OH)4].
(iii) The chlorides of both beryllium and aluminium have Cl bridged chloride
structure in vapour phase.
Both the chlorides are soluble in organic solvents and are strong Lewis acids.
They are used as Friedel
Craft catalysts.
(iv) Beryllium and aluminium ions have strong tendency to form complexes,
BeF42, AlF63.
21. Write a note on the manufacture of Cement:
Cement is an important building material. It was first introduced in
England in 1824 by Joseph Aspdin. It is also called Portland cement because it
resembles with the natural limestone quarried in the Isle of Portland, England.
Cement is a product obtained by combining a material rich in lime, CaO with
other material such as clay which contains silica, SiO2 along with the oxides of
aluminium, iron and magnesium. The average composition of Portland cement is :
CaO, 50- 60%; SiO2, 20-25%; Al2O3, 5-10%; MgO, 2- 3%; Fe2O3, 1-2% and SO3,
1-2%. For a good quality cement, the ratio of silica (SiO2) to alumina (Al2O3)
should be between 2.5 and 4 and the ratio of lime (CaO) to the total of the oxides
of silicon (SiO2) aluminium (Al2O3) and iron (Fe2O3) should be as close as possible
to 2.
The raw materials for the manufacture of cement are limestone and clay. When
clay and lime are strongly heated together they fuse and react to form cement
clinker. This clinker is mixed with 2-3% by weight of gypsum (CaSO42H2O) to
form cement. Thus important ingredients present in Portland cement are dicalcium
silicate (Ca2SiO4) 26%, tricalcium silicate (Ca3SiO5) 51% and tricalcium aluminate
(Ca3Al2O6) 11%.
22. How does the setting of cement takes place & what is the role of gypsum in
setting of cement ?
When mixed with water, the setting of cement takes place to give a hard mass.
This is due to the hydration of the molecules of the constituents and their
rearrangement. The purpose of adding gypsum is only to slow down the process of
setting of the cement so that it gets sufficiently hardened.
23.Write a note on biological importance of magnesium and calcium
An adult body contains about 25 g of Mg and 1200 g of Ca compared with
only 5 g of iron and 0.06 g of copper. The daily requirement in the human body has
been estimated to be 200 300 mg. All enzymes that utilise ATP in phosphate
transfer require magnesium as the cofactor. The main pigment for the absorption of
light in plants is chlorophyll which contains magnesium. About 99 % of body
calcium is present in bones and teeth. It also plays important roles in
neuromuscular function, interneuronal transmission, cell membrane integrity and
blood coagulation. The calcium concentration in plasma is regulated at about 100
mgL1. It is maintained by two hormones: calcitonin and parathyroid hormone. Do
you know that bone is not an inert and unchanging substance but is continuously
being solubilised and redeposited to the extent of 400 mg per day in man? All this
calcium passes through the plasma.
Subject: Chemistry
Chapter 11: P-Block Element
Decreases.
Sn+2H2O
SnO2+2H2.
B2O3+3H2O
450 K.
450 K
B2H6+NaF
c) SiCl4+H2O
SiCl4+2H2O
Si(OH)4 + 4HCl
d) C+O2+N2
1273K
2C+O2+4N2
1273K 2CO+4N2
B2H6 +6NaF
CO+H2
H2O+CO
C6H12O6+6O2+6H2O
Cu/570 K
(CH3)2SiCl2
+2H2O
-2HCl
(CH3)2Si(OH)2
***************************************
Chapter -12
Organic Chemistry-Some Basic Principles and Technique
b) sp3sp
c) sp2sp2
d) sp3sp sp
2.
3. Functional group may be defined as an atom (or) group of atoms which determine the
properties of an organic compound.
4. 1 Pentanal
5. H3C CH = CH CH CH2 CH3
|
Cl
6. The polarization of one bond caused by polarization. of adjacent bond due to
difference in electronegativity.
7. Partial displacement of sigma bond pair of electrons away from the substituent is called +I
effect.
8. Partial displacement of bond pair of electrons towards substituent is called -I effect.
9. NO2, - CN, -F, -COOH, -Cl, -Br, -I, -OCH3 etc. are electron with drawing groups
10. Alkyl groups like methyl (-CH3) & ethyl (-CH2 CH3) are usually considered as electron
donating groups.
11. The permanent polarity is produced by the interaction of lone pair & Pi electrons in
conjugate system of an organic molecule.
12. Shifting of electron pair away from substituent in a conjugate system.
13. Shifting of electron pair towards the substituent in a conjugate system.
14. Cl, -Br, -I, -NH2, -NHR, -OH, -OR, -SH, -OCH3 etc.
15. NO2,-CN, -CHO, - COOH, - COOR etc.
16. It is the complete transfer of Pi electrons of a multiple bond to one of the atom in the
presence of attacking reagent.
17. When the transfer of electrons take place towards the attacking reagent, the effect is
called +E effect attacking reagent, the effect is called +effect
Ex. Addition of H+ to ethene, H+ + CH2 = CH2 CH3 CH2
18. When the transfer of electron takes place away from the attacking reagent, the effect is
called effect.
Ex. The addition of cyanide ion (CN-) to carbonyl group
CN
+C=O
CO
|
CN
19. A series of organic compounds which can be represented by a general formula is called
Homologous series.
20. Copper sulphate acts as catalyst
21. Nitro compounds, Azo compounds & compounds containing nitrogen in ring
Ex. Pyridine.
