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IMPORTANT QUESTIONS – 2015-16
SUBJECT : CHEMISTRY
CLASS XII
1
CHAPTER 1
SOLID STATE
1mk ‘Q’
Q1. ‘Crystalline Solids are anisotropic in nature’. What does this Statement
mean?
A 1. The statement means that some of the physical properties like electrical
resistance or refractive index of Crystalline Solids show different values when measured
along different directions in the same crystal.
Q 2.
Why does the presence of excess of lithium make LiCl crystals pink?
A 2. The presence of excess of lithium makes LiCl crystals pink due to e-strapped in
anionic vacancies (F centers). These electrons absorb some energy of the white light
giving pink colour to LiCl crystal.
Q3. What is meant by anti-ferromagnetism? What type of substances exhibit anti
ferromagnetism?
A3. Substance like MnO, MnO2 in which magnetic domains are oppositely oriented
and cancel out each other’s magnetic moment exhibit anti ferromagnetism. Magnetic.
Alignment of magnetic moments in antiferromagnetic substance:-
OR
Q3
What type of substance would make better magnets, ferromagnetic or
ferromagnetic?
A3. Ferromagnetic substance would make better magnets because when
ferromagnetic substance is placed in magnetic field all the domains get oriented in the
directionof the magnetic field and a strong magnetic effect is produced eg : Co, Ni
Q4. In a compound nitrogen atoms (N) make ccp and metal atoms (M) occupy
one third of the tetrahedral voids present. Determine the formula of the compound
formed by M & N?
A4.
Let the no of nitrogen atoms (N) be x
No of tetrahedral voids = 2x
No of metal atoms = 2/3 x.
Ratio of M:N = 2/3x : x
Therefore the formula of the compound is M2 N3
2
Q 5. Classify each of the following as being either a p type or n type semi
conductor?
(i)
(ii)
A 5.
Ge doped with In.
B doped with Si.
(i)
P type semiconductor because when group 14 element is doped with group
13 element, an electron deficit hole is created.
(ii)
n type semiconductor because when group 13 element is doped wih group
14 element , free electrons will become available.
Q 6.
Write a distinguishing feature between a metallic solid and an ionic solid?
A 6.
Ionic Solids
Metallic Solids
In solid state ionic solids are Metallic Solids are good electrical
electrical insulator :- ions are not conductors in solid state because of the
free to move. Eg NaCl, CuSO4 etc
presence of free electrons , Eg :- Copper,
Iron etc
2 mk
Q 1. Analysis shows that nickel oxide has the formula Ni 0.98 0 1.00 . What fractions
of nickel exist as Ni2+ and Ni3+ ions?
A1.
The formula of the oxide is Ni 0.98 0 1.00
Let the number of O2- ions be 100.
Then number of nickel ions = 98
Let the number of Ni2+ be x
Then number of Ni3+ = 98 - x
Since total charge on cations = total charge on anions.
X x (2) + (98-x) x (3) = 100 x 2
2x + 294 – 3x = 200
x = 94
% of Ni2+ =
98
94
x 100 = 96%
% of Ni3+ = 100 -96 = 4%
Q 2. A compound forms hcp structure. What is the total number of voids in 0.5
mol of it? How many of these are tetrahedral voids?
A2.
No of atoms = 0.5 mol = 0.5 x 6.022 x 1023
= 3.011 x 1023
No of octahedral voids = no of atoms in hcp structure
= 3.011 x 1023
3
No of tetrahedral voids = 2 x no of atoms in hcp structure
= 2 x 3.011 x 1023
= 6.022 x 1023
= 6.022 x 10 23 + 3.011 x 10 23
= 9.033 x 10 23
Total no of voids
Q3.
Examine the given defective crystal:A+
BA+
B-
BBA+
A+
BA+
B-
BA+
A+
BA+
B-
A+
Answer the following questions
(i)
What type of stoichiometric defect is shown by the crystal? (1/2)
(ii)
How is the density of crystal affected by this defect?
(iii)
What type of ionic substance show such defect?
(1/2)
(1)
A 3. (i)
Schotty defect is shown by the crystals, since equal number of cations and
anions are missing from the crystal lattice.
(ii)
Due to this defect, the density of the crystal decreases.
(iii)
This defect is shown by those ionic substance in which cations and anions
are of almost similar size eg :- NaCl, KCl etc.
Q 4. If NaCl crystals are doped with 2 x 10
cation vacancies per mole?
-3
mol percent of SrCl2 ,calculate the
A 4. Doping of NaCl with 2 x 10 -3mol percent of SrCl2 means 100 moles of NaCl is
doped with 2 x 10 -3 mole of SrCl2 or 1 mole of NaCl is doped with 2 x 10-5 mole of SrCl2
Each Sr2+ will occupy the place of Na+ and displace one Na+ from crystal lattice to
create cation vacancies.
Cation vacancies = Number of Sr 2+ ion added.
= 2 x 10 -5 mol = 2 x 10 -5 x 6.022 x 10 23. = 12.046 x 10 18 mol -1
4
Q 5.
Calculate the packing efficiency of a metal crystal for a simple cubic lattice?
A 5.
Packing efficiency in simple cubic lattice
rr
r
a
Volume of one atom x 100
Volume of cubic unit cell (a3)
Since a= 2r for simple cubic
Vol of one atom =4/3π r3
=
4/3 π r3 x 100
(2 r)3
=
4 x 3.14 x r3 x 100
3 x 8 x r3
=
52.36% ≈ 52.4%
Q 6 (a) In reference to a crystal structure, explain the meaning of coordination
number?
(b) What is the number of atoms in a unit cell of
(i)
Face centered cubic structure?
(ii)
Body centered cubic structure?
A 6 (a)The number of nearest neighbours of any constituted particle in the crystal
lattice is called its coordination number.
(b)
Number of atoms in a unit cell of
(i)
Face centered cubic cell structure is 4.
(ii)
Body centered cubic structure is 2.
5
3mk
Q1.
In terms of band theory.What is the difference?
(a)
between a conductor and an insulator.
(b)
between a conductor and a semi conductor.
Empty conduction band
Filled (valence band)
_____
Partially
Over lappling band
filled band
Conductors
A1
(a)
In conductors, energy band is partially filled or it overlaps with a higher
energy unoccupied conduction band. Due to this electrons can flow easily under
an applied electric field and show conductivity
In insulators, the gap between filled valence band and the next higher unoccupied
band is large, hence electrons cannot jump to it and substance has very small
conductivity and behaves as an insulator.
Empty band
(conduction band)
Filled band
(Valence band)
(b)
In semiconductors the gap between valence band and conduction band is
small. Therefore some electrons may jump from valence band to conduction band
and show some conductivity eg ‘Si’ and ‘Ge’
6
Q2 )
A2)
Aluminum crystallizes in a cubic close packed structure. Its metallic radius
is 125 pm.
(a)
What is the length of the side of the unit cell?
(b)
How many unit cells are there in 1 cm3 of aluminum?
(a)
In a cubic close packed structure :4 r = √2
Given r = 125 x 10 -12 m
a=
√
or 2 √2 r
a = 125 x 10 -12 x 2 x 1.414 m = 354 pm.
(c)
a3 = (354)3 x (10-12) 3
= 44.21 x 10-30 m 3
No of unit cells in 1 cm3
= Total Volume
=
Vol of one unit cell
10-6
44.21 x 10-30
= 2.261x 10 22 unit cells.
Q3
How will you distinguish between the following pairs of terms :(a)
Hexagonal close packing and Cubic close packing?
(b)
Crystal lattice and unit cell?
(c)
Tetrahedral void and Octahedral void?
Hexagonal close packing
AB AB—type packing is called
hexagonal close packing. i.e
spheres of the third layer are
exactly aligned with those of the
first layer . eg : ‘Mg’ and ‘Zn’
Cubic close packing
ABC ABC—type packing is called
cubic close packing. i.e spheres of
the third layer are not aligned with
those of 1st or 2nd. When 4th layer is
placed its spheres are aligned with
those of 1st layer. eg :- ‘Cu’ and ‘Ag’
Crystal lattice
Unit cell
The
three
dimensional The smallest portion of crystal lattice
arrangement
of
constituent which when repeated in different
particles in the space which directions, generates the entire lattice.
represents how the constituent
particles
(atoms,
ions
or
molecules) are arranged in a
crystal
7
Tetrahedral void
Octahedral void
A tetrahedral void is surrounded An octahedral void is surrounded by
by 4 spheres which lie at the six spheres and formed by a
vertices of a regular tetrahedron. combination of 2 triangular voids of the
1st and 2nd layer
There are 2 tetrahedral voids per There is one octahedral void per atom
atom in a crystal
in a crystal
Q4
Account for the following :(i)
Table salt, NaCl sometimes appears yellow in colour.
(ii)
FeO(s) is not formed in stoichiometric composition.
(iii) Some of the very old glass objects appear slightly milky instead of
being transparent.
A4.
(i)
Yellow colour in NaCl is due to metal excess defect due to which unpaired
electrons occupy anionic vacancies. These sites are called F centers. These
electrons absorb energy from the visible region and transmits yellow colour.
(ii)
In the crystal of FeO, some of the Fe2+ cations are replaced by Fe3+ ions.
Three Fe2+ ions are replaced by two Fe3+ ions to make up for the loss of positive
charge. Thus there would be less amount of metal as compared to stoichiometric
properties.
(iii)
Very old glass objects become slightly milky, because of heating during the
day & cooling at nights i.e annealing. Over a number of years, glass acquires
some crystalline character.
Q5.
The density of copper is 8.95 gcm-3. It has a face centered cubic structure.
What is the radius of copper atom? (Atomic mass Cu = 63.5 g mol -1 . NA =
6.022 x 1023 mol -1) ?
A5.
Mass per unit cell
=
Atomic mass of Cu x 4
NA
=
______63.5__x 4______
6.022 x 1023
=
4.22 x 10 -22 g
Volume of unit cell =
Edge
__Mass__
Density
=
4.22 x 10 -22
8.95
=
4.7 x 10-23 cm3
=
(Volume )1/3
=
(4.7 x10 -23)1/3
=
3.61 x 10-8 cm or 361pm
8
For fcc, r
=
r
=
r
=
___a___
2√2
.
128 pm
9
CHAPTER 2
SOLUTIONS
1 mk
Q 1.
Define an ideal solution and write one of its characteristics?
A1
The solutions which obey Raoult’s law over the entire range of concentration are
known as ideal solutions. i.e PA = PoA
For ideal solutions
A
and PB = PoB
H mix = 0 and
B.
V mix = 0.
eg ;- Solution of n – hexane and n- heptane
Q 2.
What is meant by reverse osmosis ?
A2
The process in which the solvent flows from the solution into the pure solvent
through the semi permeable membrane when a pressure higher than the osmotic
pressure is applied on the solution is called reverse osmosis.
Application :- The technique is used in the desalination of sea water.
Q3.
Define mole fraction.
A3
The mole fraction of a component is the ratio of the number of moles of that
component to the total number of moles of all the components present in the solution.
For a binary solution consisting of 2 components A and B if nAis the number of
moles of A and nB is the number of moles of B then.
A=
Q 4.
__nA___
nA + nB
B
= _ nB___
nA + nB
Define the term azeotrope?
A4
The constant boiling mixture which distill out unchanged in their composition are
called azeotropes.
Eg :- A mixture of ethanol and water containing 95.4% of ethanol forms an
azeotrope with boiling point 351.15 K.
Q5. Explain boiling point elevation constant for a solvent / Define ebullioscopic
constant?
A5
Since
Tb = Kb.m where m is molality
When m = 1
Kb =
Tb
10
Therefore Ebullioscopic constant is defined as the elevation in boiling point of a
solution when 1 mole of a solute is dissolved in 1 Kg of solvent .
Q6
What are isotonic solutions ?
A6
The solutions of equimolar concentrations having same osmotic pressure at given
temperature are called isotonic solutions.
eg :- A 0.9% solution of pure NaCl is isotonic with human red blood cells.
Q7
Explain why aquatic species are more comfortable in cold water rather than
in warm water?
A7
Aquatic species need dissolved oxygen for breathing. As solubility of gases
decrease with increase of temperature, less oxygen is available in summers in lakes.
Hence they feel more comfortable in winter when the solubility is higher.
OR
Q7
Why do gases nearly always tend to be less soluble in liquid as the
temperature is raised?
A7
Dissolution of gas in liquid is an exothermic process. Gas + Solvent
Solution + heat.
Therefore as per Lechatlier’s principle if the temperature is increased, equilibrium
shifts backward i.e the solubility decreases.
2 mk
Q1
How is the vapour pressure of a solvent affected when a non volatile solute
is dissolved in it?
A1
When a non volatile solute is added to a solvent, its vapour pressure decreases
because some of the surface sites are occupied by solute molecules. Thus less space is
available for the solvent molecules to vaporize.
Q2. The depression in freezing point of water observed for the same molar
concentrations of acetic acid, trichloro acetic acid and trifluoroacetic acid
increases in the order as stated above. Explain?
A2
As depression in f.pt ( Tf) is dependent on degree of dissociation (α) and fluorine
exerts the highest – I effect. So trifluoro acetic acid is the strongest acid and ionizes to a
greater extent while acetic acid ionises to the minimum extent. Thus greater the number
of ions produced, greater is the depression in freezing point.
11
Q3. Differentiate between molarity and molality for a solution. How does a
change in temperature influence their values?
Or
Q3. State the main advantage of molality over molarity as the unit of
concentration?
Molarity
It is defined as the number of moles of
solute dissolved in one litre of the solution
Mathematically
Molarity (M) =No of moles of solute x 1000
Vol of Solution (ml)
Molality
It is defined as the number of moles of
solute dissolved in 1 Kg of the solvent
Mathematically
Molality (M) =No of moles of solute x 1000
Mass of Solvent (g)
It decreases with increase in temperature It does not change with change in
(as V α T)
temperature
Since molality does not change with a change in temperature therefore it is a
better method to express the concentration of a solution.
Q4. What is meant by colligative property. List any four factors on which
colligative properties of a solution depend?
A4. The properties of solutions which depend upon the number of solute particles and
not upon the nature of the solute are known as colligative properties eg :- Osmotic
pressure.
Factors :(i)
(ii)
(iii)
(iv)
Number of particles of solute.
Concentration of solution.
Temperature.
Association or dissociation of solute.
An aqueous solution of sodium chloride freezes below 273K. Explain the
lowering in freezing point of water with the help of a suitable diagram?
Q5.
A5. Freezing point of a substance is the temperature
at which solid and liquid phases of a substance coexist
i.e they have the same vapour pressure. As the vapour
pressure of the solution is less than that of pure solvent
For solution its vapour pressure will become equal to
that of a solid solvent only at a lower temperature.
liquid solvent
Vapo
ur …………………
pres
sure
solid solvent
Tf
Tfo
Temperature
solution
12
3 mk
Q1. A sample of drinking water was found to be severely contaminated with
chloroform CHCl3 , supposed to be carcinogen. The level of contamination was
15ppm (by mass)?
A1.
(i)
Express this in percent by mass.
(ii)
Determine the molality of CHCl3 in water sample.
15 ppm means 15 parts in 106 parts by mass in the solution.
i.e
106 parts by mass of solution= 15 parts of solute.
or 100 parts (Mass %) =
Mass of solvent
15 x 100
106
=
15 x 10 -4
=
(106 -15) g
106 g
=
nB
wA
≈
Therefore Molality
where nB = no of moles of solute
wA= Mass of the solvent
Molality
=
15/119.5X1000
106
= 1.25 x 10-4m
Q2
State Raoult’s law for a solution containing volatile components. How does
Raoult’s law becomes as special case of Henry’s law?
A2
For a solution of volatile liquids, the partial vapour pressure of each component in
the solution is directly proportional to its mole fraction?
i.e
PAα
A
or
PA
=
P0A
A
and
PBα
B
or
PB
=
P0B
B
Where PA and PB are the partial pressure of A and B, P0A , P0B are the vapour
pressure of pure component and
Aand
B
are their mole fractions.
If gas is the solute and liquid is the solvent then according to Henry’s law
PA = KH
A
Thus Raoult’s laws and Henry’s law become identical except that their
proportionality constants are different.
13
Q3
Some ethylene glycol is added to your car’s cooling system along with 5kg
of water. If the freezing point of water-glycol solution is -150c, what is the boiling
point of the solution?
Kb = 0.52 K Kgmol-1 and Kf = 1.86 K Kg mol-1
A3
Tf = 15oc Kf = 1.86 k/m
Molality =
Tf
Kf
Tb
=
15/1.86 = 8.06 m
=
=
=
Kb x m
0.52 x 8.06
4.190c
Boiling point of pure water = 100oc
Therefore Tb =
Tb+
Tbo
Tb
= 100 +4.19
= 104.19oC
Q4
Assuming complete dissociation. Calculate the expected freezing point of a
solution prepared by dissolving 6g of Glauber’s salt, Na2SO4.10 H2O in 0.1 Kg mol-1
of water.Kf of water = 1.86 K Kg mol -1.
A4
Kf
WB
MB
=
=
=
1.86 K Kgmol-1
6g
W A = 0.1 Kg mol-1
322 g mol-1
Since there is a complete dissociation
Therefore i = 3
Tf =
i K f WB
MB x W A
=
3 x 1.86 x 6
322 x 0.1
=
1.04o
14
Q5. Calculate the mass of a nonvolatile solute (molar mass 40g mol-1) which
should be dissolved in 114 g octane to reduce its vapour its pressure to 80%?
A5.
Ps=80% of Po=0.80 Po
No of moles of solute
=
W
40
No of moles of solvent
(octane)
=
114
114
Po– Ps
Po
=
XB
Po–0.8 Po
Po
=
W /40
W/40 + 1
therefore
or
0.2
W +1
40
0.8 W
40
or
=
=
1
W
40
=0.2
W = 10g
Q6
19.5g of CH2FCOOH is dissolved in 500g of water. The depression in
freezing point observed is 1oc. Calculate the Van’t Hoff factor and dissociation
constant of fluoroacetic acid .Kffor water is1.86 K Kgmol-1.
A6
W B = 19.5g W A = 500g Kf= 1.86K Kgmol-1
(ATf) obs = 1o.
MB (obs) =1000 Kf W B
W A∆Tf
=
1000 X1.86 X 19.5 =72.54 g mol-1
500 X 1
MB (cal) = 14 + 19 + 45 = 78 g mol-1
L = M(cal) = _78__
MB (Obs)
72.54
=
1.0753
15
Calculation of dissociation const (K)
Ka = _Cα2
1-α
CH2FCOO -+ H+
CH2 FCOOH
Initial C MOL L-1
C(1-α)
i = C(1+α)
C
O
Cα
=
=
O
Cα
1 + α or α = i - 1
1.0753 – 1 = 0.0753
Taking volume of solution as 500 ml
C =19.5 x_1_x 1000
78 500
Ka = Cα2
1-α
= 0.5 M
= 0.5 x (0.0753)2
1-0.0753
Ka = 3.07 x 10-3
Q7. Non ideal solutions exhibit either positive or negative deviations from
Raoult’s law. What are these deviations and why are they caused? Explain with
one example for each type?
A7
For non ideal solutions, vapour pressure is either higher or lower than that
predicted by Raoult’s law. If it is higher the solution exhibits positive deviation and if it is
lower it exhibits negative deviation from Raoult’s law.
Positive Deviation
When solute – solvent interactions are
weaker than solute – solute or solvent
solvent interactions, vapour pressure
increases which result in positive
deviation.
Eg :- Ethanol +Acetone.
for a solution showing +ve deviation
PA> PoA XA and PB > PoB XB
H mix = +ve
Vmix = +ve
Negative Deviation
When solute solvent interactions are
stronger than solute – solute or solvent –
solvent interactions, vapour pressure
decreases which result in negative
deviation.
eg :- Chloroform + Acetone.
for a solution showing –ve deviation
PA< PoA XA PB< P0B XB
H mix = - ve
Vmix = - ve
For plots of non ideal solution showing +ve and –ve deviation refer NCERT pg 56
16
5 mk
Q1. (i)
What is Van’t Hoff factor? What types of values can it have if in
forming the solution the solute molecules undergo
(ii)
(a)
Dissociation?
(b)
Association?
How many ml of 0.1M HCl solution are required to react completely with 1g
of a mixture of Na2CO3 containing equimolar amounts of both ? (Molar
Mass Na2CO3 = 106g and NaHC03 = 84g)
A1. (i)
Van’t Hoff factor (i) is defined as the ratio of the experimental value of
colligative property to the calculated value of colligativeproperty .
i
Also, i=
=
Observed Colligative property
Calculated Colligative property
Total number of moles of particles after dissociation/ association
Number of moles of particles before association/ dissociation
Therefore , For (a) dissociation i> 1
And (b) Association i< 1
(ii)
Let x g of Na2CO3 be present in the mixture
So amt of Na HCO3 = 1 –x
Moles of Na2CO3 in x g = x_
106
Moles of NaHCO3 in (1-x) = _1-x_
84
As mixture contains equimolar amounts of the two.
x_ =1-x_
106
84
106 – 106 x = 84x
X = _106__ = 0.558g
190
Moles of Na2 CO3 = __0.558__
106
= 0.00526
Moles of NaHCO3 = 1- 0.558 =0.442
84
84
= 0.00526
To calculate the moles of HCl required.
Na2CO3+ 2 HCl
2 NaCl + H2O + CO2
17
NaHCO3+HCl
NaCl + H2O + CO2
1 mole of Na2CO3 requires 2 moles of HCl
Therefore 0.00526 mole of Na2CO3 requires 2 x 0.00526 moles or 0.0152
moles of HCl.
1 mole of NaHCO3requires 1 mole of HCl
Therefore 0.00526 mole of NaHCO3 requires 0.00526 moles of HCl.
Total moles of HCl required = 0.01052 + 0.00526 = 0.01578 moles
To calculate volume of 0.1M HCl
0.1 mole of 0.1M HCl are present m 1000ml HCl
Therefore 0.01578 mole of 0.1M HCl will be present in
1000 x 0.01578
0.1
= 157.8 ml
Q2. (i)
The molecular masses of polymers are determined by osmotic
pressure method and not by measuring other colligative properties. Give two
reasons?
(ii)
At 300k, 36g of glucose C6H12O6 present per litre in its solution has a
pressure pf 4.98 bar. If the osmotic pressure of another glucose solution is
1.52 bar at the same temperature, calculate the concentration of the other
solution?
A2. (i)
because :-
The osmotic pressure method has the advantage over other methods
(a)
It uses molarities instead of molalities and it can be measured at
room temperature.
(b)
Its magnitude is large as compared to other colligative properties.
(ii)
π1 = C1 RT, π2 = C2 RT
Therefore _π1 = C1
π 2 C2
or
4.98
1.52
= 36/180
C2
or
C2
= 0.061 moll-1
= 0.061 x 180 gl-1
= 10.98 gl-1
18
Q3
(i)
Define the terms osmosis and osmotic pressure?
(ii)
An aqueous solution containing 12.48g of barium chloride in 1.0 kg of
water boils at 373.0832 k. Calculate the degree of dissociation of barium
chloride?
(Given Kb for H2O = 0.52 Km-1 molar mass of Ba Cl2 = 208.34 gmol-1)
A3. (i)
The net spontaneous flow of the solvent molecules from the solvent to the
solution or from a less concentrated solution to a more concentrated solution through a
semipermeable membrane is called osmosis.
Osmotic pressure :- The minimum excess pressure that has to be applied on
the solution to prevent the entry of the solvent into the solution through the
semipermeable membrane is called osmotic pressure.
(iii)
(iv)
Given W 2 = 12.48g , W1 = 1 Kg = 1000g
Tb (solution) = 373.0832 K
Kb for H2O = 0.52 Km-1 and M2 = (Ba Cl2) = 208.34
Tb= Tb - Tbo = 373.0832 – 373 = 0.0832 K
M2 (Observed) = Kb x W 2 x 1000
Tb x W 1
= 0.52 x 12.48 x 1000
0.0832 x 1000
= 78 g mol-1
For BaCl2
i
= M2 (calculated) =
M2 (observed)
m
= 3 as it gives 3 ions on dissociation
α
208.34 = 2.67
78
= i-1
= 2.67 -1 = 1.67 = 0.835
m-1
3-1
2
= 83.5%
Q4. (a)
What type of deviation is shown by a mixture of ethanol and acetone?
