UNIT:-1
( SOLID STATE )
STUDY MATERIAL
QUESTIONS BASED ON HIGH ORDER THINKING SKILL
1 MARK QUESTIONS
Q. 1. Name a liquefied metal which expands on solidification.
Ans. Gallium (Ga) is a silvery white metal, liquid at room temp. It expands by 3.1%
on solidifica-tion.
Q. 2. How many number of molecules per unit cell which crystallizes in the form
of end face centred (monoclinic) lattice with a molecule at each lattice.
Ans. 2.
Q. 3. What is the coordination number of carbon, in diamond ?
Ans. 4 and its unit cell has 8 atoms.
[The space lattice of diamond is FCC]
Q. 4. Name the solid which has weakest intermolecular force ?
Ans. Ice
Q. 5. Arrange the following types of interactions in correct order of their
increasing strength :
Covalent, hydrogen bonding, Vander Waals, dipole dipole
Ans. Vander Waals < dipole dipole < hydrogen bonding < covalent.
Q. 6. Give reason for the appearance of colour in alkali metal halides.
Ans. Due to F-centres.
Q. 7. Which type of defect occur in Ag Br ?
Ans. Schottky defect and Frekel defect.
Q. 8. Give one example of doping which produces p-type of semi-conductors.
Ans. Ge doped with Al.
1
Q. 9. Out of (a) Graphite and (b) Carborundum which one is harder ?
Ans. Carborundum.
Q. 10. How can a material be made amorphous ?
Ans. By melting the material and by cooling it rapidly.
2 MARKS QUESTIONS
Q. 1. Give Reason :
The energy required to vaporize one mol of copper is smaller than that of
energy required to vaporize 1 mol of diamond.
Ans. Copper is a metallic solid having metallic bonds while diamond is a covalent
solid having covalent bonds. Metallic bonds are weaker than covalent bonds
and thus less amount of energy is required to break metallic bonds than
covalent bonds.
Q. 2. The unit cube length for LiCl (NaCl) is 5.14 °A. Assuming anion-anion
contact. Calculate the ionic radius for Chloride ion.
Ans.
Cl –
Li
A
+
B
C
Interionic distance of LiCl = 5.14 / 2 = 2.57 A
AC
=
AB 2  BC 2
=
2.572  2.572
= 3.63
therefore, radius of Cl– = ½ × 3.63 = 1.81 A
Q. 3. Give reasons :
Ans.
(a)
Diamond and rhombic Sulphur are covalent solids, but the latter
has lower melting points.
(b)
Among NaCl and CsCl, CsCl is quite stable.
(a)
Due to weak Vander Waal’s Force in Sulphur molecule.
(b)
CsCl coordination number is 8. It is surrounded by 8 anion tightly.
2
Q. 4. How many unit cells are present in a cube shaped ideal crystal of NaCl of mass 1
gm ?
Ans.
Mass of 1 unit cell
= volume × density
= a³ × d
=
a3  M  Z
N0a3
=
58.5  4
6.023  1023
No. of unit cells in 1 gm
= 1/M
= 6.023 × 1023 / 58.5 × 4
= 2.57 × 1021
Q. 5. In the mineral spinal; having the formula MgAl2O4. The oxide ions are arranged
in CCP, Mg2+ ions occupy the tetrahedral voids. While Al3+ ions occupy the
octahedral voids.
Ans.
(i)
What percentage of tetrahedral voids is occupied by Mg2+ ions ?
(ii)
What percentage of octahedral voids is occupied by Al3+ ions ?
According to the formula, MgAl2O4. If there are 4 oxide ions, there will be 1 Mg2+
ions and 2 Al3+. But if the 4 O2– ions are ccp in arrangement, there will be 4 octahedral
and 8 tetrahedral voids.
(i)
Percentage of tetrahedral voids occupied by Mg2+ = (1 / 8) × 100
= 12.5%
(ii)
Percentage of octahedral voids occupied by Al3+ = (2 / 4) × 100
= 50%
Q. 6. Give reasons :
Ans.
(a)
Window glass of old building look milky.
b)
Window glass of old building is thick at bottom.
(c)
CaCl2 will introduce Schottky defect if added to AgCl crystal.
(a)
Due to annealing over a number of years glass acquires some crystalline
character.
3
(b)
Glass is not a true solid. But a super-cooled liquid of high viscosity. It has the
property to flow.
(c)
2 Ag+ will be replaced by 1 Ca2+ ions to maintain electrical neutrality. Thus a
hole is created and lattice site for every Ca2+ ion introduced.
Q. 7. Analysis shows that nickel oxide has the formula NiO.98O1.00. What fractions of
nickel exist as Ni2+ and Ni3+ ions ?
Ans.
NiO.98O1.00
Let Ni2+ be x and Ni3+ be 0.98 – x
Total charge on compd. is equal to zero.
[2 (Ni2+) + 3 (Ni3+) – 2 (O2–)] = 0
2 x + 3 (0.98 – x) – 2 = 0
x = 0.94
Therefore Ni2+ % =
0.94
× 100 = 96%
0.98
Ni3+ = 4%
Q. 8. What type of defect can arise when a solid is heated ? Which physical property is
affected by this and in what way ?
Ans.
When a solid is heated vacancy defect arises. This is because on heating some atoms
or ions leacve the lattice site completely some lattice sites are vacant. As a result of
this defect the density of the substance decreases, because some atoms leave the
structure completely.
Q. 9. (a)
(b)
What happens when a Ferromagnetic or Ferrimagnetic solid is heated ?
The ions of MgO and NaF all have the same number of electrons and
intermolecular distance are about the same (235 & 215 pm). Why are the
melting points are so different (2642 °C & 992 °C ?
Ans.
(a)
It changes into paramagnetic at hight temperature due to randomization of
spins.
(b)
The ions in MgO carry two unit charges. In NaCl only one unit charge. Hence
electrostatic forces of attraction in MgO are stronger.
Q. 10. (a)
If the radius of the Br ion is 0.182 nm, how large a cation can fit in each of
the tetrahedral hole.
4
(b)
AgI crystallizes in a cubic closed packed ZnS structure. What fraction of
tetrahedral site is occupied by Ag ion ?
Ans.
(c)
At what temp. range, most of the metals becomes super conductors ?
(a)
For a tetrahedron the limiting ratio is 0.225 – 0.414
For largest cation highest value 0.414 would be considered.
r+ / r– = 0.414
r+ = 0.414 × 0.182 = 0.075 nm.
(b)
In FCC there are 8 tetrahedral voids. Out of this ½ is occupied by Ag cation.
(c)
2 k – 5 k.
Q1. Examine the illustration of a portion of the defective crystal given below and answer the
following questions:
(i) What are these type of vacancy defect called?
(ii) How is the density of a crystal affected by these defects?
(iii) Name one ionic compound which can show this type of defect in the crystalline state.
(iv) How is the stoichiometry of the compound affected? (2 marks)
Q2. Analysis shows that a metal oxide has the empirical formula M
0.96
O . Calculate the
1.00
percentage of M and M ions in this crystal. (2 marks)
2+
3+
Q3. In an ionic compound the anion (N ) forms cubic close packing, while the cation (M )
occupy one third of the tetrahedral voids.
-
+
5
Deduce the empirical formula of the compound and the coordination number of (M ) ions. (2
marks)
+
Q4. The radius of copper atom is 118 pm. If copper crystallizes in a face-centered cubic
lattice, what is the size of the unit cell? (2 marks)
Q5. Why are amorphous solids to be considered as supercooled liquids? (2 marks)
6
Q1. What is meant by the term Co-ordination no?
Ans:- No of atoms (sphere) by which the central atom is surrounded.
Q2. What is the co-ordination no of atoms:(a) In a cubic close packet structure -12
(b) In a body centred cubic structure:-8
Q3. A cubic solid is made of two elements P and Q. Atoms of Q are at the corner of the
cube and P at the body centre. What is the formula of the compound? What are the coordination No. of P and Q.
Ans:-PQ
Q5. Analysis shows that nickel oxide has the formula Ni0.98 as I1.00. What fraction of Ni
exit as Ni2+ and Ni3+ ions?
Ans:- Let the no of Ni2+ ion = X
Ni3+ ion = 0.98 – X
As per the question
2 X + 3(0.98- X) = 2
2 X + 3 X 0.98 -3 X =2
2 X – 3 X = 2 - 3 X 0.98=2-2.94
X = 0.94
X =94
So fraction of Ni 2+ = 94%
And Ni3+ = 06%
Q4. What is difference between Frenkel and schottky defect? (any two)
Ans:Frenkel defect
Schottky defect
(a) In this defect shifting of icon (a) In this defect there is missing of equal
from its position to interstitial site.
no of cation and anion.
(b) Density remain same.
(b)Density decreases.
Q5. Silver Crystallises in fcc lattice. If edge length of the cell is 4.077 X 10-8 cmand
density is 10.5 g cm-3. Calculate the atomic mass of silver.
Ans:- density =10.5 gm/cm3, Z=4
d
zxM
4 xM