22. +E effect.
23. Fe2 [Fe(CN)6]3
24. Al2O3(alumina)
25.
1.Ans:FreshlypreparedferroussulphatesolutionisaddedtosmallportionofSFE&
warmed.Thenabout2to3dropsofFecl3solutionareadded&acidifiedwith
concentratedHcl.Theappearanceofaprussainbluecolourindicatesthepresence
ofnitrogen.
Fuse
C,N+Na NaCN
Organiccompound
Feso4+2NaOH Fe(OH)2+Na2SO4
6NaCN+Fe(OH)2 Na4[Fe(CN)6]
SodiumFerocyanide.
3Na4[Fe(CN)6]+4Fecl3 Fe4[Fe(CN)6]3+12Nacl
Ferricferrocyanide(Prussianblue)
2.Ans:a)LeadAcetatetat:FewdropsofsodiumnitroprusideisaddedtoSFE.The
appearanceofadeepvioletcolourindicatesthepresenceofSulphur
S2+[Fe(CN)5NO]2 [Fe(CN)5NOs]4
Nitroprussideionvioletcolour
b)SulphurcanalsobedetectedbyaddingleadacetatetoSFE,acidifiedwithacetic
acid.Theformationofablackprecipitate(pbs)indicatesthepresenceofsulphur.
Pb(CH3COO)2+S2 pbs +2CH3COO
3.Ans:Carbon&hydrogenpresentinanorganiccompoundcanbedetectedtogetherby
copperoxidemethod.
Thecompoundismixedwithdrycupricoxide&istakeninahardglasstesttube.
Thehardglasstesttubeisfittedtooneendofdeliverytubecontainingbulb
(containinganhydrousCuSO4)&otherendofthedeliverytubeisdippedinatest
tubecontainerlimewater.
Themixtureisstronglyheated.
CarbonpresentincompoundisoxidizedtoCO2&turnslimewatermilky.
Hydrogenpresentincompoundisoxidizedtowater&turnswhiteanhydrous
CUSO4tobluehydratedsalt.
NOTE:
C+2CUO CO2+CU
(organiccompound)
CO2+Ca(OH)2 CaCO3+H2O
Limewater(milky)
2H+CUO H2O+Cu
(organiccompound)
CuSO4+5H2O CUSO4.5H2O
Whiteblue
4.Ans:Apieceofdrysodiumistakenintoafusiontube&heatedtillitmelts.Adropoffew
crystalsoftheorganiccompoundisaddedtothefusiontube.Themixtureisheated
gently&stronglyuntilthetubebecomeredhot&plunged(added)intoamortar
containingdistilledwater.Thecontentsarecrushed&filtered.Thefilterateis
knownassodiumfusionextract.
5.Ans:Silvernitratetest:AsmallportionofSFEisboiledwithdilHNO3,cooled&silver
nitrateisadded.Awhiteprecipitatesolubleinammoniumhydroxide,showsthe
presenceofchlorine.
Apaleyellowprecipitateslightlysolubleinammoniumhydroxideinsolublein
ammoniumhydroxideshowsthepresenceofbromine&yellowprecipitate
insolubleinammoniumhydroxideshowsthepresenceofiodine
Note:Nacl+AgNO3 Agcl +NaNO3
White
NaBr+AgNO3 AgBr +NaNO3
Paleyellow
NaI+AgNO3 AgI +NaNO3
Yellow
6.Ans:Theenergydifferencebetweenconnonicalstructure&Resonancehybridisknownas
Resonanceenergy.TheResonanceenergyofbenzeneis36kcaloriesor150kJ/mol
7.Ans:Organiccompoundcontainingphosphorousisfusedwithsodiumperoxide.The
phosphorouspresentintheorganiccompoundisoxidisedtophosphate.Thefused
massisextractedwithwater&filterate.Thefilteratecontainingsodiumphosphate
isboiledwithnitricacid&thentreatedwithammoniummolybdate.Ayellow
solutionofprecipitateindicatesthepresenceofphosphorous.
8.Ans:
Inductiveeffect
Mesomericeffect
1)Itoperatesinsaturatedcompound
1)Itoperatesinconjugated
doublebondSystem
2)Itinvolvesdisplacementofsigmaelectrons
2)Itinvolvesdisplacementofpi
electrons
3)Itlastsonlyforashortdistance
3)Itlastoverlongdistance
4)Partialchargesaredeveloped
4)Complete+ve&vecharges
aredeveloped
9.Ans:
Inductiveeffect
Electromericeffect
1)ItInvolvespartialdisplacementofsigma 1)Involvescompletetransferofpi
electronselectrons
2)Permanenteffect
2)Temporaryeffect
3)Presenceofattackingreagentisnot
3)Attackingreagentisrequired
required
4)Partialchargesaredevelopedonatoms 4)Completechargeseparationtakes
place
10.Ans:1)Allthememberscanberepresentedbyageneralformula
2)EverysuccessivemembersoftheseriesdifferbyCH2group.
3)Allthememberscanbepreparedbysimilarmethods
4)Allthememberswillhavesimilarchemicalproperties
11.Ans:Theelectrondeficientspeciesorpositivelychargedionswhicharecapableof
acceptinganelectronfromsubstratemoleculearecalledelectrophiles.
Ex:Positiveelectrophiles:H+,Cl+,Br+,NO2etc.
Neutralelectrophiles:SO3,BF3,AlCl3etc.