Give reason.
(b)
What do you expect to happen when RBC’s are placed in
(i)
1% NaCl solution (ii) 0.5 % NaCl solution
(c)
Calculate the molarity of 68% (w/w) solution of nitric acid , if the
density of the solution is 1.504 g ml-1
19
A4. (a)
Ethanol and acetone shows positive deviation because on mixing the two
the forces of attraction decreases and the vapour pressure increases.
(b)
(c)
Since RBC’s are isotonic with 0.9% NaCl solution therefore
(i)
In 1% solution of NaCl they will shrink due to plasmolysis
(ii)
In 0.5% solution of NaCl they will swell or may even burst.
Molarity
=
% x density x 10
Molar mass
Mass %
=
68 , d = 1.504 Molar mass of HNO3 = 63 gmol-1
Therefore Molarity = 68 x 1.504 x 10
63
= 16.23 M
Q5.
(a)
State Henry’s law and mention two of its important applications
(b)
The partial pressure of ethane over a saturated solution containing
6.56 x 10-2g of ethane is 1bar. If the solution were to contain 5 x 10-2 g of
ethane, then what will be the partial pressure of the gas.
A5. (a)
Henry’s law states that at a constant temperature , the solubility
in a liquid is directly proportional to the pressure of the gas
of a gas
P = KH
Applications :-
To increase thje solubility of CO2 in soda water, the bottle is sealed under high
pressure
-
There is a low concentration of oxygen in the blood and tissues of the people
living at high altitudes due to which they feel weak and are unable to think
clearly (anoxia).
-
(b) M = KH x P
For the 1st case 6.56 x 10-2 = KH x 1 bar or KH = 6.56 x 10-2 g bar -1
In the 2nd case 5 x 10 -2 = (6.56 x 10-2) x P
or P = 5 x 10 -2
6.56 x 10 -2
P = 0.762 bar
20
CHAPTER 3
ELECTRO CHEMISTRY
01 MARK
Q1.
Why does the conductivity of a solution decrease with dilution?
A1. Conductivity of a solution is the conductance of ions present in a unit volume of
the solution. On dilution the number of ions per unit volume decreases. Hence, the
conductivity decreases.
Q2)
What does the negative sign the expression E0Zn2+ / Zn = -0.76 V mean?
A2) Negative sign shows that zinc is more reactive than hydrogen. This means that
when zinc electrode is connected to SHE, Zn will be oxidized to Zn2+ and H+ will be
reduced to H2.
Q3)
State Kohlrausch laws of independent migration of ions?
A3) It states that “ At infinite dilution, when the dissociation of the electrolyte is
complete, each ion makes a definite contribution to the total molar conductivity of the
electrolyte irrespective of the nature of the other ion with which it is associated.
Q4)
Write two advantages of H2O2 fuel cell over ordinary cell?
or
What advantage do the fuel cells have over primary and secondary batteries?
or
Name the type of cell which was used in Apollo space programme for providing
electrical power?
A4) (a)
Fuel cells can be run continuously so long as the reactants are supplied,
primary batteries become dead and secondary batteries take a long time for recharging .
(b)
Fuel cells do not cause any pollution
H2–O2 fuel cell was used in Apollo Space programme for providing electrical
power.
Q5) Express the relation among the conductivity of a solution in the cell, the cell
constant and the resistance of solution in the cell?
A5)
K= 1xl
R A
Where K = Conductivity
l/A
= Cell constant
R
= Resistance.
21
Q6)
What is electrode potential?
A6) It is defined as the tendency of an electrode to lose or gain electrons when it is in
contact with solution of its own ions.
Q7)
Is it safe to stir AgNO3 solution with a copper spoon? Why or Why not ?
Given :- Eo Ag+/Ag = 0.8 volt and Eo cu2+ /Cu = 0.34 volt
A7) The given values of reduction potentials show that Cu is more reactive than Ag. i.e
Cu reacts with AgNO3 solution. Hence it is not safe to stir AgNO3 solution with copper
spoon.
Q8)
What is the role of ZnCl2 in a dry cell?
A8) ZnCl2 combines with the NH3 produced to form the complex salt [Zn (NH3) 2 Cl2] as
otherwise the pressure developed due to NH3 would crack the seal of the cell.
Q9)
State faraday’s first law of electrolysis ?
A9) The mass of any substance (w) deposited or liberated at any electrode is directly
proportional to the quantity of electricity (Q) passed through the electrolyte or W α Q
W = ZQ
Z = electrochemical equivalent
02 MARKS
Q1. Calculate the potential of hydrogen electrode in contact with a solution whose pH
is 10.
A1.
H+ + e-
½ H2
Applying Nernst eqn
EH+/1/2 H2 = EoH+ ½ H2 -0.0591 log _1_
n
[H+]
= 0 – 0.0591log _1_
1
10-10
because pH = 10 means [H+] = 10 -10 M.
= - 0 .0591 x -10
= + 0.591V
Q2. Mention the reactions occurring at (i) Anode (ii) Cathode, during the working of a
mercury cell. Why does the voltage of a mercury cell remain constant during its
operation?
22
A2.
At Anode :Zn (Hg) + 2OH-
ZnO (s) + H2O + 2e-
At Cathode
HgO + H2O + 2e-
Hg (l) + 2OH-
______________________________
Zn(Hg) + HgO (s)
ZnO (s) + Hg(l)
The voltage of a mercury cell remains constant during its life as the overall
reaction does not involve any ion in solution whose concentration can change
during its life .
Q3. Explain why electrolysis of aqueous solution of NaCl gives H2 at cathode & Cl2 at
anode? Write overall reaction?
Given EoNa+ /Na = - 2.71 V
Eo H2O/H2 = -0.83 V
Eo Cl2/Cl- = + 1.36 V
Eo H+/H2/H2O = + 1.23V
A3. At Cathode the reaction with higher value of Eois preferred, therefore H2 is
released at cathode, which is produced by dissociation of H2O.
Therefore at Cathode H2O (l) + e-
½ H2 (g) + OH – (aq)
At anode the reaction with lower value Eo is preferred, therefore water should
oxidised to O2 but due to over potential of oxygen, chlorine is released at anode.
At anode :- Cl- (aq)
½ Cl2 (g) + e-
Net reactions :NaCl (aq) + H2 O (l)
Na+ (aq) + OH- (aq) + ½ H2 (g) + ½ Cl2 (g)
Q4.
Calculate
G for the reaction, Mg (s) + Cu2+ (aq)
Given Eo cell = + 2.71V
1F = 96500 (Cmol -1)
A4.
For the reaction
Mg(s) + Cu2+ (aq)
n=2
Go= -nFEo cell
= -2 x 96500 x 2.71
= - 523030 J mol-1
Mg 2+ (g) + Cu (s)
Mg2+ (aq) + Cu (s)
23
Q5.
Calculate the degree of dissociation of acetic acid at 298K given that
λmo (CH3 COOH) = 11.7 Scm2 mol -1
λmo (CH3 COO-) = 40.9 Scm2 mol -1
λmo (H+)
A5.
= 349.1 Scm2 mol -1
Degree of dissociation
α = λm
λ 0m
λ Mo
= λMo (CH3 COO-) + λM (H+)
= 40.9 + 349.1
= 390 Scm2 mol-1
α = λm
λm
= 11.7
390
o
α = 3 x 10-2
Q6. A solution of Ni(NO3)2 is electrolyzed between platinum electrodes using a current
of 5 A for 20 min. What mass of nickel will be deposited at the cathode?
Given : Atomic mass of Ni = 58.7 gmol-1 1F = 96,500 (mol-1)
A6.
Given :- I = 5A
Time (t) = 20 x 60 = 1200s
Q = I x t = 5 x 1200 = 6000C.
M=ZxIxt
M = Eq. wtx I x t
96,500
M = __58.7__ x 6000 = 1.82g
2 x 96500
Q7.
(a)
Define electrochemical series .
(b)
Given that the standard electrode potentials (Eo) metals are :K+/K = -2.93V, Ag+/Ag = 0.8V, Cu 2+/Cu = 0.34V, Mg 2+/Mg = -2.37V,
Cr 3+/Cr = - 0.74, Fe2+/Fe = -0.44V.
Arrange these metals in an increasing order of their reducing power.
24
A7.
(a)
The arrangement of the various electrodes in order of their increasing
values of standard reduction potentials is called electrochemical series.
(b)
Greater the negative value of the standard electrode potential (Eo) greater
is the reducing power of the electrode . Thus the increasing order of reducing
power is Ag+/Ag < Cu 2+/Cu < Fe2+/Fe < Cr 3+/Cr < Mg 2+/Mg < K+/K
03 MARKS
Q1. Three electrolytic cells A, B,C containing solutions of ZnSO4 , AgNO3 and CuSO4
respectively are connected in series . A steady current of 1.5A was passed through them
until 1.45 g of Ag were deposited at the cathode of cell B. How long did the current flow?
What mass of copper and what mass of zinc were deposited in the concerned cells?
(Atomic mass of Ag = 108 , Zn = 65.4 u and Cu = 63.5u)
A1.
Ag+ + e(1F)
Ag
108g
108 g of Ag is deposited by 96500C of charge
Therefore 1.45g of Ag will deposited by
96500 x 1.45 =1295.6 C
108
Q =It
t
= 1295.6
1.5
Cu2+ + 2e-
=
863.73 s
Cu
2 x 96500 c deposits Cu = 63.5 g
Therefore 1295.6c charge will deposit
Cu
=
63.5 x 1295.6
2 x 96500
=
0.426g
Zn2+ + 2e-
Zn
Amt of zinc deposited =
=
65.4 x 1295.6
2 x 96,500
0.439g
25
Q2. What type of battery is lead storage battery? Write the anode and cathode
reactions and the overall cell reaction occurring in the operation of a lead storage
battery?
A2. Lead storage battery is a secondary cell i.e it can be recharged by passing direct
current through it and can be reused.
Anode :-
Spongy lead.
Cathode :-
Grid of lead packed with PbO2
Electrolyte :- 38% H2 SO4
At Anode :- Pb (s) + SO42- (aq)
Pb SO4 + 2e-
At Cathode :- Pb O2 (s) + SO42- + 4H+ (aq) + 2e-
Pb SO4 (s) + 2H2O (l)
Overall cell reaction
Pb (s) + PbO2 (s) + 4H+ + 2SO42-
2 PbSO4 (s) + 2H2O(l)
Q3. The resistance of 0.01M NaCl solution at 25oc is 200 Ω. The cell constant of the
conductivity cell used is unity. Calculate the molar conductivity of the solution.
A3.
Given Resistance ® = 200 Ω
Molarity of NaCl solution = 0.01M
Cell constant (l/A) = 1cm-1
Conductivity (K) = _1_ x _l_
R
A
= __1__
200
= 5 x 10-3 Ω -1 cm-1
Molar conductivity λM = 1000K
Molarity
= (5 x10-3) x 1000
0.01
= 500 Scm2 mol-1
26
Q4. A strip of nickel metal is placed in a 1 molar solution of Ni(NO3)2 and a strip of
silver metal is placed in one molar solution of AgNO3. An electrochemical cell is created
when the two solutions are connected by a salt bridge and the two strips are connected
by wires to a voltameter?
(i)
Write the balanced equations for the overall reaction occurring in the cell
and calculate the cell potential.
(ii)
Calculate the cell potential, E at 25oC for the cell, if the initial concentration
of Ni(NO3)2 is 0.100 molar and initial concentration of AgNO3 is 1.00 molar.
[Eo Ni2+/Ni = - 0.25V, Eo Ag+/Ag = 0.8V log 10-1 = -1]
A4.
Ni2+ (aq) + 2e-
At anode :- Ni (s)
At Cathode :- 2Ag+ (aq) + 2e-
2Ag (s)
Overall rn = Ni (s) + 2 Ag+ (aq)
Ni2+ (aq) + 2 Ag(s)
Eo cell = Eo cathode – Eo Anode
= + 0.8V – (-0.25V)
= 1.05V
Ecell
= Eo cell – 0.0591 log [Ni2+]
2
[Ag+]2
Eo cell – 0.0591
2
=
log 0.1
(1)2
=
1.05 – 0.0591 x -1
2
1.05 + 0.0295
=
1.0795V
Q5. Determine the values of equilibrium constant (Kc) and
reaction.
Ni (s) + 2 Ag+ (aq)
Ni2+ (aq) + 2Ag (s)
Eo= 1.05V (1F = 96500 cmol-1)
A5.
Given Eo = 1.05V
n =2
Therefore
Go = -nf Eo
Go = -2 x 96500 x 1.05 = -202.65 KJ mol-1
-nfEo = -2.303RT log Kc (because
Go = -2.303 RT log Kc)
G0 for the following
27
Log Kc =
nEo
0.0591
=
2 x 1.05 = 35.5329
0.0591
Kc
=
antilog 35.5329
Kc
=
3.411 x1035
Q6. Set up Nernst equation for the standard dry cell. Using this equation show that the
voltage of a dry cell has to decrease with use?
A6. Dry cell (Leclanche cell) consists of a zinc container which acts as anode and
cathode is a graphite rod surrounded by powdered MnO2 and ‘C’ The space between the
electrodes is filled by a moist paste of NH4Cl and ZnCl2.
Zn 2+ (aq) + 2e-
At anode :- Zn (s)
At cathode :- 2MnO2(S) + 2NH4+(aq) + 2e-
Mn2O3 (s) + 2NH3 (g) + H2O
NH3 formed combines immediately with Zn2+ ions to form complex ion [Zn(NH3)2] 2+
Zn 2+(aq) + 2NH3(g)
[ Zn (NH3)2] 2+ (aq)
Overall reaction :Zn2+ (aq) + 2NH4+ (aq)+ 2MnO2(s)
[ Zn(NH3)2] 2+ (aq) + Mn2O3 (s) + H2O
Nernst equation for dry cell
E cell = Eocell – 2.303 RT
nF
log [Zn (NH3)2]2+
[NH4+]2
Due to the presence of ions in the overall reaction, its voltage decreases with time.
Q 7. What is the relationship between Gibbs free energy of the cell reaction in a
galvanic cell and emf of the cell. When will the maximum work be obtained from a
galvanic cell?
A7.
In a galvanic cell –
Electrical work done = Electrical energy produced =
= Quantity of electricity flowing x EMF
for n moles of electrons transferred in any cell reaction. The
quantity of electricity flowing = nf faradays.
Therefore Electrical work done = nF Ecell.
Also electrical work done = Decrease in free energy
Therefore -
`
r
G = nF Ecell. Or -
Go = nF E o cell
28
ΔrGo = –2.303 RT log Kc
Hence, knowing Eocell, ΔG o can be calculated which in turn can be used for the
calculation of the equilibrium constant Kc.
–ΔG = Wmax. Hence, the decrease in free energy is equal to the maximum work that can
be obtained from the cell.
5 MARKS
Q1. (a)
The chemistry of corrosion of iron is essentially an electrochemical
phenomenon. Explain the reactions occurring during the corrosion of iron in the
atmosphere?
(b)
Give reasons :(i)
Alkaline medium inhibits the rusting of iron.
(ii)
Rusting of iron pipe can be prevented by joining it with a piece of
magnesium.
A1. (a)
The chemistry of corrosion of iron is essentially an electrochemical
phenomenon. At a particular spot of an object made of iron, oxidation takes place and
that spot behaves as anode.
At anode
2 Fe(s)
2 Fe2+ + 4e- ;
EoFe2+/ Fe(s) = -0.44 V
Electron released at anodic spot go to another spot on the metal and
reduce oxygen in the presence of H+ ions (H+ ions obtained from H2CO3 which is
formed due to dissolution of CO2 from air into water). This spot behaves as
cathode for the reduction.
At cathode
O2 (g) + 4H+ (aq) + 4e-2H2O(l);
EoH+/O 2/H20 = 1.23V
Overall reaction
2Fe(s) + O2 (g) + 4H+ (aq)
2 Fe2+ (aq) + 2H2O (l) , E0cell = 1.67V
29
The ferrous ions are further oxidized to ferric ions by atmospheric oxygen .
Reaction is as follows :2 Fe2+ (aq) + 2H2O (l) + ½ O2(g)
Fe2O3 + x H2O
Fe2O3 (s) + 4H+ (aq)
Fe2O3 . xH2O
Hydrated ferric oxide
(b) (i)
Rusting of iron takes place in presence of H+ and as alkaline
medium neutralizes H+ , the rusting is inhibited.
(ii)
It is due to cathodic protection in which Mg metal is oxidized in
preference to iron and acts as anode.
Q2.
(i)
Predict the product of electrolysis in each of the following :(a)
An aqueous solution of AgNO3 with platinum electrodes.
(b)
An aqueous solution of H2SO4 with platinum electrodes.
(ii)
Estimate the maximum potential difference needed to reduce Al2O3 at
5000C . The Gibbs energy change for the decomposition reaction
4/3 Al +O2 is 960 kJ. (F = 96500 C mol-1)
2/3 Al2O3
A2.
(i) (a) Electrolysis of AgNO3 (aq) using Pt electrodes
Ag+ (aq) + NO-3 (aq)
AgNO3 (aq)
At cathode Ag+ (aq) + e-
Ag (s)
O2 (g) + 4H+ (aq) + 4e-
At Anode 2H2O (l)
(1 ½ )
(b) Electrolysis of H2SO4 (aq) using Pt electrodes.
2H2O (l)
2H2(g) + O2 (g)
At Cathode H2O + e-
O2 + 4H+ + 4e-
At Anode 2H2O
(iii)
½ H2 + OH-
Reaction involved
2/3 Al2O3
4/3 Al + O2
Here n = 4
Given ,
G = 960K J
-960 kJ = - 4 x 96500 x E0
G = - nFE0
(1 ½ )
30
E0 = ___960000 J____
4 x 96500
Q3.
=
2.48V ≈ 2.5V
(2)
(i)
Define molar conductivity of a solution and explain how molar conductivity
changes with change in concentration of solution for a weak and a strong
electrolyte.
(ii)
The resistance of a conductivity cell containing 0.001M KCI solution at 298
K is 1500Ω. What is the cell constant if the conductivity of 0.001M KCI solution at
298K is 0.146 x 10-3 S cm-1?
A3. Molar conductivity.
It is defined as the conductivity of the solution which
contains one mole of the electrolyte such that entire solution is in between the two
electrodes kept one centimeter apart.
Molar conductivity,
λm = __K___
C
where k = conductivity.
c is concentration of solution
Variation of conductivity and molar conductivity with concentration.
CH3COOH
(weak electrolyte)
λm
2
(S cm mol-1)
KCI
(strong electrolyte)
C1/2 (mol/L) ½
Conductivity and molar conductivity change with change in concentration of
electrolyte. Conductivity always decreases with decrease in concentration for both
weak as well as strong electrolytes.
But molar conductivity increases with decreases in concentration. For strong
electrolytes λm increases slowly with dilution but for weak electrolytes λm increases
steeply on dilution, especially near lower concentrations.
Molar conductivity increases with decrease in concentration because both the
number of ions as well as the mobility of increase with dilution.
(ii)
Given, conductivity k = 0.146 x 10-3 S cm-1
Resistance , R = 1500Ω
31
Therefore Cell constant , G* = K x R
= 0.146 x 10-3 x 1500
= 0.219 cm-1
Q4) (a)
What is standard hydrogen electrode? Give reactions that occur at this
electrode when it acts as a positive electrode in an electro chemical cell.
(b)
What is the function of salt bridge in an electrochemical cell.
(c)
Cu2+ + 2e-
Cu,
Eo= + 0.34 V
Ag + + e-
Ag,
Eo = + 0.80 V
(i)
Construct a galvanic cell using the above data.
(ii)
For what concentration of Ag+ ions will the emf of the cell be zero at
25oc , if the concentration of Cu2+ is 0.01 M?
A4)
(a)
It is a reference electrode used to measure the electrode potential of other
electrode. The electrode potential of SHE is taken as zero.
(b)
(c)
Salt bridge performs the following functions :(i)
It completes the inner circuit by flow of ions.
(ii)
It maintains the electrical neutrality in the solution of half cells.
(i)
Cu (s) I Cu2+ (aq) I I Ag + (aq) I Ag (s)
At Anode :
Cu 2+ (aq) + 2e-
Cu (s)
At Cathode 2 Ag+(aq) + 2eCu (s) + 2Ag+ (aq)
2Ag (s)
Cu2+ (aq) + 2Ag (s)
Thus , n = 2
E cell = EoAg+/Ag - EoCu2+ /Cu = + 0.80 V – 0.34V = 0.46V
Using Nernst equation to calculate concentration of Ag+
Ecell = Eo cell – 0.0591 log [ Cu2+]
2 [ Ag+] 2
or
0 = .46 V – 0.0591 log _0.01_
2
[ Ag+] 2
or
log 0.01_
[Ag+] 2
= +0.46 V x 2 =
0.0591
_0.92_
0.0591
= 15.567
32
Q5A)
= Antilog 15.567 = 3.690 x 1015
or
log 0.01_
[Ag+] 2
or
[Ag+] 2 = ___0.01___ =
3.688 x 1015
or
[Ag+] = 1.65 x 10-9 mol L-1
(i)
___1__ x 10-17 = 0.271 x 10—17 = 2.71 x 10-18
3.688
State two advantages of H2 –O2 fuel cell over ordinary cell.
(ii)
Silver is electrodeposited on a metallic vessel of total surface area
900 cm3 by passing a current of 0.5A for two hours. Calculate the thickness
of silver deposited. (Given : Density of silver = 10.5 g cm-3, Atomic mass of
silver = 108 u, 1 F = 96500 C mol-1]
A5A)
(i)
It has high efficiency and is eco –friendly.
(ii)
Use the following relation to calculate mass of silver deposited
m=ZxIxt
or
m = _108 x 0.5 x 2 x 60 x 60 = 108 x 5 x 2 x 6 x 6 = 4.03g
96500
965x 10
Also, Mass = Volume X Density
or
4.03 g = Volume x 10.5g cm-3
or
Volume = 4.03 cm3 = 0.38 cm3
10.5
Volume = Area x Thickness
Q5B)
or
0.38 cm3 = 900 cm2 x Thickness
or
Thickness = 0.38 cm3 = 4.22 x 10-4 cm.
900 cm2
Give reasons for the following :(i)
Aluminium metal cannot be produced by the electrolysis of aqueous
solution of aluminium.
(ii)
A 5B)
Equilibrium constant is related to Eo cell but not to E cell
(i)
It is because aluminium metal is more reactive that hydrogen and
will react with H2O.
(ii)
When equilibrium is reached in the cell reaction. E
equal to zero . However Eo cell is a constant quantity .
Therefore, Eocell =RT ln Kc
nF
cell
becomes
33
CHAPTER – 4
CHEMICAL KINETICS
01 MARK
Q1)
State a condition under which a bimolecular reaction is kinetically first order?
A1) Bimolecular reaction becomes kinetically first order when one of the reactants is in
excess.
Q2)
Write the rate equation for 2A + B
A2)
Rate = k [A]o [B]o or Rate = k.
C if the order of the reaction is zero?
Q3) The reaction between H2 (g) an O 2 (g) is highly feasible yet allowing the gases to
stand at room temperature in the same vessel does not lead to the formation of water.
Explain?
A3) Activation energy of this reaction is very high. Therefore the reaction does not
take place.
or
Q3) Oxygen is available in plenty in air yet fuels do not burn by themselves at room
temperature. Explain?
A3) Activation energy for combustion reactions of fuels is very high at room
temperature therefore they do not burn by themselves.
Q4) For a certain reaction large fraction of the molecules has energy more than the
threshold energy, yet the rate of the reactions very slow. Why?
A4) Apart form energy considerations, the colliding molecules should have a proper
orientation for effective collision. It appears that this condition is not fulfilled in this case.
That is why the reaction is very slow.
Q5)
Why is the probability of reaction with molecularity higher than three vey rare?
A5) The probability of more than three molecules colliding simultaneously is very very
small. Hence the probability of reaction with molecularity higher than three is very rare.
Q6)
Why does the rate of reaction generally decrease during the course of reaction?
A6) Rate of the reaction depends upon the concentration of reactants. With the
progress of the reaction, reactants convert into products. Thus the concentration of the
reactant decreases and hence the rate of reaction decreases.
Q7) Thermodynamic feasibility of a reaction alone cannot decide the rate of the
reaction. Explain with the help of example?