3
23
N 0 xa
6.022 x10 x(4.077 x108 )3
6.022 x1023 x10.5 x(4.077)3 x10 24
4
 107.14 gm
Hence atomic mass of silver=107.14 gm.
We Know M 
Q6. Classify each of the following as being either a p-type or an n-type Semiconductor.
(a) Ge dopped with In----p-type.
(b) B dopped with Si----n-type.
Q7. Zinc oxide is white but it turns yellow on heating. Explain?
Ans:- When Zinc oxide is heated it loses oxygen.
ZnO-------- Zn2+ + 1/2 O2 + 2eZn2+ adopt the interstitial void and electron in the neighbouring voids. Due to
presence of electron in void, the colour is yellow.
7
Q8. If NaCl is doped with 10-3 mole % SrCl2.What is the concentration of cation
vacancies?
Ans:- Since 100 mole of NaCl are doped with 10-3 mol of SrCl2
 1 mole of NaCl are doped with =10-3/100 mol=10-5 mol
As Sr2+ ion introduced one cation vacancy
Conc. Of cation Vacancy=10-5 X 6.022X1023 mol-1
=6.022X1018 mol-1
Q9. What type of defect arise when a solid is heated? Which physical property is affected
by it and what way ?
Ans:- Vacancy defect is created, it is because on heating some atoms leaved the lattice site
completely also the density decreases.
Q10. What is the affect of pressure on NaCl type crystals ?
Ans:- The coordination No. of NaCl increases from 6 to 8 .
Q11. Al crystallizes in a CCP structure. Its metallic radius is 125 pm.
(i)
What is the length of side of the unit cell?
(ii)
How many unit cells are there in 1cm3 of Al?
Ans:- (i) In CCP
a= 2 2 r
=2 X 4.14 X 125 pm = 354 pm
(ii) Volume of 1 unit cell = (3.54X10-10)3 = 4.44 X 10-23 cm3
1
 2.25 X 1022
unit cell in 1cm3 =
4.44 x10 23
Q12. What happens when ferrimagnetic Fe3O4 is heated to 850 K and why?
Ans:- Fe3O4 on heating to 850 K becomes paramagnetic. This is due to greater alignment
of domains in one direction on heating.
FOR BELOW AVERAGE
01 Mark Question
1.
2.
What is the number of atoms per unit cell in a
i)
Simple cube/primitive cube.
ii)
Body centered cube.
iii)
Face centered cube.
Name the element with which silicon can be doped to give a (i) p-type semi conductor
(ii) n-type semi conductor
3.
A cube solid is made of two elements P and Q. Atoms of Q are at the corner of the
cube and P at the body centre. What is the formula of the compound? What are the coordination number of P and Q.
8
4.
What makes the crystal of KCl appear some time violet?
5.
What is the effect of Schottky and Frenkel defects on the density of crystalline solid?
6.
Name a substance which on addition to AgCl causes cation vacancy in it.
7.
What happens when ferromagnetic Fe3O4 is heated at 850 K and why?
8.
What is curie temperature?
9.
What is ferromagnetism different from paramagnetism?
10.
What structural changes will occur if NaCl crystal is subjected to high pressure?
FOR AVERAGE
1.
An element (at. Mass=60) having fcc structure has a density of 6.23 g cm -3. What is
the edge length of the unit cell? (Avg. const. N = 6.02 x 1023 mol-1)
2.
The density of CsBr, which has a BCC structure, is 4.4 g cm-3. The edge length of the
unit cell is 400 pm. Calculate the interionic distance in crystal of CsBr (NA= 6.023 x
1023, At. Mass of Cs= 133, Br= 80)
3.
The composition of a sample wustite is Fe0.93 O1.00, what percentage of the iron is
present in the form of Fe(III)?
4.
NaCl crystallizes in FCC structure. Its density is 2.165 gm cm-3. If the distance
between Na+ and its nearest Cl- is 281 pm. Find out the Avog. No. (Na=23 g mol-1, Cl
= 35.5 g mol-1)
5.
The compound CuCl has ZnS structure and the edge length of its unit cell is 500 pm.
Calculate the density. (At. Mass, Cu=63, Cl=35.5, NA=6.02 x 1023)
FOR BELOW AVERAGE
02 or 03 Marks Questions
1.
Calculate the efficiency of packing in a case of metal crystals for simple cube or body
centered cube or face centered cube (with the assumption that atoms are touching
each other)
2.
How can you determine the atomic mass of an unknown metal if you know its
density and the dimension of its unit cell? Explain.
3.
If the radius of the octahedral void is r and radius of the atoms in close packing is R.
derive relationship between r and R.
4.
Explain the following with suitable examples:
(a) Ferromagnetism
(b) Paramagnetism
9
(c) Ferrimagnetism
(d) Anti ferromagnetism
(g) Schottky defect
(j) Doping
(e) 12-16 groups
(h) Frenkel defect
(k) n-type semiconductor
(f) 13-15 groups
(i) F-centre
(l) p-type semiconductor
(m) Conductor/ Insulator/ Semiconductors.
NUMERICAL PROBLEMS
1.
If NaCl is doped with 10-3 mol%. What is the concentration of cation vacancies?
2.
Aluminum crystallizes in a cubic close packed structure. Its metallic radius is 125pm.
(a) what is the edge length of unit cell? (b) How many unit cells are there in 1.00 cm3
of aluminum?
3.
Silver forms ccp lattice and x-ray studies of its crystal show that edge length of its
unit cell is 408.6pm. Calculate the density of silver (Atomic mass = 107.9 u)
4.
Niobium crystallizes in body centre cubic structure. If density is 8.55 gm/cc.
Calculate atomic radius of niobium using its atomic mass 93.
5.
Silver crystallizes in face lattice. If edge length of the cell in 4.07 x 10-8 cm and
density is 10.5 gm/cc. Calculate the atomic mass of silver.
6.
Gold (atomic radius = 0.144 nm) crystallizes in a fcc unit cell. What is the length of a
side of the cell?
7.
An element (Atomic mass = 60) having fcc structure has a cell edge of 400 pm. What
is its density? NA = 6.023 x 1023.
FOR ABOVE AVERAGE
1.
Analysis shows that nikel oxide has the formula Ni.98O1.00. What fraction of nickel
exists as Ni+2 and Ni+3 ions?
2.
An element (atomic mass = 60) having face centred cubic unit cell has a density of
6.23 gm/cc. what is edge length of unit cell?
3.
Metallic gold crystallizes in a fcc structure and has a density of 19.3 gm/cc. Calculate
the radius of gold atom. [atomic mass of Au=197, NA= 6.023 x 1023]
4.
Unit cell of an element (Atomic mass = 108 and density 10.5 gm/cc) has an edge
length 409 pm. Deduce the type of crystal lattice.
5.
An element has a body centred cubic structure with a cell edge of 288 pm. The
density of the element in 7.2 gm/cc. How many atoms are present in 208 gm of the
elements?
10
UNIT:-2
SOLUTION
STUDY MATERIAL
1. Solution : Homogeneous mixture of two or more pure substances.
2. Types of solutions :
a) Solid in solid solution. Eg: Bronze, Brass.
b) Liquid in solid solution. Eg: Zinc-amalgam.
c) Gas in solid solution. Eg: Solution of hydrogen in palladium.
d) Solid in liquid solution. Eg: Aqueous Sodium Chloride sol.
e) Liquid in liquid solution. Eg: Ethanol dissolved in water.
f) Gas in liquid solution. Eg: Carbondioxide dissolved in water, ammonia dissolved in
water.
g) Solid in gas solution. Eg: Iodine vapour in nitrogen .
h) Liquid in gas solution. Eg: Water vapoursin air, Chloroform vapour in nitrogen.
i) Gas in gas solution. Eg: Water gas, Producer gas.
3. Concentrations of solution :
a) Mole fraction ( X):
Xsolvent = nsolvent /nsolvent + nsolute
Xsolute = nsolute /nsolvent + nsolute
b) Normality (N) :
N = WB/EB x 1000/V
c) Molarity (M) :
M = WB/MB x 1000/V
d) Molality (m) :
m = WB/MB x 1000/WA
WB = Mass of solute in gram.
WA = Mass of solvent in gram.
MB = Molar mass of solute.
V = Volume of the solution.
EB = Equivalent weight of the solution.
4. Henery’s Law : Mass of gas dissolved per unit volume of the solvent is directly
proportional to the pressure of the gas in equilibrium with the solution.
Limitations : Not applicable when pressure is very high.
11
Not applicable when the temperature is too low.
Not applicable when the gas is highly soluble.
Not applicable when the gas react chemically with solvent and dissociate and associate in
the solvent.
5. Colligative Properties : Properties of solution which depends upon the number of
solute particles dissolved in it. Eg: R.L.V.P, Osmotic pressure, Elevation of boiling
point.
6. Vapour Pressure : Pressure exerted by vapours in equilibrium with its liquids at a
given temperature.
7. Lowering of vapour Pressure :
Δ P = PA0 – P (PA0 = v.p of pure solvent, p = v.p of sol.)
8. Relative lowering of vapour pressure : Ratio of lowering of vapour pressure of pure
solvent.
R.L.V.P = (PA0 – P)/ PA0
9. Rault’s Law :
a) Partial pressures of components in the
solutions are directly proportional to their mole fractions.
b) R.L.V.P. of solution containing non volatile solute is equal to mole fraction of the
solute.
10. Limitations of Raults Law;
a) Applicable only to dilute solution.
b) Applicable only to homogeneous solution.
c) Applicable to the solution in which solute doesnot undergo association or dissociation.
11. Boiling Point : Temperature at which vapour pressure of the liquid is equal to
atmos[heric pressure.
12. Elevation of boiling Point :
Δ Tb = T0 – T = kb x m.
Here T0 and T are Boiling points of pure solvent and solution.
m = molality , kb = Ebullioscopic constant.
1. Freezing points : Temperature at which liquids and its solid phase have same vapour
pressure.
2. Depression of freezing point :
Δ Tf = T - T0 = kf x m.
Here T0 and T are freezing points of pure solvent and solution
m = molality , kf = Cryoscopic constant.
3. Osmotic pressure : External pressure applied to the solution to just prevent osmosis.
4. Osmosis : Spontaneous flow of solvent molecule from lower concentrated solution to
higher concentrated solution through semi permeable membrane.
5. Reverse Osmosis : When the applied pressure is greater than osmosis pressure, the
reverse osmosis take place. (used in desalination plants to meet potable water)
6. Isotonic solution : The solutions having same osmotic pressure.
7. Azeotropic mixture : Binary mixture of liquids having same composition in liquid and
vapour phase and boils at the constant temperature.
12
a) Maximum Boiling Azeotropes : Azeotropes boils at the temperature higher than the
boiling point of its components. Eg: Mixture of HNO3 (68 %) and water (32%)
b) Minimum Boiling Azeotropes : Azeotropes boils at the temperature lower than the
boiling point of its components. Eg: Mixture of Ethanol (95%) and water .
8. Ideal Solutions: Solutions obeying Rault’s Law and solvent – solute molecular forces
are same as that of solvent – solvent and solute – solute molecular forces.
Eg: Solution of N- Hexane and N- Heptane, Solution of Benzene and Toluene etc.
9. Positive Deviation from Rault’s Law : Solutions having solvent – solute molecular
forces are weaker than that of solvent – solvent of solute – solute molecular forces .
Δ Hmix = + ve, ΔVmix = +ve and P > PA + PB .
Eg: solution of ethanol and acetone, solution of carbondisulphide and acetone.
10. Negative Deviation from Rault’s Law : Solutions having solvent – solute molecular
forces are stronger than that of solvent – solvent or solute – solute molecular forces.
Δ Hmix = - ve, ΔVmix = - ve and P < PA + PB
11. Abnormal molecular mass : Molecular mass calculated by measurement of colligative
properties differ from normal value due to association or dissociation of solutes in
solution.
Modified equation for colligative properties are
a) Δ P/ PA0 = i XB
b) Δ Tb = i kb x m.
c) Δ Tf = i kf x m.
d) π = i C R T.
12. Vanthoff factor :
i = M normal / M observed
ratio of observed value of colligative properties and normal value of colligative properties
is called vanthoff factor.
13. Degree of dissociation (α) = (i -1) / (n-1).
n = no. of particles formed from dissociation of 1 molecule.
i = van’t hoff factor.
14. Degree of association (α) = (i -1) / (1/n-1).
n = no. of simple molecule to form an associated molecule.
15. a) i >1 (solute undergoes dissociation).
b) i <1 (solute undergoes association).
c) i =1 (solute undergoes neither dissociation nor association).
13
Important Questions:
1.Define the Mole Fraction of a substances in a solution. What is sum of mole fraction
of all components in a three component system?
2.
How is the molality of a solution different from its molarity?
3.
Calculate the density of H2SO4 solution if molality and molarity are 94.5 and 11.5
respectively.
4.
How many ml of 0.1 (M) HCl required to react completely with a 1 gm mixture of
sodium carbonate and sodium bicarbonate containing equimolar amount of both.
5.
State Henery’s Law for a solubility of a gas in a liquid. Give its applications
(three) and limitations?
6.
Explain the significance of Henery’s Law constant KH . At the same temperature
hydrogen is more soluble in water than Helium. Which of them will have a higher
value of KH and Why?
7.
What concentration of nitrogen should be present in a glass of water at room
temperature? Assume a temperature of 250 celcius. A total pressure of 1 atm and
mole fraction of nitrogen in air is 0,78 (KH for nitrogen is 8.42 x 10-7 M/mm Hg).
8.
Henery’s Law constant for CO2 in water is 1.67 x 108 Pa at 298 K. calculate the
quantity of CO2 in 500 ml of soda water when packed under 2.5 atm CO2 pressure
at 298 K.
9.
Mention some limitations of Rault’s Law.
10. Urea form ideal solution of water. Determine the vapour pressure an aqueous
solution containing 10 % by mass of urea at 400 C. (V.P of water at 400 C = 55.3
mm Hg.).
11. Derive the equation to express the relative lowering of vapour pressure for a
solution is equal to its mole fraction of the solute in it when the solvent alone is
volatile.
12. Two liquids X and Y boil at 1100 C and 1300 C, which one of them has higher
vapour pressure at 500 C.
13. Benzene and Toluene forms ideal solution over entire range of composition. The
vapour pressure of pure Benzene and Toluene at 300 K are 50.71 mm Hg and
32.06 mm Hg respectively. Calculate the mole fraction of Benzene in vapour
phase if 80 gm of Benzene is mixed with 100 gm Toluene.
14.
An aquous solution of 2 % non – volatile solute exerts a pressure of 1.004 bar at
normal boiling point of solvent. What will be the molar mass of solute. (V.P of
pure water = 1.013 bar).
14
15. A person suffering from high blood pressure is advised to take minimum quantity
of salt.
16. Write the name of two inorganic substance (compounds) which can be used as
semipermiable membrane.
17. 100 mg protein is dissolved in enough water to make 10 ml of solution if this
solution has an osmotic pressure of 33.3 mm Hg at 250 C . what is the molar mass
of protein. ( R = 0.0821 L atm mol-1 K-1 and 760 mmHg = 1 atm).
18. State condition resulting in reverse osmosis. Give one of its application.
19. Define the following with a suitable example :
a) Maximum and minimum boiling azeotropes.
b) Desalination
c) Ideal solution
d) Abnormal molecular mass
20.
a) Define Vanthoff factor.
b) What is the Vanthoff factor for a compound which
undergoes tetramerisation in an organic solvent.
c) What would be the value of Vanthoff factor for
1) dilute solution of K2SO4 in water
2) Na2SO4.10H2O in water
3) Ethanoic acid in Benzene
d) When is the value of Vanthoff factor
1) more than one
2) less than one
3) equal to one
Hints:
3. m = 1000 M / (1000d – MMB) ,
M=molarity, MB = molar mass of solute, d= density of sol.
d = 1.24 gm/cm3
4. 157.8 ml
7. Partial pressure of nitrogen in atmosphere (PN2) = Ptotal x XN2
PN2 = 592.8 mm Hg
Solubility of N2 = KH x PN2 = 4.99 x 10-4 (M)
8. KH = 1.67 x 108 Pa , PCO2 = 2.5 atm = 2.5 x 1.01325 x 10-5 Pa
PCO2 = KH x XCO2 , XCO2 = 1.517 x 10-3
nCO2/(nCO2 + nH2O) = 1.517 x 10-3
nCO2 = 42.14 x 10-3 mole.
15
10. From Raults Law : (PA0 –P)/PA0 = XB = nB/nA
P = 53.46 mm Hg.
12. X has higher vapour pressure than Y.
13. nbenzene = 80/78 = 1.026 moles
ntoluene = 100/92 = 1.087 moles
Xbenzene = 0.486 , Xtoluene = 0.514
From Rault’s Law : PBen = PBen0 x XBen = 24.65 mm Hg
PTol = PTol0 x XTol = 16.48 mm Hg
Mole fraction of Benzene in vapour phase =
PBen/PBen + PTol = 0.60.
0
0
14. (PA –P)/PA = XB = nB/nA
MB = 41.34 gm mol-1
16. Calcium phosphate , Copper Ferrocyanide
17.
=CxRxT
= WB x R x T/MBV
MB = 13980.45 gm Mol-1
18. External pressure greater than osmotic pressure
Desalination plant .
20. a) Ratio of observed value of colligative properties to the
normal value of colligative properties is called Vanthoff factor.
b) i = ¼
c) 1) 3,
2) 3,
3) ½
d) 1) solute undergoes dissociation in the solution
2) solute undergoes association in the solution
3) solute undergoes neither dissociation nor association
1 MARK QUESTIONS
Q. 1.The vapour pressure of deliquescent substance is less or more than that of water
vapours in air ?
Ans. Less than that of water vapours in air.
Q. 2. If
2SO4 then write the Vant Hoff factor used
for calculating the molecular mass.
–1/m–
– 1) = i –
–
– 1), =>
i
Q. 3. If 6.023 × 1020 molecules of urea are present in 10 ml of its soln. then what is the
conc. of urea soln.?
Ans. N0 = 6.023 × 1023
= 1 mol
6.023 × 1020 molecules = 0.001 mol in 10 ml
.001  1000
N  1000
=
= 0.01 M.
10  1000
v
Q. 4.Why camphor is used in molecular mass determination ?
Ans. Because it has very high cryoscopic constant.
It has large depression in m. p. when an organic solute is dissolved in it.
Q. 5.0.004 M soln of Na2SO4 is isotonic with 0.01 M soln of glucose at the temp. What is
the apparent degree of dissociation of Na2SO4 ?
Ans. 75%
Q. 6.What happen when mango is placed in dilute aqueous soln. of HCl?
16
M=
Ans. When mango is placed in dilute aqueous soln. of HCl it swells.
Q. 7. Out of (a) 200 ml of 2 M NaCl Soln. and (b) 200 ml of 1 M glucose Soln. which one
has higher osmotic pressure?
Ans.: (a) 200 ml of 2 M NaCl Soln.
NaCl is an electrolyte which dissolve to give ions. Glucose and urea are non electrolytes.
Thus glucose has minimum conc. and hence minimum osmotic pressure.
Q. 8.Out of (a) 0.01 M KNO3, (b) 0.01 M Na2SO4 which aqueous soln. will exhibit high
B. P.?
Ans. (a) 0.01 M Na2SO4
Q. 9.Out of (a) 1 M CaCl2 (b) 1 M AlCl3 which aqueous soln. will show max. vapour
pressure at 300 K ?
Ans. (a) 1 M CaCl3, if we assume 100% dissociation, i for CaCl2 = 3 and AlCl3 = 4 and
relative lowering of V. P. is directly proportional to i.
Q. 10. Out of (a) HNO3 + H2O and (b) C6H6 + C6H5CH3 which will form max. boiling
azeotrope ?
Ans. (a) HNO3 + H2O.
2 MARKS QUESTIONS
Q. 1.Two solns of a substance (non-electrolyte) are mixed in the following manner – 480
ml of 1.5 M (First Soln) + 520 ml of 1.2 M (Second Soln). What is the molarity of the
final mixture ?
M V  M2 V2
1.5  480  1.2  520
Ans. Total molarity = 1 1
=
= 1.344 M
V1  V2
480  520
Q. 2.To get the hard boiled eggs, why common salt is added to water before boiling the
eggs?
Ans. Due to addition of common salt the B. P. of the salt containing water elevated, hence
the egg at high temperature becomes hard.
Q. 3.Equimolar Soln. of NaCl and BaCl2 are prepared in H2O. D. F. pt. of NaCl is
found to be – 2 °C. What freezing point do you expect from BaCl2 soln ?
Ans. i for NaCl = 2
i for BaCl2 = 3
F (BaCl2) =
32
 3,
2
F for BaCl2 = 3 °C ,
Freezing Point of BaCl2 Solution is TF = – 3 °C.
Q. 4.Why water cannot be separated completely from ethyl alcohol by fractional
distillation?
Ans. Ethyl alcohol and water (95.4% ethyl alcohol and 4.6% water) form constant boiling
mixture (azeotrope) boiling at 351.1 K. Hence, further water cannot be separated completely
from ethyl alcohol by fractional distillation.
Q. 5.Why a person suffering from high blood pressure is advised to take minimum
quantity of common salt ?
Ans. Osmotic pressure is directly proportional to the conc. of solutes. Our body fluid contains
a number of solutes. On taking large amount of salts, ions enter the body fluid there by
raising the conc. of the solutes. As a result osmotic pressure increases which may rapture the
blood cells.
17
Q. 6.Chloro acetic acid is a monoprotic acid and has Ka = 1.36 × 10–3. Calculate b. p. of
0.01 M aqueous solution? (Kb = 0.51 k kg/mol)
Ans. Kb = 0.51 k kg/mol
= 1 + 0.3687 = 1.3687
= Ka / C
b = i × Kb m
= 1.36 × 10–3 / 0.01
= 1.36 × 10–2 × .51
= 0.3687
= 0.0069 °C
Q. 7.Which colligative property is preferred for the molar mass determination of macro
molecules ? Why ?
Ans. Osmotic pressure is preferred over all other colligative properties because :
(a)
even in dil. soln the o. p.values are appreciably high and can be measured accurately.
(b)
o. p.can be measured at room temp. on the other hand elevation in B. P. is measured
at high temp. where the solute may decompose. The depression in freezing point is measured
at low temp.
Q. 8. How much ethyl alcohol must be added to 1 litre of water so that the solution will
freeze at 14 °F ?
(Kf for water = 1.86 °C/mol)
Ans. (14 – 32) / 9 = C / 5
C
= 5 × (– 18) / 9
= – 10 °C
K  1000  Wb
= F
Wb = mass of solute
F
Wa  Mb
Mb = molar mass of solute
Wa = mass of solvent
1.86  1000  Wb
1000  46
Wb
= 247.31 g
Q. 9. 75.2 g of phenol is dissolved in solvent of KF = 14, if the depression in freezing
point is 7 k. What is the % of phenol ?
10
Ans.
Mb =
=
2 C6H5OH ——
6H5OH)2
1
0
1–
Total = 1 –
=1–
i
=1–
=1–
= 0.375
= 0.75
% of association = 75%
KF = 14
1000  KF  W2
W1  TF
Taking the solvent as 1 kg
1000  14  75.2
Mb =
1000  7k
= 150.4 g per mol
phenol (molar mass) — 94 g/mol
Calculated molar mass
i=
Observed molar mass
= 94 / 150.4 = 0.625
18
Q. 10. How many ml of 0.1 M HCl are required to react completely with 1 gm mixture
of Na2CO3 & NaHCO3 containing equimolar amounts of both ?
Ans.
Let the amount of Na2CO3 be x
Let the amount of NaHCO3 be 1 – x
Since no. of moles of both are equal
x
1 x