12.Ans:theanegativelychargedionswhicharecapableofdonatinganelectronpair
Eg.Cl,Br,OHetc
13.Ans:Symmetricalbreakingofacovalentbond,inwhicheachofthetwospeciescontain
oneelectronofsharedelectronpairiscalledhemolyticfissionorhomolysis.
Exclcl . +cl.
Chlorinefreeradicals
14.Ans:Unsymmetricalbreakingofacovalentbond,inwhichonethespeciescarrybonded
electronpairiscalledheterolyticfissionorheterolysis.
15.Ans:Freeradicalscanbedefinedasanatomorgroupofatomshavinganunpaired
electron.
homolysis
Ex.clcl
cl.+cl.(Chlorinefreeradicals)
16.Ans:Areactionintermediateformedbyheterolyticfissionofacovalentbondwhich
containsonepositivelychargedcarboniscalledcarbocation.
Ex.CH3Br +CH3+
17.Ans:Areactionintermediateformedbyheterolyticfissionofacovalentbondwhich
containsonenegativelychargedcarboniscalledcarbonion.
Ex.CH3MgI
+
I
18.Ans:Thesearesaturatedhydrocarbonsjoinedbycovalentbondtoformringstructure.
Ex.,etc.
19.Ans:Thesearethecompoundscontainingonebenzenering
Ex.Benzene,Napthaleneetc.
20.Ans:Thesearethecompoundscontainingringstructureinwhichoneormorecarbon
atomsarereplacedbyheteroatomssuchasN,S,Oetc.
21.Ans:ItisdonesoastodecomposeNaCNtoHCN&Na2StoH2S
22.Ans:a)2,2,4trimethylhexane
b)6chloro3methyl2hexanone
CH3
|
23.Ans:a)H3CCCHCH2CH3b)clH2CC=CHCH2OH
||
CH3CH3
CH3
24.Ans:Afunctionalgroupisanatomorgroupofatomspresentinamoleculewhich
determinesthepropertiesoftheorganiccompounds.
Ex:C2H5OH,OHisF.Gpresentinethylalcohol
CH3COOH,COOHisF.G.presentinAceticacid.
25.Ans:Twoormorecompoundshavingsamemolecularformulabutdifferinpositionof
thesamefunctionalgroup
Ex:CH3CH2CH2OH&CH3CHCH3
1propanolOH
2propanol
26.Ans:Twoormorecompoundshavingsamemolecularformulabutdifferinthe
functionalgroup
Ex.Alcoholandether
CH3OCH3
CH3OC3H7
Dimethylethermethylpropylether.
27.Ans:Thereactioninwhichonegroupreplacesanother.
Uv
CH4+Cl2 CH3cl+Hcl
28.Ans:Thereactioninwhichthereagentaddsuptothesubstratemoleculewithout
eliminationofanymoleculeiscalledadditionreaction
Ni
Ex.CH2=CH2+H2 CH3CH3
29.Ans:Sp2,Sp2Sp3,Sp2,Sp2
Method:
1. The apparatus is arranged as shown in the diagram.
2. It contains a combustion tube connected to a U shaped tube containing anhydrous
calcium chloride and a glass bottle containing Potassium hydroxide solution in series [Which
is then connected to guard tube containing anhydrous Calcium chloride to avoid the entry of
moisture and CO2 into the apparatus.]
3. A known mass of organic compound is mixed with cupric oxide and placed in the
combustion tube which is heated strongly.
4. Carbon present in organic compound is oxidized to carbon dioxide and absorbed in
potassium hydroxide solution.
5. Hydrogen present in the organic compound is oxidized to water and absorbed in anhydrous
calcium chloride.
6. The U tube and glass bottle are weighted before and after the experiment.
Calculation:
I. Estimation of hydrogen :
a) Mass of organic compound = Wg
b) Mass of U tube before experiment = m1g.
c) Mass of U tube after experiment = m2g
d) Increase in mass of CaCl2 = Mass of water = (m2-m1g)
e) 18 grams of water contains 2 grams of hydrogen
(m2 m1)g of water contains
= 0.1111(m2-m1)grams
f) W grams of organic compound contains 0.1111(m2-m1)g of hydrogen
.
100 grams of organic compound contains
percentage of hydrogen =
= 0.2727(m2-m1)g of carbon
f) W grams of organic compound contains 0.2727(m2-m1)grams of carbon
.
100 grams of organic compound contains
.
percentage of Carbon=
2. Ans: Principle: A known mass of organic compound is heated with concentrated sulphuric
acid. Nitrogen is converted into ammonium sulphate. Which is treated with sodium hydroxide
solution to liberate ammonia. This ammonia is absorbed in excess of standard sulphuric acid.
The unreacted acid is estimated by titration with standard alkali. From which the amount of
ammonia is determined and the percentage of nitrogen in the compound is calculated.
Percentage of Nitrogen =
3. Ans: Principle: The organic compound containing nitrogen when heated with excess of
copper oxide in the atmosphere of carbon dioxide, gives nitrogen in addition to carbon
dioxide and water.
Traces of nitrogen oxides formed during combustion of organic compound are reduced to
nitrogen by passing the gaseous mixture over a heated copper gauze. The percentage of
nitrogen present in a given organic compound is calculated from the volume of nitrogen
collected over potassium hydroxide solution from a known mass of organic compound.
Procedure: The apparatus used for the estimation of nitrogen by this method is shown in the
figure.