A7) For a reaction to take place, appropriate activation must be supplied to the
reactant. If this not available, the reaction does not take place. For example
thermodynamically, conversion of diamond to graphite is possible, but it requires an
34
activation energy which is not available at room temperature. Hence the reaction does
not take place.
Q8) Why in the redox titration of KMnO4 vs oxalic acid, we heat oxalic acid solution
before titration ?
A8) The reaction between KMnO4 and oxalic acid is slow. On raising the temperature,
we provide the required activation energy and the rate of reaction is increases.
Q9) Write the difference between instantaneous rate of a reaction and average rate of
reaction?
Instantaneous Rate of Reaction
Average Rate of Reaction
Instantaneous Rate of Reaction at It is the appearance of product or
any instant of time is defined as the disappearance of reactants over a long
rate of change in concentration of time interval
any one of the reactants or
products at that particular instant of
time.
For a reaction :A+2B
C
[ ]
=
=-
[ ]
=
[ ]
=
[ ]
=-
[ ]
=
[ ]
02 MARKS
Q1)
(a)
For a general reaction A
B , plot of concentration of A vs time is
given in the figure below. Answer the following questions on the basis of this
graph?
(i)
What is the order of the reaction?
(ii)
What is the slope of the curve?
(iii)
What are the units of the rate constant?
A
Co
nc
.of
A
t
B
35
(b)
For the reaction A + B
Products,
the
rate
law
Rate = k A] [B] 3/2. Can the reaction be elementary reaction? Explain.
A 1)
is
:
(a)
(i)
For a zero order reaction, a plot of [A] with t gives the above type of
graph. Hence this is a zero order reaction.
(ii)
Slope of the curve = - k
(iii)
Unit of rate of constant for zero order reaction are mol L-1 s-1
(b)
Had this been an elementary reaction, the order with respect to B would
have been 1. But it is given by 3/2 in the problem. Hence it is not an elementary
reaction.
Q2)
(a)
Why molecularity is applicable only for elementary reactions and order is
applicable for elementary as well as complex reactions?
(b)
Why can we not determine the order of a reaction by taking into
consideration?
A2)
(a)
A complex reaction proceeds through a number of elementary reactions.
Number of molecules involved in each elementary reaction may be different i.e,
the molecularity of each step may be different. Therefore, molecularity of overall
complex reaction has no sense. On the other hand, order of a complex reaction is
determined by the slowest step in its mechanism and is valid even in the case of
complex reactions.
(b)
Balanced chemical equation often leads to incorrect order or rate law as
the rate of reaction may not depend upon all the molecules of a reactant present
in the balanced chemical equation.
For eg :- Decomposition of NH3 on Pt surface is actually a zero order
reactions which is only determined experimentally & cannot be predicted
from the chemical equation.
2NH3
Q 3)
Pt
N2 + 3H2
(a)
Illustrate graphically the effect of catalyst on activation energy.
(b)
Catalyst have no effect on the equilibrium constant. Why?
36
A 3)
(a)
---------Pot
enti
al
ene
rgy
-------Reactants
Reaction
path with
catalyst
Energy
activation
with
catalyst
Energy of
activation
without
catalyst
-----------Products
Reaction coordinate
The catalyst provides an alternate pathways or reaction mechanism by
reducing the activation energy between the reactants and products and hence
lowering the potential energy barrier as shown in the figure.
A catalyst does not change the equilibrium constant of a reaction, rather it
helps in attaining the equilibrium faster. That is, it catalyses the forward as well as
backward reactions to the same extent so that the equilibrium state remains same
but is reached earlier.
(b)
Q4)
How can you determine the rate law of the following reaction ?
2NO(g) + O2(g)
A4)
2NO2(g)
The rate law can be determined by initial rate method. Keeping the concentration
of one of the reactant constant and changing the concentration of the other, the
effect on the rate of reaction is determined.
for eg : for the given reaction
(i)
Keeping [O ] constant, if [NO] is doubled rate is found to become four
times. This shows that. Rate α [NO]2
(ii)
Keeping [NO] constant, if [O ] is doubled, rate is also found to become
doubled . This shows that Rate α [O ]
(iii)
Hence overall rate law will be Rate = k[NO]2[O ]
Q5) How does a change in temperature affect the rate of a reaction? How can this
effect on the rate constant of a reaction be represented quantitatively?
A 5) Rate of reaction increases with temperature. Temperature coefficient is the ratio
of rate constants of reaction at two different temperatures differing by 10o.
Temperature coefficient =
It is observed that for a chemical reaction with rise in temperature by 10 o, the rate
constant is nearly doubled.
37
O3 MARKS
Q1. The half life of decay radioactive 14C is 5730 years. An archeological artifact
containing wood had only 80% of 14C activity as found in a living tree. Calculate the age
of the artifact?
A1.
Radioactive decay is a first order reaction.
or
k
=
t
=
.
log [ ]
[ ]
.
log [ ]
[ ]
For 80% activity, [ ] = 0.8[ ]
Substituting in the above equation, we get
t
or
.
=
t
⁄
.
.
=
x
.
log [ ]o
0.8[ ]o
x 5730 x 0.0969 = 1845 years
.
Thus, the artifact in 1845 years old.
Q2. During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If
1 µg of 90 Sr was absorbed on the bones of a newly born instead of calcium, how much of
it will remain after 10 years and 60 years if it is not lost metabolically?
A2)
Determine k, using the following relation :
.
k=
=
.
.
/
.
k=
log
Substituting the values in the equation of first order, we get
After 10 years,
or
log =
or
N=
.
.
=
.
.
.
.
=
log
.
.
= 0.1071 or
= Antilog 0.1071 = 1.279
= 0.7818 µg
Thus, 0.7818 µg of 90Sr will remain after 10 years
38
.
After 60 years, 60 =
.
orlog =
or
x 28.1 log
.
=
.
N=
.
= 0.6427 or
.
= Antilog 0.6427 = 4.392
= 0.227 µg.
.
Thus , 0.227 µg of 90Sr will remain after 60 years.
Q3. For a decomposition reaction, the values of rate constant at two different
temperatures are given below :k1= 2.15 x 10
L/(mol.s) at 650 K
k2= 2.39 x 10
L/(mol.s) at 700 K
Calculate the value of
for the reaction.
(log 11.11 = 1.046) (R = 8.314J K-1mol
A3.
Using the formula, log
=
= 2.15 x 10 L/mol
at 650K.
= 2.39 x 10 L/mol
at 700K.
Given,
=
log
log
−
.
.
/
.
/
log 11.12 =
=
.
.
=
.
.
1
.
.
x
.
.
/
650
−
1
700
.
.
= 182.25 kJ
Q4. The following data were obtained during the first order thermal decomposition of
SO Cl at a constant volume :SO Cl (g)
SO (g) + Cl (g)
Experiment
1
2
Time
0
100
Total pressure/atm
0.4
0.7
39
Calculate the rate constant.
(Given , log 4 = 0.6021, log 2 = 0.3010)
A4)
SO Cl (g)
SO (g) + Cl (g)
Initial pressure
After time, t
–p
0
0
p
p
Total pressure aftertime t,
i.e.
=
– p+ p+ p =
p=
Thus, a =
-
and a –x =
=
+p
– p=
– pt+
=2
-( -
)
-
Substituting the values of a and (a-x) in equation,
.
k =
log
.
k =
log
(
)
Calculation of rate constant (k), when t = 100 s
Given, p = 0.4 atm and
.
Then, k =
=
.
=
=
log
log
=
= 0.7 atm
.
.
(
log
.
.
.
.
. )
=
.
log 4
x 0.6021 = 0.01387
1.387 x10
40
Q 5) Nitrogen pent oxide decomposes according to equation, 2N O (g)
4NO (g) +
O (g). This first order reaction was allowed to proceed at 40 C and the data below were
collected.
[
Time (min)
0.00
20.0
40.0
60.0
80.0
]M
0.400
0.289
0.209
0.151
0.109
A5)
(i)
Calculate the rate constant. Include units with your answer.
(ii)
Calculate initial rate of reaction.
(iii)
After how many minutes will [N O ] equal to 0.350 M?
(i)
k=
.
log
[ ]
[ ]
[ ] = 0.400 M
[ ] = 0.289 M, t = 20 min
k=
k=
k=
.
log
.
[ .
[ .
]
]
x log 1.384
.
x log 0.1414
0.01625 min
=
When t = 40 min, [ ] = 0.209M
.
k=
.
=
.
log
.
=
.
log 1.914
x 0.2819 = 0.01623 min
Similarly, when t = 60 min, [ ]= 0.151
k=
k=
.
.
log
.
.
x 0.4231 = 0.01623min
k =0.01625 + 0.01623 + 0.01623
3
= 0.016236 min
41
(ii)
Initial rate = k [N O ] = 0.016236 x 0.4
= 0.00649 mol L s
(iii)
.
k=
t=
t=
t=
log
[ ]
[ ]
.
.
.
.
.
.
.
log
.
log 1.143
log 0.0581
t = 8.24 min.
05 MARKS
Q1)
(a)
Explain the following terms :
(i)
(ii)
(b)
Rate of reaction
Activation energy of reaction
The decomposition of phosphine, PH , proceeds according to the following
equation : 4PH (g)
P (g) +6H (g)
It is found that the reaction follows the following rate equation : Rate = k[PH ].
The half life ofPH is 37.9 sat120 C.
A1)
(i)
(ii)
How much time is required for 3/4th of PH to decompose?
(a)
(i)
Rate of a reaction : Rate of a reaction may be defined as change
in concentration of reactant or product in unit time.
What fraction of the original sample of PH remains behind after 1 minute?
(ii)
Activation energy of a reaction : Difference in energy of the
reactants and the activated complex in the reaction is called activation
energy. Reactants must gain energy to reach the state of activated
complex in order to give the products.
(b)
4PH (g)
P (g) +6H (g)
Half – life of PH = 37.9 s
Rate = k[PH ]
This means that the reaction is of first order. For a first order reaction
k=
.
or
/
.
.
= 0.01828496
42
(i)
Here R =
.
Apply the reaction k =
-
log
[ ]
[ ]
=
Substituting the values in the equation above, we have
0.01828496 =
or
t
=
or
t
=
.
log
[ ]
[
.
]
/4
log 4
.
.
x 0.60206
.
= 75.83 s
(ii)
After 1 minute or 60 seconds k =
or
log
[ ]
[ ]
=
.
log
[ ]
[ ]
.
Substituting the values, we havelog
[ ]
[ ]
=
.
.
= 0.4768
Taking antilogarithms , we have
[ ]
[ ]
or
[ ]
= 2.992
=
.
= 0.3342
Q2) (a)
All energetically effective collisions do not result in a chemical change.
Explain with the help of an example?
(b) Show that for a first order reaction, the time required for half the change (half –life
period) is independent of initial concentration.
A2)(a) All energetically effective collisions do not result in a chemical change because
even if colliding molecules have energy greater than the threshold energy, they may not
have proper orientation at the time of collision and no breaking of bonds in the reactant
molecules and formation of new bonds to form product molecules may occur.
For eg :The reaction between bromomethane and an alkali may or may not
lead to the formation of methanol depending upon whether they have a proper
orientation or not at the time of collision.
43
H
Proper orientation OH + H
H
-
H
H
H
H
C – Br HO C   -- Br  
H
H
Intermediate



C   – Br   + OH–
CH3 OH + Br–
Methanol
Improper orientation
H
H
H
C   – Br   HO  No product
(Repel)
A2)(b) For a first order reaction, rate constant,
k=
When, t =
.
log
/ ,
/ =
/ =
=
/ =
.
.
.
[ ]
[ ]
[ ]=
log
[ ]
[ ]
[
]
log 2
.
.
Thus for a first order reaction,half life period is constant & is independent of
initial concentration of the reacting species.
Q3)(a) What are pseudo first order reactions? Give one example of each reactions.
(b) Hydrogen peroxide,H O (aq) decomposes to H O (l) and O (g) in a reaction that
is first order in H O and has a rate constant. k = 1.06 x 10 min-1.
(i)
How long it will take for 15% of a sample of H O to decompose?
(ii)
How long it will take for 85% of the sample to decompose?
A3)(a) Pseudo first order reaction
The reaction which is bimolecular but has order
one, is called pseudo first order reaction, e.g acidic hydrolysis of ester.
H+
CH COOC H + H O
CH COOH(l) + C H OH(l)
44
(b)
(i)
When 15% of a sample of H O is decomposed.
For a first order reaction, k =
.
log
[ ]
[ ]
Given, k = 1.06 x 10 min−1
[ ] = 100M,
[ ] (after time, t) = 100 – 15 = 85M
t =
.
.
t=
=
(ii)
=
=
(a)
log 1.176
.
.
.
= 152.9 min
.
When 85% of a sample of H O is decomposed, [R] = 100-85 = 15M
t =
Q4)
log
.
.
log
.
.
log 6.667
.
.
.
= 1790.25 min
.
Explain the following terms :
(i)
Order of a reaction.
(ii)
Molecularity of a reaction
(b)
The rate of a reaction increases four times when the temperature changes
from 300 K to 320 K. Calculate the energy of activation of the reaction, assuming
that it does not change with temperature. (R = 8.314 J K -1 mol -1).
A4)
(a)
(i)
Order of a reaction : The sum of the powers of the concentration of
the reactants in the rate law expression is called the order of that chemical
reaction.
(ii)
Molecularity of a reaction : The number of reacting species (atoms, ions or
molecules) taking part in an elementary reaction, which must collide
simultaneously in order to bring about a chemical reaction is called molecularity of
a reaction. Thus there are unimolecular, bimolecular, trimolecular reactions when
the numbers of molecules taking part are one, two or three.
(b)
Apply Arrhenius equation
log
=
.
45
Substituting the values, we have
=
log
or
log 4 =
or
log 4 =
or
0.6021 =
or
=
(a)
.
.
.
x
x 1.088 x 10
.
.
x 1.088 x 10
x 10
= 5.53 x 10 J = 55.3 kJ/mol.
or
Q5)
.
A reaction is first order in A and Second order in B.
(i)
Write the differential rate equation.
(ii)
How is the rate affected when concentrations of both A and B are
doubled?
(b) In a reaction between A and B , the initial rate of reaction( ) was measured
for different initial concentrations of A and B as given below :
A/mol L-1
B/mol L-1
/mol L-1
0.20
0.30
5.07 x 10
0.20
0.10
5.07 x 10
0.40
0.05
7.16 x 10
What is the order of reaction with respect to A & B ?
A5
(a)
= K [A] [B]2
(i)
(Rate)1
(ii)
If both [A] = [B] are doubled
(Rate)2 = k [2A] [2B]2
= 8K [A] [B]2
(Rate)2 = 8K [A] [B]2 = (Rate)2 = 8 (Rate)
(Rate)1
k [A] [B]2
i.e The rate increases 8 times
(b)
= k [A]α[B]β
( )1 = k x 5.07 x 10-5 = (0.20) α (0.30) β
( )2 = k x 5.07 x 10-5 = (0.20) α (0.10) β
( )3 = k x 1.43 x 10-4 = (0.40) α (0.05) β
Dividing (i) by (ii)
( )1/( )2 = 1 = (0.30/0.10)β = (3)β
…..(i)
…..(ii)
…..(iii)
46
Therefore
β =0
Dividing (iii) by (ii)
( )3/( ) =
.
.
= (0.40/0.20)α (0.05/0.10)β
or
1.412 = 2α
= 2α
or
log 1.412 = α log 2
or
α = 0.1523/0.3010 = 0.5
Thus, order w.r.t A = 0.5 , order w.r.t B = 0
Because [β] = 0
47
CHAPTER 5
SURFACE CHEMISTRY
01 MARK
Q1).
What is meant by shape selective catalysis?
A1) The catalytic reaction that depends upon the pore structure of the catalyst and the
size of the reactant and product molecules is called shape selective catalysis.
eg:-
zeolites (eg ZSM 5) converts alcohol directly into gasoline.
Q2)
What are the dispersed phase and dispersion medium in milk?
A2)
Dispersed phase : Fat
Dispersion medium : water
Q3)
What is Collodion?
A3)
Collodion is 4% solution of nitro cellulose in a mixture of alcohol and ether.
Q4)
Define Kraft temperature?
A4) The formation of micelles from the ionic surfactant can take place only above a
certain temperature which is called Kraft temperature.
Q5)
Why is adsorption always exothermic ?
Or
What is the sign of
A5)
H&
S when a gas is adsorbed by adsorbent ?
When a gas is adsorbed in the surface of solid, its entropy decreased i.e.
S =-ve. From Gibbs Helmholtz equation :
G=
H–T
S for the process
to be spontaneous ,
G must be negative, which is possible only
when
H = -ve . Hence adsorption is always exothermic.
Q6) Why is it essential to wash the precipitate with water before estimating it
quantitatively?
A6) Some amount of the electrolytes mixed to form the precipitate remains adsorbed
on the surface of the particles of the precipitate. Hence it is essential to wash it with
water before it quantitatively.
Q7)
What happens when persistent dialysis of colloidal solution is carried out?
A7) The stability of a colloidal sol is due to the presence of a small amount of the
electrolyte. On persistent dialysis, the electrolyte is completely removed, so the colloidal
sol becomes unstable and gets coagulated.
Q8)
What causes Brownian movement in a colloidal solution?
48
A8) Brownian movement ie zigzag movement of the colloidal particles is due to hitting
of these particles by the molecules of the dispersion medium with different forces from
different directions.
Q 9)
Why it is important to have clean surface in surface studies?
A 9)
It facilitates the adsorption of species on the adsorbent.
Q 10) What happens when gelatin is mixed with gold sol?
A 10) Gold sol is a lyophobic sol. On addition of gelatin, the sol is stabilized..
Q 11) Gelatin which is a peptide is added in ice – creams. What can be it s role?
A 11) Ice –creams are emulsion which get stabilized by emulsifying agents like gelatin.
Q 12) What is the role of activated charcoal in gas mask used in coal mine?
A 12) Activated charcoal adsorbs poisonous gases present in coal mine.
Q 13) In what way is a sol different from a gel?
A 13) Sols are colloidal solutions of solid dispersed in liquid while gels are colloidal
solutions of liquid dispersed in solid.
Q 14) Of NH3 and N2,which gas will be adsorbed more readily on the surface of charcoal
and why?
A 14) NH3 is adsorbed more readily as it is more easily liquefiable compared to N2,
Moreover, NH3 molecule has greater molecular size.
Q 15) How does it become possible to cause artificial rain by spraying silver
iodide on the clouds?
A 15) Clouds are colloidal in nature and carry a charge. On spraying silver iodide which
is an electrolyte, the charge on the colloidal particles is neutralized. Clouds coagulate to
form rain.
Q 16) What is the role of diffusion in heterogeneous catalysis?
A 16) The gas molecules diffuse onto the surface of the catalyst and get adsorbed. After
the chemical change, the products formed diffuse away from the surface of the catalyst
setting the surface free for other reactant molecules to adsorb on the surface and give
the product.
02 MARKS
Q1)
Write the differences between physisorption and chemisorption with respect to
the following :(i) Specificity
(ii)
Temperature dependence
(iii) Reversibility
(iv)
Enthalpy change
Criteria
Physisorption
Chemisorption
49
Specificity
It is not specific in nature
It is highly specific in nature
Temperature
dependence
It decreases with increase in
temperature.
Thus,
low
temperature is favorable for
physisorption
Reversible in nature
It increases with increase in
temperature. Thus, highly
temperature is favorable for
chemisorption
Irreversible in nature
Low enthalpy of adsorption
High enthalpy of adsorption
Reversibility
Enthalpy
change
Q2)
Distinguish between homogeneous and heterogeneous catalysis.
A2)
When reactants and the catalysts are in the same phase (i.e. liquid or gas) the
catalysis is known as homogeneous catalysis.
3
(g)
O2 ( g ) NO

 O3 ( g )
2
When reactants and the catalysts are in the different phase (i.e. liquid or gas) the
catalysis is known as heterogeneous catalysis.
(s )
4NH3(g) + 5O2(g) Pt

 4 NO ( g )  6 H 2 O( g )
Q3)
What is meant by coagulation of a colloidal solution? Describe briefly any three
methods by which coagulation of lyophobic sols can be carried out?
A3)
The process of aggregating together the colloidal particles is called coagulation
of the sol. It is also known as precipitation. Following are the three methods by which
coagulation of lyophobic sols can be carried out.
Q4)
(i)
Electrophoresis. In this process, the colloidal particles move towards
oppositely charged electrodes and get discharged resulting in
coagulation.
(ii)
Mixing of two oppositely charges sols.
When equal proportions of
positively charges sols are mixed, they neutralize each other resulting in
coagulation.
(iii)
Dialysis. By this method, electrolytes present in sol are removed
completely and colloid becomes unstable resulting in coagulation colloid
is a heterogeneous system, e.g gold sol, sulphur sol, soap, etc.
What are emulsions ? Discuss the role of an emulsifier in forming emulsion.
A4)
Emulsions are one of the types of colloidal system, in which both the dispersed
phase and dispersion medium are liquids, e.g milk.
50
Role of emulsifier Emulsifying agents are added to emulsions to stabilize them. The
emulsifying agent forms an interfacial film between suspended particles and the medium.
For oil in water emulsions, the principal emulsifying agents are gums, proteins, natural
and synthetic soaps.
Q 5)
A5)
How are the following colloidal solutions prepared?
(i)
Sulphur in water
(ii)
Gold in water
(i)
Sulphur sol is prepared by the oxidation of H2S with SO2.
SO2 + 2H2S
oxidation
3S + 2H2O
(Sol)
(ii) Gold sol is prepared by Bredig’s arc process or by the reduction of AuCl3
with HCHO.
2AuCl3 + 3HCHO + 3H20
Reduction
2Au + 3HCOOH + 6HCI
(Sol)
Q6)
What are enzymes? Write in brief the mechanism of enzyme catalysis.
A6)
Enzymes are biochemical catalysts whih are globular proteins and form
macromolecular colloidal solution in water.
The mechanism of enzyme catalysis may be explained on the basis of lock
and key theory. Acc to the theory , Enzymes are highly specific due to the
presence of active sites on their surface. The shape of active site of any given
enzyme is such that only a specific substrate can fit in to it just as one key can fit
into a particular lock enzyme catalyzed reactions takes place in two steps as
follows :Step I :-
E
Formation of Enzyme – substrate complex
+
Enzyme
Step II :-
S
Substrate
(fast and reversible)
Enzyme substrate complex
Dissociation of enzyme substrate complex to form the products
ES
Enzyme
Substrate
Complex
Q7)
ES
[EP]
E
+
P (slow and rate determining)
Enzyme
Enzyme
Product
(Regenerated)
association
What do you mean by activity and selectivity of catalysts?
Product
51
A7) Activity : Ability of a catalyst to accelerate chemical reactions is known as its
activity . For example, Pt catalyses the combination of H2 and O2 to form water. It has
been found that for hydrogenation reactions, the catalytic activity increases from Group 5
to Group 11 metals with maximum activity being shown by Group 7-9 elements of the
periodic table.
Selectivity : The ability of a catalyst to direct a reaction to yield a particular
product is called its selectivity. Combination of CO and H2 yields different products with
different catalysts as given below :CO (g) + 3H2(g)
Ni
CH4 (g) + H2O (g)
CO (g) + H2(g)
Cu
CO (g) + H2(g)
Cu I ZnO – Cr O
2 3
HCHO (g)
CH3OH (g).
Q8)
Describe some features of catalysis by zeolites?
A8)
(i)
Zeolites are hydrated alumino-silicates which have a three- dimensional
network structure containing water molecules in their pores.
(ii)
On heating, water of hydration present in the pores is lost and the pores
become vacant to carry out catalysis.
(iii)
The size of the pores varies from 260 to 740 pm. Thus, only those
molecules can be adsorbed in these pores and catalyzed whose size fits these
pores. Hence, they act as molecular sieves or shape selective catalysts.
An important catalyst used in petroleum industry is ZSM -5 (Zeolite sieve of
molecular porosity – 5). It converts alcohols into petrol by dehydrating them to
form a mixture of hydrocarbons.
ZSM -5
Alcohols
Hydrocarbons
Dehydration
03 MARKS
Q1)
Account for the following :(a)
Medicines are more effective in the colloidal form/ colloidal gold is used for
intramuscular injection.
(b)
Sky appears blue in colour.
(c)
Alum is added to purify water.