N(Na2CO3 ) M(NaHCO3 )
x
1 x

106
84
84 x = 106 – 106 x
x
= 0.5578
xNa2CO3 = 0.5578 / 106
= 0.00526
xNaHCO3 = 0.00526
Na2CO3 + 2 HCl ———
2 + H2
NaHCO3 + HCl ———
2+
H2O
M1V1 = 2 M2V2 + M3V3
0.1 × V1 = 2 × 0.00526 + 0.00526
V1
V
0.01578
0.1
= 0.1578 L
= 157.8 ml.
=
Q.11 Given below is the sketch of a plant for carrying out a process.
(i) Name the process occurring in the
above plant
(ii) To which container does the net flow
of solvent take place
(iii) Name one SPM which can be used
in this plant
(iv) Give one practical use of the plant
Ans. (i) The process is called Reverse Osmosis. [Reverse Osmosis: If the pressure applied
on solution side exceeds the osmotic pressure then the osmosis can be reversed (i.e. pure
solvent can be forced out of the solution to pass through the pores of membrane in opposite
direction). This is called Reverse osmosis].
(ii) The solvent moves from sea water container to fresh water container.
(iii) Cellophane (It is one of the semi-permeable membranes that can be used)
(iv) This process is used for desalination of sea water.
Q.12 On increasing the temperature the solubility of most of the gases in water
decreases. An important consequence of this effect is the thermal pollution of water, such
as occurs around power plants cooled by a river or stream. How is the survival of fish
affected by thermal pollution of water?
Ans.: When the water temperature of the river or stream is increases the solubility of oxygen
in water decreases. Due to this decrease in dissolved oxygen the survival of fish becomes
difficult.
Q.13 How much NaOH is required to preapre 50 ml of aqueous solution with 70mg of
Na+ ions per ml. of the solution?
Ans. : Na+ ions in 1 ml = 70 mg
Na+ ions in 50 ml = 70 × 50 = 3.5g
19
Q.4 Volatile hydrocarbons are not used in the brakes of automobile as lubricants, but
non-volatile hydrocarbons are used as lubricants.
Ans.: The vapour pressure of volatile hydrocarbons is very high and they get evaporated
leaving behind the system. Due to this they are not used as lubricants in automobiles. Nonvolatile hydrocarbons having low vapour pressure are used as lubricants.
Q.5 When fruits and vegetables that have dried are placed in water, they slowly swell
and return to the original form. Explain why ? Would a temperature increase accelerate
the process ? Explain.
Ans.: The cell walls of the fruits and vegetables have semi-permeable membrane. When they
have dried, concentration inside is higher. On placing in water, water enters into the cells, i.e.,
osmosis takes place. Hence, they swell and return to the original form. The process will be
accelerated with increase of temperature because osmosis becomes faster with increase of
temperature.
Q1. Define osmotic pressure.
Ans:- The extra pressure that must be applied to the solution side to prevent the flow of
solvent molecules into solution through a semi permeable membrane is called
osmotic pressure.
Q2.
State Raoult’s Law for a binary solution containing volatile components.
Ans:- At a particular temperature the partial vapour pressure of each component of a
solution containing volatile liquid is directly propositional to its mole fraction.
Q3. Define Vant Hoff factor.
Ans:-It is the ratio of experimental value of a colligative property to the theoretical value.
Q4. What do you understand by colligative properties?
Ans:- Properties of solutions which depend on the number of solute and solvent particles
but not on the nature of solute are called colligative properties.
Q5. State any two characteristics of ideal solution.
Ans:- Characteristics of an Ideal solution .
(i)  mix H  0
(ii) p A  P A  A , pB  pB  B
0
0
Q6. Mention a large scale use of the phenomenon called reverse osmosis
Ans:- In desalination of sea water.
Q7. When is the value of Van’t Hoff factor less than 1?
Ans:- Value of Vant Hoff factor is less than 1 in case of association of solute in solution.
Q8. When is the value of Van’t Hoff factor more than 1?
Ans:- Value of vant Hoff factor is more than 1 in case of dissociation of solute in solution.
Q9. Why does molality of a solution remain unchanged with change in temperature while
its molarity changes?
20
Ans:- Molality of a solution depends on the mass of solvent and man does not vary with
change in temp. where as molarity depends on volume of solution and volume
change with change in temp.
Q10. What are azotropes?
Ans:- A liquide mixture which distills over without changes in composition is called
ozeotropes or ozeotropic mixture.
Q11. What are maximum boiling azeotropes? Give one example?
Ans:- A mixture which boils at temp. brighter than the boiling point of its components the
mixture is known as maximum boiling azeotropes. Eg- a mixture of HCl and H2O
containing 20.2% of HCl by weight.
Q12. What are minimum boiling azeotropes? Give one example?
Ans:- A mixture which boils at temp. lower than boiling point of its components the
mixture is known as minimum boiling azeotropes. E.g water and Benzene.
Q13. What is expected value of ‘i’ for K4[Fe(CN)6] in dilute solution?
Ans:- Expected value of n in K4(Fe(CN)6) is 5.
Q14. What are isotonic solutions?
Ans:- Those solutions are said to be isotonic which have the same osmotic pressure.
Q15. Define molal elevation constant or ebullioscopic constant.
Ans:- It is equal to the elevation in boiling point of1 molar soln. i.e one mole of a solute
dissolved in 1 kg of the solvent.
Q16. What are non-ideal solutions? Explain as to why non-ideal solutions deviate from
Raoult’s Law.
Ans:- Non ideal solution are those which do not obey Raoult’s law.
When the molecular interaction between solute molecules solvent. Solvent molecules
is not same as the molecular interaction between solute & solvent molecules then
deviotion occurs.
Q17. What are colligative properties? Mention them.
Ans:-these are the colligative properties.
(a) Osmotic pressure (b) Relative lowering of vapour pressure
(b) Elevation in B.P (d) Depression in F.P
NUMERICALS
Q1. Calculate the mole fraction of ethylene glycol (C2H6O2) in solution containing 20 %
of ethylene glycol by mass.
Ans:Mole fraction of ethylene glycol =
no.of .mole.of .thelene.glycol
No.of .moles.of .water  No.of .moles.ofC2 H 6O2
21
i.e
 glycol  nglycol / nwater  nglycol 

20 / 62
20 / 62  80 / 18
20 X 62 X 18
9

 0.0676
62 X 5320
133
So  H 2 O  1   glycol  1  0.0676  0.9323
Q2. An aqueous solution of glucose made by dissolving 10 g of glucose (C6H12O6) in 90 g
of water at 303 K. if vapor pressure of pure water at 303 K be 32.8 mmHg. What
would be the vapor pressure of solution
Ans:Hence wB= 10 g, MB=180, WA=90 g, MA=18
P0A=32.8 mmHg.
PS=?
10
P A  PS
P A  PS
180  1  p S  1 / 90




B
90 / 18
P0 A
P0 A
P0 A
PS
1
89
89
0

1


P

P

32
.
8

 32.43mmHg
A 
S
90
90
90
P0 A
0
0
Q3. A solution 3.8 g of sulphur in 100 g of CS2 (Boiling point=46.3 0 C) boils at 46.66 0c.
what is the formula of sulphur molecules in the solution .(Atomic mass of sulphur is
32 g mol-1 and Kb= 2.4 K kg mol-1)
Ans:We have :- M B 
WB  1000  K B
Tb  WA
Here
Tb  (273  46.66)  (273  46.3)  0.36 K
MB 
3.8  1000  2.4 760

 253.3
0.36  100
3
No of atoms of sulphur S=Molecualar wt./Atomic Wt. = 253.33/32  8
Hence the formula of sulphur = S8
22
Q4. Calculate the osmotic pressure of a solution obtained by mixing 100 cm3 of 0.25 M
Solution of urea and 100 cm3 of 0.1M soln of cane sugar at 293 K (R=0.082 L atm
mol -1K-1)
Ans:We know V  nRT
0.25  0.1
)  0.0821  293
2
Or
0.35  0.0821  293
 
 21.022atm
2  0.2
  0.2  (
Q5. 2 g of benzoic acid (C6H5COOH) dissolved in 25 g of benzene shows a depression in
freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol1
. what is the percentage association of acid if it forms dimmer in soln.?
Ans:Given that
WB=2g , Kf=4.9 KKgmol-1, WA=25 g, T f  1.62 K
We have
MB 
WB  1000  K f
T f  WA