A known mass of organic compound is mixed with copper oxide and placed in the
combustion tube. The carbon dioxide gas is passed through the combustion tube to displace
air present in the tube. The combustion tube is now heated in the furnace. The nitrogen
evolved collects in the nitrometer. The volume of the nitrogen collected is recorded after
adjusting the levels of potassium hydroxide solution in the two limbs are equal. Room
temperature and atmosphere pressure are recoded.
Calculation: Mass of organic Compound
=
mg
3
Volume of nitrogen in nitrometer
=
V cm
Room temperature
= toC = (273 + t)K
Atmosphere pressure = P1mm
'
Aqueous tension at room temperature = P mm
Pressure of dry nitrogen gas formed = P = (P-P')mm .
cm3
Volume of nitrogen at STP (V0) =
22,400 cm3 of nitrogen of STP = 28g of nitrogen
Mass of V0 cm3 of nitrogen =
g
,
Percentage of Nitrogen =
nitric acid does not enter the weighing tube. The Carius tube is now sealed and heated in a
furnace at 3000 C for about six hours.
The tube is than cooled and its narrow end is cut off and the contents are completely
transferred to a beaker by washing with water. The precipitate of silver halide formed is
filtered through a weighed sintered glass crucible. It is washed, dried and weighed.
Observation and calcualteion:
i) Mass of organic compound taken =
W 1g
ii) Mass of silver halide obtained
=
W2g
a) For chlorine: AgCl Cl
143.5g 35.5g
143.5g of AgCl contains 35.5g of chlorine
.
g of chlorine
= a grams (say)
This amount of chlorine was present in w1g of the compound.
%Cl2 =
g of bromine.
g of Iodine
***************************************************************************
CHEMISTRY
CHAPTER- HYDROCARBONS (I PUC)
One mark questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
3) Why is Wurtz reaction not preferred for the preparation of alkanes containing
odd number of carbon atoms? Illustrate your answer by taking one example.
4) State Huckels rule. Draw the structure of Pyridine and Furan. Are these
aromatic?
5) Explain the mechanism involved in the chlorination of methane.
(i)
(ii)
(iii)
(iv)
8) Addition of HBr to propene yields 2-bromopropane, while in the presence of
benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give
mechanism.
9) Write chemical reactions for the following conversions:
(i) Phenol to benzene
(ii) Benzene to ethyl benzene
10) How do you convert the following?
(i) Benzene to p-nitrotoulene
(ii) Benzene to acetophenone
(iii) Benzene to m-chloronitrobenzene
11) Discuss the orbital structure of benzene.
CHEMISTRY
CHAPTER- HYDROCARBONS (I PUC)
a. Bayers test
b. Bromination
2-methyl-propane
3) X is Cis-but-2-ene ; Y is trans-but-2-ene
4)
5)
Due to the occurrence of both methyl groups on the same side of the C=C
bond, the combined effect of the two polar bonds makes cis-but-2-ene
much more polar than trans-but-2-ene.
8)
9)
Ca(OH)2 + C2H2
11) Ketone
12) Markovnikov's rule states that, negative part of the addendum (adding molecule)
gets attached to that carbon atom which possesses lesser number of hydrogen
atoms.
13) Sodium salt of butanoic acid is required for the preparation of propane.
CH3CH2CH2 COO-Na+ + NaOH
CH3CH2CH3 + Na2CO3
14) Alkyl halides on treatment with sodium metal in dry ether solution give
higher alkanes. This reaction is known as Wurtz reaction.
This reaction is used for the preparation of higher alkanes containing even number
of carbon atoms.
15) Sodium salts of carboxylic acids on heating with soda lime (mixture of sodium
hydroxide and calcium oxide) give alkanes containing one carbon atom less than
the carboxylic acid. This elimination of carbon dioxide from a carboxylic acid is
known as decarboxylation.
CH3COO-Na+ + NaOH
CH4 + Na2CO3
16) Alkanes undergo free radical substitution. The examples of this category are
halogenation, nitration and sulphonation.
17) The spatial arrangements, which are obtained by free rotation around sigma
bonds, are called conformation or conformational isomers.
18) Ethene is obtained.
C2H5OH
20)
23) Any Lewis acid like anhydrous FeCl3, SnCl4, BF3 etc. can be used during the
ethylation of benzene.
24)
25) The repulsive interaction between the electron clouds, which affects stability of a
conformation, is called torsional strain. Magnitude of torsional strain depends upon
the angle of rotation about C-C bond. This angle is called dihedral angle or torsional
angle. Of all the conformations of ethane, the staggered form has the least torsional
strain and the eclipsed form has the maximum torsional strain.
26)
a) Unusual stability of benzene.
b) According to Kekule, two ortho disubstituted products are possible. But in
practice only one ortho disubstituted product is known.
c) Heat of hydrogenation of benzene is 49.8 kcal/mole, whereas theoretical value of
heat of hydrogenation of benzene is 85.8 kcal/mole. It means resonance energy is
36 kcal/mole.
d) C - C bond length in benzene are equal, (although it contains 3 double bonds and
3 single bonds) and are 1.39 .
27)
H
H
H
H
H
H
H
H
Eclipsed
H
H
Staggered
28)
10
The boiling points of alkanes (obtained in the mixture) are very close. Hence, it becomes
difficult to separate them.