OR
Bleeding stops on rubbing moist alum on the cut surface.
A1)
(a)
Medicines are more effective in the colloidal form because they have a
large surface area and are easily assimilated in this form.
(b)
There are dust particles in the atmosphere. These dust particles are of
colloidal size and scatter the light. Blue light coming from the sun is scattered and
the sky appears blue.
52
(c)
Alum coagulates colloidal impurities present in water.
or
Alum brings about and coagulation of blood and stops further bleeding.
Q2)
A 2)
Explain clearly how the phenomenon of adsorption finds application in
(i)
Production of vacuum in a vessel
(ii)
Heterogeneous catalysis
(iii)
Froth floatation process in metallurgy.
(i)
Production of high vacuum. The traces of air can be adsorbed by
charcoal from a vessel evacuated by a vacuum pump to give a very high vacuum.
(ii)
Heterogeneous catalysis.
Adsorption of reactants on the solid
surface of the catalyst increases the rate of reaction.
(iii)
Froth floatation process. A low grade sulphine ore is concentrated by
this method using pine oil and frothing agent. The mineral particles become wet
by oils while the gangue particles by water.
Q3)
What is an adsorption isotherm? Describe Freundlich adsorption isotherm.
A 3) Adsorption isotherm. It is the variation in the amount of gas adsorbed by the
adsorbent with pressure at constant temperature.
Freundlich’s adsorption isotherm. It is an empirical relationship between the
quantity of gas adsorbed by unit mass of solid adsorbent and pressure at a particular
temperature.
1
x
 kp n ( n  1)
m
When, n  1,
…. (i)
x
x
 kp or  p
m
m
Where x is the mass of gas adsorbed on mass m of the adsorbent at pressure p.k
and n are constants which depend on the nature of the adsorbent and the gas at a
particular temperature.
Taking log in Eq. (i), gives
log
x
1
 log k  log p
m
n
x
on Y-axis and
m
log p on X-axis. If, it comes to be a straight line, the Freundlich isotherm is valid.
The validity of Freundlich isotherm can be verified by plotting log
53
Q4)
Explain the terms with suitable examples :(i)
A4)
Alcosol
(ii)
Aerosol
(iii)
Hydrosol
(i)
Alcosol :- It is a colloidal dispersion having alcohol as the dispersion
medium. For eg :- Collodion, which is a colloidal sol, of cellulose nitrate in ethyl
alcohol.
(ii)
Aerosol :-
It is a colloidal dispersion of a liquid in a gas eg:- fog.
(iii)
Hydrosol :- IT is a colloidal dispersion of a solid in water as the dispersion
medium.eg :- starch sol.
Q5)
Explain the following observations :(a)
Cotrell’s smoke precipitator is fitted at the mouth of the Chimneys use is
factories.
(b)
Physical adsorption is multilayered, while chemisorption is monolayered.
(c)
A white precipitate of silver halide becomes coloured in presence of the dye
eosin.
A5)
(a)
The charged collodial particles of carbon after coming in contact with
oppositely charges electrode in Cottrell precipitator lose their charge and settle
down at the bottom.
(b)
In physical adsorption, there are Weak Vander Waals forces. Therefore it
forms multilayers. In Chemisorption, adsorbate is attached by chemical bond.
There is a strong force of attraction. Therefore, only one layer is obtained.
(c)
Eosin is adsorbed on the surface of silver halide precipitate making it look
coloured.
54
CHAPTER – 6
GENERAL PRINCIPLES OF EXTRACTION
01 MARK
Q1.
Name the chief ores of aluminium and zinc.
A1.
Chief ores of – Aluminium – Bauxite. A 0 (OH) 3- 2x[Where O<X<1]
Zinc :- zinc blende (ZnS)
Q2. What is the function of collectors in the froth floatation process for the
concentration of ores?
or
What is the role of collectors in froth floatation process?
A2. Collectors such as pine oils, fatty acids, xanthates etc, enhance non – wettability
of the mineral particles.
Q3.
What is the role of flux in metallurgical processes?
A3. Flux is a substance that chemically combines with gangue (earthy impurities)
which may still be present in the roasted or the calcined ore to form an easily fusible
material called the slag.
Q4.
What types of ores can be concentrated by magnetic separation method?
A4. Those ores which are magnetic in nature and associated impurities are nonmagnetic in nature or vice-versa, are concentrated by magnetic separation method.
Q5.
Differentiate between a mineral and an ore?
A5. Naturally occurring substances from which metal may or may not be extracted
profitably is known as mineral. The minerals from which metal can be extracted chiefly,
easily and profitably are known as ores.
Q6.
Why is it that only sulphide ores are concentrated by froth floatation process?
or
Why is the froth floatation method selected for the concentration of sulphide ores?
A6. Only sulphide ores are concentrated by froth floatation method because sulphide
ores are wetted preferentially by pine oil and impurities (gangue) are wetted by water.
Q7.
Why are sulphide ores converted to oxide before reduction?
A7. It is more easy to reduce an oxide ore than a sulphide ore. This is why sulphide
ores are roasted and first converted to its oxide form.
55
Q8. Although carbon and hydrogen are better reducing agents, but they are not used
to reduce metallic oxides are at high temperature. Why?
A8. This is because carbon and hydrogen react with metals to form carbides and
hydrides respectively at high temperature.
Q9.
What is the role of CO in the extractive metallurgy of aluminium from its ore?
or
What role is played by CO in the getting pure alumina (Al O ) in the extraction of
aluminium?
A9. The aluminate in solution is neutralized by passing CO and hydrate Al O is
precipitated
2 Na [Al (OH)4] (aq) + CO (g)
Al O .xH O( ) + 2NaHCO (aq)
Q10. Why is electrolytic reduction preferred over chemical reduction for the isolation of
certain metals?
A10. More reactive metals, i.e. alkali metals and alkaline earth metals are usually
extracted by electrolysis of their fused salts because in electrolytic reduction, by applying
external potential from outside source, reduction can be brought easily. Further, more
reactive metals have more affinity with oxygen than that of carbon. That’s why electrolytic
reduction is preferred over chemical reduction.
02 MARKS
Q1.
How is copper extracted from a low grade ore of it?
A1. Copper is extracted by hydrometallurgy from low grade ores. It is leached out
using acid or bacteria. The solution containing Cu is treated with scrap iron or H .
Cu
(aq) + H .(aq)
Cu(s) + 2H (aq)
Cu
(aq) + Fe(s)
Cu(s) + Fe
(aq)
Q2. How do we separate two sulphide ores by Froth floatation method? Explain with
an example?
Or
State the role of depresent in froth floatation process?
A2. Two sulphide ores can be separated by adjusting properties of oil to water or by
using depressants which prevent one type of sulphide ore particles from forming the froth
with air bubbles.
Eg :-In case of an ore containing Zns and Pbs, the depresent NaCN is used. It
forms complex with ZnS and thus prevents it from forming a froth while Pbs forms
the froth and hence can be separated from ZnS.
4NaCN + ZnS
Na [Zn CN)4] + Na S
sodiumtetracyanidozincate(II)complex
56
Q3.
State Write the reactions involved in the following process :
(i)
Leaching of bauxite ore to prepare pure alumina.
(ii)
Recovery of gold after gold ore has been leached with NaCN solution.
A3. Leaching of alumina from bauxite ore Bauxite usually contains SiO , iron oxide
and titanium oxide (TiO ) as impurities. Powered ore is digested with conc NaOH solution
at 473-523Kand 35-36 bar pressure.Al O and SiO are dissolved in the solution while
impurities do not.
Al O (s) + 2NaOH (aq) + 3H O(l)
2Na [Al(OH)4](aq)
Filtrate is neutralized by passing CO gas and hydrated Al O is precipitated.Some fresh
alumina is also added to solution to induce precipitation.
2 Na [Al (OH)4] (aq) + 2CO (g)
Al O .xH O (s) + 2NaHCO (aq).
ppt
Precipitate is filtered, washed, dried and heated to give back pure Al O .
Al O .xH O (s) 1470KAl O ( ) +.xH O (g)
Alumina
Recovery of gold
4Au(s)+ 8CN (aq) +2H O (aq) + O (g)
4[Au(CN)2]2-(aq) + 4OH (aq)
22[Au(CN)2] (aq) + Zn (s)
[Zn(CN)4 ] (aq) +2Au(s)
Soluble gold complex
In this reaction, zinc acts as a reducing agent.
Q4.
A4.
Define the following terms :
(i)
Roasting
(ii)
Calcination
(i)
Roasting. The process of heating of metal ore below its melting point in the
presence of air is called roasting.
(ii)
Calcination. The process of heating of metal ore in the absence of air is
called calcination.
Q5. Why is an external emf of more than 2.2V required for the extraction of CL from
brine?
A5.
For the reaction 2Cl (aq) + 2H O (l)
2OH (aq) + H
(g) + Cl (g)
Value of
is + 422kJ. Using the equation
= -n
, the value of
comes
out to be -2.2V. Therefore an external emf of more than 2.2V is required for the
extraction of
from brine.
57
Q6.
A6.
(a)
What is the composition of copper matte?
(b)
What is the function of SiO in the metallurgy of copper?
(a)
Copper matte contains Cu s and FeS.
(b)
Silica(SiO ) is added in the reverberatory furnace during the extraction of
Cu to remove impurities of iron oxide (FeO) present in the ore. Silica here acts as
flux and reacts with iron oxide gangue to remove it as slag, iron silicate.
FeO + SiO
Iron oxide
Q7.
A7.
Silica
FeSiO
Iron silica slag
Give reason for the following :(i)
Alumina is dissolved in cryolite instead of being electrolysed directly.
(ii)
Zinc oxide can be reduced to the metal by heating with carbon
but
.
(i)
Melting point of alumina is very high and it is a bad conductor of electricity.
Cryolite is added to alumina to lower its melting point and to make it conductor of
electricity.
(ii)
Reduction of zinc oxide is done by using coke ZnO +C
1673K
Zn + CO
Chromium oxide is reduced through alumina (thermic reduction process}.
Cr O + 2Al Al O + 2Cr
This is because, the free energy change for the formation of Cr O is more
negative than that of ZnO.
Q8. State the basis of refining a substance by chromatographic method. Under what
circumstances is this method specially useful?
A8. Chromatographic Method.
This method is based on the principle that
different components of a mixture are differently adsorbed on an adsorbent. Adsorbed
components are removed by using suitable solvent (eluent).
This method is very useful for the purification of the elements which are available
in minute quantities and the impurities are not different in chemical properties from
the element to be purified.
Q9. What chemical principle is involved in choosing a reducing agent for getting the
metal from its oxide ore? Consider the metal oxides, Al O and Fe O , and justify the
choice of reducing agent in each case?
A9. Thermodynamic factor helps us in choosing a suitable reducing agent for the
reduction of a particular metal oxide to metal. From Ellingham diagram, it is evident that
metals which have more negative f of their oxide can reduce those metal oxide for
58
which
is less negative. Since, the free energy change for the formation of
f
less negative than CO, CO is a good reducing agent for it.
+ 3CO 823K 2Fe + 3
,G = -ve [because
f
G(
) is less negative then
f
is
G (CO)]
The free energy change for the formation of Al O is highly negative (~ –1000 to
–1100), thus, no reducing agent is suitable for its reduction. It is thus,reduced by
electrolysis of its oxide.
Q10. How is wrought iron different from steel?
A10. Pig iron contains 4% C and trace elements like, S, P,Si and Mn as impurity. It is
heated strongly (in excess of oxygen) in Bessemer converter. By this, volatile impurities
like C,P,S are removed as CO, P O and SO repectively. Si amd Mn are also oxidised
into their oxides which combine to form MnSiO (slag). The iron so obtained is called
wrought iron. When, it is heated with 0.5% C, it gives steel. Thus, steel is less pure3 than
wrought iron.
03 MARKS
Q1.
Explain the basic principles of following metallurgical operatins :
(i)
Zone refining
(ii)
Vapour phase refining
(iii)
Electrolytic refining
A1. (i)
Zone refining It is based on the fact that the impurities are more soluble in
the melt than in the solid state of the metal. When one end of the impure metal rod is
heated with the help of mobile heater, the molten zone moves forward along with
impurities.
In this way impurities are concentrated at the other end of the rod. This end is cut off.
The process is repeated several times to obtain ultrapure metals for producing
semiconductors. Ge,Si,Ga and In are purified by this method.
Noble gas atmosphere
Metal rod
Induction –coil heaters moving as shown
Molten zone
Zone refining process
59
(ii)
Vapour phase refining In this method, the metal is converted into its
volatile compound and is collected elsewhere. It is then decomposed to give pure
metal. So the two requirements are
(a)
the metal should form a volatile compound with an available reagent.
(b)
The volatile compound should be easily decomposable, for the
recovery of metal.
(iii)
Electrolytic refining
In this method, the impure metals is made anode
and a strip of pure form of same metal is made cathode. Aqueous solution of salt
of same metal is taken as electrolyte. On passing electric current, metal ions from
the electrolyte are deposited at the cathode in the form of pure metal while an
equivalent amount of metal dissolves from the anode and goes into the electrolyte
solution as metal ions, i.e. pure metal is dissolved from anode and deposited at
cathode through electrolyte in equivalent amount.
Q2. Write down the reactions which occur in upper, middle and lower zones in the
blast furnace during the extraction of iron from iron ore?
A2.
Reduction of iron oxide in blast furnace
(i)
Lower zone of the blast furnace
C + O CO + Heat
C + CO
2CO
Coke is burnt to give temperature upto 2200K at lower part of the blast
furnace.
(ii)
Middle zone of the blast furnace
CO and heat move up in the furnace. The temperature range on the middle
zone of the blast furnace is 900 – 1500 K.
FeO + CO
Fe + CO
Limestone is also decomposed to CaO which removes silicate impurity of
the one as slag.
CaCO
1100K
C + CO
CaO + SiO
CaO + CO
2CO
CaSiO
Slag
60
(iii)
Upper zone of the blast furnace
Temperature range in this zone is 500-800K. Following reactions take
place in this zone.
3Fe O + CO
2Fe O + CO
Fe O + 4CO
3Fe + 4CO
Fe O + CO
2FeO + CO
The iron obtained from blast furnace is known as pig iron. It contains about
4% carbon.
Q3. (a)
The reaction : Cr O + 2Al Al O
+ 2Cr ( Δ
= - 421kJ) is
thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not
take place at room temperature?
(b)
The value of Δf for formation of Cr O is -540kJ mol
is - 827 kJ mol . Is the reduction Cr O possible with AI?
and that of Al O
A3. (a) In the given redox reaction, all the reactants and the products are solids at
room temperature, therefore, there does not exist any equilibrium between the reactants
and the products and hence the reaction does not occur at room temperature. The
interpretation of
is based on K (
= -RT in K) where K is the equilibrium
constant. Where there is no equilibrium in the solid state and the value of K becomes
insignificant. However, at high temperature, when chromium melts, the value of
TSincreases. As a result, the value of
proceeds rapidly.
(b).
r
becomes more –ve and hence the reaction
The two thermochemical equations can be written as :
Al (s) + O (g)
Al2 O3 (s) ;
Cr (s) + O (g)
Cr2 O3 (s):
AI, Al O
f
f
= -827kJmol
= -540 kJ mol
(i)
(ii)
Subtracting equation (ii) from equation (i), we get
Al (s) + Cr2 O3 (s) Al2 O3 (s) + Cr (s) ;
f
= -287 kJ mol
As f of the combined redox reaction is –ve, therefore , reduction of Cr O by
AI is possible.
61
Q4.
Write chemical reactions taking place in the extraction of zinc from zinc blende.
A4.
The various steps involved in the extraction of zinc from zinc blende are :
(a)
Concentration.
floatation process.
The ore is crushed and then concentrated by froth
(b)
Roasting. The concentrated ore is roasted in presence of excess of air
at about 1200 K when zinc oxide (ZnO) is formed and SO is evolved.
2ZnS + 3O
Zinc blende
1200K
2ZnO + 2SO
Zinc oxide
(c)
Reduction Zinc oxide obtained in step (b) is mixed with powdered coke
and heated to 1673 K in a fire clay retort when it is reduced to zinc metal.
ZnO + C 1673K Zn + CO
At 1673 K, zinc metal being volatile (b.p 1180K), distils over and is condensed.
(d)
Electrolytic refining
The metal obtained as above is impure. It is
purified by electrolytic method. Impure zinc is made the anode while cathode
consists of a sheet of pure zinc. The electrolyte consists of ZnSO solution
acidified with dil. H SO . On passing electric current, pure Zn is deposited on the
cathode.
Q5.
(a)
(b)
Describe the principle involved in each of the following process :(i)
Mond process for refining of nickel.
(ii)
Liquation method
What is the role of graphite rod in the electrometallurgy of aluminium?
A5. (a)
(i)
Mond process for refining of nickel. (vapour phase refining) is
based upon the principle that nickel is heated in the presence of carbon monoxide to
form nickel tetra carbonyl, which is a volatile complex.
Ni+ 4CO
330-350 K
Ni(CO)4
Nickel tetra carbonyl
Then, the obtained nickel tetra carbonyl decomposed by subjecting it to a higher
temperature (450 – 470K) to obtain pure nickel metal.
Ni(CO)4
Ni + 4CO
Nickel tetra
Nickel
carbonyl
(ii)
Liquation method In this method alow melting metal like tin can be
made to flow on a sloping surface. In this way, it is separated from higher
melting impurities.
62
(b)
In the electrometallurgy of alumina, a fused mixture of alumina, cryolite and
fluorspar (CaF2) is electrolysed using graphite as anode and graphite lined as
cathode. During electrolysis, Al is liberated at the cathode while CO and CO2 are
liberated at the anode. The relevant equations are as under :
At cathode :
At anode :
Al 3  3e  Al (l )
C (s) + O
C (s) + 2O
(melt)
CO(g) + 2e
(melt) CO (g) + 4e
Oxygen evolved at the anode burns, the anode material (graphite).
If instead of graphite, some other metal is used as the anode, then O
liberated will oxidise the metal of the electrode. Since graphite is much cheaper
than any metal, therefore, graphite is used as the anode.
63
CHAPTER – 7
P-BLOCK ELEMENTS
01 MARK
1)
Ans:
Why does PCl3 fume in moisture?
PCl3 hydrolyses in the presence of moisture giving fumes of HCl.
PCl3 + 3H2O - H3PO3 + 3HCl
2)
PH3 forms bubbles when passed slowly in water but NH3 dissolves. Explain why?
Ans:- Due to high electronegativity & small size of N, NH3 forms H- bonds with water and
hence it is water soluble. On the other hand, due to its lower electronegativity and its bigger size.
PH3 does not form H-bonds with H2O. As a result, it does not dissolve in H2O and hence escapes
as bubbles.
3)
Solid Phosphorus pentachoride behaves as an ionic compound. Explain why?
Ans In the solid state PCl5 exists as an ionic solid [PCl4]+ [PCl6]-in which the cation [PCl4]+
tetrahedral and the anion [PCl6]-is octahedral.
4)
Sulphur in vapour state exhibits paramagnetic behavior. Give reason.
Ans In vapour state, sulphur partly exists as S2 molecule which has two unpaired electrons in
the antibonding π-orbitals like O2 and hence, exhibits paramagnetism.
5)
In solution of H2SO4 in water, the second dissociation constant, Ka2 is less than the first
dissociation constant, Ka1. Explain.
Ans Ka2 is less than Ka1 because the negatively charged HSO4- ion has much less tendency to
donate a proton to H2O as compared to neutral H2SO4.
6)
Draw the structure of SO2 molecule. Comment on the nature of two S-O bonds formed in
it. Are the two S-O bonds in this molecule equal?
Ans Structure of SO2molecule is given below:
In SO2, sulphur is sp 2 hybridised. The molecule of SO2 is angular. It is resonance hybrid of the
above two canonical forms. In SO2, both the S-O bonds are covalent and have equal strength due
to resonance.
64
7)
Ans
Write the balanced chemical equation for the following reaction :
Excess of SO2 reacts with sodium hydroxide solution.
Chemical equation for sodium hydrogen sulphite.
2NaOH + SO2 - Na2SO3 + H2O
Na2SO3 + H2O + SO2 - 2NaHSO3
Sodium hydrogen sulphite
8)
Name two poisonous gases which can be prepared from chlorine gas.
Ans Phosgene (COCl2), tear gas (CCl3NO2), mustard gas (ClCH2CH2SCH2 CH2Cl) all are
obtained from chlorine gas..
9)
Ans
Write two uses of ClO2
i)
ClO2 is a powerful oxidizing agent & chlorinating agent
ii)
It is an excellent bleaching agent.
10)
What inspired N Bartlett for carrying out reaction between Xe and PtF6?
Ans Bartlett found that the first ionization enthalpy of molecular oxygen is almost similar with
that of xenon. Thus, after preparing red coloured compound O2+[PtF6]-, he got inspired for
carrying out reaction between Xe and PtF6 and made efforts to prepare Xe+[PtF6]- by mixing Xe
and PtF6.
02 MARKS
1)
Arrange the following NH3, PH3, A3H3, SbH3 BiH3
in order of –
a)
Increasing basic strength
b)
Increasing reducing character.
Ans a)
On moving down the group, size of the elements increases. As the size increases,
tendency to attract a proton decreases and thus, the basic character decreases. Hence the
increasing order of basic strength is BiH3< Sbh3< AsH3< Ph3< NH3
b)
The stability of hydrides decreases from NH3 to BiH3 which can be observed from
their bond dissociation enthalpy. Thus the reducing character of hydrides increases in the order –
NH3< PH3< AsH3< SbH3< BiH3
2)
Explain the following :
i)
NO2 readily forms a dimer.
ii)
BiCl3 is more stable than BiCl5.
Ans
i)
NO2 is an odd molecule due to the presence of odd number of valence electrons.
Hence, it readily dimerises to give more stable N2O4 molecule with even number of electrons.
ii)
Due to inert pair effect, +3 oxidation state of Bi is more stable than its +5 oxidation
state. Thus, BiCl3 is more stable than BiCl5.
3)
Draw the structures of white phosphorus and red phosphorus. Which one of these two
types of phosphorus is more reactive and why?
65
Ans
Structure of white phosphorus is given below :
Structure of red phosphorus is given below:
White phosphorus is more reactive because of its discrete tetrahedral structure and angular
strain.
4)
Ans
Write the conditions to maximize the yield of H2SO4 by contact process.
The key step in the production of H2SO4 is the oxidation of SO2 to SO3.
SO2(g) + O2(g)
2SO3 (9) ∆ƒ Ho = - 196.6 KJmol-1
The reaction is exothermic & reversible and the forward reaction proceeds with decrease in
volume. Therefore in accordance with Le chatliers; principle, to maximize the yield of SO3 &
hence of H2SO4, a low temperature (720 K), a high pressure (2 bar) & V2O5 is used as a catalyst.
5)
Assign reasons for the following.
a)
H2S is more acidic than H2O.
b)
Sulphur has a greater tendency for catenation than oxygen.
Ans a)
Due to decrease in (E-H) bond dissociation enthalpy down the group, acidic
character increases. Thus, H2S is more acidic than H2O.
b)
Bond energy of S-S bond (213 kJ mol-1) is greater than O-O bond (138 kJ mol-1).
Due to small size of oxygen atom, there is greater lp-bp repulsion in O-O, resulting in weakening
of O-O bond more than in S-S bond. Therefore, the tendency of catenation in oxygen is lower
than sulphur.
6)
Draw the structure of each of the following.
a)
H2S2O8
b)
H2SO4
66
Ans
7)
Ans
8)
Ans
a)
Structure of H2S2O8 is given below:
b)
The structure of H2SO4
Complete the following chemical equations:a)
NaOH (hot & Conc.) + Cl2 →
b)
Cl2 + F2 (excess) →
a)
3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O
(hot & conc)
b)
Cl2 + 3F2 (excess) → 2ClF3
a)
b)
a)
b)
Which neutral molecule would be isoelectronic with ClO-?
How are interhalogen compounds formulated and how are they prepared?
ClO- has 26 electrons. A neutral molecule with 26 electrons is OF2.
Interhalogen compounds are formulated as XX | XX 3| XX 5| , etc.
The interhalogen compounds can be prepared by direct combination or by the action of
halogen on lower interhalogen compounds. The product formed depends upon some specific
conditions, eg:
K
Cl2 + F2 437

 2ClF
(Equal volume)
K
Cl2 + 3F2 573

 2CIF3
(Excess)
9)
Ans
a)
i)
ii)
i)
ii)
How are the following compounds prepared from XeF6?