2  1000  4.9
 241.98 gmol 1
1.62  25
Also 2C6H5COOH  (C6H5COOH)2
Initially
1
0
( 1-x)
x/2 mole
Total no. of particles at equilibrium. =
i  1
(1  x) 
x
x
 1
2
2
x
2
Now i=(Normal molecular mass/Abnormal molecular mass)
=122/241.98
Thus 122/241.98=1- x/2
Or x/2 =1- (122/241.98) =0.4958
So x = 0.9916
The degree of association of benzoic acid in benzene is 99.16 %
***************
23
FOR BELOW AVERAGE
01 Mark questions
1.
State Henry’s law.
2.
Define osmotic pressure.
3.
How pressure effect the solubility of a solid in a liquid.
4.
State Raoult’s Law for a binary solution containing volatile components.
5.
Define Henry’s law about solubility of a gas in a liquid.
6.
Define Vant Hoff factor.
7.
What do you understand by colligative properties?
8.
Define an ideal solution.
9.
State any two characteristics of ideal solution.
10.
Distinguish between molarity and molality.
11.
What is a non- ideal solution?
12.
Mention a large scale use of the phenomenon called reverse osmosis.
13.
What is antifreeze? Give one example.
14.
When is the value of Van’t Hoff factor less than 1?
15.
Name the two factors on which the vapour pressure of the liquids depend.
16.
Define mole fraction of a substance in a solution
17.
When is the value of Van’t Hoff factor more than 1?
FOR AVERAGE
1.
Why is the elevation in b.p. of water different in the following solutions? (i) 0.1 M
NaCl solution (ii) 0.1 M Sugar solution.
2.
What are azotropes?
3.
What happens when blood cells are placed in pure water?
4.
Why is the cooking temperature in pressure cooker higher than in open pan?
5.
Why does molality of a solution remain unchanged with change in temperature while
its molarity changes?
6.
Why is ether not miscible in water?
7.
What are maximum boiling azeotropes? Give one example?
8.
What are minimum boiling azeotropes? Give one example?
9.
Why do doctors advice gargles by saline water in case of sore throat?
10.
Why is boiling point elevated when a non volatile solute is added?
11.
A person suffering from high blood pressure should take less common salt, why?
12.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
24
FOR ABOVE AVERAGE
1.
How mole fraction of a solute and molality are related?
2.
Two liquids A and B boil at 145C and 190C respectively. Which of them has a
higher V.P. at 80C?
3.
What is expected value of ‘i’ for K4[Fe(CN)6] in dilute solution?
4.
How molarity and molality of solute are related.
5.
What possible value of ‘i’ will it have if solute molecules undergo association in
solution.
6.
Define molal elevation constant or ebullioscopic constant?
7.
Define molal depression constant or cryoscopic constant?
8.
Why is osmotic pressure considered as a colligative property?
9.
On mixing equal volumes of water and ethanol what type of deviation would you
expect from Raoult’s Law?
10.
What are isotonic solutions?
FOR BELOW AVERAGE
02 Marks question
1.
Under what condition Van’t Hoff factor is (i) equal to one, (ii) greater than 1, (iii) less
than 1.
2.
Two liquid A and B on mixing produce a warm solution. Which type of deviation
from Raoult’s Law does it show?
3.
What are non-ideal solutions? Explain as to why non-ideal solutions deviate from
Raoult’s Law.
4.
What are colligative properties? Mention them.
5.
Differentiate between molarity and molality of a solution. When and why is molality
preferred over molarity in handling solutions in chemistry?
FOR AVERAGE
1
Name the factors which affect the vapour pressure.
2.
Amongst the following compounds, identify which are insoluble, partially soluble and
highly soluble in water.
(a) Phenol
(e)Chloroform
3.
(b)Toluene
(c) Formic Acid
(d)Ethylene Glycol
(f) Pentanol
What do you mean by relative lowering of vapor pressure ? How is relative lowering
of vapor pressure related with mole fraction of non volatile solute in a solution?
25
4.
With the help of a suitable diagram show that the vapour pressure of a solution is
lower than the pure solvent, causes a lowering of freezing point for the solution
compared to that of the pure solvent.
5.
Carbon tetra chloride and water are immiscible where as ethanol and water are
miscible in all proportions. Correlate this behaviour with molecular structures of these
compounds.
FOR ABOVE AVERAGE
1.
What do you mean by abnormal molecular mass? What are its cause?
2.
With the help of suitable diagrams, illustrate the two types of non ideal solutions.
3.
State Raoult’s law for solutions of non volatile solutes in volatile solvents. Derive a
mathematical expression for this law.
4.
What is osmotic pressure and how is it related with the molecular mass of non-volatile
solute?
5.
What is meant by abnormal molecular mass of solute? Discuss the factors which bring
abnormality in the experimentally determined molecular masses of solutes using
colligative properties.
6.
State Raoult’s Law. Discuss the factor responsible for the deviation from this law.
7.
State Henry’s Law and mentions some important application.
8.
Explain the significance of Henry’s Constant (KH). At the same temperature,
hydrogen is more soluble in water than helium. Which of them will have a higher
value of KH and why?
9.
Explain the difference between osmotic pressure and vapour pressure of a solution.
10.
Give one example each of miscible liquid pairs showing positive and negative
deviation from Raoult’s Law. Give one reason each for such deviations.
FOR AVERAGE/BELOW AVERAGE
02 Marks question (Numerical)
1.
The vapour pressure of pure liquid A and B are 450 and 700 mm Hg respectively at
350 K. Find out the composition of the liquid mixture if total vapour pressure is 600
mm Hg.
2.
Calculate the mass percentage of aspirine(C9H8O4) in acetonitrile(CH3CN) when 6.5 g
of C9H8O4 is dissolved in 450 g of CH3CN.
3.
H2S, toxic gas with rotten a like smell, is used for qualitative analysis. If the solubility
of H2S in water at STP is 0.19 m. Calculate Henry’s law constant.
4.
18 g of glucose is dissolve in 1 kg of water in a saucepan. At what temperature will
water boil at 1.03 bar? Kb for water is 0.52 K kg mol-1.
26
5.
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the
normal point of the solvent. What is the molar mass of the solute?
6.
The boiling point of benzene is 353.23 K. When 1.08 gm of non-volatile solute was
dissolved in 90 gm of benzene. The boiling point is reached to 354.11 K. Calculate
the molar mass of solute. Kb for benzene is 2.53 K kg mol-1.
7.
45 g of ethylene glycol is mixed with 600 g of water. Calculate (a) the freezing point
depression and (b) the freezing point of the solution.
8.
200 cm3 of an aqueous solution of a protein contains 1.26 g of protein. The osmotic
pressure of such a solution at 300 K is found to be 2.57 x 10-3 bar. Calculate the molar
mass of the protein.
9.
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate
the freezing point of 5% glucose in water if the freezing point of pure water is 273.15
K.
FOR ABOVE AVERAGE
03 Marks Questions (Numerical)
1.
If N2 gas is bubbled through water at 293K, how many millimoles of nitrogen gas
would dissolved in 1 ltr. of water. Assume that N2 exerts a partial pressure of 0.987
bars. Given that Henry’s law constant for N2 at 293 K is 76.48 kbar.
2.
0.6 mL of acetic acid having density 1.06 g mL-1, is dissolve in 1 litre of water. The
depression in freezing point observed for this strength of acid was 0.0205°C.
Calculate the Van’t Hoff factor and the dissociation constant of acid.
3.
100 g of liquid A (molar mass 140 g/mol) was dissolved in 1000 g of liquid B (molar
mass 180 g mol-1). The vapour pressure of pure liquid B was found to be 500 torr.
Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution
if the total vapour pressure of the solution is 475 Torr.
4.
2 g of benzoic acid dissolved in 25 g of benzene shows a depression in freezing point
equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol-1. What is the
percentage association of acid if it forms dimmer in solution?
5.
Determine the osmotic pressure of solution prepared by dissolving 25 mg of K2SO4 in
2 litre of water at 25°C. Assuming that it is completely dissociated.
6.
Two elements A and B from compounds having formula AB2and AB4. When
dissolved in 20 g of benzene, 1 g of AB2 lowers the freezing point by 2.3 K where as
1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K
kg mol-1. Calculate atomic mass of A and B.
27
7.
An antifreeze solution is prepared from 222.6 g of ethylene glycol and 200 g of water.
Calculate the molality of solution. If the density of the solution is 1.072 g/mol, then
what shall be the molarity of the solution?
*************************
28
UNIT :-3
ELECTRO CHEMISTRY
STUDY MATERIAL
(ELECTR
O CHEMISTRY)
Q1. Define molar conductivity. How does it varies with dilution ?
Ans:- Molar conductivity:It is the product of specific conductivity and volume of solution containing 1 g molecular
mass of the electrolyte.
 mol 
K  1000
C
Unit= Sm2 mol-1
* Variation :-
Strong Electrolyte:-Molar conductivity increases slowly with decrease in concentration.
Weak Electrolyte:- Molar Conductivity increases sharply for weak electrolyte on dilution
.Q2. States Koahlrausch law.
Ans:- Limiting molar conductivity of an electrolyte is the sum of molar conductivity of cation
and anion at concentration approaches zero.
mol    
0
0
0
  =Molar Conductivity of cation
0
  =Molar Conductivity of anion
0
Q3. Wat do you mean by fuel cell? Write cathode and anode reaction in fuel cell.
Ans:- Fuel Cell:- These are electrical cells which can convert the energy of combustion of
a fuel like (H2, CO etc.) directly into electrical energy
Ex:-H2-O2 fuel cell.
29
Reaction :Anode reaction:- H2(g)+2OH-(aq) --- 2H2O+2e….X 2
Cathode reaction:- O2(g) +2H2O+4e ---- 4OHOverall reaction :-2H2(g) + O2(g) ------ 2H2O(l)
Q3. What do you mean by Corrosion? How it is a electrochemical phenomena?
Ans:- in corrosion a metal is oxidized by a lose of electrons to oxygen and metal oxide is
formed. It is an electrochemical phenomenon.
Chemistry of corrosion of iron:The sport where iron under goes oxidation is considered as anode.
At anode: 2Fe ---- 2Fe2+ + 4eAt another sport on the metal these electrons reduce oxygen in presence of
H+.
At anode :O2(g) + 4H+ + 4e- ---- 2H2O(L)
The overall reaction is
2Fe + O2 + 4H+ ----- 2Fe2+ + 2H2O
The ferrous ions formed are further oxidized to ferric ions by atmospheric oxygen and
products comes out as rust in the form of Fe2O3XH2O(hydrated Ferric oxide)
Q4. A solution of CuSO4 is electolysed for 10 minutes with a current of 1.5 amperes. What is
the mass of copper deposited at cathode?
Ans:- t= 10X60=600 Sec.
I=1.5 amperes.
Q=nF
(Cu2+ +2e --- Cu)
=2X96500C
Q=It=1.5X600
=900 C
 2 X 96500 C charge deposited 63 g of copper
 900 C charge deposited
63  900
gm of copper
2  96500
=0.2938 gm.
Q5. The resistance of a conductivity cell containing 0.001 M Kcl solution at 298 K is 1500
30
 . What is the cell constant If conductivity of 0.001 M Kcl solution 298 K is 0.148 X
10-3 S cm-1.
Ans:- R=1500 
K=0.148 X 10-3 S cm-1
l
 Cell Constant = ?
a
l
 RK
a
=1500 X 0. 148 X 10-3
=0.222 Cm-1
Q6. Conductivity of 0.00241 M acetic acid is 7.896 X 10-5 Scm-1. Calculate its molar
conductivity and if  0 for acetic acid 390.5 Scm2 mol-1. what is its dissociation
constant?
Ans:K=7.896 X 10-5 S cm-1
 0  390.5Scm 2
C 
K  1000
C
7.896  105  1000 7896


 32.93
0.00241
241
C 32.93
 0 
 0.0838

390.5
K   2C  (0.0838) 2  0.00241  1.85  10 5
Q7. Represent the cell in which the following reaction takes place.
Mg(S) + 2Ag +(0.0001 M) ----- Mg2+(0.130M) + 2Ag(s)
Calculate its E.M.F when E0=3.17V
Ans:- Cell reaction
Mg ----- Mg2+ + 2e -- - - - - Oxidation at anode
2Ag+ + 2e --- 2 Ag --- - - - - - -Reduction at cathode
Representation of cell
Mg│Mg2+ (0.130 M) ││ Ag+(0.0001 M) │Ag
C1
C2
31
E.M.F of the cell E= E

0.591
C
log 2
n
C1
 
 
2
0.0591
Ag 
=3.17 +
log
2
Mg 2 
0.0591
(0.0001)2
=3.17 +
log
0.130
2
=3.17 -0.21
E=2.96 V
Q8.  0 mol for NaCl, HCl and Na AC(CH3COONa) are 126.4, 425.9 and 91.0 Scm2 mol-1
respectively. Calculate  0 mol for HAC(CH3COOH, Acetic Acid).
Ans: 0 HAC   0 CH # COO    0 Na    0 H    0 Cl    0 Na    0 Cl
 91.0  425.9  126.4  516.9  126.4

=390.5 Scm2 mol -1
Q9. States Faraday’s laws of electrolysis.
Ans: 1st law:- The amount of substance deposited during electrolysis is directly
proportional to quantity of electricity passed.
m  Q, m=Z Q
=Zit
Where Z=electrochemical equivalent.
2nd law:- If same change is passed through different electrolytes, the mass of substance
deposited at each electrodes will be propositional to their equivalent weights.
W1/E1=W2/E2
Where W is mass of substance and E is its equivalent weight.
Q10. Calculate ▲G0 for Zn-Cu cell at standard conditions.
Given E 0 Zn 2  / 2 n  0.76V , E 0Cu 2  / Cu  0.34V , F  96500C
Ans:E 0 cell  E 0Cathode  E 0 anode
 0.34  (0.76)  0.34  0.76
 1.10V
G 0  nE 0 F  2  1.10V  96500c
 212.2kjmol1
32
Q11. Depict the galvanic cell in which the reaction
Zn(s)+2Ag+(aq) --- Zn2+(aq) + 2Ag(s) takes place.
(a) Which of the electrode is negative chage.
(b) The carries of the current in the cell.
(c) Individual reaction at each electrode.
Ans:
Cell Reaction
AT anode
Zn  Zn 2   2e      oxidation
AT cathode
2 Ag   2e  2 Ag     Re duction