4) Huckels rule states that, compounds that have (4n + 2)
be Aromatic compounds, where n = 1,2,3,4.etc.
Pyridine is aromatic because it folllows Huckel's rule and has 6 pi electrons where n=1.
Furan is also aromatic compound because one of the lone pair of electrons at the oxygen
delocalise towards benzene ring and then it follows Huckel's rule where n=1.
5) Chlorination of methane proceeds via a free radical chain mechanism. The whole
reaction takes place in the given three steps.
11
Step 1: Initiation:
The reaction begins with the homolytic cleavage of Cl Cl bond as:
Step 2: Propagation:
In the second step, chlorine free radicals attack methane molecules and break down the
CH bond to generate methyl radicals as:
These methyl radicals react with other chlorine free radicals to form methyl chloride
along with the liberation of a chlorine free radical.
Hence, methyl free radicals and chlorine free radicals set up a chain reaction. While HCl
and CH3Cl are the major products formed, other higher halogenated compounds are also
formed as:
Step 3: Termination:
Formation of ethane is a result of the termination of chain reactions taking place as a
result of the consumption of reactants as:
12
6) (i)
For the given compound, the number of -electrons is 6.By Huckels rule,
4n + 2 = 6; 4n = 4; n = 1
For a compound to be aromatic, the value of n must be an integer (n = 0, 1, 2). Since
the value of n is an integer, the given compound is aromatic in nature.
(ii)
By Huckels rule,
4n + 2 = 4; 4n = 2 ;
(iii)
For the given compound, the number of -electrons is 8.
By Huckels rule, 4n + 2 = 8; 4n = 6;
For a compound to be aromatic, the value of n must be an integer (n = 0, 1, 2). Since
the value of n is not an integer, the given compound is not aromatic in nature.
7) (i) Pen-1-ene-3-yne
(ii) Buta-1,3-diene
(iii) 4-Phenyl but-1-ene
13
8)
This reaction follows Markovnikovs rule where the negative part of the addendum is
attached to the carbon atom having a lesser number of hydrogen atoms.
In the presence of benzoyl peroxide, an addition reaction takes place anti to
Markovnikovs rule. The reaction follows a free radical chain mechanism as:
14
Secondary free radicals are more stable than primary radicals. Hence, the former
predominates since it forms at a faster rate. Thus, 1 bromopropane is obtained as the
major product.
In the presence of peroxide, Br free radical acts as an electrophile. Hence, two different
products are obtained on addition of HBr to propene in the absence and presence of
peroxide.
9) (i) Phenol into benzene.
(ii)Benzene into ethyl benzene.
15
16
11) The structure of benzene molecule is best described in terms of molecular orbital
treatment theory. According to this theory, all the C-atoms in benzene are sp2hybridized. Two sp2-hybrid orbitals of each C-atom overlap with two sp2-hybrid orbital
of two other C-atoms to form sigma bonds. In this way there are six sigma bonds are
formed between six C-atoms which are 120o apart. Remaining six sp2-orbital of six Catoms overlap with 1s orbital of six H-atoms individually to form six sigma bonds. Since
sigma bond results from the overlap of above said planar orbital, all H and C atoms are in
the same plane and their generate a hexagonal ring of C-atoms.
Each C-atom in benzene also has an unhybrid 2pz-orbital containing one electron. These
2pz-orbital are perpendicular to the plane of sigma bonds.
OR
Actually these 2pz-orbital produce a pi-molecular orbital containing six electrons. One
half of this pi- molecular orbital lies above the plane of hexagonal ring and remaining
half below the ring like a sandwich.
18
CHAPTER14
ENVIRONMENTALCHEMISTRY
I)Onemarkquestionsandanswers:
1.Nameoneinsecticide?
A. DDT
2.Whichacidisnotpresentinacidrain?
HNO3,H2SO4,CH3COOH,H2CO3?
A. CH3COOH
3.Definethetermpollution?
A. Itisasubstancepresentintheenvironmentingreat
proportionthanitsnaturalabundanceandresultingin
harmfuldamageeffect.
4.Nametwogaseswhichareresponsibleforgreenhouse
effect?
A. CO2andCH4gases.
5.Whichpartoftheatmospherecontainsozone
layers?
A. Stratospherecontainsozonelayers.
6.WhatisfullformBODandDDT?
A. BODBiochemicaloxygenDemandandDDTDichchloro
DiphenylTrichloroethane.
7.WhatarePCBs?
A. Polychlorinatedbiphenyls(PCBs)areusedascleansing
solvent,detergentsandfertilizerscausewaterpollution
anditiscarcinogeniccompound.
8.WhatisPAN?
A. Peroxyacetylnitrate(PAN)isoneofthecomponentsof
photochemicalsmoganditispowerfuleyeirritant.
9.WhatisdesirableconcentrationofF ionsandPHof
drinkingwater?
A. DesirableconcentrationofF ionsis1ppmor1mgdm3
andPHis5.5to9.5
10.Nametheoxidesofnitrogen?
A. Nitricoxide(NO)andNitrogendioxide(NO2).
11.WhichgascausedBhopalgastragedy?Giveitsformula.
A.Methylisocyanate(MIC)anditsmolecularformula
CH3N=C=0
12.Writeanytwocommonchemicalsofphotochemicalsmog?
A.Acroleinandformaldehyde
13.Whichcandamagethegreathistoricalmonument
Tajmahal?