XeOF4
XeO3
XeF6 + H2O → XeOF4 + 2HF
XeF6 + 3H2O → XeO3 + 6HF
67
9)
b)
XeF2 is linear molecule without a bent. Explain.
Ans XeF2 is linear molecule. According to VSEPR theory, the three lone pairs will occupy the
equalorial positions and two bond pairs will occupy axial positions to minimize lp-lp and lp-bp
repulsions.
10)
Explain Why –
a)
Noble gases have comparatively large atomic sizes.
b)
Noble gases form compounds with fluorine & oxygen only.
Ans a)
It is due to the reason that noble gases have only vander waals radii while others
have covalent radii & by definition, Vander waals radii are larger than covalent radii.
b)
Fluorine and oxygen are the most electronegative elements and hence, are very
reactive. Therefore, they form compounds with noble gases particularly with xenon.
03 MARKS:
1)
What happens when –
a)
White phosphorus is heated with conc. NaOH solution in an inert gas atmosphere?
b)
Orthophosphorous acid is heated?
c)
PCl5 is heated.
Ans a)
When white phosphorus is heated with Conc.NaOH solution in an inert gas
atmosphere, phosphine gas is produced.
P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2
2)
Ans
b)
On heating, H3PO3 disproportionate into H3PO4 and phosphine.
4H3PO3 → 3H3PO4 + PH3
c)
When PCl5 is heated, it decomposes into PCl3 and Cl2.
PCl5 → PCl3 + Cl2
Complete the following chemical reaction –
i)
I2 + conc.HNO3 →
ii)
HgCl2 + PH3 →
iii)
AgCl(s) + NH3 (aq) →
i)
I2 + 10HNO3 → 2HIO3 + 10NO2 +4H2O
ii)
3HgCl2 + 2PH3 → Hg3P2 + 6HCl
iii)
AgCl(s) + 2NH3 (aq) → [Ag(NH3)2] cl (aq)
68
3)
On heating compound (A) gives a gas (B) which is a constituent of air. This gas when
treated with 3 mol of hydrogen (H2) in the presence of a catalyst gives another gas (C) which is
basic in nature. Gas C on further oxidation in moist condition gives a compound (D) which is a
part of acid rain. Identify compounds (A) to (D) and also give necessary equations of all the steps
involved.
Ans Compounds (A) to (D) are as follows :
(A) = NH2NO2
(B) = N2
(C) = NH3
(D) = HNO3
The reactions are given as under :

i)
NH4NO2 
N2 + 2H2O
ii)
N2 + 3H2 → 2NH3
iii)
4NH3 + 5O2 → 4NO + 6H2O
iv)
2NO + O2 → 2NO2
v)
3NO2 + H2O → 2HNO3 + NO
4)
On reaction with Cl2, phosphorus forms two types of halides A and B. Halide A is
yellowish-white powder but halide B is colourless oily liquid. Identify A and B and write the
formulas of their hydrolysis products.
Ans A is PCl5 (It is yellowish-white powder)
P4 + 10Cl2 → 4PCl5
B is PCl3 (It is a colourless oily liquid)
P4+6Cl2 → 4PCl3
Hydrolysis products are formed as follows:
PCl3 + 3H2O → H3PO3 + 3HCl
PCl5 + 4H2O → H3PO4 + 5HCl
5)
An amorphous solid “A” burns in air to form a gas “B” which turns lime water milky. The
gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises
acidified aqueous KMnO4 solution and reduces Fe3+ to Fe2+. Identify the solid “A” and the gas
“B” and write the reactions involved.
Ans The solid A is sulphur, S8 and the gas B is SO2.
The reactions are given as under:

S8 + 8O2 
8SO2
2MnO 4 + 5SO2 + 2H2O → 5SO42- + 4H+ + 2Mn2+
2Fe3+ + SO2 + 2H2O → 2Fe2+ + SO42- + 2H+
SO2 is produced as a by product during roasting of sulphide ore. The sulphides are converted into
oxides with the evolution of SO2.
6)
Ans
Complete the following chemical equation
i)
C + H2SO4 (conc.) →
ii)
SO3 + H2SO4 (conc.)→
iii)
O3(9) + I-(aq) + H2O (e) →
i)
C + 2H2SO4 (conc.) → CO2 + 2SO2 + 2H2O
ii)
SO3 + H2SO4 (conc.) → H2S2O7
Oleum
iii)
O3(g)+ 2I (aq) + H2O(l) → 2OH-(aq) + I2(s) + O2(g)
69
7)
Account for the following:
a)
Iron on reaction with HCl forms FeCl2& not FeCl3 .
b)
O2& F2 both stabilize higher oxidation states of metals but O2 exceeds F2 in doing
so.
c)
ICl is more reactive than I2 ?
Ans a)
HCl reacts with Fe and produces H2.
Fe + 2HCl → FeCl2 + H2
Liberation of hydrogen prevents the formation of ferric chloride
Ans b)
O2 and F2 both stabilize higher oxidation states of metal but O2 exceeds F2 in doing
so due to ability of oxygen to form multiple bonds with metals.
Ans c)
Interhalogen compounds are more reactive than halogens (except fluorine) because
|
XX bond (I-Cl bond in question) in interhalogens is weaker than X-X bond (I-I bond) in
halogens, except F-F bond. In other words, I-Cl bond is weaker than I-I bond. That’s why, ICl is
more reactive than I2.
8)
Ans
(ii)
a)
i)
ii)
iii)
i)
Draw the structure of
XeOF4 Molecule.
HClO4 Molecule
BrF3
70
(iii)
9)
How are xenon fluorides XeF2, XeF4 and XeF6 obtained ?
Ans These xenon fluorides are obtained by direct reaction between Xe and F2, under different
conditions as shown below:
,
Xe (g) + F2 (g) ⎯⎯⎯⎯⎯⎯ XeF2 (s)
(excess)
,
Xe (g) + 2F2 (g) ⎯⎯⎯⎯⎯⎯ XeF4 (s)
(1:5 ratio)
,
Xe (g) + 3F2 (g) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ XeF6 (s)
(1: 20 ratio)
10)
Account for the following –
a)
Helium is used in diving equipments.
b)
Structure of xenon fluorides cannot be explained by valence bond approach.
c)
Bleaching of flowers by chlorine is permanent while that by sulphur dioxide is
temporary.
Ans a)
Helium is used as a diluent for oxygen in modern diving apparatus because of its
very low solubility in blood.
b)
According to the valence bond approach, covalent bonds are formed by the
overlapping of half-filled atomic orbitals. But xenon has fully-filled electronic configuration.
Hence, the structure of xenon fluorides cannot be explained by VBT.
c)
Cl2 bleaches coloured material by oxidation. Cl2 + H2O → 2HCl + [O]
hence bleaching is permanent.
On the other hand, SO2 bleaches coloured material by reduction & hence bleaching is
temporary SO2 + 2H2O → 2H2SO4 + 2[H]
71
05 MARKS
1)
On heating lead (II) nitrate gives a brown gas “A”. The gas “A” on cooling changes to
colourless solid “B”. Solid “B” on heating with NO changes to a blue solid ‘C’. Identify ‘A’, ‘B’
and ‘C’ and also write reactions involved and draw the structures of ‘B’ and ‘C’.
Ans The gas A is NO2
The reactions are explained as under :

2Pb(NO3)2 
2PbO + 4NO2 + O2
A
(brown colour)
2NO2 ⎯⎯⎯⎯⎯⎯ N2O4
(Colourless solid)
2NO + N2O4 ⎯⎯⎯ 2N2O3
B
C
(Brown coloursolid)
Structure of N2O4
Structure of N2O5
2)
i)
How would you account for the following ?
a)
NF3 is an exothermic compound whereas NCl3 is not. Explain.
b)
All the P-Cl bonds in PCl5 molecule are not equivalent. Explain why?
c)
Why nitrogen gas is very unreactive ?
Ans a)
In case of nitrogen, only NF3 is known to be stable. N-F bond strength is greater
than F-F bond strength, therefore formation of NF3 is spontaneous. In case of NCl3, N-Cl bond
strength is lesser than CL-Cl bond strength. Thus, energy has to be supplied during the formation
of NCl3.
72
Ans b)
PCl5 has a trigonal bipyramidal structure and the three equatorial P-Cl bonds are
equivalent while the two axial bonds are different and longer than equatorial bonds. This is
because the axial bond pairs suffer more repulsion as compared to equatorial bond pairs.
Ans c)
Nitrogen is chemically less reactive. This is due to the presence of a more stable
triple bond in N2 molecule whereas phosphorus forms only P-P single bond. Therefore,
phosphorus is more reactive than nitrogen.
ii)
Ans
Draw the structures of the following
a)
(HPO3)3
b)
H4P2O7
a)
Structure of (HPO3)3 is given below:
b)
Structure of H4P2O7 is given below:
3)
a)
P4O6 reacts with water according to equation P4O6 + 6H2O → 4H3PO3, Calculate
the volume of 0.1 M NaOH solution required to neutralize the acid formed by dissolving 1.1 g of
P4O6 in H2O.
73
P4O6 + 6H2O → 4H3PO3
Ans
H3PO3 + 2NaOH → Na2HPO3 + 2H2O] x 4
---------------------------------------------------------------------------P4O6 + 8NaOH →4Na2HPO3 + 2H2O
1 mol 8 mol
No. of moles in P4O6 =
.
1 mol P4O6 requires = 8 mol NaOH
.
mol P4O6 requires = 8 ∗
=
.
.
mol NaOH
=0.04 mol NaOH
Molarity of NaOH solution is 0.1 M
0.1 mol NaOH is present in = 1 litre
0.04 mol NaOH is present in =
.
x 0.04 litre
= 400 ml.
b)
Give an example to show the effect of concentration of nitric acid on the formation
of oxidation product.
Ans Dilute and concentrated nitric acid gives different oxidation products on reaction with
copper metal.
3Cu + 8HNO3 (dil.) → 3Cu(NO3)2 + 2NO + 4H2O
Cu + 4HNO3 (conc.) → Cu(NO3)2 + 2NO2 + 2H2O
NO and NO2 are the oxidation products obtained wilh dil. HNO3 and conc. HNO3 respectively.
4)
Give reasons for the following.
i)
Decomposition of O3 molecule is a spontaneous process.
ii)
SF6 is inert towards hydrolysis.
iii)
(CH3)3 P=O exists but (CH3)3 N=O does not.
iv)
Oxygen has less electron gain enthalpy with negative sign than sulphur.
v)
SO2 is an air pollutant.
Ans i)
Decomposition of O3 is an exothermic process (ΔH = -ve) and occurs with increase
in entropy (ΔS = +ve). These two effects reinforce each other which results in large negative
Gibbs energy change. It favours its decomposition into oxygen.
ii)
In SF6, S is surrounded by 6 F- octahedrally. Therefore, attack of water molecule
on S is sterically hindered.
iii)
Due to absence of d-orbitals,N cannot form pπ-dπ multiple bonds.As a result,N
cannot expand it’s covalency more than four,but in R3N=O,N has it’s covalency 5,therefore,the
compound R3N=O doesnot exist.On the other hand,due to the presence of d-orbitals in P,it forms
pπ-dπ multiple bonds,therefore can expand it’s covalency more than four.As a result P can form
(CH3)3 P=O.In this compound the covalency of P is 5.
iv)
The electron gain enthalpy of oxygen is less negative than sulphur due to its
compact size. As a result of which, the electron repulsions in the relatively compact 2p subshell
74
are comparatively large hence the incoming electrons are not accepted with the same ease as in
case of sulphur.
v)
SO2 is water soluble, therefore it dissolves in rainwater causing acid rain.
Moreover, when released in air, it mixes with it and leads to several diseases like eye irritation,
redness, asthma, bronchitis, etc. Thus, it is considered as an air pollutant.
5)
a)
Name the two most important allotropes of sulphur. Which one of the two is stable
at room temperature? What happens when the stable form is heated above 370 K?
Ans Two most important allotropes of sulphur are
i)
Rhombic sulphur (α-sulphur)
ii)
Monoclinic sulphur
The stable form at room temperature is rhombic sulphur which transforms to monoclinic sulphur
when heated above 369K
370K
Rhombic sulphur
Monoclinic Sulphur
b)
i)
Ans
i)
ii)
iii)
Draw the structures of –
H4S2O7
ii)
O3
iii)
S8
75
Crown Shape
6)
a)
i)
ii)
iv)
How would you account for the following –
The oxidizing Power of oxoacids of chlorine follows the order –
HClO4< HClO3< HClO2< HClO
The halogens are coloured.
F2 is a stronger oxidizing agent than Cl2
b)
Complete the following chemical equations.
i)
I2 + NaClO3→
ii)
I2 + H2O + Cl2→
Ans a)
As the oxidation number of halogen atom in oxoacid increases, its oxidizing power
decreases. Therefore, HClO is least stable and gives [O] most easily, so its oxidizing power is
greater than HClO4.
b)
All halogens are coloured due to the absorption of raditions in visible region which
results in the excitation of outer electrons to higher energy level. By absorbing different quanta of
radiation, they display different colours.
c)
Since Eo for F2/F- eleclrode is higher than that of Cl2/Cl- electrode  F2 is more
easily reduced than Cl2& is a stronger oxidizing agent than Cl2.
Ans
b)
i)
I2 + 2NaClO3→ 2NaI+2ClO2 + O2
ii)
I2 + 6H2O + 5Cl2→ 2HIO3 + 10HCl
Iodic acid.
2)
a)
How can you prepare Cl2 from HCl & HCl from Cl2. Write reactions only.
b)
Write the reactions of F2& Cl2 with water.
Ans a)
HCl can be oxidized to Cl2 by a number of oxidizing agents such as MnO2,
KMnO4, K2Cr2O7, etc.
Reaction with MnO2 is given below:
MnO2 + 4HCl → MnCl2 + Cl2+ 2H2O
Cl2 can be reduced to HCl by reaction of H2 in presence of diffused sunlight.
H2 + Cl2
> 2HCl
b)
F2 being a stronger oxidizing agent oxidises H2O to O2 or O3. The reactions are
given as under:
2F2 (g) + 2H2O (l) → 4H+ (aq) + 4F- (aq) + O2 (g)
2F2 (g) + 3H2O (l) → 6H+ (aq) + 6F- (aq) + O3 (g)
76
Cl2, on the other hand, reacts with H2O to form hydrochloric acid and hypochlorous acid as per
the following equation :
Cl2(g) + H2O (l) → HCl (aq)
Hydrochloric acid
8)
+
HOCl (aq)
Hypochlorous acid
a)
i)
ii)
iii)
Give reason for the following.
F2 is more reactive than ClF3 but ClF3 is more reactive than Cl2.
H3PO2 is a stronger reducing agent than H3PO3.
PCl5 is more covalent than PCl3.
b)
Complete the following chemical equation.
i)
XeF4 + O2F2→
ii)
XeF4 + SbF5 →
Ans a) i) Fluorine due to its small size, high electronegativity and low bond energy is more
reactive than ClF3 but bond energy of C-Cl bond is higher than Cl-F bond, therefore ClF3 is more
reactive than Cl2.
ii)
The structure of H3PO2 ,H3PO3&H3PO4 are as follows:
The acids which contain P-H bond, have strong reducing properties. Hypophosphorous
acid (H3PO2) contains two P-H bonds, whereas orthophosphorous acid (H3PO3) has one P-H
bond. Hence, H3PO2 is a stronger reducing agent than H3PO3.
iii)
Since, pentavalent metal ion has higher polarizing power than trivalent metal ion.
Thus, PCl5 is more covalent than PCl3.
b) i)
ii)

XeF4 + O2F2 
XeF6 + O2
XeF4 + SbF5→ [XeF3]+ [SbF6]-
77
CHAPTER – 8
d AND f BLOCK ELEMENTS
01 MARK
Q1.
Why EO values for Mn, Ni and Zn are more negative than expected?
Ans. Negative values for Mn2+ and Zn2+ are related to the stabilities of half-filled and fully filed
configurations respectively.
For Ni2+, EO value is related to the highest negative enthalpy of hydration.
Q2.
Transition metals show high melting points. Why?
Ans. The high melting points of transition metals are attributed to the involvement of greater
number of electrons in the interatomic metallic bonding from (n – 1)d orbitals in addition to ns
electrons.
Q3. When Cu2+ is treated with KI, a white precipitate is formed. Explain the reaction with the
help of chemical equation.
Ans. Reduction of Cu 2+ to Cu+ takes place due to reaction with I- ions. This is given by the
equation
2Cu2+ + 4I-
Q4.
Cu 2I2 + I2
Out of Cu2Cl2 and CuCl2, which is more stable and shy?
Ans. CuCl2 is more stable than Cu 2Cl2. Greater stability of Cu2+ (aq) is due to more negative
o
2+
+
∆hyd H of Cu (aq) than that of Cu (aq).
Q5. Although fluorine is more electronegative than oxygen, but the ability of oxygen to
stabilize higher oxidation states exceeds that of fluorine. Why?
Ans. The electronic configuration of fluorine is 1s22s22p 5. Thus it can form only one bond as it
has only one unpaired electron. Electronic configuration of oxygen is 1s22s22p63s23p4.
It may be mentioned that oxygen also has vacant d-orbitals along with two 3p orbitals
containing single electrons.Thus oxygen has greater ability to stabilize higher oxidation states.
Q6. Although +3 oxidation state is the characteristic oxidation state of lanthanoids but cerium
shows +4 oxidation state also. Why?
Ans. The electronic configuration of Ce is -4f15d16s2. Usually 5d 1 and 6s2 electrons are lost by
the lanthanoids in their reaction i.e., they exhibit +3 oxidation states. But Ce exhibit +4 oxidation
state also because it gains extra stability by losing 4f1 electron because it will give rise to
completely filled orbitals.
Q 7. Explain why does the colour of KMnO4 disappear when oxalic acid is added to its solution
in acidic medium.
78
Ans. KMnO4 acts as oxidizing agent. It oxidises oxalic acid to CO2 and is itself changed to
Mn2+ ions which are colourless. The reaction is given as under:5
+2
+ 16
(Coloured)
→2
+8
+ 10
(Colourless)
Q8. The halides of transition element becomes more covalentwith increassing oxidation state
of the metal. Why?
Ans. As the oxidation state of the element increases, its charge increases. According to Fajan’s
Rules, as the charge of the metal ion increases covalent character increases because the positively
charged cation attracts the electron cloud on the anion towards itself.
Q9. While filling up of the electrons in the atomic orbitals, the 4s orbital is filled before the 3d
orbital but reverse happens during the ionization of the atom. Explain why?
Ans. Filling of electrons is explained by (n+l) rule. Electrons will go to that level where the
value of (n+l) is minimum.
For 3d orbital n + 1 = 3 + 2 = 5
For 4s orbital n + 1 = 4 + 0 = 4
So the electrons will be filled first in 4s orbital and then in 3d orbital.
When it comes to ionization of the atom, we need to compare the ionization enthalpy.
Removal of electron from 4s requires less energy than removal of electron from 3d orbital.
Q10.
Reactivity of transition elements decreases almost regularly from Sc to Cu. Explain.
Ans. Reactivity of an element is dependent on the value of ionization enthalpy. In moving from
Sc, the first element to Cu, the ionization enthalpy increases regularly. Therefore, the reactivity
decreases as we move from Sc to Cu.
02 MARKS
Q 1. When a brown compound of manganese (A) is treated with HCl it gives a gas (B). The
gas taken in excess reacts with NH3 to give an explosive compound (C). Identify compounds A,
B and C.
Ans.
The compounds A, B and C are as follows:A = MnO2
B = Cl2
C = NCl3
The reaction are explained as under:+4
→
+
[A]
3
+2
[B]
+
→
+3
Excess [C]
Q2.
Complete the following chemical equations:(i)
(
)+
(
)+
( )→
79
(ii)
Ans.
(i)
8
(
)+
(
)+3
8
(ii)
(
2
(
)+
(
+6
)+6
+6
(
)+
)→
()→
+2
(
) + 14
(
)→
+7
Q3. When orange solution containing
ion is treated with an alkali, a yellow solution is
formed & when H+ ions are added to yellow solution, an orange solution is obtained. Explain
why does this happen?
Ans.
This is due to the following inter conversion:OHCr2O72
CrO42
H+
Dichromate (orange) Chromate (Yellow)
Q4.
Assign a reason for each of the following observations:(i)
The transition metals (with the exception of Zn, Cd and Hg) are hard and have high
melting and boiling points.
(ii)
The ionization enthalpies (first and second) in the first series of the transition
elements are found to vary irregularly.
Ans. (i)
The transition metals (except Zn, Cd and Hg) are hard and have high melting and
boiling metallic points. They show strong metallic bonding. Greater the number of valence
electrons stronger is the resultant bonding.
(ii)
Ionisation enthalpy increases with increase in nuclear charge along each series.
However, its value for Cr is lower because of the absence of any change in d-configuration (d5)
and the value for Zn is higher because it represents an ionization from the completely filled 4s
level..
Q5.
Assign reason for each of the following:(i)
Transition elements exhibit paramagnetic behavior.
(ii)
Co 2+ is easily oxidized in the presence of a strong ligand.
Ans. (i)
Transition metal ions have unpaired electrons in d-orbitals (d1 to d9).
Paramagnetism arises due to the presence of unpaired electrons. Therefore, they exhibit
paramagnetic behaviour.
(ii)
In Co 2+, electronic configuration is 3d 7 there is one unpaired electron even after
pairing occurs in the presence of a strong ligand. Hence, Co 2+ is oxidized to more stable Co 3+
80
3 marks
Q1. (a)
Transition metals can act as catalysts because these can change their oxidation
state. How does Fe(III) catalyse the reaction between iodide and persulphate ions?
(b)
Mention any three processes where transition metals act as catalysts.
Ans.
(a)
Reaction between iodide and persulphate ions is:
2 I   S 2 O82 Fe
( III
)  I 2  2SO42
Role of Fe(III) ions:2Fe3  2I   2Fe 2  I 2
2 Fe 2  S 2 O82  2 Fe3  2SO42
Q2.
(b)
(i)
Vanadium (V) oxide in contact process for oxidation of SO2 to SO3.
(ii)
Finely divided iron in Haber’s process in conversation of N2 and H2 to NH3.
(iii)
MnO2 in preparation of oxygen from KClO3.
Assign suitable reasons for the following:(i)
The Mn2+ compounds are more stable than Fe2+ towards oxidation to their +3 state.
(ii)
In the 3d series from Sc (Z=21) to Zn (Z=30), the enthalpy of atomization of Zn in
the lowest.
Ans.
(iii)
Sc3+ is colourless in aqueous solution whereas Ti3+ is coloured.
(i)
Electronic configuration of Mn2+ = [Ar] 3d5.
Electronic configuration of Fe2+ = [Ar] 3d 6.
It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn
in (+2) state has a stable d5 configuration. Due to this Mn2+ is resistant towards
oxidation to Mn3+.
Also Fe2+ has 3d6 configuration and by losing one electron, its configuration
changes to a more stable 3d5 configuration. Therefore, Fe2+ easily gets oxidized to
Fe3+ oxidation state.
(ii)
Transition metals have high enthalpies of atmoisation due to strong metallic
bonding and addition covalent bonding. Metallic bonding is due to their smaller size
while covalent bonding is due to d-d overlapping.
But zinc has completely filled d-orbitals so it cannot overlap with other atom to show d-d
overlapping.
(iii)
Sc3+ is colourless because of d o – orbitals configuration here d-d transition is
forbidden. But in Ti3+ due to presence of one electron in d-orbital (d1) d-d transition are
possible (allowed).
Q3.
How would you account for the following?
(i)
Cr2+ is reducing in nature while with the same d-orbital configuration (d4) Mn3+ is
an oxidizing agent.
81
(ii)
In a transition series of metals, the metal which exhibits the greatest number of
oxidation states occurs in the middle of the series.
(iii)
Ans.
What is misch metal. Mention its two important uses.
(i)
Cr2+ is a reducing agent as its configuration changes from d4 to d 3, the latter having
a half-filled t2g level. On the other hand, the change from Mn2+ to Mn3+ results in the halffilled (d5) configuration which has extra stability. Thus, it behaves as oxidizing agent.