Zn  2 Ag  Zn 2   2 Ag
Cell representation
Zn Zn
2  ( aq)
Ag  Ag
E  E  E  Ecathode  Eanode
(c) Ag electrode is negative charged.
(d) Ag+ and Zn2+ ions carries the current in the cell. Current flow from Ag
electrode to Zn electrode.
(e) At anode
Zn  Zn 2   2e      Oxidation
At cathode.
2 Ag   2e  2 Ag     Re duction
1 MARK QUESTIONS
Q. 1. Which solution will allow greater conductance of electricity, 1 M NaCl at 293 K
or 1 M NaCl at 323 K and why ?
Ans.
1 M NaCl at 323 K as the ionic mobilities increase with increase in temperature.
Q. 2. What does the negative value of E°cell indicate ?
Ans.
Q. 3. Why is the equilibrium constant K, related to only E°cell and not Ecell ?
Ans.
This is because Ecell is zero at equilibrium.
33
Q. 4.
Ans.
Positive.
Q. 5. Rusting of iron is quicker in saline water than in ordinary water. Why is it so ?
Ans.
In saline water, NaCl helps water to dissociate into H+ and OH–. Greater the number
of H+, quicker will be rusting of Iron.
Q. 6. What would happen if the protective tin coating over an iron bucket is broken in
some places ?
Ans.
Iron will corrode faster as the oxidation potential of Fe is higher than that of tin.
Q. 7. Can a nickel spatula be used to stir a solution of Copper Sulphate ? Justify your
answer.
(E°Ni²+/Ni = – 0.25 V
Ans.
E°Cu²+/Cu = 0.34 V)
Reduction potential of Ni is less than Cu. Ni will replace the Cu from CuSO4. Thus Ni
spatula cannot be used to stir a solution of CuSO4.
Q. 8.
m
and
why ?
Ans.
m
because H+ (aq) being smaller in size than Na+ (aq)
and have greater mobility.
Q. 9. Three iron sheets have been coated separately with three metals A, B, C whose
standard electrode potentials are given below :
A
E°value – 0.46 V
B
C
Iron
– 0.66 V
– 0.20 V
– 0.44 V
Identify in which rusting will take place faster when coating is damaged.
Ans.
Rusting of iron will take place when coated with metal C as it is placed above iron
more than other metal.
Q. 10. Which will have greater molar conductivity ? Solution containing 1 mol KCl in
200 cc or 1 mol of KCl in 500 cc.
Ans.
1 mol of KCl in 500 cc.
2 MARKS QUESTIONS
34
Q. 1. (a)
How will the value of Ecell change in an electrochemical cell involving the
following reaction of the concentration of Ag+ (aq) is increased ?
(b)
What will be e. m. f. when the cell reaches equilibrium :
Mg (s) + 2 Ag+ (aq) ——
Ans.
(a)
2+
(aq) + Ag (s)
Mg2 
0.059
Ecell = E°cell –
log
2
2
 Ag 
As the concentration of [Ag+] ion increases, Ecell increases.
(b)
Q. 2. (a)
e.m.f. = 0
In a cell reaction, the equilibrium constant K is less than one. Is E° for the
cell positive or negative ?
(b)
Ans.
What will be the value of K of E°cell = 0 ?
For a cell E° =
0.0591
log K
n
K<0
i.e. log K is – ve.
Then E°cell will be negative.
(b)
If E°cell = 9 then 0 =
0.0591
log K
n
Q. 3. Knowing that :
Cu2+ (aq) + 2 e– ———
E° = + 0.34 V
2 Ag+ (aq) + 2 e– ———
E° = + 0.80 V
Reason out whether, 1 M AgNO3 solution can be stored in Copper Vessel or 1 M
CuSO4 solution in Silver Vessel.
Ans.
A solution of an electrolyte can be stored in a particular vessel only in case there is no
chemical reaction taking place with the material of the vessel.
Cu is a strong reducing agent and can lose electrons to Ag+ as E° of Cu is less than
that of Cu. So AgNO3 cannot be kept in Copper Vessel.
35
CuSO4 solution can be stored in Ag Vessel as no chemical reaction will take place as
Ag is placed above Cu in the activity series and Ag is less reactive than Copper.
Q. 4. What is the number of electrons in one Coloumb of electricity ?
Ans.
Charge on one mole of electrons = 1 F = 96500 C
96500 C of Charge is present on electrons = 6.022 × 1023
1 C of Charge is present on electrons =
6.022  1023
×1C
96500 C
= 6.24 × 1018
Q. 5. Which of the following pairs will have greater conduction and why ?
Ans.
(a)
Copper wire at 25 °C and Copper wire at 50 °C.
(b)
0.1 M acetic acid solution or 1 M acetic acid solution ?
(a)
Copper wire at 25 °C because with increase in temperature metallic
conduction decreases due to vibration of kernels.
(b)
0.1 M acetic acid solution because with dilution degree of dissosciation
increases and hence no. of ions.
3 MARKS QUESTIONS
Q. 1.
) is plotted against
m
the square root of concentration for 2 electrolytes A and B.
(a)
What can you say about the nature of the two electroyltes A and B ?
(b)
H
m
for the
electrolytes A and B on dilution ?
400
lm
200
A
0.2 ½
C
Ans.
B
0.4
(a)
A is a strong electrolyte and B is a weak electrolyte.
(b)
Molar conductivity of a strong electrolyte (A) increases with dilution as ionic
mobility increases. In a weak electrolyte molar conductivity increases steeply
36
with dilution as degree of dissociation increases and hence no. of ions
increases.
Q. 2. Iron and nickel are used to make electrochemical cell by using a salt bridge to
join a half cell containing 1 M Fe2+ (aq) in which a strip of iron has been
immersed to a second half cell which contains 1 M Ni2+ (aq) in which a strip of Ni
has been immersed ? A voltmeter is connected between the two metal strips :
E°Fe²+/Fe = – 0.44 V
E°Ni²+/Ni = – 0.25 V
(a)
Write the name of the cathode and anode.
(b)
Write the half reactions involved ?
(c)
What would be the effect on the Voltmeter reading if Fe2+ concentration
were increased ?
Ans.
(a) Anode : Fe
Cathode : Ni
(b)
Reaction at anode : Fe ———
2+
+ 2 e–
Reaction at cathode : Ni2+ + 2 e– ———
(c)
Voltmeter reading decreases.
Q. 3. Consider the electrochemical cell :
Zn (s) / Zn2+ (aq) // Cu2+ (aq) / Cu. It has an electrical potential of 1.1 V when
concentration of Zn2+ and Cu2+ ions is unity.
State the direction of flow of electrones and also specify if Zinc and Copper are
deposited or dissolved at their respective electrodes. When :
Ans.
(a)
an external opposite potential of 0.8 V is applied.
(b)
an external opposite potential of 1.1 V is applied.
(c)
an external opposite potential of 1.4 V is applied.
(a)
Electrons flow from Zn rod to Cu rod.
Zinc dissolved and Copper gets deposited.
(b)
No flow of electrons and current.
No change observed at Zinc and Copper electrodes (system is at equilibrium).
(c)
Electrons flow from Cu rod to Zn rod.
Zinc is deposited and Copper gets dissolved.
Q. 4. Given that :
CO3+ + e– ———
2+
E° = 1.82 V
37
2 H2O ———
E° = – 1.23 V
+ 4 H+ +4 e–
2
Explain why CO3+ is not stable in aqueous solution ?
Ans.
The E°cell can be calculated as :
4 [CO3+ + e– ———
2 H2O ———
2
] E° = 1.82 V
2+
E° = – 1.23 V
+ 4 H+ +4 e–
——————————————————————
Cell reaction : 4 CO3+ + 2 H2O ———
2+
O2 + 4 H+
E°cell = 1.82 V – (– 1.23 V) = 3.05 V
Since E°cell is positive, the cell reaction is spontaneous. CO3+ iron will take part in the
reaction and hence unstable in aqueous solution.
Q. 5. For the reaction :
Ag+ + Hg ———
E° = 0.80 V
2+
2
E° = 0.79 V
Predict the direction in which the reaction will proceed if :
[Ag+] = 10–1 mol/h [Hg2+] = 10–3 mol/h
Ans.
Cell reaction is :
2 Ag+ + 2 Hg ———
Ecell
= E°cell –
2+
2
Hg22 
0.0591
log 
2
2
 Ag 
103
0.0591
= (0.80 V – 0.79 V) –
log
2
2
101

= 0.01 V –

0.0591
(– 1) = 0.01 + 0.0295
2
= 0.0395 V
Since Ecell is positive, the reaction will be spontaneous in the forward direction.
FOR ABOVE AVERAGE
01 Marks Questions
1. Express mathematical relationship among resistance(R), Specific conductivity (K)
and cell constant.
2. Write relation between specific conductance and molecular conductivity of an
electrolytic.
3. State the Kohlrausch law.
38
4. Name a metal that can be used for the cathodic protection of iron against rusting.
5. How are secondary cells different from primary cells?
6. Write the reaction occurs in fuel cell.
7. What do you mean by standard electrode potential?
8. What is galvanization of iron?
9. Define strong electrolytes.
10. Define molar conductivity.
02 Marks Questions
1.
What is corrosion? How is rusting of iron protected by cathodic protection.
2.
How is cathodic protection different from galvanization in protection of iron from
rusting?
3.
What do you understand by equilibrium constant? Calculate the equilibrium
constant of the reaction
Cu (s ) + 2Ag + (aq) → Cu 2+ (aq) + 2Ag (s) Eº = 0.46 v .
4.
Represent Zn –Cu cell and write Nernst equation for the calculation of emf of the
cell.
5.
The conductivity of 0.2M solution of NaCl at 298 K is 0.0248 S cm-1. Calculate its
molar conductivity.
6.
The standard electrode potential for Daniell cell is 1.1v. Calculate the standard
Gibb’s energy for the reaction
Zn (s) + Cu 2+ (aq) → Zn 2+ (aq) + cu (s)
7.
State and explain the Faraday’s laws of electrolysis. What is the value of
Faraday’s constant?
8.
State reasons for the following –

Rusting of iron is said to be an electrochemical phenomena.

Rusting of iron is quicker in saline water than in ordinary water.

Arrange the following metals in the order in which they displace each from
the solution of their salts.
Al, Mg, Fe, Cu, Zn
03 Marks Questions
1.
Write the Nernst equation and calculate the emf of the following cell at 298 K.
Cu(s) / Cu 2+ ( 0.130 M ) // Ag+ ( 1 x 10-4 M ) / Ag (s). Given that E0 Cu+2 / Cu =
0.34 V and E0 Ag+ / Ag = 0.80 V,
F = 95500 C Mol-1
2. What is standard hydrogen electrode? How it measures the electrode potential of an
electrode? Explain.
39
3. What are secondary cells? Explain lead storage battery with electrode reactions.
4. A galvanic cell is constructed in which the cell reaction is
Zn(s) + 2 Ag+ (aq) → Zn2+ (aq) + 2 Ag (s),
Now write:
(a) Which of the metal electrode is negatively charged?
(b) The direction of the current (either anode to cathode or cathode to anode)
(c) Individual half electrode reaction for cathode
5. What are electrode potential and emf of a cell?
Calculate the emf of the cell
Mg(s) / Mg 2+ ( 0.1 M ) // Ag+ ( 0.01 M ) / Ag at 250 C
Given that E0 Ag+ / Ag = 0.8 V, E0 Mg2+ / Mg = - 2.37 V
05 Marks Question
1.
What is EMF of a cell?
Calculate the emf of the cell
Mg(s) / Mg2+ ( 0.1 M ) // Ag+ ( 1 x 10-4 M) / Ag (s) at 250 C,
Given that E0 Ag+ / Ag = 0.8 V , E0 Mg2+ / Mg = -2.37v
2. What will be the effect on EMF of the cell if concentration of Ag+ is increased to 1 x 103
M?
3. Explain the product of electrolysis of molten NaCl and aqueous NaCl. Explain the
difference between these two phenomena and list two other application of electrolysis.
4. A cell is formed as
Ni (s) / Ni2+ ( 0.01 M ) // Cu2+ ( 0.1 M ) / Cu(s)
The E0 values for Ni2+ / Ni and Cu2+ / Cu electrodes are -0.25 V and 0.34 V
respectively.
Calculate the cell potential, equilibrium constant and work done by the cell.
5. Electrolysis of aqueous sodium chloride gives H2 gas at cathode instead of sodium metal
and Cl2 gas anode while electrolysis of molten NaCl gives sodium metal at cathode.
How is this difference in results explained? Give electrode reactions for anode and
cathode.
FOR BELOW AVERAGE
01 Mark Questions
1.
What is over voltage in an electrolytic reaction?
2.
What is Nickel- Cadmium cell?
3.
How is unit of molar conductivity arrived at?
40
4.
How many faradays of electricity are required to liberate 2 moles of hydrogen gas in
electrolysis of a solution ?
5.
Why does cu not displace Fe from FeSO4 Solution?
6.
Why does an alkaline solution enhance the rusting of iron?
7.
What is concentration cell? Give an example.
8.
In function of a galvanic cell, one of its electrodes does oxidation reaction. What is
the name of the electrode and what is its polarity?
9.
What is the electrolyte used in a fuel cell?
10.
Suggest one method to operate galvanic cell if the salt bridge is absent.
11.
The standard reduction potential for the Zn2+ aq/Zn (s) half cell is -0.76 V. Write the
electrode reactions of the cell when it is coupled with standard hydrogen electrode
(SHE). Also give the standard cell potential.
02 Marks Questions
1. Why is it not allowed to determine the molar conductivity at infinite dilution of a weak
electrolyte by extrapolating is allowed in the graph of strong electrolytes. Explain.
2. If E0 for copper electrode is +0.34V; how will you calculate its emf when the solution in
contact with it is 0.1 M in copper ions? How does emf of copper electrode change when
concentration of Cu2+ ions in solution is decreased?
3. Calculate the potential of a zinc - zinc ion electrode in which the zinc ion activity is
0.001M
[E0 Zn2+ /Zn = -0.76 V, R= 8.314 JK-1mo1-1, F=96500 Cmo1-1]
4. The molar conductance of sodiumacetate, hydrogenchloride and sodiumchloride at
infinite dilutions are 91, 426 and 126 Simen cm2 mol-1. Calculate the molar conductance
of acetic acid at infinite dilution.
5.
Estimate the minimum potential difference needed to reduce Al2O3 at 500°c. The ΔG
for the decomposition reaction 2/3 Al2O3 → 4/3 Al + O2 is ΔG = 960 KJ.
6.
The E0 Values for two metal electrodes are given below
(i) Cr3+/Cr2+= -0.4V (ii) Fe3+/Fe2+ = 0.8V
Comment on the result of treating a solution of Cr (II) with a solution containing
Fe (III) ions
7.
Calculate the standard free energy change for the reaction occurring in the cell
Zn|Zn2+ (IM) // Cu2+ (IM) / Cu
Given E0zn2+/zn = -0.76V and E0 cu2+/Cu= 0.34V.
41
19.
Explain the electrolysis of aqueous NaCl and molten NaCl with their Chemical
reactions. Give reasons for this difference and deduce criteria for product formation.
20.
Explain the electrode reaction of a fuel cell. Discuss the advantage of fuel cell in
space programme.
03 Marks Question
1.
Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO3 With Ag electrodes.
(ii) A Dilute solution of H2SO4 With platinum electrodes.
(iii) An aqueous solution of CuCl2 with platinum electrodes.
2.
The resistance of a conductivity cell containing 0.001 M KCl Solution at 298 K is
1500 ohms. What is the cell constant if conductivity of 0.001 M KCl Solution at 298
K is 0.146X10-3 S cm-1.
3.
Conductivity of 0.00241 M acetic acid is 7.896X10-5 S cm-1, calculate its molar
conductivity and if ^0m for acetic acid is 390.5 S cm2 mol-1. What is its dissociation
constant?
4..
“Rusting is an electrochemical phenomena.” Explain this phenomena with the help
of rusting of iron with reactions involved.
5.
Silver is electrodeposited on a metallic vessel of surface area 800 cm2 by passing
current of 0.2 ampere for 3 hours. Calculate the thickness of Silver deposited.
(Density of Ag =10.47 g cm-3, Atomic mass of Ag=108 amu)
05 Marks Question
6. Calculate the potential of following cell reaction at 298 K
Sn4+(1.50 M)+ Zn → Sn2+(0.5M)+ Zn2+(2M)
The standard potential E0; of the cell is 0.89V, whether the potential of the cell
increase or decease (R=8.314 JK-1mol-1, F=96500 C mol-1) if the concentration of
Sn2+ is increased in the cell.
7. The conductivity of NaCl at 298 K has been determined at different concen-tration and the
results are given below:
Concn / M
0.001
102 x K/S m-1
0.010
1.237 11.58
0.020
23.15
0.050
55.53
0.100
106.74
Calculate ^m for all concentrations and draw a plot between ^m and C1/2. Find the
value of ^°m.
8.
(a)
State the reasons for the following :a. Iron does not rust even if Zinc coating is broken in a galvanized iron
pipe.
42
b. Electrolysis of KBr (aq) gives Br2 at anode, but that of KF (aq) does
not give F2.
(b)
Write the electrode reaction of H2-O2 Cell.
(c)
Calculate the EMF of the following cell at 298 K
Sn/Sn2+(0.1M) // Ag+(0.1M) / Ag
Given E0 Sn2+/Sn = -0.14V, E0 Ag+/Ag = 0.8V
9.
(a)
State Kholrausch’s law for electrical conductance of an electrolyte at infinite
dilution.
(b)
Give the composition and reaction of cathode and anode in a mercury cell?
Give one use.
(c)
Silver is electrodeposited on a metallic vessel by passing a current of 0.2
ampere for 3 hours. Calculate the weight of silver deposited.
(Atomic mass of Ag=108 amu)
FOR AVERAGE
01 Mark Question
1.
Define the molar conductivity of an electrolytic solution.
2.
Express the relation between degree of dissociation of an electrolyte and its molar
conductivities.
3.
What does the standard electrode potential of a metal being negative E0
Zn2+/Zn = - 0.76 v shows?
4.
What happens when the protective coating of galvanized iron is broken?
5.
How is electroplating different from galvanization?
6.
How does a fuel cell operate?
7.
How much charge is required for the following reduction of 1 mol of Cu2+ ions to cu
(s)?
8.
Which type of metal can be used for the cathodic protection of iron?
9.
Give one difference between primary and secondary cells.
10.
What are uses of salt bridge in a galvanic cell?
02 Marks Question
1.
What is mercury cell? Give an electrode reaction?
2.
Make difference between molar conductivity and equivalent conductivity.
3.
Suggest a cell to determine molar conductivity experimentally.
4.
How will you explain the sharp increase in molar conductivity of a week electrolyte
on dilution? How is molar conductivity related to the degree of dissociation?
5.
The conductivity of 0.20 M solution of potassium chloride 298 K is 2.48 x 10-2 ohm-1
cm-1 calculate the molar conductivity ?
43
6.
A solution of Ni (NO3)2 is electrolyzed between platinum electrodes using a current
of 5 A for 20 minutes? What mass of Nickel is deposited at the cathode?
7.
The Zinc - silver oxide cell has the following reactions.
Zn → Zn 2+ + 2 e- E0= 0.76 v
Ag2O + H2O + 2e- →2Ag+ + 2OH-
E0=0.34V
Calculate the standard free energy in joules
8.
Give reason for the following phenomena:
(i) Iron does not rust if coating is broken in a galvanized iron pipe.
(ii) Rusting is said to be an electrochemical phenomena
9.
Molar conductance of a 1.5 M solution of an electrolyte is found to be 138 simen cm2.
What would be specific conductance of this solution?
10.
What is rusting? Explain how galvanization gives better protection than electroplating
for an iron object from rusting?
11.
Describe the characteristics of variation in molar conductivities (^m) strong and weak
electrolytes on dilution.
03 Marks Question
1.
Given E0 Cu2+ / Cu = 0.34 volt and E0 Ag+ /Ag = 0.80 volt for a cell
a) Calculate cell potential for the cell containing 0.1 M Ag+ and 4 M Cu2+ at 25°C.
b) How many hours does it take to reduce 3 moles of Fe2+ with 2 amp current.
2.
Calculate the standard cell potential, standard free energy charge ∆rG0 and equilibrium
constant for the Galvanic cell 2Cr(s) + 3 Cd2+  2Cr3+ + Cd
Given that E0Cr3+ /Cr = -0.74 v, E0 Cd2+ /Cd= -0.40 v
3.
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss
their variation with concentration.
4.
The resistance of a conductance cell containing 7.5 x 10-3 M solution of KCl at 250 C.
was 1005 ohms.
Calculate the specific conductance and molar conductance of the solution if cell
constant is 1.25 Cm-1.
5.
What happens during the corrosion of a metal? Explain the electro-chemical basis of
corrosion of iron with chemical equation involved?
05 Marks Question
1
What is EMF of a cell? How is it different from cell potential (Ecell)? Calculate the
EMF of the cell reaction
Sn4+ (1.5M) + Zn → Sn2+ (0.5 M) + Zn2+ (2M )
The standard cell potential of the cell is 0.89 v.
44
2.
Explain the primary and secondary batteries. How are secondary cells different from
primary cells? Give the composition and reaction of cathode and anode in a mercury
cell?
3..
What do you mean by molar conductivity and equivalent conductivity? Explain the
experimental Determination of molar conductivity with help of conductivity cell.
4
In the electrolysis of aqueous solution of sodium chloride, there are two possible
anode reactions.
2Cl- (aq) → Cl2 + 2e- E0 = 1.36 v
2 H2O (l) → O2 (g) + 4H+ (aq) + 4e- E0= 1.23 v
Which of the above reaction actually takes place and why? Also give other utilities of
electrolysis?
********************
UNIT – 04
CHEMICAL KINETICS
STUDY MATERIAL
Chemical Kinetics : Study of chemical reactions w.r.t. reaction rate, effect of various
variables, rearrangement of atoms and formation of intermediates.
Rate of Reaction : It is concerned with decrease in concentration of reactant per unit time.
Rate = - d[c]/ dt
45
It can also be defined as increase in concentration of products per unit time.
Rate= d[p]/ dt
Average Rate: It is rate of reaction measured over a long time interval.
Average Rate = ΔX/ Δt
Lt
Instantaneous Rate
t  0
X dX