A.Acidrain
(CaCO3(marble)+H2SO4CaSO4+H2O+CO2)
14.WhatiseffectofexcessofSO42ionindrinkingwater?
A.ExcessofSO42ionindrinkingwatercauseslaxativeeffect
(>500ppm)
15.Whatisthecauseofmethemeglobinemia?
A.Excessofnitrateion(>50ppm)indrinkingwatercause
methemeglobinemia(bluebabysyndrome).
16.Whatistroposphere?
A.Thelowerregionsofatmosphereinwhichthehumanbeings
along with other organisms live are called troposphere. It
extendsuptothehighof~10KMfromsealevel
17.Whatisstratosphere?
A. Above the troposphere, between 10 and 50km above sea
levelliesiscalledstratosphere.
18.Nametheharmfulradiationemittedfromsun?
A.UVradiation
19. Which type of harmful radiations absorbed in ozone
layers?
A.UVradiation
20.Namethetypesofpollutantscausetropospherepollution?
A.1.Gaseousairpollutants
2.Particulatepollutants
21.Whatarethesourcesofdissolvedoxygenin
water?
A. In water, the source of oxygen is either atmospheric
oxygenorphotosynthesiscarriedinplantsduringdaylight.
22.Whichofthefollowinggasesisnotagreenhousegas?
CO,CO3,CH4,H2Ovapours
A. COisnotagreenhousegas.
II)Twomarksquestionsandanswers:
1.Nametwoherbicides?
A.NaClO3(sodiumchlorate)and
Na3AsO3(SodiumArsenite)
2.Listanytwoharmfuleffectsofsmog?
A.1.OzonePANActaspowerfuleyeirritants
2.Ozoneandnitricirritatethenoseandthroatandtheir
highconcentrationcausesheadache,chestpain,anddrynessof
thethroat,coughanddifficultyinbreathing.
3.Writeanytwoachievementofgreenchemistry?
A.1.Developmentofpolystyrenefoamsheetpackaging
materialthisTechnologyallowseliminationsCFCSwhich
contributetoozonedepletion,globalwarmingandground
levelsmog.
2.Catalytichydrogenationofdiethanolamineinwhich
environmentalfriendlyherbicideisproducedinlessdangerous
ways.
4.Defineenvironmentalchemistry?
A.Itisthestudyofchemicalandbiochemicalprocessoccurring
innature
(OR)
Itdealswiththestudyoforigin,transportrelation,effectsand
fatesofvariouschemicalspeciesintheenvironment
5. Whatdoyoumeanbyozonehole?Whatareits
Consequences
A. Ozone hole implies distribution of the ozone layer by the
HarmfulUVradiationsthedepletionwillvirtuallyresult
Increatingsomesortofholesintheblanketofozonewhich
surround us. As a result, the harmful radiations cause skin
cancer,lossofsightandalsoaffectourimmunesystem
6.WhatdoyoumeanbyBiochemicaloxygendemand?
A. Biochemical oxygen demand is the amount of oxygen
requiredbybacteriatodecomposeorganicmatterinacertain
volumeofsampleofwater.CleanwaterwouldhaveBODvalue
oflessthan5ppm,whereashighlypollutedwaterhasaBODof
17ppmormore
7.Howdoesdetergentcausewaterpollution?
A. Tetrachlorothene (Cl2C=CCl2) was used as solvents for dry
cleaningofclothes.Thiscompoundissuspectedtocarcinogenic
and contaminated the suspected to carcinogens for bleaching
cloths in the laundry, hydrogen peroxide (H2O2)is being used
whichgivesbetterresultandisnotharmfuldetergentsproduce
pathogens which are diseases causing bacteria and result in
gastrointestinaldiseases.
8.Writethemethodsformanagementofwastematerial?
1.Recycling:materialsarerecycledwhichsavesthecastofraw
materialandwastedisposal.
2.Sewagetreatment
3.BurningandIncineration
4.Digesting
5.Dumping
9. Write the chemical reaction take place during acid rain in
theatmosphere?
A.1.H2O+CO2>H2CO3
+
2.H2CO3>H +HCO3
3.2SO2+O2+2H2O>2H2SO4
10.Foryouragriculturefieldorgardenyouhavedevelopeda
compostproducingpit.Discusstheprocessinthelightofbad
odour,filesandrecyclingaswastesforagoodproduce.
A.Itisessentialtotakepropercareofthecompostproducing
pit in order to protect ourselves from bad odour and files. It
should be kept covered to minimize bad odour and prevent
filesfromenteringit.
The recyclable waste should not be dumped in the compost
producing pit. It should be sent to the industries through
vendorsforrecycling.
11.Alargenumberoffisharesuddenlyfoundfloatingdeadon
a lake. There is no evidence of toxic dumping by you find an
abundanceofphytoplankton,suggestareasonforthefishkill.
A.Theamountofdissolvedoxygenpresentinwaterislimited.
The abundance of phytoplankton causes depletion of this
dissolvedoxygen.Thisisbecausephytoplanktonsaredegraded
by bacteria present in water. For their decomposition, they
require a large amount of oxygen. hence, they consume the
oxygen dissolved in water. As a result, the BOD level of water
drops below 6ppm, inhibiting the growth of fish and causing
excessivefishkill.
12.Whatareharmfuleffectsasdepletionofozonelayer?
A. 1) The ozone layer protects the earth from the harmful UV
radiation of the sun, with the depletion of the layer, more
radiation will enter the earths atmosphere. UV radiations are
harmfulbecausetheyleadtotheskincancerandsunburns.