(ii)
This is due to large number of unpaired electrons in d-orbitals in the middle of the
series.
(iii)
Misch metal consists of lanthanoids (95%) iron, traces of S, C, Ca & Al.
It is used to produce bullets.
It is used in shells & lighter flints.
Q4. What is meant by disproportionation? Give two examples of disproportionation reactions
in aqueous solution.
Ans. The disproportionation reactions are those in which the same substance gets oxidized as
well as reduced. When a particular oxidation state becomes less stable relative to other oxidation
states, one lower and one higher, it undergoes disproportionation.
e.g. Mn (VI) becomes unstable relative to Mn (VII) and Mn (IV) in acidic solution.
6
7
4
3 Mn O42  4 H   2 Mn O4  Mn O2  2 H 2 O
Many copper (I) compounds are unstable inaqueous solution and undergo disproportionation.
2Cu   Cu 2 Cu
Q5.
Account for the following:(i)
Oxidising power in the series VO 2  Cr2 O72  MnO4 .
(ii)
Actinoid contraction is greater from element to element than lanthanoid
contraction.
Ans.
(iii)
Oxoanions of a metal show higher oxidation state.
(i)
The oxidizing power of the species varies in the order:-
VO 2  Cr2 O72  MnO4
This is explained on the basis of increasing stability of the lower species to which they are
reduced.
(ii)
Actinoid contraction is greater from element to element than lanthanoid
contraction due to poor shielding of 5f electrons compared to that of 4f electrons in
lanthnoids.
82
(iii)
Oxoanions of a metal shows higher oxidation state due to the ability of oxygen to
form multiple bonds and its high electronegativity.
05 MARKS
1)
When a chronite ore (A) is fused with sodium carbonate in free excess of air and the
product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of
this yellow solution with sulphuric acid, compound (C) can be crystallised from the solution.
When compound (C) is trated with KCl, orange crystals of compound (D) crystallise out. Identify
(A) to (D) and also explain the reactions.
Ans The compounds A, B, C and D are given as under:
A = FeCr2O4
B = Na2CrO4
C=Na2Cr2O72H2O
D = K2Cr2O7
The reactions are explained as under :
4FeCr2O4 + 8Na2CO3 + 7O2→8Na2CrO4 + 2Fe2O3 + 8CO2
(A)
(B)
2NaCrO4 + 2H+→ Na2Cr2O7 + 2Na+ + H2O
Na2Cr2O7 + 2KCl →
(C)
K2Cr2O7 + 2NaCl
(D)
2)
When an oxide of manganese (A) is fused with KOH in the presence of an oxidizing agent
and dissolved in water, it gives a dark green solution of compound (B). Compound (B)
disproportionates in neutral or acidic solution to give purple compound (C). An alkaline solution
of compound (C) oxidises potassium iodide solution to a compound (D) and compound (A) is also
formed. Identify compounds (A) to (D) and also explain the reactions involved.
Ans. The compounds (A), (B), (C) and (D) are given as under :
(A)
MnO2
(B)
K2MnO4
(C)
KMnO4
(D)
KlO3
The reactions are explained as under:
2MnO2 + 4KOH + O2→ 2K2MnO4 + 2H2O
(A)
(B)
3MnO42- + 4H+→ 2MnO-4 + MnO2 + 2H2O
(C)
2MnO-4 + H2O + KI → 2MnO2 + 2OH- + KIO3
(A)
(D)
3)
A violet compound of manganese (A) decomposes on heating to liberate oxygen and
compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the
presence of potassium nitrate to give compound (B). On heating compound (C) with conc. H2SO4
and NaCl, chlorine gas is liberated and a compound (D) of manganese along with other products
is formed. Identify compounds (A) to (D) and also explain the reactions involved,
Ans The compounds A, B, C and D are given as under:
A = KMnO4
B = K2MnO4
C = MnO2
D = MnCl2
The reactions are explained as under

KMnO4 
K2MnO4 + MnO2 + O2
(A)
(B)
(C)
83
MnO2 + KOH + O2→2K2MnO4 + 2H2O
MnO2 + 4NaCl + 4H2SO4→ MnCl2 + 2NaHSO4 + 2H2O + Cl2
4)
On the basis of Lanthanoid contraction explain the following –
i)
Nature of bonding in La2O3 & Lu2O3
ii)
Trends in the stability of oxosalts.
iii)
iv)
v)
Stability of the complexes of lanthanoids.
Radii of 4dand 5dblock elements.
Trends in acidic character of lanthanoid oxides.
A 4) (i) As the size decreases covalent character increases. ThereforeLa2O3is more ionic
andLu2O3is more covalent.
5)
(ii)
(iii)
(iv)
(v)
As the size decreases from La to Lu, stability of oxosalts alsodecreases.
Stability of complexes increases as the size of lanthanoids decreases.
Radii of 4dand 5dblock elements will be almost same.
Acidic character of oxides increases from La to Lu.
i)
Complete the following equations:
a)
Cr2O72 + 2OH-→
MnO4 + 4H+ + 3e-→
Account for the following :
a) Zn is not considered as a transition element.
b) Transition metals form a large number of complexes.
c) The Eo value for the Mn3+ / Mn2+ couple is much more positive than that for
Cr3+ / Cr2+ couple.
b)
ii)
Ans
i)
a)
Cr2O72- + 2OH-→ 2CrO42-+ H2O
b)
MnO4- + 4H+ + 3e-→ MnO2 + 2H2O
ii)
a)
Zn has completely filled d-oritals in its atomic as well as in its common
oxidation states (Zn2+ state). Therefore it is not regarded as transition element.
b)
Due to the comparatively smaller size of the metal ions, the high ionic
charges and the availability of vacant d-orbitals for bond formation, transition metals form
a large number of complex compounds.
c)
Mn3+ (3d4) is less stable than Mn2+ (3d 5), which has stable half-filled
configuration. Cr3+ has stable t2g configuration, therefore, Cr3+ cannot be reduced to Cr2+.
So, Eo vaule for the Mn3+ / Mn2+ couple is much more positive.
84
CHAPTER – 9
COORDINATION CHEMISTRY
01 MARK
Q1.
Define ambidentate ligand with an example ?
A1. Ligand that can ligate through two different atoms eg NO - ion can ligate through N
or through oxygen.
Q2.
Draw structures of geometrical isomers of [Fe (NH )2 (CN)4]NH
A2.
NH
NH
NC
NC
CN
Fe
NC
Fe
CN
NC
CN
NH
CN
cis isomer
trans isomer
Q3.
Calculate the magnetic moment of Mn
A3.
Mn
Q4.
Give the IUPAC name of [NiCl ]2-
A4.
Tetrachloridonickelate (II)
Q5.
Out of [Fe(C O )3]3- and [FeCL ]3-, which is more stable and why?
A5.
[Fe(C O )3]3-, as it is a chelate complex.
3d 45
μ=
ion?
( + 2) =
5(7) = √35BM
02 MARKS
Q1. What is spectro chemical series? What is the difference between a strong field
ligand and a weak field ligand?
A1. Arrangement of ligands in increasing orde of field strength is called spectro
chemical series.
A strong field ligand produces greater splitting in the d-orbitals, the complex is
likely to be low – spin. A weak field ligand, on the other hand produces lesser splitting,
and complex is likely to be high – spin.
Q2. Using valence bond, approach, deduce the shape and magnetic character of [Co
(NH ) 6]3+ ion. (Atomic no of Co = 27)
85
A2.
27Co
CO
3d74s
3d64s o
CO
3d
4s
4p
Co3+ assumes d2sp3 hybridization and accepts six pairs of e-from 6NH
moleculesShape – octahedral Magnetic nature – Diamagnetic
Q3. How is coordination chemistry useful in medicine? Where is Wilkinson’s catalyst
used?
A3. Chelates are being used to beat metal poisoning. E.g EDTA is used to treat lead
poisoning. Wilkinson’s catalyst [(Ph P)3 Rh Cl] is used for hydrogenation of alkenes.
Q4. Write the structure of the complex Pentaamminenitrito – N – Cobalt (III). Which
isomerism can this complex exhibit?
A4.
[Co (NH )5 (NO )] 2+, Linkage isomerism.
Q5.
A solution of [Ni(H2O)6]2+ is green , but a solution of [Ni(CN)4]2- is colorless. Why?
A5.
Ni
3d 8 4 s o
H O is a weak ligand, so high-spin complex is formed (sp3d2), leaving two
unpaired e-.
CN- being a strong ligand, produces a low spin square planar (dsp2)complex
leaving no unpaired e-.
03MARKS
Q1. List the postulates of Werner’s Theory for coordination compounds on the basis of
the theory, assign primary and secondary valencies to Pd in [Pd (NH )4]Cl2
A1.
(a)
In coordination compounds metals show two types of linkages (valences) –
primary and secondary.
(b)
Primary valencies are ionisable and satisfied by –ve ions.
(c)
Secondary valencies are non-ionisable and satisfied by neutral or –ve ions.
Secondary valency is equal to coordination number and is fixed for a metal.
(d)
The ions/groups bound by secondary linkages to the metal having
characteristic spatial arrangements corresponding to different coordination
numbers.
[Pd (NH )4]Cl2
Primary valence - +2
Secondary valence – 4
86
Q2.
Discuss the nature of bonding in metal carbonyls?
A2.
The M-C bond in metal carbonyls possess both s and p character. M-C sigma
bond is formed by donation of lone pair of e on the carbonyl carbon into a vacant
orbital of the metal. The M-C π bond is formed by donation of a pair of e from a
filled d-orbital of metal into the vacant anti bonding π∗ orbital of CO. The metal –
ligand bonding creates a synergic effect which strengthens the bond between CO
and the metal
Q3.
Explain the following :-
A3.
(a)
Low spin octahedral complxes of Ni aren’t known.
(b)
π- complexes are known only for transition elements.
(c)
CO is a stronger ligand than NH for many metals.
(a)
For octahedral complex, coordination number should be 6. Since Ni has
8
[Ar]3d 4s configuration, i.e only one d orbital is empty, it can form only high spin
complexes.
(b)
Transition metals have empty d –orbitals to accommodate
ligands.
(c)
Q4.
donated by
Due to small size and ability to form π-bonds alogwith normal sigma bonds.
For the complex [Fe(en)2 Cl2]Cl, identify the following :(i)
Oxidation number of iron.
(ii)
Hybrid orbitals and shape of the complex.
(iii)
Magnetic behaviour of the complex.
(iv)
Number of its geometrical isomers.
(v)
Whether there may be optical isomers
(vi)
Name of the complex.
87
A4.
(i)
x + 2(0) + 2 (-1) + (-1) = 0
oxidn no +3
(ii)
d sp octahedral geometry
(iii)
paramagnetic (1 unpaired e-)
(iv)
Two geometrical isomers (CIs, trans)
(v)
Yes, due to polydentate ligends
(vi)
Dichloridobis – (ethane -1,2- diamine) iron (III) chloride.
Q5. What will be correct order of wavelengths of absorption in the visible region for the
following :[Ni(NO )6]4- , [Ni(NH )6]2+, [Ni(H O)6]2+
A5. The strength of ligands according to sepctrochemical series NO ->NH >H O
Therefore splitting produced bythem in d-orbitals will also be in the same oreder (or O
Will be maximum in NO -, followed by NH and H O). Hence , energy absorbed will be in
the same order. Since E α , wavelength absorbed will be in the opposite order. i.e
[Ni(NO )6]4-< [Ni(NH )6]2+< [Ni(H O)6]2+.
88
CHAPTER – 10
HALOALKANES AND HALOARENES
1 Mk Questions:
1)
Write the IUPAC name of the following compound :
CH3
H3C – C – CH2Cl
CH3
Ans
1 – Chloro – 2, 2 – dimethylpropane
2)
Ans
Draw the structure of 2-bromopentane
H3C – CH2 – CH2 – CH – CH3
Br
3)
Write a chemical reaction in which iodide ion replaces the diazonium group in a
diazonium salt.
N2+ Cl- + KI(aq) . >
Ans
I + N2 + KCl
4)
Out of CH3 CH Cl CH2 CH3 and CH3 CH2 CH2 CH2 Cl which one is hydrolysed more
easily by aq. KOH?
Ans CH3 CH – CH2 CH3 , as it’s a secondary halide
|
Cl
5)
How can you convert methylbromide to methylisocyanide in a single step?
Ans
CH3Br
.
> CH3NC + Ag Br
2 Mks Questions
1)
Ans
What are ambident nucleophiles? Explain with an example.
Ambident nucleophiles have two nuclophilic sites through which they can attack.
Eg. [ΘC  N ↔: C = NΘ] linking through C results in alkyl cyanides and through N results
in isocyanides.
2)
Write chemical equations when
i) ethylchloride is treated with aq.KOH
ii) chlorobenzene is treated with CH3COCl in presence of anhyd.AlCl3.
89
Ans
i) C2H5Cl
.
> C2H5OH + KCl
COCH3
ii)
>Cl
Cl
2-chloroacetophenone
3)
Cl + HCl
4-chloroacetophenone
Write the mechanism of the following reaction:
N – BuBr + KCN
Ans
+ H3COC
,
> n – BuCN
CN- is an ambident nucleophile  it can attack through C and N
C-C bond is stronger than C-N bond, cyanide is formed.
4)
Ans
5)
Ans
Give reasons:
a)
The order of reactivity of haloalkanes is RI>RBr>RCl
b)
Neopentyl chloride, (CH3)3C – CH3 Cl doesn’t follow SN2 mechanism.
a)
Larger the size of halogen atom, weaker the C-X bond,
R group remaining same i.e; bond strength follows the order
R-I <R-Br<R-Cl, hence reactivity is opposite.
b)
R – group being bulky, provides steric hindrance to the incoming nucleophile.
Why does Wurtz reaction fail in case of tertiary halides?
3o halides prefer to undergo dehydrohalogenation in presence of a base like Na metal.
CH3
CH3
|
(CH3)3 C-Na+ + H – CH2 – C – Br
> (CH3)3 CH + CH2 = C – CH3
|
CH3
3 Mk Questions:
1)
Ans
Write a chemical test to distinguish between
a)
Chlorobenzene and benzyl chloride
b)
Chloroform and carbontetrochloride
c)
n-propyl bromide and isopropyl bromide
a)
Benzyl chloride gives a white ppt with AgNO3 Soln., chlorobenzene does not.
b)
Chloroform gives offensive smelling isocyanides, on heating with 1o amine and
KOH, CCl4 doesn’t give this test.
c)
Isopropyl bromide, on treating with aq.KOH, gives Propan-2-ol which gives a
yellow ppt. on heating wiith NaOH and I2 n- propyl bromide doesn’t answer this test
90
2)
Complete the following reactions:
i)
ii)
iii)
Ans
i)
ii)
iii)
3)
Give reasons:
a) p-dichlorobenzene has higher melting point than o- and m- isomers.
b) Haloarenes are less reactive than haloalkanes towards nucleophilic substitution
reactions.
c) Alkyl chloride gives alcohol with aq.KOH, but alkene with alc. KOH.
Ans
a)
p-isomer has more symmetric structure, so molecules can pack closely.
b)
The C-X bond has partial double – bond character due to resonance. Also, in
haloalkanes, X is bonded to sp3 – C while in haloarenes, X is bonded to sp2 –C. sp3-C is less
electronegative, releases electrons to halogen more easily.
c)
In aq.solution, KOH is almost completely ionized to give OH- ions, which being a
strong nucleophile, gives a substitution product. An alcoholic soln. of KOH contains alkoxide
ions (RO-), which being a strong base, preferentially eliminates an HCl molecule to form alkenes.
4)
Convert:
a)
Ethene to ethanol
b)
Chlorobenzene to toluene
c)
Chlorobenzene to diphenyl
Ans
a)
b)
c)
CH2 = CH2 + H2O
.
.
> CH3CH2OH
91
5)
What are enantiomers? Identify the asymmetric carbons in the following molecule:a b
c
d
HOOC CH (OH) – CH (OH) COOH
Ans
Enantiomers are non-superimposable mirror images of an optically active compound.
b and c are asymmetric carbon atoms as they are bonded to 4 different groups.
6)
Write short notes on :
a) Gatterman reaction
b) Wurtz reaction
c) Peroxide effect or Kharasch effect
Ans a) Aromatic 1o amines produce benzene diazonium salts with HNO2 (produced in situ) at
273 K. Chloro arenes and bromoarenes can be prepared using Cu/HCl or Cu/HBr from these
diazonium salts.
b)
Alkyl halides, when treated with sodium in presence of dry ether, produce an
alkane with even number of C atoms.
2 R-X + Na
E.g. 2 CH3 Br + Na
> R _ R + 2NaX
.
> CH3 – CH3 + 2NaBr
If a mix of two different alkyl halides is taken, then a mixture of alkanes is obtained which is
difficult to separate.
E.g. CH3 Br + C2H5Br
> CH3-CH3 + C2H5 + CH3-C2H5
c)
The addition of Hydrogen bromide to unsymmetrical alkenes in presence of a
peroxide takes place in such a way that H goes to that C which has lesser H atoms and Br goes to
the C with greater number of H atoms.
CH3 – CH = CH2 + H Br
(
)
> CH3 – CH2 – CH2 Br
This rule doesn’t apply to addition of HCl or HI
7)
Explain the following with an example each
a)
Swarts reaction
b)
Finkelstein reaction
c)
Hunsdiecker reaction
Ans a) Flourination of hydrocarbons directly with F2 occurs explosively due to the large
amount of energy released. Hence, they can be conveniently prepared indirectly by halogen
exchange with chloro and bromoalkanes.
CH3Br + AgF → CH3F + AgBr
2 C2H5Cl + .Hg2F2 → 2C2H5F + Hg2Cl2
92
b) Iodoalkanes can be easily prepared from chloroalkanes or bromoalkanes by heating
with NaI in acetone.
R-Cl + NaI
R –Br + NaI
> R – I + NaCl
> R – I + NaBr
c)
Bromoalkanes can be prepared by refluxing the silver salt of a carboxylic acid with
Bromine in CCl4.
CH3CH2 COOAg + Br2
> CH3CH2Br + CO2 + AgBr
The yield of alkyl bromide is 1o>2o>3o
8)
Explain why.
a)
Vinyl chloride is unreactive in nuceophilic substation reaction
b)
3-Bromocyclohexene is more reactive than 4-Bromocyclohexene in hydrolysis
with aq.NaOH
c)
tert-butyl chloride reacts with aq.NaOH by SN1 mechanism, while n-butyl chloride
reacts with aq.NaOH by SN2 mechanism.
Ans a)
Vinyl chloride is unreactive in nucleophikic substitution reactions due to
reasanance which results in a partial double bond character of C-Cl bond which is difficult to
break.
b)
3-bromocyclohexene forms allyl carbocation which is more stable than carbocation
formed by 4-Bromocycohexene
c)
Tertiary carbocation is stable, so tert-butyl chloride follows SN1 mechanism. Nbutyl chloride would form 1o carbocation which isn’t that stable. Hence it undergoes SN2
mechanism through formation of transition state.
9)
Identify the products A and B formed in the following reaction:
CH3CH2 CH = CH –CH3 + HCl → A + B
Ans Since both the doubly-bonded C atoms have same number of H atoms, Markonikov’s rule
becomes irrelevant.
So, products will be → CH3 – CH2 – CH – CH2 – CH3
CH3CH2 CH2CH CH3
|
(A) and
|
(B)
Cl
Cl
3-chloropentane
2-choropentane
10)
Ans
What happens when:
a)
n-butyl chloride is treated with alc.KOH
b)
Bromobenzene is treated with Mg in presence of dry ether.
c)
Chlorobenze is subjected to hydrolysis.
a)
CH3CH2 CH2CH2Cl - alc.KOH → CH3-CH2-CH=CH2+KCl + H2O
But – 1 – ene
93
a) Br + Mg
a)
>Mg Br
Cl2 + aq.NaOH → no reaction.
CHAPTER – 11
ALCOHOLS, PHENOLS AND ETHERS
1 Mark Questions
1)
A)
Write the IUPAC name of CH3 CH= CH – CH- CH2 – CH3
|
OH
Hex – 4 –en – 3 –ol
2)
A)
Why is the boiling point of C2H5OH higher than that of CH3OH ?
Due to more number of C atoms, van der waal’s forces increase.
3)
A)
Name the reagent used in bromination of phenol to 2,4, 6 – tribromophenol.
Aqueous solution of bromine (Bomine water)
4)
A)
Name the alcohol used to prepare the ester CH3 – C OOCH(CH3)CH3
Propan-2-ol CH3-CH(OH)-CH3
5)
A)
Which is more volatile – o-nitrophenol or p-nitrophenol?
o-nitrophenol,, as it has intramolecular H-bonding, whereas p-nitrophenol has
intermolecular H-bonding.
O-Nitrophenol with intra molecular hydrogen bonding
94
02 MARKS
1)
A)

How will you convert:
a)
Propene to propan – 2 – ol
b)
Ethyl chloride to ethanol
a)
CH3-CH = CH2 H2O/H+ CH3 – CH – CH3
|
OH
Propan – 2 – ol
b)
2)
CH3 -CH2- Cl aq – NaOH CH3 CH2 OH [o] CH3 - CHO
-----------
------- Ethanal
CrO3/PCC
Write the mechanism for the following reaction:
CH3CH2OH
A)
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ CH3CH2Br + H2O
1 o alc undergoes the reaction by SN2 mechanism
Protonated Alcohol
3)
A)
How would you obtain
i)
Picric acid from phenol?
ii)
2- Methylpropan-2-ol from 2- Methylpropene?
i)
95
ii)
4)
A)
5)
A)
Give reasons :
a)
Boiling points of alcohols decrease with increase in branching of the alkyl chain.
b)
Phenol does not give protonation reaction readily
a)
Because increase in branching of the alkyl chain reduces surface area, so
intermolecular forces of attraction decrease.
b)
Because the lone pair on oxygen is delocalized over the benzene ring due to
resonance, hence not available for protonation easily.
Explain why cleavage of phenyl alkyl ether with HBr always gives phenol and alkyl
bromide.
Due to resonance, the O-C bond in phenyl alkyl ether has a partial double bond character,
hence it is difficult to break.
Also, phenoxide ion is stabilized by resonance, hence we get phenol and alkyl bromide.
03 MARKS
1)
A)
Write a short note on Williamson’s Synthesis.
An alkl halide, on treating with a suitable sodium alkoxide gives an ether
RONa + RX - R– O – R + NaX
Symmetric and unsymmetric ethers can be prepared by this method
Aryl alkyl ethers can be prepared as shown
But there is no reaction if arylhalide and sodium alkoxide are taken
Best yields of unsymmetrical ethers are obtained when alkyl halides are primary with 2o
and 3 o alkyl halide, dehydrohalogenation occurs to give alkene.
96
2)
Convert:
a)
Methyl magnesium chloride --2-methyl propan-2-ol
b)
Benzyl chloride --- Benzyl alcohol
c)
Phenol -- benzoquinone
A)
a)
propanone
b)
c)
3)
A)
Give reasons:
a) The boiling point of ethanol is higher than methoxymethane.
b) Phenol is more acidic than ethanol
c) O- and p-nitrophhenols are more acidic than phenol.
a)
Ethanol has intermolecular H-bonding while methoxymethane only has dipoledipole forces.
b)
In phenols, O atom acquires a partial positive charge due to resonance, this
weakens O-H bond, release of H+ is easy. Also, phenoxide ion is stabilised by
resonance. On the other hand, alkoxide ion is destabilized due to the +I effect of
alkyl group.
c)
-NO2 group, being electron-withdrawing, stabilizes the o- and p- nitrophenoxide
ions.
97
4)
A)
Describe the mechanism of hydration of ethene to yield ethanol.
Protonation of Alkene
Nucleophilic attack by water on the carbocation
Deprotonation
5)
Give equations for Reimer Tiemann reaction and Kolbe’s reaction.
A)
Reimer Tiemann reaction – Phenol reacts with chloroform in the presence of alkali to give
o- and p-hydroxy aldehyde.
Kolbe’s reaction – Phenoxide ion generated by treating phenol with NaOH undergoes
electrophilic substitution with CO2
98
6)
A)




7)
A)
8)
Arrange the following compounds in increasing order of their acid strength.