T dT
i.e. rate of reaction when the average rate is taken over a very small interval of time.
Rate law or Rate equation : Rate law is relation between Rate of reaction and concentration
of reactants.
x
y
Rateofreaction  K A B 
Mind always, it is not theoretical but it is written when order of reactions w.r.t. A & order of
reaction w.r.t. B are known experimentally.
Rate Constant :
unity.
It is defined as rate of reaction when molar concentration of reactant is
Molecularity :
Total number of molecules of the reactants in an elementary reaction
(Single step reaction) is called molecularity of reaction.
Unimolecular reaction : When molecularity is one.
NH4NO2
N2 + 2 H2O
BiMolecular Reaction : When molecularity is two.
2HI
H2+ I2
Trimolecular or Termolecular reactions : When three species collide
2NO + O2
2NO2
Note that Molecularity greater than three is not observed? It is due to the probability
that more than three molecules can collide and react simultaneously is very small. Hence, the
molecularity greater than three is not observed.
Order of Reaction : The sum of the powers of the concentration of reactants in the rate law
is termed as order of the reaction. It can be in fraction. It can be Zero.
Order of a reaction is an experimental quantity. It is applicable to elementary as well
as complex reactions where as molecularity is applicable only for elementary reactions.
For complex reaction, Order is given by the slowest step and generally, molecularity
of the slowest step is same as the order of the overall reaction.
Zero order Reaction :
Rate = K[A]o
The rate of a reaction does not change with the concentration of reactants.
First Order Reaction : The reaction in which the rate of reaction is directly proportionate to
the concentration of reacting substance.
Rate = dx = k[A]
dt
46
Rate constant of first order reaction is
2.303
a
K
log
t
ax
or K 
A 
2.303
log 0
A
t
Where a is initial concentration, (a-x) is the concentration of reactants after time ‘t’.
The unit of K is s-1 or Min-1.
Half Life of a reaction : The time taken for a reaction when half of the starting material has
reduced is called half life of a reaction.
t1 
2
0.693
K
Where K= rate constant
Second Order Reaction :
dX
1
1
Rate 
 K A B 
dT
The reaction in which sum of powers of concentration terms in rate law or rate equation is
equal to 2.
Third Order Reaction :
The reaction in which sum of powers of concentration terms in
rate law is equal to 3 i.e.
dX
x
y
Rate 
 K A B  ,
Where x+y =3
dT
Pseudo First Order Reaction : The reaction which is bimolecular but order is one is called
Pseudo first order reaction.
H
CH 3COOC 2 H 5  H 2 O 
 CH 3COOH  C 2 H 5 OH
Large Excess
Factors affecting rate of reactions :
Rate of reaction is affected by
Nature of
Reactants
Physical
State
Surface
Area
Reactions of fraction order :
H 2  Br2  2HBr
Rate  K H 2  Br2 
1
1
2
Half life of Reaction of nth Order :
47
Concentration Temp.
In general, t 1 
2
1
A0 n1
where n is order of reaction.
For t 1  a for Zero order
2
t½ is independent for first order.
1
t 1  for second order.
2
a
1
t 1  2 for 3rd order.
2
a
Activation Energy (Ea) : It is extra energy which must be possessed by reactant
molecules so that collision between reactant molecules is effective and leads to formation
of product molecules.
Arrhenius equation of reaction rate :

Ea
K  A.e RT
Where K = rate constant
A= frequency factor,
Ea= Energy of activation,
R = gas constant,
T= temperature in Kelvin
lnK = lnA – Ea/ RT
log K = log A- Ea / 2.303 RT
Rate determining step :
The slowest step in the reaction mechanism is called rate determining step.
Temperature coefficient :
It is the ratio of rate costant at temperature 308 K to the rate constant at
temperature 298 K,
Temperature Coefficient = Rate Constant’K’ at 308 K / Rate Constant’K’ at 298 K
48
CHEMICAL KINETICS
Below Average Students :
01 Mark
Q1.
Ans.:
Define rate of chemical reaction?
It is defined as change in concentration of reactants or products per Unit time.
Q2.
Ans.
What is the unit of rate of chemical reaction.
ML-1S-1.
Q3.
Ans.:
Defineaverage rate of chemical reaction?
When change in concentration is measured over bigger interval of time. It is called
average rate. It is denoted by ΔX/ ΔT where ΔT is bigger time interval.
Q4.
Ans.:
Define instantaneous rate of chemical reaction ?
When change in concentration is measured over very small interval of time. It is
called instantaneous rate. It is denoted by dx/dt where ΔX is small change in
concentration and dt is small interval of time.
Q5.
Write the expression of rate constant for first order reaction
K
Ans.:
R 
2.303
log 0
R
t
Where [R0] is the original concentration of reactants
[R] is the concentration of reactant after time t.
Q6.
Ans.
Write the half life time expression for first order.
t1 
2
0.693
K
Q7.
Ans.:
Is half life time independent to initial concentration in first order reaction.
Yes
Q8.
Write the Arrhennius equation of reaction rate.
Ans.:
K  A.e
Q9.
Ans.:
What is elementary reaction.
When reaction is compelled in single step.
Q10.
Ans.:
What do you mean by molecularity of reaction.
It is the number of molecules taking part in each act of leading chemical reaction.
Example : (1) decomposition of NH4NO2.
NH 4 NO2  N 2  2H 2 O i.e. Unimolecular
 Ea
RT
where A is frequency factor,
Ea is activation Energy.
(2) 2HI  H 2  I 2 i.e. Bimolecular reaction
Q11. What do you mean by order of reaction.
Ans. : Sum of powers of the comcentarion of the reactants in the rate law expression is
called the order of that chemical reaction.
Suppose, Rate = K[A]x[B]y
Hence, Order of reaction = x+y
Q.12.
Ans.
Calculate the overall order ofn a reaction which has the rate expression.
Rate = K[A]½[B]3/2
Hence, Order of reaction = ½ + 3/2 = 2 ( i.e. Second Order)
49
Q13.
Ans.:
Write the Unit of K in Zero Order reaction.
M0
Q14.
Ans.
What is the Unit of K in first order reaction.
K=S-1
Q15.
Ans.
What is the unit of K in Second Order reaction?
K=Mol-1.L.S-1
Q16.
Ans.
Write the determination of second order reaction?
Molecularity is two & order of reaction is one.

H
Ex : CH 3COOC 2 H 5  H 2 O 
CH 3COOH  C 2 H 5 OH
Q17.
Ans.
Identify the reaction order if the unit of rate constant is Sec-1.
First order reaction.
Q18.
Ans.:
What is the order of reaction of Nuclear reaction.
First Order reaction
For Average Students :01 Mark
Q1.
Ans.:
State any one condition under which a bimolecular reaction may be kinetically of first order?
CH 3  COOC 2 H 5  HOH  CH 3COOH  C 2 H 5 OH
Rate  K [CH 3COOC 2 H 5 ]1[ H 2 O]0
i.e. H2O is in excess.
Q2.
Ans.:
Define activation energy.
It is defined as extra energy to be supplied to the reactants so that they can
change into products.
Q3.
Ans.
A reaction is 50% complete in 2 Hrs and 75% complete in 4 Hrs.What is the order
of the reaction?
First Order.
Q4.
Ans.
The plot of log K versus X is linear with a slope = -Ea/ 2.303 R. What is X?
X is 1/T.
The rate constant of a reaction is 1.2 x 10-2 L.Mol-1S-1. What is the order of the
reaction ?
Ans.: Order =2
For Above Average Students :
01 Mark
Q5.
Q1.
What is temperature coefficient ?
Ans.:
R308
R298
Q2.
Fill up the blanks :
Ans.:
Ea
T .T
 1 2  ..............
R T2  T1
R
Ln 2
R1
50
Q3.
Ans.
A catalyst provides a path of …………………. Activation energy.
Lower
Q4.
Ans.:
In a photochemical reaction, the energy of activation is provided by ……………
Radiation
Q5.
Which of the following graphs correspond to first order reaction ?
(a)
Rate
(b)
(c)
Rate
Rate
C
Ans.:
1/c
(D)
Rate
C2
C
(a)
Q6.
Photochemical reaction between hydrogen & Chlorine on the Surface of water is a reaction of
(a) Zero Order, (b) First Order, (c) Second Order, (d) Third Order
Ans. : Zero Order
Q7.
For a chemical reaction XA
yM.
The rate law is r=K[A]3. If the concentration of A is doubled the reaction rate will be (a)
Doubled,
(b) Quadrupled, (c) Increased by 8 times, (d) Unchanged
Ans.: (c)
CHEMICAL KINETICS
Below Average Students :02 Marks
Q1.
Differentiate between order of reaction and molecularity of reaction.
Ans.
Order of reaction
Molecularity
1. It is defined as sum of powers to which
1. It is the number of molecules taking part in
concentration terms are raised in rate law
chemical reaction.
2. It is determined experimentally
2. It is determined theoretically
3. It can be zero or even in fraction
3. It is always in whole number except Zero.
4. Order of complex reaction can be determined, 4. Molecularity of complex reaction can not be
determined.
Q2.
For a reaction A+H2O B, rate ∞ TA. What is its
i) Molecularity,
ii) Order of reactions ?
Ans.
I) Its molecularity is 2
ii)
Its order is 1.
Q3.
Ans.:
Define pseudo unimolecular reaction. Give two examples.
The reaction which is molecular but its order is one is called pseudo unimolecular
reaction. Eg.:

H
1.)CH 3 COOC 2 H 5  H 2 O 
CH 3 COOH  C 2 H 5 OH
2.)C12 H 22O11  H 2 O invertage

 C 6 H 12O6  C 6 H 12O6
Glucose
Q4.
Name four factors affecting rate of reduction.
51
Fructose
Ans.:
1)
Nature of reactants
2)
Temperature
3)
Concentration
4)
Catalyst
Q5.
When could order an molecularity of a reaction i) be Same ii) be different
Ans.: i) Order and molecularity will be same in single step reaction.
ii)
They will be different if reaction is of two or more steps.
For Average Students :
02 Marks
Q1.
Plot a graph between log cone Vs time and give expression value of K.
Ans. Slope= -K/2.303

Log
(Conc.)
time
Q2.
The rate constant for a first order reaction is 0.005 min-1. Calculate its half life.
Ans.:
t1 
2
0.693 0.693