2) They cause death of many phytoplanktons which lead to a
decreaseoffishproductivity.
3)IncreaseinUVradiation,decreasesthemoisturecontentof
thesoilanddamagesbothplantsandfibres.
13.Whatarepathogens?Mentionitsharmfuleffect?
A. Pathogens are water pollutants include bacteria and other
organism.Theyenterwaterfromanimalexcretaanddomestic
sewage.
14.Whatareharmfuleffectsofacidrain?
A.1.Itisharmfulforcrops
2.Itdamagesbuildingsmadeupofmarble.
14.Writebytwoharmfuleffectsofoxidesofnitrogen?
1. Damage the leaves of plants and retard the rate of
photosynthesis.
2. Nitrogen dioxide is a lung irritant that can lead to an acute
respiratorydiseaseinchildren.
15.Writeanytwoharmfuleffectsofoxideofsulphur?
1. It causes respiratory diseases e.g. Asthama bronchitis in
humanbeings
2.Itcausesirritationtotheeyes,resultingintearsandredness.
16. Write the harmful effects of hydrocarbons pollutants?
Mentionitssources?
Harmfuleffects:
1.Hydrocarbonsarecarcinogenici.e.theycausecancer
2.Theyharmplantsandsheddingofleavesflowers&twinges
Sources:Incompletecombustionoffuelusedinautomobiles.
17.WhataretheharmfuleffectsofCO?Mentionitssources?
Harmfuleffects:
1. Itishighpoisonoustolivingbeings
2. It causes, headache, weak eyesight, nervousness and
cardiovasculardisorder
CO2presentintheaciddissolvesinwatertogivecarbonic
acid.
III)Threemarksquestionsandanswers
1.Explaintroposphericpollution?
A. Troposphere pollution occurs due to the presence of
undesirable gases and the solid particles in the air the major
gaseous and the Particulate pollutants presents in the
troposphereasfollows.
1. Gaseous air pollutants: These include mainly oxides of
sulphur (SO2&SO3), oxide of nitrogen (NO&NO2) and oxides of
carbon (CO&CO2) in addition to hydrogen sulphide (H2S),
hydrocarbonsandotheroxidants.
2. Particulate pollutions: These include dust, mist, fumes,
smoke,smogetc..
Classicalsmogphotochemicalsmog
ItoccursincoolhumidItoccursinwarmdryand
Climatesunnyclimate
ItiscalledreducingsmogItiscalledoxidizing
smog
Itisamixtureofsmoke,Itisamixtureof
Fogandsulphurdioxideunsaturated
hydrocarbonsand
Oxidesofnitrogen
3.StatuesandmonumentsinIndiaareaffectedbyacidrain,
how?
A.TheairaroundthestatuesandmonumentsinIndiacontains
fairlyhighlevelsoftheoxidesofsulphurandnitrogen.
Thisisduetoalargenumberofindustriesandpowerplantsin
thenearbyareas.
Theproblemhasbeenfurtheraggravatedduetouseofpoor
qualityofcoal,keroseneandfirewoodasfuelfordomestic
purposes.
Thestatesacidrainaffectsformarbleofthesestatuesand
monuments.
CaCO3+H2SO4>CaSO4+H2O+CO2
Asaresult,thisawayandmarbleisgettingdiscoloredand
lusterless.
6.Defineenvironmentalpollution?Nameitstypes?
A. The addition of any undesirable material to air, water and
soilbyanaturesource
or
Due to human activity to such a level of concentration warm
adversely affects the quality of environment is called
environmentalpollution.
Types
1.Waterpollution
2.Soilpollution
3.Airpollution
8.Howcandomesticwastebeusedasmanure?
A.Dependinguponthenatureofthewastedomesticwastecan
besegregatedintotwocategories.i.e.biodegradableandnon
biodegradables. Biodegradables waste such as leaves, rotten
food etc. should be deposited in land hills, where they get
decomposed aerobically and anaerobically in to manure. Non
biodegradable waste (Which cannot be degraded) such as
plastic,glass,metalscrapesetc.shouldbesentforrecycling.
11.Whatarethereactionsinvolvesforozonelayerdepletion
inthestratosphere?
A. CFCS (chlorofluorocarbons) that are released in the
atmosphere mix with the other atmospheric gases and when
eventually reach the stratosphere, gets broken down by UV
radiationsasfallows
CF2Cl2>Cl +CF2Cl
Thechlorineradicalreactswithozoneandbreaksdownozone
moleculeasfollows
Cl +O3>ClO +O2
ClO radicalfurtherreactswithatomicoxygenandproduces
morechlorineradicalsasfollows
3. NO(g)+O3(g)>NO2(g)+O2(g)
Sources: 1. At high attitude when lightning strokes, dinitrogen
anddioxygencombinetoformoxidesofnitrogen.
2. Burning of fossil fuels in an automobile engine at high
temperature, dinitrogen and dioxygen combine to yield
significantquantitiesofnitricoxideandnitrogendioxide.
13.Writethechemicalreactionsfortheformationofoxidesof
sulphur?Mentionitssources?
A.1.2SO2+O2>2SO3
2.SO2+O3>SO3+O2
3.SO2+H2O2>2H2SO4
Sources:Burningoffossilfuelscontainingsulphur
14.Writedownthechemicalequationofreactionsinvolved
duringtheformationofphotochemicalsmog.Howcanitbe
controlled?