Propan-1-ol, 2, 4, 6-Trinitrophenol, 3 – Nitrophenol,
3, 5- Dinitrophenol, phenol, 4-Methylphenol, 4-Methoxyphenol
Propan-1-ol <4-Methoxyphenol<4-Methylphenol<phenol< 3-Nitrophenol
<3, 5 – Dinitrophenol<2, 4, 6 – Trinitro phenol
Propan – 1 ol has R group, which destabilizes the propoxide ion due to +I effect, hence
weaker than all phenols.
All e- donating groups reduce acidity of phenols by destabilizing phenoxide ion and ewithdrawing groups increase acidity.
Methoxy group is stronger e- - donating groups compared to methyl group.
Larger the number of e- withdrawing groups, greater the acidity. 2, 4, 6 – Trinitrophenol
is strongest, followed by 3, 5 – Dinitrophenol and 3 – Nitrophenol.
Give a simple chemical test to distinguish between:
a)
Phenol and cyclohexanol
b)
Propan-2-ol and Benzyl alcohol
c)
Propan-1-ol and 2-Methylpropan-2-ol
a)
Phenol produces a deep violet colour with neutral FeCl3 solution., cyclohexanol
doesn’t.
b)
Propan-2-ol gives yellow ppt of iodoform with NaOH and I2, benzyl alcohol
doesn’t.
c)
On treating with conc. HCl and anhydrous ZnCl2 at room temperature, 2methylpropan-2-ol produces turbidity immediately while propan-1-ol doesn’t
produce turbidity at room temperature.
(
)
Give the mechanism for the reaction: 2C2H5OH ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ C2H5O C2H5
A)
Protonation of Alcohol
Nucleophillic attack by Alcohol
Deprotonation
99
9)
An organic compound A (C6H6O) gives a characteristic colour with aq. FeCl3 Solution.
when (A) is treated with CO2 and NaOH at 410 K under pressure, it gives compound (B) which on
acidification gives compound (C). Compound (C) reacts with acetyl chloride to give (D), which is
a popular pain killer. Deduce the strutures of A,B,C and D and explain all reactions
A)
(A) C6H6O must be phenol
Phenol gives sodium salicylate with CO2 which on acidification gives salicylic acid.
Salicylic acid forms aspirin with acetyl chloride.
10)
A
(a)
Explain hydroboration-oxidation reaction with an example
Diborane reacts with alkenes to give trialkyl boranes as addition product. This is
oxidized by H2O2 to alcohol, in the presence of aq.NaOH
(b)
Identify A B C in the following reaction.
CH3Br
A
B
C
(
)
⎯⎯⎯ A ⎯⎯⎯⎯ B ⎯⎯⎯⎯⎯⎯⎯⎯ C
CH3CN
CH3CH2NH2
CH3CH2N2Cl (Unstable, so forms CH3CH2OH + N2 + HCl with Water
100
CHAPTER – 12
ALDEHYDES – KETONES AND CARBOXYLIC ACIDS
01 MARK
1)
Write the structure of p-methylbenzaldehyde.
A)
2)
Complete the reaction :
.
+ CH3COCl ⎯⎯⎯⎯⎯
A)
Acetophenone
3)
Arrange the following in increasing order of their acid strength (CH3)2 CHCOOH,
CH3 CH2 CH(Br) COOH, CH3 CH(Br) CH2 COOH.
A)
(CH3)2 CHCOOH < CH3 CH(Br)CH2COOH<CH3CH2CH(Br)COOH
4)
Write IUPAC name of
A)
5 – Bromo -3- chlorobenzoic acid
5)
A)
Give a chemical test to distinguish between acetaldehyde and benzaldehyde.
On heating with Na OH and I2, acetaldehyde gives a yellow ppt of iodoform,
benzaldehyde doesn’t
101
02 MARKS
1)
a)
A)
Convert:
Ethanol to 1,2 Ethanediol
a)
(b) Phenol to acetophenone
(
)
C2H5OH ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ CH2 = CH2 + H2O ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
Ethane – 1, 2 - diol
b)
3)
Give reasons:
a)
b)
Monochloroethanoic acid is a weaker acid than dichloroethanoic acid.
Benzoic acid is stronger than ethanoic acid
A) a)
Cl has – I effect, making release of protons easier. Also, it stabilises the carboxylate ion.
Monochloroacetic acid has one Cl atom. While dichloroacetic acid has two Cl atoms.
Hence it is a stronger acid.
b) Benzene ring is e-withdrawing, while CH3 group is e- donating. Hence, release of
protons is easier in benzoic acid than ethanoic acid. Also, the benzoate ion is much more
stable than ethanoate ion due to e- - withdrawing benzene ring.
4)
Describe Cannizaro’s reaction with example.
A)
Aldehydes without α-H undergo self-oxidation and reduction (disproportionation) on
treatment with conc. Alkali. One molecule is reduced to alcohol while the other is oxidized to salt
of corresponding carboxylic acid
102
5)
Complete the equations
A)
a)
a)
(
CH3CHO ⎯⎯⎯
CH3CH2OH
.
b) CH3COCH3 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
b) CH3 CH2CH3
03 MARKS
1)
Explain Aldol condensation with an example.
A)
Aldehydes or ketones having at least one α-H undergo condensation to produce β-hydroxy
aldehyde (aldol) in the presence of a dil. Base.
.
⎯⎯⎯⎯⎯⎯
∆(
)
CH3CH(OH)CH2CHO ⎯⎯⎯⎯⎯⎯⎯ CH3CH=CH-CHO
3-Hydroxybutanal
Eg.
If two different molecules of aldehydes and/or ketones undergo aldol condensation,
the reaction is called CROSS ALDOL REACTION.
Eg.2CH3CHO
2)
Give reasons
a)
There are two –NH2 groups in semicarbazide, but only one is involved in the
formation of semicarbazones:
b)
Cyclohexanone
forms
cyanohydrins
in
good
yield,
but
2,2,6Trimethylcyclohexanone doesn.t.
c)
Aldehydes are more reactive than ketones towards nucleophilic addition.
A)
a) One of the NH2 groups is involved in resonance, hence it cannot act as a nucleophile.
Other one is free,  it can act as a nucleophile.
b)
CN-can attack easily without any steric hindrance in cyclohexanone.
Lot of CH3 groups offer too much steric hindrance.
103
c) Aldehydes have one R group and ketones have two.Two R groups cause more steric
hindrance and greater +I effect reduces the electrophilicity of the carbonyl carbon.
3)
How will you convert:
a)
Propanone to propane
b)
Benzoyl chloride to benzaldehyde
c)
Ethanal to but-2-enal
A)
a)
b)
c)
4)
How will you bring about the following conversions?
a)
Benzene to Benzophenone
b)
Propyne to acetone
c)
p-nitrotoluene to p-nitrobenzaldehyde
Ans 4 a)
b)
104
c)
5)
An organic compound A contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The
molecules mass of the compound is 86. It doesn’t reduce Tollen’s reagent but forms an addition
compound with sodium hydrogen sulphite and gives positive iodoform test. On vigorous
oxidation, it gives ethanoic abd propanoic acids. What is the structure of compound A?
Ans
Element
C
H
O
%
Molar Mass
69.77
12
11.63
1
18.6
16
Moles
5.81
11.63
1.16
Simple ratio
5
10
1
Empirical formula C5H10O
Empirical formula mass = 5(12) + 10(1) + 1(16) = 86
n=
=1
 Moecules formula C5H10O
Since it doesn’t give Tollen’s test but gives positive iodoform test. It is a methyl ketone i.e it has –
C – CH3 group. Since on oxidation it gives ethanoic and propanoic acid, it must be pentan-2-one.
CH3 CH2 CH2 CO CH3→CH3COOH + CH3 CH2 COOH
Pentan-2-one
05 MARKS
1)
Ans
a)
i)
ii)
Draw the structures of:
5-Chloro-3methyl-pentan-2-one
p-nitropropiophenone
b)
i)
ii)
iii)
Give simple chemical tests to distinguish between:
Ethanal and propanal
Phenol and benzoic acid
Benzaldehyde and acepopheone
a) i)
ii)
105
b)
2)
i) CH3 CHO gives a yellow ppt of iodoform with NaOH/I2 (positive iodoform test)
CH3CH2 CHO doesn’t answer this test.
ii) Phenol produces deep violet colour with aq-neutral FeCl3 Soln., benzoic acid
doesn’t answer the test.
iii) Acetophenone produces yellow ppt on heating with NaOH+I2, benzaldehyde
doesn’t.
Identify A,B,C,D and E in the following sequence:
A ⎯⎯
⎯⎯⎯⎯ B ⎯⎯⎯⎯⎯⎯⎯⎯
/
C ⎯⎯⎯⎯⎯⎯⎯⎯⎯ D + E
A) A → CH4, B → HCOONa, C → C6H5COOOCH,
D →
CHO E → C6H5COOH
3)
An organic compound A (C3H2O) on treatment with Cu at 573 K gives B. B doesn’t
reduce Fehling’s solution. But gives a yellow ppt. of compound C with I2/NaOH. Deduce the
structures of A, B and C.
A)
Since compound B gives iodoform test, it must be having –CO-CH3 group i.e it is a
ketone.
Since B is obtained by dehydrogenation of A,  A is a 2o alcohol
 A is R-CH – CH3
|
OH
Comparing with the given molecules formula, R is –CH3
 A is CH3-CH-CH3
|
OH
Propan-2-ol
The reactions are:
O
CH3-CH-CH3
> CH3-C-CH3
.
> no reactions
OH
Acetone
(A)
(B)
CH3 COCH3 + 3I2 + 4NaOH → CH3COONa + 3NaI+CHI3 + 3H2O
(C)
So, A = CH3CHCH3 Propan-2-ol
|
OH
B = CH3COCH3 Propan-2-one
C = CHI3 Iodoform
106
4)
a)
An organic compound A (C3H6O) is resistant to oxidation but forms a compound B
(C3H8O) on reduction. B reacts with H Br to form a bromide ‘C’ which on treatment with
alcoholic KOH forms an alkene D (C3H6). Deduce the structures A,B,C and D.
b)
Carboxylic acids contain a carbonyl group, but do not show the nucleophilic
addition reaction like ketones or aldehydes. Why?
Ans
a)
CH3-CO-CH3
.
> CH3-CH-CH3
OH
Propan-2-ol
‘B’
Acetone
‘A’
H Br
CH3-CH=CH2<
.
CH3-CH-CH3
.
‘D’
Propene
b)
Br
2-Bromo propane
‘C’
This is due to resonance (for structures refer NCERT)
The carboxyl carbon of the resonance hybrid is less positive and hence less
electrophilic than carbonyl ‘C’ of aldehydes and ketones
5)
a) Explain the following :
i) Cyanohydrin
ii) Acetal
iii) 2,4 – DNP derivative
b) Complete the following:
i)
CH2CH3
.
>
COOH
ii)
iii)
Ans
COOH
>
C6H5CHO
>
a) i) Formed addition of HCN to aldehydes or ketones
C  O  H   CN  
C
OH
CN
Cyanohydrin
ii) Formed on addition of two moles of alcohol to an aldehyde or ketone.
C = O + 2 R OH DryHC lg as
Acetal
C
OR
OR
+ H2O
107
iii) Addition product of aldehydes or ketones with 2,4 – Dinitro phenyl hydrazine in
weakly acidic medium.
NO2
NO2
C = O + H2N NH
b)
NO2
i)
ii)
iii) C6H5CH = NNHCONH2
Benzaldehyde semicarbazone
C = NNH
NO2
108
CHAPTER – 13
AMINES
01 MARK
1)
Ans
Draw the structure of prop-2-en-1-amine
H2C = CH – CH2 – NH2
2)
Give a chemical test to distinguish between ethylamine and aniline.
Ans They can be distinguished by Azo dye test.
On treating with HNO2 (NaNO2+HCl) followed by alkaline soln. of 2-naphthol (Temp – O-5 oC)
Aniline forms an orange dye, while ethylamine only gives ethanol and N2.
3)
Arrange the following in increasing order of their solubility in water : C6 H5NH2,
(C2H5)2NH, C2H5NH2
Ans C6H5NH2<(C2H5)2NH<C2H5NH2
4)
Out of CH3NH2 and (CH3)3 N, which one has higher boiling point?
Ans CH3NH2 due to its ability to form Intermolecular H-bonds (CH3)3N has no H atoms
bonded to N so cannot participate in H bonding.
5)
Why is the pkb of aniline greater than methylamine?
Ana In aniline, lone pair of N is delocalized over the benzene ring. reducing basicity. In
CH3NH2, +I effect of –CH3 group increases basicity of CH3NH2.
02 MARKS
1)
Explain Gabriel Pthalimide synthesis.
Ans It is used to prepare pure 1o amine. Aromatic 1 o amines can’t be prepared by this method
because aryl halides do not undergo nucleophilic substitution with the anion formed by pthalimide
2)
Ans
Give reasons:
a) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
b) Amines are more basic than alcohols of comparable molecular masses.
a) Resonance stabilization of diazonium salts of aromatic amines
109
b) N is less electronegative than O,  lone pair of electrons are easily available Such
resonance stabilization is not possible in diazonium salts of aliphatic amines.
3)
Identify A and B in each of the following:
a) C2H5Cl
>A
b) C6H5NH2
Ans
>A
a) A → C2H5CN,
Prppanenitrile
b) A → C6H5N2+ ClBenzene diazonium
Chloride
4)
Ans
/
>B
> B
B → C2H5CH2NH2
propanamine
B→
Complete the reactions
a)
C6H5N2Cl + H3PO2 + H2O →
b)
C6H5NH2 + Br2 (aq) →
a)
C6H6. + N2 + H3PO3 + HCl
b)
2, 4, 6 – Tribromoaniline
5)
Ans
b)
Convert
a)
Nitrobenzene to phenol
a)
b)
aniline to chlorobenzene
110
03 MARKS
1)
Ans
Give the structures of A, B, C in the following:
a)
CH3Br
>A
>B
>C
b)
CH3COOH
>A
a)
A → CH3CN,
B → CH3CH2NH2, C → CH3CH2OH
b)
A → CH3CONH2,
>B
B → CH3NH2
>C
C → CH3NC
2)
Write short notes on :
a)
Coupling reaction
b)
Ammonolysis
Ans a) Benzene diazonium salts react with e- rich aromatic compounds such as phenols and
amines to form azo compounds, which are often coloured and are used as dyes.
E.g
b) An alkyl or benzyl halide, on reaction with ethanolic soln. of NH3 undergoes
nucleophilic substitution reaction in which the halogen atom is replaced with –NH2 group. This
cleavage of C-X bond by ammonia molecule is called AMMONOLYSIS.
NH3 + R – X → RN+H3 XSubstituted ammonium salt
RNH2
.
> R2NH
> R4N+X-
> R3N
Quarternary Ammonium salt
Disadvantage : Mixture of 1 , 2 , 3 amines and quarternary ammonium salt is obtained.
Order of reactivity of halides with amines is RI > R Br > RCl
o
3)
o
Write the main products of the following reactions:
a) CH3CH2NH2
b)
a) CH3CH2OH
b) SO2NH C2H5
d) CH3NH2
>
SO2Cl + C2H5NH2
c) CH3CONH2
Ans
o
>
111
4)
Ans
Give reasons:
a) Amines are basic while amides are neutral.
b) CH3NH2 in water reacts with ferric choride to precipitate Fe (OH)3
c) Reactivity of –NH2 group gets reduced in acetanilide.
a) In amines, R group increase e- density on N due to +I effects, whereas in amides,
O
R – C – group is electron withdrawing.
b) CH3NH2 + H2O → CH3NH3 + + OHFeCl3 + 3OH- → Fe(OH)3(↓) + 3ClReddish - brown
O
c) The lone pair on N is involved in conjugation with R-C-group
O
OC6H5NH – C – CH3 ↔ C6H5 N+H = C – CH3
5)
An aromatic compound ‘A’ on treatment with ammonia followed by heating forms
compound ‘B’, which on heating with Br2 and KOH forms a compound ‘C’ (C6H7N). Give the
structures of A, B and C and write the reactions involved.
Ans ‘A’ must be benzoic acid
A
6)
Ans
B
C
Convert
i) Benzene diazonium chloride to nitrobenzene
ii) Nitrobenzene to aniline
iii) Aniline to benzonitrile
N 2Cl  HBF4 
i)
Sn
ii)
iii)
NO2  NH2
HCl
NaNO2
N 2 BF4 
Cu
NO2
112
7)
Give reasons
a) Although amine group is O, P- directly in aromatic substitution reaction, aniline on
nitroation gives some amount of m-nitro aniline.
b) Aniline doesn’t undergo Friedel-Craft reaction.
Ans a) Nitration is usually carried out with conc. HNO3+ con.H2SO4 In presence of these acids,
some amount of aniline undergoes protonation to form anilinium ion, so the reaction mixture
consists of aniline and anilinium ion, -NH2 group in aniline is o- and p- directing and activating
while the –NH3 group in anilinium ion is m-directing and deactivating. Nitration of aniline mainly
gives p-nitroaniline (due to steric hindrance at o- position) nitration of anilinium ion gives mnitroaniline.
b) Aniline being a lewis base reacts with Lewis acid AlCl3 to form a salt.
Cl3. As a result N acquires a positive charge
8)
C6H5NH2+Al-
Describe Hinsberg’s test to distinguish between 1 o,2 o and 3o amines.
Ans The amine is shaken with benzenesulphonyl chloride (Hinsberg reagent) in presence of
aq.KOH soln.
Ans
113
9)
Ans
How will you convert 4-Nitrotoluene to 2-Bromobenzoic acid?
10)
a) Predict, with reasons, the order of basicity of the following compounds is gases phase.
(CH3)3 N, (CH3)2 N H, CH3NH2 , NH3
b) Describe Carbylamine reaction.
Ans a) In gaseous phase, solvation effects are missing, Hence, greater the number of alkyl
groups, greater +I effect and stronger the base.
 (CH3)3 N > (CH3)2 NH > CH3NH2> NH3
b) Aliphatic and aromatic primary amines on heating with CHCl3 and ethanolic KOH form
isocyanides or carbylamines. which are foul-smelling substances. Secondary and tertiary amines
do not show this reaction. Hence this reaction is used as a test for primary amines.
R-NH2 + CHCl3 + 3KOH ⎯ RNC + 3KCl + 3H2O
Carbylamine
or
Isocyanide
114
CHAPTER – 14
BIOMOLECULES
01 MARK
1)
What are polypepticles?
A)
Polymers of amino acids having peptide linkage (– CONH –)
2)
Which vitamin’s deficiency causes pernicious anemia?
A)
Vitamin B12
3)
What is the biological effect of denaturation of proteins?
A)
The protein molecule uncoils from an ordered and specific conformation into a
more random conformation. Primary structure remains undisturbed.
4)
In what sense are the two strands of DNA not identical but complementary to each
other?
A)
If one strand has the bases A T C G, the other strand has T A G C, i.e. A can only
bond with T and C can pair with G. Hence the strands are not identical, but
complementary.
5)
Show the reaction of Glucose with Tollen’s reagent.
02 MARKS
Q1.
What is meant by
(a)
Pyranose structure of glucose
(b)
Glycosidic linkage
115
Ans 1
a) Six-membered cyclic structure of glucose, in analogy with pyran
b)
Two monosaccharide units are linked through oxygen atom accompanied
by loss of a water molecule. This linkage is called glycosidic linkage.
Q2)
What is the difference between α – form of glucose and β-form of glucose?
Ans These two forms differ from each other in orientation of –OH group at C-1.
α – form is obtained by crystallization from concentrated solution of glucose at 303K while
β – form (melting point 423 K) is obtained by crystallization from hot saturated soln. at 371 K.
Q3)
Mention the type of linkages responsible for the formation of the following:
i)
Primary structure of proteins
ii)
Cross-linking of polypeptide chains
iii)
α – helix formation
iv)
β – sheet structure.
Ans
i)
ii)
iii)
iv)
Q4)
Ans
Why are carbohydrates generally optically active?
Due to presence of chiral or asymmetric carbon atom and absence of plane of symmetry.
Peptide linkage
H-bond, sulphide linkage, van der waal’s forces
H-bond
Intermolecular H-bonds.
Q5) Give reactions characteristic of –CHO group, but not by glucose, as in it, free – CHO
group is absent.
116
Ans
i) No reaction with Schiff’s reagent
CHO
|
(CHOH)4 + Schiff’s reagent → no reaction
|
CH2OH
ii) No reaction with NaHSO3 and NH3
CHO
|
(CHOH)4 + NaHSO3/NH3 → no reaction
|
CH2OH
03MARKS
Q1)
Ana
Define :
a) Invert sugar
b) Vitamins
c) Nucleosides
a) Hydrolysis of sucrose brings about a change in sign of rotation from dextro (+) to laevo
(-), hence it is called invert sugar.
b) Organic compounds which can’t be produced by the body and must be supplied in
small amounts in diet to perform specific biological functions for normal health, growth
and maintenance of body.
c) A unit formed by attachment of a base to I’ position of sugar
Q2)
Ans
Differentiate between fibrous and globular proteins.
FIBROUS PROTEINS
1. Fibre – like structure
2. Water – insoluble e.g.keratin, myosin,
fibrin etc.
3. Stable to moderate changes in
temperature and pH.
GLOBULAR PROTEINS
1. Polypeptide chains coil around to give a
spherical shape.
2. Water soluble e.g.insulin, haemoglobin,
enzymes, hormones.
3. Very sensitive even to small changes in
temperature and pH.
117
Q3)
What happens when D-Glucose is treated with
a) HI
b) Br2 water
c) HNO3
Ans 3
(a) C H
(b)
(c)
6 +HI
CH – (CH )6- CH
n-Hexane
CHO
COOH
(CHOH)4Br2 H2O
(CHOH)4
CH OH
Glucose
CH OH
Gluconic acid
CHO
HNO3
CH OH
(a)
(Mild Oxidation)
COOH
(CHOH)4
Q7.
(Reduction)
(CHOH)4
(Strong Oxidation)
COOH (Saccharic acid)
Define Zwitter ions with examples?
(b)
What is the difference between essential amino acids and non-essential
amino acids?
A7.
(a)
Amino acids contain -NH and –COOH groups. These two groups interact
by transferring a proton from carboxyl group to amino group within the molecule.
Hence, a dipolar ion called ZWITTER ION is formed.
NH
R
(b)
NH
CH
S No
1.
2.
COOH
R
Essential Amino Acids
Cannot be synthesized by the
body.
Therefore should be
supplemented through diet.
Eg Valine, leucine
+
CH
COO-
Non-Essential Amino Acids
Can be synthesized by the
body.
E.g Glycine, alanine
Q8. Which is the sugar present in milk? How many monosaccharaides are present it?
What are such oligosaccharides called?
A8. Lactose, two monosaccharaide units
oligosaccharides are called DISACCHARIDES.
(glucose
and
galactose)
such
05 MARKS
Q1. Discuss the structure of proteins in detail? What is the difference between
helix and β-pleated sheet structures of proteins?
A1.
α-
(a)
Specific sequence in which various amino acids in a protein are linked to
one-another – primary structures.
118
(b)
The conformation adopted by these polypeptide chains as a result of Hbonding – SECONDAY STRUCTURES.
(i)
α- Helix.
Formed by intra molecular H – bonds, causing the
polypeptide chain to coil – up into a spiral structure or right handed helix
Eg . Fibrous protein like keratin and myosin.
(ii)
β-pleated sheet structures
Polypeptide chains lie side by side
held by intermolecular H-bonds , forming sheets. These sheets can then be
stacked one over the other forming a 3-D structure. This structure
resembles pleated folds of drapery, hence also called β-pleated sheet
structures.
(c)
Tertiary Structure. The secondary structure is further arranged, leading to
flowing two possibilities.
(i)
FIBROUS PROTEIN.The long linear protein chains form thread like
structure. These are insoluble in water and have β-pleated structure.
(ii)
GLOBULAR PROTEINS. Different segments of the protein fold up
to give the entire molecule a spherical shape. The folding involves various
interactions between the side-chains-such as, vander waal’s interactions,
disulphide bridges ,hydrogen bonding etc.
(d)
Quaternary Structure.
Some proteins exist as assembly of two or more
polypeptide chains called SUBUNITS or PROMOTERS. These subunits may be
identical or different and are held together by H-bonds, electrostatic and van der
waal’s interactions.
The quaternary structure refer to the determination of number of subunits
and their spatial arrangement w.r.t each other in an aggregate protein
molecule.