 1386Min
K
0.05
Q.3.
Ans.:
What is meant by effective collision?
The collision which leads to formation of product. Molecules is called effective
collision.
Q4.
When do molecules undergo effective collision?
Ans.: Molecules undergo effective collision if they cross energy barrier and orientation
barrier, that is they possess activation energy and collide in proper orientation.
Q5.
How does concentration effect the rate of reaction?
Ans,: When we increase concentration of reactants, total no. of collision will increase, therefore,
probability of effective collision will increase, hence the rate of reaction will increase.
For Above Average Students :
02 Marks
Q1.
An increase of 10 K in temp. rarely doubles the kinetic energy of particles but
doubles the rate of reaction, why?
Ans.: When temp. is increased 10 K , kinetic energy of molecules increases, no. of
molecules possessing activation energy become double, therefore no. of effective
collision doubles hence rate of reaction.
Q2.
The rate of constants of a reaction at 500 K & 700 K are 0.02 Sec-1 and 0.07 Sec1
respectively. Calculate the values of Ea & A.
log
K2
Ea  T2  T1 



K 1 2.303R  T1 .T 2 
log
0.07
Ea

0.02 2.303  8.314 JK 1 mol 1
 700  500 
 700  500 


10  4
19.15
19.15
Ea  0.544 
 18230.8 J
5.714  10 14
0.544  Ea  5.714 
Ans.
Since, K  Ae
 Ea
RT
18230.8
0.02  A.e 8.314500
0.02
A
 1.61
0.012
52
FOR BELOW AVERAGE STUDENTS :03 Marks
Ro
2.303
log
R
t
Q1.
Drive the relation K 
Ans.:
We know that rate of reaction is proportional to the first power of concentration of
the reactant R. For Ex.
RP
Rate  
or ,
d R 
 K R 
dt
d R 
  K .dt
R 
Integrating this equation, we get
Ln[R]=-Kt+ I -------------------(1)
Again;
I is the constant of integration and its value can be determined easily,
When t=0, R=[R0], where [R0] is the initial concentration of the reactant.
Ln[R0]=-K x 0 +I
So, Ln[R0]=I
Substituting the value of I in Equation (1)
Ln[R]= -Kt + Ln [R0]
Rearranging this equation,
Ln
R 
R 0
  Kt
1 R 
or , K  Ln 0
t
R
R 
2.303
or , K 
Log 0
t
R
Q2.
Derive the expression t½ = 0.693/ K for first order reaction.
R 
2.303
Log 0
t
R
R 
2.303
K
log O
RO
t1
2
2
2.303
K
log 2
t1
K
Ans,:
2
t 1 
2
0.693
K
FOR AVERAGE STUDENTS :03 Marks
Q1.
Derive the relationship between activation energy & rate constant.
Ans.
Arrhenius equation K  A.e RT where ‘A’ is frequency factor, Ea is activation
energy, R= 8.314 JK-1mol-1, T is temp. in Kelvin.
E
53
Ea
RT
EG
ln K  ln A 
RT1
ln K  ln A 
ln k  ln A 
 ln
Ea
RT 2
K 2 Ea  1 1 

  
K1
R  T1 T2 
 log
K2
Ea  T2  T1 



K 1 2.303R  T1 .T2 
For Above Average Students :
03 Marks
Q1.
The following results have been obtained during the kinetic studies of the reaction.
2A + B C + D
Experimental
1
2
3
4
[A] M
0.1
0.3
0.3
0.4
Initial rate of formation of D/M min-1
6.0 x 10-3
7.2 x 10-2
2.88 x 10-1
2.4 x 10-2
[B] M
0.1
0.2
0.4
0.1
Determined the rate law and the rate constant for the reaction.
Ans.:
dx
 K [ A] x [ B] y          (1)
dt
7.2  10  2  K [0.3] x [0.4] y    (2)
Dividing (1) /( 2), weget ,
1
1

 2 y  22
4 2Y
 y2
6.0  10 3  K [0.1] x [0.1] y      (3)
2.40  10  2  K [0.4] x [0.1] y      (4)
Dividing (3) by (2), we get,
1
1

x 1
4 4x
dx
 K [ A] x [ B ] y  K [ A]1 [ B ] 2
dt
6.0  10 3  K [0.1]1 [0.1] 2
K  6.0 M  2 S 1
Q2.
Calculate the half life of a first order reaction from their rate constants given below :
(a) 200 S-1
(b) 2 Min-1
(c) 4 year-1
Ans.:
(a)
t1 
0.693
0.693

 3.465  10 3 Sec.
K
200 S 1
t1 
0.693
 3.465 min  0.35 min ..
2Min 1
2
(b)
2
54
(c)
t1 
2
Q3.
0.693
 0.173 years
4
Give the mechanism of the reaction
H 2 ( g )  Cl2 ( g ) Sunlight
 2 HCl ( g )
Ans.:
Chain initiation step :
Cl  Cl HN
 2Cl
Chain propagation step
H 2  Cl  H  HCl
H  Cl2  HCl  Cl
Chain termination step
Q4.
State the role of activated complex in a reaction and state the relation with energy of
activation.
Ans.: When the colliding reactant molecules possess the kinetic energy equal to energy of activation
then the configuration of atoms of the species at this state is different from the reactants as well as the
products.This state is called activated or transition state and specific configuration of this state is
called activated complex.
The reactant molecules do not change directly into products, first they absorb energy
equivalent to the energy of activation and form an activated complex. The activated complex formed
has a very short life spam and splits into the products. The activated complex is at the top of the
energy barrier graph and possess the highest energy. In this activated complex, all the bonds are in the
transition state.
Low energy activation barrier means the activated complex is readily formed and it is fast
reaction.
High energy of activation barrier means the activated complex is difficult to form and it is a
slow reaction.
A……….B
Activated complex
Eaf
H o S  E R  E P
Energy
ER
JHoS
Reaction condition
55
UNIT – 05
SURFACE CHEMISTRY
STUDY MATERIAL
1. Adsorption:or liquid .
It is the process of attracting molecular species on the surface of solid
2. Adsorbate:The molecular species which concentrates or accumulates on the
surface known as Adsorbate.
3. Adsorbent:as adsorbent .
The solid or liquid substance on which adsorption takes place known
4. Desorption:The process of removal of adsorbed substance from a surface of solid
or liquid known as desorption.
5. Absorption:The uniform distribution of molecular throughout the bulk of the solid
known as absorption.
6. Sorption:The process in which both adsorption and absorption takes place
simultaneously known as sorption.
7. Enthalpy of Adsorption:The amount of energy released by the attraction of one
mole of adsorbate on the adsorbent.
8. Types of adsorption :- There are two types of adsorption :
(a) Physisorption or Physical adsorption:- When the gas molecules are attracted or
accumulated on the solid by Vander waal’s forces.
(b) Chemisorption or Chemical adsorption:- When the gas molecules are accumulated on the
solid by chemical bonds.
9. Freundlich adsorption isotherm:He gave a relation between the quantity of gas
adsorbed by unit mass of solid adsorbent and pressure at a particular temperature
x/m=k P1/n
(n>1)
Where x be the mass of adsorbate ,m be the mass of adsorbent k & n be the constants
P be the pressure .This relation can be changed at different Pressure:At low pressure x/m=k P 1
At high pressure x/m=k P 0
At intermediate pressure x/m=k P 1/n
10. Adsorption from solution phase:It has similar relation with Freundlich adsorption
isotherm .Only in place of pressure ,concentration of solution is taken.
Therefore x/m=k C 1/n
11. Catalyst:A substance which increase or decrease the rate of chemical reaction
and quantitatively unchanged after the reaction known as catalyst. There are two types of
catalyst:- (i) Positive Catalyst and (ii) Negative Catalyst.
12. Promters:A substance enhances the activity of catalyst.
13. Poisons:A substance decreases the activity of catalyst.
14. Homogeneous catalysis:When reactants and catalysts are in same phase ,the
process is known as Homogeneous catalysis.
56
Eg.
(g)
2SO2 ( g )  O2 ( g ) NO