A. Photochemical smog is formed by absorption of sunlight
by oxides of nitrogen to form free radicals which are highly
reactive
a)NO2>NO +O
b)O +O2>O3
c)NO +O3>NO2+O2
d)3CH4+2O3>3HCHO+3H2O
Itcanbepreventedbysprayingchemicalswhichwilldestroy
freeradicalsintheatmosphere.
IV)Fourmarksquestionsandanswers
1. Explain the reactions involved during the formation of
photochemicalsmog?
A. Photochemicalsmogisformedasaresultofthereaction
ofsunlightwithhydrocarbonsandnitrogenoxides.
Ozone, nitricoxide, Acrolein, formaldehyde and
peroxyacetyl nitrate are common compounds as
photochemicalsmog.
Theformationofphotochemicalsmogcanbeburningasfossil
fuelsleadstotheemissionofhydrocarbonandnitrogendioxide
in atmosphere .High concentration of these pollutants in air
resultsintheirinteractionwithsunlightasfallows.
1. NO2(g)>NO
(g)+O (g)
2. O
(g)+O2(g)>O3(g)
3. O3(g)+NO
(g)>NO2(g)+O (g)
Controlofphotochemicalsmog
1. Use of catalysts converters in automobiles Which prevents
the release of oxides of nitrogen and hydrocarbons to the
atmosphere
2.Certainplantse.g.pines,Juniparus,quercuspyrusandvitis
can metabolize oxides of nitrogen and therefore, their
plantationcouldhelpinthismatter.
3.Whatarethemajorcausesofwaterpollution?Explain
A. 1. Pathogens: These are water pollutants include bacteria
and other organisms.They enter water from animal excreta
and domestic Sewage. Bacteria presents in human excreta
causes gastrointestinal diseases (Excreta contains, escherichia
Coilandstreptococcusfaecalis)
2.Organicwastes:Thesearebiodegradablewaterthatpollutes
water as a result run off. The presence of excess of organic
wastesinwaterdecreasestheamountofoxygenheldbywater.
This decrease in the amount of dissolved oxygen inhibits
aquaticlife.
3.Chemicalpollutants:Thesearethewatersolublechemicals
likeheavymetalssuchascadmium,mercury,nickeletc
The presence of these chemicals (above the tolerance
limit) can damage the kidneys, central nervous system
andliver.
4.Whatdoyoumeanbygreenchemistry?Howwillithelpin
decreasingenvironmentalpollution?
A. It is a production process that aims at using the existing
knowledge and principles of chemistry for developing and
implentingchemicalproductsandprocessestoreducetheuse
andgenerationifsubstanceshazardoustotheenvironment.
Thereleaseofdifferentharmfulchemicals(particulates,gases,
organic and inorganic wastes) causes environmental pollution.
In green chemistry, the reactants to be used in chemical
reactions are chosen in such way that the yield of the end
products is up to 100%. This prevents or limits chemical
pollutantsfrombeingintroducedintotheenvironment
For example, through the efforts of green chemists, H2O2 has
replaced tetrachloromethene and chlorine gas in drying and
bleachingofpaper.
CO2 has replaced CFCs as blowing agents in manufacture of
polystyrenefoamsheet.
5.Namethepollutantswhichcausesoilpollution?
A.1.Pesticides
2.Insecticides
3.Herbicides
4.Fungicides
5.Industrialwastes
6.Urbanwastes
7.Agriculturepollutants
8.Fertilizers
6. What is green house effect? How does it affect the global
climate?
A.Itisthephenomenoninwhichearthsatmospheretrapthe
heatfromthesunandpreventitformescapingintotheouter
space.
GreenhousegasessuchasCO2,CH4,ozone,chlorofluorocarbon
compounds and water vapour in the atmosphere result in
climate changes sunlight intense a green house through. The
transparent glass or plastic panes and heats the plants by the
heat emitted by the plants in the form of infrared radiation
cannot puss through the glass or plastic panes. As a result of
inside temperature increases. Increased CO2 levels in the
atmospherecanalsocauseplants,undergoingphotosynthesis,
to take use the gas at a greater rate so that plants in warmer
climatewithadequaterainfallwouldgrowfaster.Anincreasein
average global temperature to increase the incidence of
infectious diseases like malaria, sleeping sickness, dengue and
yellow fever CFCs are also damaging ozone layer. The average
globaltemperatureincreasetoalevelwhichmayleadmelting
polar ice caps and flooding of lying areas all over the earth.
Theremaybelessrainfallintemperaturezonesandmorerain
fallinthedrierareasoftheworld.(CO2inthemajorcontribute
toglobalwarming)
7.Whatisacidrain?Howisitharmfultotheenvironment?
A. When the PH of the rain water below 5.6 due to the
presenceofoxidesofsulphur&nitrogenandcarbondioxidein
theatmosphereiscalledacidrain.
Harmfuleffectsofacidrain
1. Itistoxictovegetationandaquaticlife
2. Itdamagesbuildingandstatesanddissolvesheavymetals
from soils, rocks and sedimentals. Tajmahal has been
damagedbyacidrain.
3. Theheavymetalionssuchascopper,leadandaluminum,
leached from the soil, enter well water and produced
varietyoftoxiceffects.
4. Acid rain also corrodes water pipes resulting in the
leachingasheavymetalssuchasiron,leadandcopperin
todrinkingwater.
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