Primary
Structure
Secondary
Structure
Tertiary
Structure
Quarternary
Structure
119
Q2. (a)
RNA?
Write important structural and functional differences between DNA and
S
Structural Differences
No
1. Sugar present is 2 – deoxy –D-(-) Sugar present is D-(-) ribose
ribose
2. Cytosine and thymine are the Cytosine and Uracil are the
pyrimidine bases
pyrimidine bases
3. Double- stranded α- helix
Single- stranded α- helix
4. Very large molecules ; molecular Smaller molecules with
mass may vary from six – sixteen molecular mass ranging
million
from 20,000- 40,000µ
S No
1.
2.
Q2.
Functional Differences
Unique property of replication
Usually doesn’t replicate
Responsible for transmission of Responsible
hereditary characteristics
synthesis
protein
(b)
What deficiency diseases are caused due to lack of vitamins A,B ,B and
K in human diet?
S No
Q3.
for
Vitamins
Deficiency disease
1.
A
Xerophthalmia , night blindness
2.
B
Beriberi, loss of appetite
3.
B
Anaemia, general weakness
4.
K
Hemorrhage , slow blood clotting
(a)
An optically active amino acid (A) can exist in three forms depending on the
of the medium. If the molecular formula of (A) is C H NO , write
(i)
Structure of (A) in aqueous medium. What are such ions called?
(ii)
In which medium will the cationic form of (A) exist?
(iii)
In alkaline medium, towards which electrode will compound (A)
migrate?
A3.
(a)
(i)
CH
CH
NH2
(A)
COOH In aqueous medium, CH
NH
CH
COO-
+
ZWITTER ION
120
(ii)
In acidic medium, CH
NH
(ii)
COO-+ H+ CH
CH
In alkaline medium CH
+
NH
+
+
COOH
(cationic
form)
COO- + OH-CH
CH
NH
CH
+
COO-
CH
NH2
Anionic form
Therefore the compound in alkaline medium will migrate towards anode.
(b)
What are reducing sugars ? How are proteins related to amino acids?
A3
(b)
Carbohydrates with free aldehyde group that can reduce Fehling’s solution
and Tollen’s reagent are called reducing sugars. All monosaccharaides are
reducing sugars. Also, those disaccharides where the linkage between two
monosaccharaides is not through aldehydic group are also called reducing
sugars.
Q4.
(a)
Write four characteristics features of enzymes. Name a disease caused by
the deficiency of a particular enzyme?
A4.
(b)
What is a peptide bond? Explain its formation with an e.g?
(a)
(i)
Enzymes are specific in their action.
(ii)
Work at specific pH & moderate temperature.
(iii)
Activity is high.
(iv)
Catalyse biochemical reactions and their deficiency can cause
diseases.
Eg
Phenylketane urea is caused by deficiency of phenylalanine
hydroxylase and Albinism is caused by deficiency of tyrosinase.
(b)
Proteins are polymers of α-amino acids connected to each other by
peptide linkage. Chemically peptide linkage is an amide formed between –
COOH group and –NH2 group.
Q5.
(a)
Two samples of DNA, A and B have melting points 340K and 350K
respectively. What conclusion can you draw from these data regarding their base
content?
(b)
Discuss the composition of starch.
121
A5.
(a)
CG base pair has 3 H- bonds and AT base pair has two H – bonds.
Therefore CG base pair is more stable than AT base pair. Since sample B has
higher melting point than sample A, therefore sample B has higher CG content
than sample A.
(b)
Starch is made of two components
(i)
Amylose (15 – 20%) – Water – soluble, linear polymer of α-D
Glucose in which C of one glucose unit is attached to C of the other
through α – glycosidic linkage. Aqueous solution of amylose gives blue
colour with iodine solution.
(ii)
Amylopectin (30-85%) – Highly branched polymer, large number of
short chains containing 20-25 glucose units joined through α – glycosidic
linkage involving C of one glucose unit and C of another. The C of terminal
glucose unit in each chain is further linked to C of some other glucose unit
in the next chain through C - C α – glycosidic linkage.
122
CHAPTER – 15
POLYMERISATION
01 MARK
Q1.
Define polymerization?
A1.
The process of formation of a macromolecule from its repeating units (monomers).
Q2.
Is (CH – CH)na homo polymer or copolymer?
A2.
Homopolymer with
Q3.
Write the distinguishing feature between homo polymers and copolymers?
CH = CH as monomer.
A3. Homopolymers. These polymers are formed by the polymerisation of single
monomeric species, e.g. formation of polythene from ethene.
Copolymers. These polymers are formed by the polymerisation of different monomeric
species, e.g. formation of buna-S from styrene and buta-1, 3-diene.
Q4. Write the name and structure of a commonly used initiator in free – radical
addition polymerization.
o
o
A4.
Benzoyl peroxide C H -C – O-O-C-C H
Q5.
Based on molecular forces, what type of polymer is neoprene?
A5.
Elastomer (weakest intermolecular forces).
02 MARKS
Q1.
What is a biodegradable polymer? Give an example of biodegradable polyester?
A1. The natural polymer disintegrates by itself or by microorganisms within a certain
period of time. Eg PHBV (Poly-β-Hydroxy butyrate-co-β- hydroxy valerate).
Q2.
What are the differences between thermoplastic and thermosetting polymers?
A2. Thermoplastic polymers are linear, can be softened by repeated heating and
hardened on cooling, so can be used again and again without change in chemical
composition. E.g polythene.
Thermosetting polymers: When heated in a mould, they under permanent change in their
chemical composition to give a hard mass. Thus, they can be heated only once and then
set into a solid which can’t be remoulded e.g. Bakelite.
Q3.
(a)
What does 6,6 in nylon – 6,6 mean?
(b)
On free radical polymerization of chloroprene, which polymer is obtained?
123
A3) (a)
The monomers – adipic acid and hexamethaylene diamine – both have 6 C
atoms each.
(b)
Neoprene
Q4. A natural linear polymer of 2- methyl – 1,3 butadiene becomes hard on treatment
with sulphur between 373 to 415K and –S-S bonds are formed between chains. Write the
structure of product of this treatment?
A4.
CH3
CH
C
CH
S
S
H C C
CH
CH
CH
CH
Q5. What is the repeating unit in the condensation polymer obtained by combining
HO C CH2CH CO H (succinic acid) and H NCH CH NH ?
A5.
-(NH CH CH NH CO CH CH CO---) n
03 MARKS
Q1. How are polymers classified into different categories on the basis of
intermolecular forces? Give one example of a polymer of each of these categories?
A1.
There are 4 types of polymers based in intermolecular forces.
(i)
Fibres
–
Strongest
intermolecular
forces,
thread-like,
crystalline
; Nylon
(ii)
Elastomers – Weakest intermolecular forces, stretchable e.g natural
rubber.
(iii)
Thermoplastics – No cross – linkages become soft on heating, eg
polyethene.
(iv)
Thermosetting polymers – undergo extensive cross –linking on melting
eg Bakellite.
Q2. Distinguish between addition (chain- growth) and condensation (step-growth)
polymers?
124
A2.
S No
1.
2.
3.
4.
Q3.
Addition Polymerisation
Condensation Polymerisation
Formed by repeated addition of Formed by condensation reactions
monomer units
between bi-functional componds
No removal of any molecules
Removal of small molecules like
H O, NH
Also known as chain-growth Also known as step- growth
polymers and the chain grows by polymers as each step is
free – radical mechanism
independent of the other.
Eg :- Polythene, PVC
Nylon 6,6 , Dacron
Write chemical equations for the synthesis of
(i)
Nylon – 6
(ii)
Nylon – 6,6 (iii)
O
A3. (i)
[O]
Polythene
o
NH OH
H2N(CH )5COOH
cyclohexane
cylohexanone
cylohexanone
caprolactum
polymer
risation
o
(NH - (CH )5 C)n
Nylon – 6
(ii)
(iii)
n(CH2=CH2)
- (CH2 – CH2–)n
Q4.
What is the purpose of vulcanization of rubber?
A4.
The main purpose of vulcanization of rubber is to improve upon its properties.
Natural rubber becomes soft at high temperature (.335K) and brittle at low temperature
(<283) and shows high water absorption capacity. It is soluble in non-polar solvent and
resistant to oxidizing agents. To improve upon these properties, raw rubber is mixed with
an additive such as zinc oxide and heated between 373K and 415K. Sulphur forms
cross-links at the reactive sites of double bonds and rubber gets stiffened. The probable
structures of vulcansied rubber.
125
CH
CH
C
H C
CH
CH
CH
S
S
C
CH
CH
C = CH - CH
S
CH
CH
CH = CH - CH2
CH
Q5.
A5.
Write the structure and one use of each of the following :(i)
PVC
(ii)
Urea – formaldehyde resin.
(ii)
Bakelite
(i)
(CH
CH)n
use : pipes and raincoats
Cl
(ii)
(NH
C
NH
CH )n use : making unbreakable crockery
O
OH
(iii)
CH
OH
CH
O
OH
CH
CH
O
O
CH
CH
CH
H2O
CH
CH
O
O
O
OH
OH
OH
Use : electrical switches, handles of utensils
126
CHAPTER – 16
CHEMISTRY IN EVERY DAY LIFE
01 MARK
Q1.
Define narrow spectrum antibiotics with an example?
A1. Antibiotics that kill or inhibit a short range of gram positive or gram negative
bacteria are called narrow spectrum antibiotics. For example – Penicillin G.
Q2.
Name a substance that can be used as an antiseptic as well as disinfectant?
A2. Phenol can be used as both .A 0.2% soln of phenol is an antiseptic, while its 1% is
disinfectant.
Q3.
Why is bithional added to soap?
A3. Bithional acts as an antiseptic agent and reduces the odour produced by bacterial
decomposition of organic matter.
Q4.
What makes detergents non-biodegradable?
A4.
Branching in the hydrocarbon chain.
02 MARKS
Q1.
A1.
Explain the following with an example?
(a)
Antifertility drugs
(b)
Antibiotics
(a)
Drugs which help in birth-control eg norethindrone
(b)
Antibiotics are substances produced wholly or partially by chemical
synthesis, which in low concentrations inhibit the growth or destroy
microorganisms by intervening in their metabolic processes.
Q2.
Give reasons :(a) Soaps don’t work well in hard water.
(b) Synthetic detergents are better than soaps.
A2.
(a)
Mg and Ca ions of hard water react with soap to form magnesium and
calcium salts which are insoluble in water and form scum.
(b)
Q3.
A3.
Synthetic detergents can work even with hard water.
Mention one important use of:
(i)
Equanil
(ii)
Sucralose.
(i)
Equanil is a tranquilizer.
(ii)
Sucralose is an artificial sweetener.
127
Q4.
A4.
Q5.
(i)
Why is the use of aspartame limited to cold foods and drinks?
(ii)
Give two examples of macromolecules that are chosen as drug targets.
(i)
It is unstable at high temperature.
(ii)
Nucleic acids/proteins/carbohydrates/lipids.
Classify the following into antihistamine, antacid, tranquiliser, antibiotic.
Drug: penicillin, meprobamate, terfenadine, and ranitidine.
A5. Pencillin – antibiotic, Meprobamate – tranquiliser, Terfenadine – antihistamine,
Ranitidine – Antacid.
03 MARKS
Q1. What are analgesics? How are they classified? When are they recommended for
use ?
A1.
Chemical substances used to relieve pain are called analgesics.
Types of analgesics :
(i)
Non-narcotic drugs – Non-habit forming, effective in relieving or preventing
heat attack, skeletal pain, viral inflammation.
(ii)
Narcotic drugs – Habit forming produce sleep, unconsciousness,
convulsions. Useful in post – operative pain, cardiac pain, terminal cancer.
Q2.
What are antipyretics? Given an example. Can it play any other role?
A2.
Antipyretics reduce body temperature under conditions of fever. E.g Paracetamol.
They can also act as analgesics.
Q3.
Define the following and give an example of each?
(a)
Antacids
(b)
Sulpha drugs
(c)
Antioxidants
A3. (a)
Antacids are drugs that are administered into the body to neutralize excess
acid produced in the stomach. E.g ranitidine.
(b)
Derivatives of sulphanilamide have great antibacterial powers, e.g sulpha
diazine, and sulphapyridine.
(c)
Antioxidants retard the action of oxygen on food and thus prevent the
decomposition of foods. Eg (BHA)
Q4.
Explain the mechanism of drug – enzyme interaction.
128
A4.
Drugs can block activities of enzymes in the following two manners.
(a)
Drugs can block activities of enzymes in the following two manners
(i)
Drug might compete with the substrate for attachment on the active
sites of enzymes. Such drugs are called competitive inhibitions.
(ii)
Some drugs don’t bind to the active site instead they bind to a
different site on the enzyme called allosteric site. This action changes the
shape of the active site to an extent that the substrate doesn’t recognize it.
Q5.
How do receptors transfer message to the cell?
A5. Message between two neurons or between neurons and muscles is
communicated through chemical substances called chemical messengers. They are
received at the binding site of the receptor protein. To accommodate these chemical
messengers shape of receptor protein changes a little and messenger gives the
message. After the message is given, chemical messenger departs and active site of
receptor protein returns to its original shape.
129
VALUE – BASED QUESTIONS
1)
Mr Arora and family were having a tough time. It had been raining all night and the sewer
water was moving back in their toilets. Mr Arora thinks it is due to polythene bags choking the
sewer lines. His wife wonders why these bags don’t dissolve in water. Mr Arora explains that
these bags are non-biodegradable. He says that cloth bags must be used for shopping to minimise
use of polythene bags.Answer the following questions:
(a)
Is there any biodegradable polymer? What are its uses?
(b)
What values were displayed by Mr. Arora?
A)
a)
PHBV [Poly-β-Hydroxybutyrate-Co-β-hydroxyvalerate] is biodegradable. It is a
co-polymer of 3- Hydroxybutanoic acid and 3-Hydroxypentanoic acid. It is used in speciality
packing, orthopaedic devices and in controlled release of drugs.
b)
Mr. Arora displays concern for environment. He finds the polythene bags a serious
threat to the environment. He must follow his ideals.
2)
Shyam wants to wash his woolen clothes. His grandmother suggested that he washes the
clothes in warm water. Shyam follows her advice and finds that the clothes become cleaner that
how they were, when washed in cold water!
Answer the following questions:
a)
b)
Why is it easier to wash the clothes in lukewarm water?
What values are associated with this information?
A)
a)
Soaps work through micelle-formation. But the optimum temperature (KRAFT
TEMPERATURE) at which this happens is attained in lukewarm water.
b)
Be humble to accept a scientific fact form an elderly person.
3)
During Chemistry practical’s, the students were told to add reagents very carefully and not
to unnecessarily touch or play with the chemicals. Even then, a student quietly removed the labels
from bottles of 2-propanol and 1-propanol. Nikhil, a responsible student, reported the matter to
the teacher. Answer the following questions:
a)
What values are expressed by Nikhil?
b)
The teacher easily found out the correct alcohols by performing a simple test.
Write the chemistry of the test performed.
A)
a)
Nikhil shows very responsible behavior. He not only obeys the teacher, but also
reports the matter, showing concern for others.
b)
.
(
,
)
CH3-CH(OH)-CH3 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ CH3-CH(Cl)-CH3
Propan-2-ol
turbidity after 5-mins
(Secondary Alcohol)
.
,
(
)
CH3CH2 CH2OH ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ CH3CH2CH2Cl
Primary Alcohol
turbidity appears only on heating
130
4)
Fermentation of sugars gives a mixture containing an alcohol A and water. The alcoholwater mixture on fractional distillation gives an azeotropic mixture. The azeotropic mixture is
composed of 95.6% alcohol and 4.4% water by mass. The mixture is commonly called rectified
spirit. To refrain people from drinking rectified spirit, industrial alcohol is denatured. Some
countries now use alcohol A as an additive in gasoline since it is a cleaner fuel.Answer the
following questions:
a)
Identify alcohol A. Write reactions involved in fermentation of a sugar.
b)
What is denaturation of alcohol? Why is industrial alcohol denatured?
c)
Would you support the use of alcohol ‘A’ as an additive in gasoline in India,
although it may result in decrease in production of sugar?
d)
What are the values associated with your decision?
A)
a)
A  Ethanol C2 H5OH
C12H22O11+ H2O ⎯⎯⎯⎯⎯⎯⎯⎯ C6H12O6 + C6H12O6
Glucose
Fructose
C6H12O6 ⎯⎯⎯⎯⎯⎯ 2C2H5OH + 2CO2
Glucose or Fructose
b)
Making industrial alcohol unfit for drinking by adding CuSO4 (to give it colour)
and pyridine ( a foul smelling liquid) is called Denaturation of Alcohol. Ethanolis
required for various chemical industries. To stop people from drinking it, it is denatured.
c)
Gasoline need of our country is met by importing it which increase our fiscal
deficit. To meet energy requirements and reduce oil-consumption, ethanol is used as an
additive to gasoline. But it must be regulated by the Government.
d)
Concern for energy crisis and economy of the country.
5)
There is growing interest in the use of chelate therapy in medicinal chemistry. An example
is the treatment of problems caused by the presence of metals in toxic proportions in plant/animal
systems. Detection of cations through coloured complex formation is done in qualitative
analysis.Answer the following questions:
A)
a)
Name the chelating agents that can remove copper, iron and lead from water.
b)
Name the compound that inhibits the growth of tumours.
c)
Recent studies show that cis-platin can cause serious side-effects, including severe
kidney damage. What is the alternative medicine?
a)
Copper, iron - D – penicillamine desferrioxime D
lead  EDTA
b)
131
c)
cis-platin is being replaced by trans-isomer.
6)
Vitamin C is water-soluble and can’t be synthesized by humans, monkeys and guineapigs. It is also required for the synthesis of collagen, which is the structural protein of skin,
tendons, connective tissue and bone.Answer the following questions:
a)
b)
c)
A)
Write structure of vitamin C and the functional groups therein.
Is it acidic in nature? What is its common name?
What is the oxidation product of vitamin C?
a)
b)
It is acidic due to enolic -OH group. Its common name is ascorbic acid.
c)
7)
Deepti and her father went to the market to buy some mosquito-killer chemical. Deepti
stopped her father from buying DDT.Answer the following questions:
a)
b)
A)
Why is Deepti’s action justified?
What are the values shown by Deepti?
a) i) Many species of insects have developed a resistance to DDT.
ii) It is highly toxic for aquatic life.
iii) Stability of DDT makes it difficult to disintegrate or degrade.
b)
Concern for environment, application of knowledge to everyday situations,
awareness and willingness to educate others.
132
8)
Dentist Dr Verma opened a clinic in a village and found many people suffering from
bleeding gums. He spread awareness amongst villagers about some inexpensive steps to prevent
the problem.Answer the following questions:
a)
b)
c)
d)
A)
What disease were the villagers suffering form?
What causes this disease?
What is the cure for this disease?
What values are displayed by Dr Verma?
a)
Scurvy
b)
Deficiency of vitamin C
c)
Vitamin C in the form of citrus fruits like orange, lemon, amla should be taken on
a regular basis.
d)
Dr Verma displays sensitivity to the villagers, who may not be able to afford
expensive medicines, but simple fruits should be easily available. He also shows social
responsibility, a duty of a doctor.
9)
At a chemist shop, you see a person pleading with the shopkeeper to give him sleeping
pills, without a doctor’s prescription. The chemist refuses, stating that the pills can be misused.
Answer the following questions:
a)
Should the shopkeeper give the customer the pills without a doctor’s prescription?
Why?
b)
What alternative approach can the shopkeeper take to help the customer without
giving him the pills? Which values can be reflected in this alternative approach?
A)
a)
No, sleeping pills should not be given without prescription. An overdose
of these pills can even cause death.
b)
He can suggest some good doctor nearby, convincing the customer to seek
a doctor’s advice.Value : Professional ethics, responsibility, human approach.
10)
Rohini wanted to give her baby some medicine. She added boiled and cooled water as per
instructions and shook the bottle well before giving the medicine to the baby.Answer the
following questions:
A)
a)
b)
What values are associated with selling medicine in this form?
Why did she shake the contents? Name the process.
a)
The medicines sold in anhydrous form have higher shelf life and can be stored for
a longer time. This is a way of being thrifty by not wasting the available
resources.
Rohini shook the contents of the bottle to change the content into the form of Sol
because adsorption of medicines are easier in the colloidal form. The process
involved is Peptisation
b)
133
11)
I work in a lab where aldehydes are prepared by the following method :
/
RCOCl + H2 ⎯⎯⎯⎯⎯⎯⎯⎯
+
My mother, an organic chemistry professor, advised me against this method and
suggested another one.Answer the following questions:
c) Why did my mother advise me against this method?
d) What could be the other method? Can you judge it for environment and economically?
A)
12)
a)
BaSO4 harms the environment and can be toxic for us.
b)
CH2 = CH2 + O2 ⎯⎯⎯⎯⎯⎯⎯⎯
It is economical, but heavy metals like Pd and Cu drain underground and
contaminate underground water.
( )/
( )
Mrs Sharma lives near a lake along with 5 neighbours. Initially the lake was clear and
beautiful, but now it has a lot of phytoplanktons. One day Mrs Sharma’s daughter, after
returning from school, advised the ladies to use natural soaps for washing clothes, not
detergents. Mrs Sharma and the others follow her advice and get the lake cleared
gradually!
Answer the following questions:
a)
What values are displayed by Mrs Sharma and her daughter?
b)
Why did the phytoplanktons grow excessively whie they were using detergents?
A)
a)
b)
Mrs Sharma – Environmental awareness, giving due importance to educated
person’s advice.
Daughter – Knowledge of chemistry, environmental awareness, application of
knowledge to real-life situation.
Synthetic detergents contains phosphates which lead to overgrowth of vegetation
or phytoplanktons.
13)
Rajiv saw policemen stopping vehicles and asking drivers to blow air into an instrument.
His father told him that they were checking whether the driver had consumed alcohol or not.
Answer the following questions:
A)
a)
b)
c)
What is the instrument called?
What values are the police trying to inculcate?
How does the instrument test whether or not a driver is drunk?
a)
b)
c)
Alcometer
Social responsibility, awareness of risks involved due to drunken driving
Alcometer uses K2Cr2O7 which changes colour on reacting with alcohol.
Depending on the amount of alcohol, the colour range varies.
2Cr2O7-2 + 3C2H5OH + 16 H+ 4Cr3+ + 3CH3COOH + 11 H2O
(orange)
blue
green
134
14)
These days, people watch “Solar Eclipses” through UV-protected sun-glasses. Earlier,
people avoided eclipses, as they were believed to be evil.Answer the following questions:
a)
b)
A)
c)
Which transition metal oxide is used in making UV protected lens?
By watching the eclipse through UV protected lens, which value are people
displaying?
Which rays can damage the eyes while watching solar eclipse with naked eyes?
a)
b)
c)
Neodymium and praseodymium oxides (f-block)
Scientific attitude
UV rays
15)
Sameer and his friends decide to play Holi with natural colours. When some people try to
play with synthetic colours, Sameer explains how they could be harmful, cause skin allergies etc.
Answer the following questions:
A)
a)
b)
c)
Mention the values shown by Sameer.
Write the reaction for preparation of an azo (synthetic) dye.
Name two pigments present in natural colours. How are natural colours prepared?
a)
Sensibility, awareness and courage
b)
c)
Chlorophyll and carotene. Natural colour can be prepared from leaves and flowers.
16)
Zahira’s brother took cough syrup one day, even though he was healthy. After sometime
his head started aching and skin started itching. Zahira insisted on her parents to take him to a
doctor.Answer the following questions:
a)
b)
c)
A)
Mention the values shown by Zahira?
Why did her brother’s body start itching?
Why was it necessary to go to the doctor?
a)
Critical thinking, keeping calm in crisis
b)
Due to an allergic reaction
c)
Only a doctor can prescribe the correct medicine. Incorrect drug or dosage can be
fatal.
135
17)
Nita’s mother got easily tired and was lethargic. Nita took her to a doctor, who diagnosed
her with pernicious anemia. Nita took good care of her mother till she recovered.
Answer the following questions:
A)
a)
b)
c)
Which vitamin-deficiency causes pernicious anemia?
Name the sources that can provide this vitamin
Mention the values shown by Nita.
a)
b)
c)
Vitamin B12
Cheese, eggs, meat, milk and yogurt.
Caring, helpful, keen observation