 2SO3 ( g )
15. Heterogeneous catalysis :When reactants and catalysts are in different phases,
this process is known as Heterogeneous Catalysis.
Eg.: N2(g) + 3 H2(g)
Fe(S)
2 NH3(g)
16.
Mechanism of heterogeneous catalysis :- The steps are as follows :
(i) Diffusion of reactants on the surface of catalyst.
(ii)
Adsorption of reactants on the surface of catalyst.
(iii)
Chemical reaction between reactants on the solid surface to form intermediate
product.
(ii)
Desorption of product from solid surface.
(iii)
Diffusion of products away from solid surface.
17.
Shape-selective catalysis : A catalytic reaction that depends upon the pore structure
of catalyst as well as the size of reactant & product molecules is called Shape-selective
catalysis.
18.
Enzyme Catalysis : Enzymes are biological catalyst which catalyse specific
biochemical reactions. They are complex nitrogenous organic compounds which are
produced by living plants and animals having high molecular mass. They are highly efficient
because they increase the rate of reaction by 108 to 1020 times.
NH 2 CONH 2  H 2 O Urease
 2 NH 3  CO2
Eg.
Urease is an enzyme used for the decomposition of Urea only.
19.
Colloids or Colloidal Solution :
phase mixed in dispersion medium.
A heterogeneous solution in which dispersed
20.
Lyophilic Colloids : Those colloids which are solvents loving or attracting they are
reversible Sols.
21.
Lyophobic Colloids :irreversible Sols.
Those colloids which are solvent hating. It is
22.
Multimolecular Colloids: A colloid in which large no. of atoms or smaller
molecules aggregate together to form species having size in colloidal range for eg. A sulphur
sol consists of particles containing about a thousand of S8 molecule.
23.
Macromolecular colloids : A colloid in which the size of dispersed particles are in
a colloidal range. Such dispersed phase in dispersion medium is called Macromolecular
colloids. For Eg. Proteins, Starch and Cellulose form macromolecular colloids.
24.
Associated Colloids : Those colloids which behave as strong electrolyte at low
concentration but at high concentration behave like colloids due to formation of aggregated
particle of colloidal dimensions. For eg. Micelles are associated colloids.
25.
Kraft Temperature : The formation of Micelles takes place only above a particular
temp. is called Kraft temp.
26.
Peptization : The process of converting precipitate into colloidal sol by mixing
small amount of electrolyte into dispersion medium known as peptization and the electrolyte
known as peptizing agent.
57
27.
Dialysis :
It is the process of purification of colloids by diffusion through a semi
permeable membrane.
28.
Ultra filtration:
It is the process of separating the colloidal particles from the
solvent and the solute present in colloida solution by ultra filter paper.
29.
Tyndal effect :
The process of scattering of light in colloidal solution by
colloidal solute known as Tyndal effect. The bright cone of light is called Tyndal cone.
30.
Brownian Movement
:
The continuous zig zag motion of dispersed
phase in dispersion medium is called Brownian Movement. It is due to the unbalanced
bombardment of the particles by the molecules of the dispersion medium.
31.
Helmholtz Electrical double layer :
The combination of the two layers of
opposite charges around the colloidal solute is called Helmholtz Electrical double layer.
When the colloidal particles acquire +ve or –ve charge by selective adsorption of one of the
ions, It attracts counter ions from the medium forming a second layer like.
Ag I / I - K+
AgI / Ag+ I –
32.
Zeta potential or Electro kinetic potential :
The potential difference between
the fixed layer and the diffused layer of colloidal solution having opposite charges.
33.
Electrophoresis :
The process of migration of charged colloidal solute towards
the oppositely charged electrode in colloidal solution is known as Electrophoresis.
34.
Coagulation : The process of converting colloidal solution into precipitate by mixing
small amount of oppositely charged electrolyte known as Coagulation.
35.
Coagulative Value : The minimum no. of a milli moles of electrolyte requires to
coagulate one litre colloidal solutions.
36.
Hardy-Schultz Rules:
It has two rules :
a)
Oppositely charged ions are effective for coagulation.
b)
The coagulative power of electrolyte increases with increase in charged on the
ion used for coagulation. For Eg. :
Al3+> Ca++ > Na+ for negatively charged colloids. Similarly
[ Fe(CN)6]4- > PO43- >SO42- > Cl - for positively charged colloids.
37.
Emulsion :
A colloids contain dispersed phase and dispersion medium both in
liquid state is known as Emulsion.
There are two types of emulsion.
(i)
Oil in water & (ii) water in oil.
38.
Cottrell Smoke precipitator :
Smoke is a colloidal solution of solid particles
like C, As compounds and dust in air. It comes out through the chimneys of industrial plants.
It consists of two metal discs charged to high potential. The charged dust and C , As particles
get discharged towards oppositely charged metal disc in form of precipitate while gases come
out through chimney.
58
1 MARK QUESTIONS
Q. 1. What kind of adsorption is represented by the following graph :
x
M
T
Ans.
Chemisorption.
Q. 2. In the titration of oxalic acid by acidified KMnO4, the oxidation of oxalic acid is
slow in the beginning but becomes fast as the reaction progresses. Why ?
Ans.
Autocatalysis by Mn+2.
Q. 3. Out of PO43–,, SO42–,, Cl–,, which wil act as the best coagulating agent for for Fe
(OH)3 ?
Ans.
PO43–.
Q. 4. Arrange the following in correct order of their coagulating power :
Na+, Al3+, Ba2+
Ans.
Na+ < Ba2+ < Al3+
Q. 5. Which type of charged particles are adsorbed on the surface of As2S3 during its
preparation ?
As2O3 + 3 H2S ——
Ans.
S3 + 3 HOH
2
S2–.
Q. 6. Which type of metals act as effective catalysts ?
Ans.
Transition metals.
Q. 7. The colloidal solution of gold prepared by different methods have different
colours. Why ?
Ans.
Due to difference in the size of colloidal particles.
Q. 8. At high pressure, the entire metal surface gets covered by a mono molecular
layer of the gas. What is the order of the process ?
59
Ans.
Zero order.
Q. 9. What is the term used for minimum concentration of an electrolyte which is able
to cause coagulation of a sol ?
Ans.
Flocculation value.
Q. 10. A liquid is found to scatter a beam of light but leaves no residue when passed
through the filter paper. What can the liquid be described as ?
Ans.
Colloid.
Q. 11. If an electric field is applied to a colloidal sol, the dispersed phase particles are
found to move towards the electrode of opposite charge. If however, the
dispersed phase is made stationary, the dispersion medium is found to move in
the opposite direction. What is the term used for such movement of dispersion
medium ?
Ans.
Electro osmosis.
Q. 12. Out of glucose, urea and dodecyl trimethyl ammonium chloride, which one
forms micelles in aqueous solution above certain concentration ?
Ans.
Dodecyl trimethyl ammonium chloride.
Q. 13. A plot of log versus log p for the adsorption of a gas on a solid gives a straight
line. What is the slope equal to ?
Ans.
1
n
Q. 14. The formation of micelles occurs only beyond a certain temperature. What is the
temperature called ?
Ans.
Kraft temperature
Q. 15.
Ans.
–ve
–ve
–ve
Q. 16. Out of CO and NH3 which is adsorbed on activated charcoal to a large extent
and why ?
Ans.
Ammonia; because more easily liquefiable gas undergoes adsorption to a greater
extent.
Q. 17. On passing H2S through dilute HNO3 the colourless solution becomes turbid. Why ?
Ans.
Due to formation of colloidal sol of Sulphur.
60
Q. 18. A sol is prepared by addition to excess AgNO3 solution in KI solution. What
charge is likely to develop on the colloidal particles ?
Ans.
Positive.
Q. 19. If we add equimolar amounts of ferric hydroxide sol and arsenic sulphide sol,
what will happen ?
Ans.
Both the sols will get coagulated.
Q. 20. What happens when freshly precipitated Fe (OH)3 is shaken with a little amount
of dilute solution of FeCl3 ?
Ans.
It causes peptization leading to the formation of a positively charged sol of Fe (OH)3.
Q. 21. What happens to a gold sol if gelatin is added to it ?
Ans.
It causes stabilisation of gold sol.
Q. 22. Out of NaCl, MgSO4, Al2 (SO4)3, K4[Fe(CN)6], which one will bring about the
coagulation of a gold sol quickest and in the least of concentration ?
Ans.
Al2 (SO4)3.
Q. 23. What is the unit for expressing flocculation value ?
Ans.
millimole per litre.
Q. 24. Out of PO43–, SO42–, Al3+ and Na+, which will have the highest coagulating power
for As2S3 colloid ?
Ans.
Al3+.
2 MARKS QUESTIONS
Q. 1. Bleeding is stopped by the application of alum to a wound. Why ?
Ans.
Blood is a colloid alum being an electrolyte, makes the blood to coagulate and form
clot.
Q. 2. What is the purpose of adding gelatin to ice cream ?
Ans.
Ice cream is a colloid. Gelatin imparts stability to it because gelatin is a protective
colloid.
Q. 3. Dialysis is a method of purification of sols. But prolonged dialysis of the sol
makes it unstable. Why ?
Ans.
Traces of electrolytes in the sol, impart charge to dispersed phase particles making it
stable. Prolonged dialysis removes all electrolytes thus making the sol unstable.
Q. 4. What is the function of gum arabic in the preparation of Indian ink ?
Ans.
Gum arabic is a protective colloid and thus provides stability to Indian ink.
61
Q. 5. What is collodion ? What is its use ?
Ans.
Cellulose dispersed in ethanol, is called collodion. It is used for making membranes
for ultrafiltration.
Q. 6. Why the sun looks red at the time of setting ? Explain on the basis of colloidal
properties.
Ans.
At the time of setting, the sun is at the horizon. The light emitted by the sun has to
travel a longer distance through the atmosphere. As a result, blue part of the light is
scattered away by the dust particles in the atmosphere. Hence the red part is visible.
Q. 7. Addition of H2 to acetylene gives ethane in presence of palladium but if BaSO4
and quinoline or sulphur are also added, the product is ethane. Why ?
Ans.
BaSO4 + quinoline / s poison the catalyst. Hence, the efficiency of the catalyst
decreases and the reaction stops at the first stage of reduction.
Q. 8. SnO2 forms a positively charged colloidal sol in acidic medium and a negatively
charged sol in the basic medium. Why ?
Ans.
SnO2 is amphoteric in nature. It reacts with acid eg. HCl to form SnCl4 in the solution.
The common Sn4+ ions are adsorbed on the surface of SnO2 particles giving them a
positive charge.
SnO2 reacts with a base eg. NaOH to form Sodium Stannate in the solution. The
stannate ions are adsorbed on the surface of SnO2 particles giving them a negative
charge.
Q. 9. Why physical adsorption is multimolecular whereas chemisorption is
unimolecular ?
Ans.
Chemisorption takes place as a result of reaction between adsorbent and adsorbate.
When the surface of the adsorbent is covered with one layer, no further reaction can
take place.
Physical adsorption is simply by Vander Waal’s forces. So any number of layers may
be formed one over the other on the surface of the adsorbent.
Q. 10. What is meant by induced catalysis ? Give an example.
Ans.
It is a phenomenon in which a chemical reaction increases the rate of another reaction
which otherwise may not occur in similar conditions.
Eg. Sodium arsenite (Na3AsO3) is not oxidised in air but if air is blown into a solution
containing Na3AsO3 and Na2SO3, then both AsO33– and SO32– ions are oxidised.
Q. 11. What type of colloidal sols are formed in the following ?
(i)
Sulphur vapours are passed through cold water.
(ii)
White of an egg is mixed with water.
62
Ans.
(iii)
Concentration of soap solution is increased.
(i)
Multimolecular colloid
(ii)
Macromolecular colloid
(iii)
Associated colloid.
Q. 12. What is common to aquasol and aerosol ? In what respect do they differ ?
Ans.
Both are colloids.
In aquasol, water acts as dispersion medium.
In aerosol, air acts as dispersion medium.
Q. 13. Explain as to why SnO2 forms a positively charged sol in solutions with pH < 7
and negatively charged sol in solutions with pH > 7.
Ans. Refer Ans. 8.
For average
Q.1
Ans.:
Why does physisorption decrease with increase of temperature ?
Because the attraction between gas molecules and solid surface is very weak
which easily over come by increase of temperature.
Q2.
Why are powdered substances more effective absorbents than their crystalline
forms?
Due to increase of surface area of powered substances.
Ans.
Q.3.
Ans.:
What do you mean by activation of adsorbant? How is it achieved?
Activation of adsorbent means increase the extent of adsorption. It is achieved by
the increase of surface area.
Q4.
Discuss the effect of pressure and temperature on the adsorption of gases on
solids.
Effect of pressure : By the increase of temp., extent of adsorption decreases
uniformly for physisorption but for chemisorption initially increases then
decreases.
Ans:
Q5.
Ans.:
What is observed
(i)
When a beam of light is pressed through a colloidal sol
(ii)
An electrolyte, NaCl is added to hydrated ferric oxide sol.
(iii)
Electric current is passed through colloidal sol.
(i)
Tyndal Effect
(iv)
Coagulation
(v)
Electrophoresis
Q6.
Ans.
What do you mean by activity and selectivity of catalysis?
Activity of catalyst means to increase the extent of chemisorption.
Selectivity of Catalyst means to direct a reaction to form selective product.
Q7.
Ans.:
What is ZSM-5? What is its formula?
ZSM-5 is a zeoli8te seine of molecular porosity5. Its formula is
Hx[(AlO2)x(SiO2)96-x ] . 16H2O
Q8.
Ans.:
Why adsorption is always exothermic?
In Adsorption, change of entropy ΔS is –ve i.e. non favourable condition for exothermic
change where as change of enthalpy ΔH is –ve i.e. favourable condition for exothermic
change. Overall ΔG becomes –ve in adsorption; therefore adsorption is always exothermic.
63
Q9.
Ans.:
Explain the following terms :
(i) Electrophoresis,
(ii) Coagulation,
(iii) Dialysis
(iV) Tyndal Effect
(i) Electrophoresis :- The immigration of colloidal solute towards oppositely charged
electrode under an electric potential is called Electrophoresis.
(ii) Coagulation :The process of settling down of colloidal partcles is known as
coagulation.
(iii) Dialysis :- It is a process of removing dissolved impurities from colloidal solution by
means of diffusion through a suitable membrane.
(iv) Tyndal effect :The scattering of light in colloidal solution by colloidal solute is
known as Tyndal effect.
Q10.
Ans.
Comment on the statement that “ colloid is not a substance but state of a substance” .
A substance shows different physical properties in different medium. It may exist as colloid
or crystalloids under certain conditions. Eg. NaCl in water behaves like crystalloid while in
benzene behaves like colloid. Similarly dilute soap solution behaves like crystalloid whereas
concentrated soap solution like colloids. It is the size of particles which matters i.e. the state
in which the substances exists. If the size of the particles lies in the range 1 to 1000 no., it is
in colloidal state.
Q11.
Why are substances like Platinum and Palladium offer used for carrying out electrolysis in
aqueous solution ?
Platinum and palladium is inert electrodes. They are not react with the ions of electrolyte and
products of electrolysis. Hence they are used as electrodes during electrolysis. Hence they are
used as electrodes during electrolysis. Hence they are used as electrodes during electrolysis in
aqueous solution.
Ans.:
Q12.
Ans.:
Why it is necessary to remove CO when ammonia is obtained by Haber’s process?
Because CO acts as poison for the catalyst in the manufacture of NH3 by Haber’s process.
Q13.
Ans:
Which will be adsorbed more readily on the surface of charcoal and why NH3 or CO2?
The critical temperature NH3 is more CO2, therefore NH3 is liquefied more easily than CO2.
Hence NH3 has higher intermolecular forces of attraction and hence it adsorbed more readily.
Q14.
Ans.
What is similarity and dissimilarity in aqua solution and solid aerosols?
Aqua Sols and solid aerosols both have solid as dispersed phase and have different dispersion
medium. Aqua solution contain water and aerosol contain air as dispersion medium.
Q15.
How can a colloidal solution and true solution of the same colour be distinguished from each
other?
By Tyndal effect( colloidal solution will scatter light and path becomes lighted where as no
such phenomena is observed in true solution.)
Ans.:
Q16.
Ans.:
Why is ferric chloride preferred over potassium chloride in case of a cut leading to bleeding?
Blood is negatively charge colloidal particles. It is coagulated by +ve ions. As Fe ++ ions has
greater number of charges than K+ ions. Therefore on the basis of Hardy-Schulze rule,
Coagulation with Fe+++ is faster and hence it is preferred.
BELOW AVERAGE STUDENTS :01
Mark Questions
1.
Define Adsorption with one example.
2.
What is physisorption and chemisorption ?
3.
What is Freundlich adsorption isotheorem ?
4.
Write the relation between quantity of gas adsorbed by unit mass of Solid adsorbent
and pressure at a particular temperature.
64
5.
Draw a graph between rate of adsorption and pressure at different temperature like
T1>T2>T3.
6.
Define catalysis with one example.
7.
Write the use of ZSM-5.
8.
What is Biochemical catalysis?
9.
What is Colloids?
10.
Define Dispersed phase and Dispersion medium.
11.
Define Micelle.
12.
What is Peptization?
13.
What is Tyndal effect?
14.
Define Coagulation and Precipitation of Sol.
15.
What is Gel?
16.
Define Peptizing agent.
AVERAGE STUDENTS :01 MARK QUESTIONS
1
Why the pressure of gas is decreased when a gas like O2, H2, CO, Cl2, NH3 or
SO2 is taken in closed vessels containing powdered charcoal ?
2.
Why does physisorption decreases with increase of temperature?
3.
Why are powdered substances are more effective adsorbents than their crystalline
forms?
4.
Write the process by which vegetable oil converted into vegetable Ghee.
5.
What is shape-selective catalysis?
6.
Write two examples of heterogeneous catalysis.
1.
01 Mark
Define Enthalpy of Adsorption.
2.
What is the role of desorption in the process of catalysis?
3.
Define Tyndal Cone.
4.
What is Electrokinetic Potential or Zeta Potential?
5.
Define Coagulating Value.
ABOVE AVERAGE STUDENTS :
SURFACE CHEMISTRY
Below Average Students :02 Marks Questions
1.
Explain Desorption and Sorption.
2.
Write two differences between physisorption and chemisorption.
3.
What is homogeneous and Heterogeneous catalysis? Give one example each.
4.
Define Promoters and Poisons.
65
5.
Write two applications of catalyst in Industry.
6.
Explain cleansing action of soap.
7.
What are the methods for the preparation of colloids.
8.
What is purification of colloidal solution? Name the process of purification of
colloids.
9.
Write brief notes on
(a) Brownian Movement and
(b) Electrophoresis.
10.
State Hardy-Schulze Rule. What is its application?
11.
Write the example of colloids which is being used in our daily life.
AVERAGE STUDENTS :
02 Marks Question
1.
Write the mechanism of Adsorption.
2.
Why are substances like Platinum and Palladium often used for carrying out
electrolysis of aqueous solution?
3.
What is function of promoters and Poisons?
4.
Write the name of process and catalyst used for the preparation of NH3
5.
How can be Lyophobic colloids prepared?
6.
Give two examples for preparation of colloids by chemical methods.
7.
What is demulsification? Name two demulsifiers .
ABOVE AVERAGE STUDENTS :
02 Marks Questions
1.
Write the mechanism of Homogeneous catalysis reaction
g 
2 SO2 g   O2  g  NO

 2SO3  g 
2.
Define Kraft temperature (TK) and critical Micelle concentration (CMC).
3.
What is the role of electrolyte in coagulation or precipitation?
4.
Write the cause of formation of delta in brief.
5.
Define Intrinsic and Extrinsic colloids with example.
BELOW AVERAGE STUDENTS :
03 Marks Questions
1
Define Adsorption, Adsorbent and Adsorbate.
2.
Write three characteristics of physisorption and chemisorption.
3.
Write at least three factors on which rate of adsorption depends.
66
4.
Write three applications of Adsorption.
5.
What are the characteristics of Enzyme catalysis?
6.
Write three differences between Lyophilic and Lyophobic colloids.
7.
Define Multimolecular, Associated and Macro molecular colloids.
8.
Define Dialysis, Electro dialysis and Ultra filtration.
9.
What is Emulsion. Write the types of Emulsions. How can be tested?
10.
Write at least three applications of colloids.
AVERAGE STUDENTS :
03 Marks Questions
1.
Draw a graph between log
x
and log P.
m
Write the value of intercept and slope.
2.
Write a brief notes on activity and selectivity of solid catalyst with example.
3.
What is Zeolite? Write its structure and use.
4.
What are Enzymes. Write the mechanism of Enzyme catalysis.
5.
Write at least three differences between true solution , colloidal solution and
suspension.
6.
Write the causes for the creation of charge on colloidal particles.
7.
Explain the terms with suitable examples
(a) Alcosol,
8.
(b) Aerosol,
(c) Hydrosol
Name the catalyst for Haber’s Process, Ostwald’s Process and Contact Process.
ABOVE AVERAGE STUDENTS :
03 Marks Questions
1.
Write the steps involve for mechanism of Heterogeneous catalysis?
2.
Write a brief note on mechanism on Micelle formation.
3.
What do you mean by Cottrell smoke precipitator.
67
UNIT – 06 :
GENERAL PRINCIPLES AND
PROCESS OF ISOLATION
OF ELEMENTS
STUDY MATERIAL
Important Points :
Mode of occurrence of elements : Elements are found in nature either in the free state is also called
the native state, or in the combined state i.e. found in the form of compounds. This is mainly due to
the reason that different elements have different chemical reactivities.
Native States : Elements or metals which are not attacked by moisture, oxygen and carbon dioxide of
the air occur in the native state. Examples: Carbon, Sulphur, Gold, Platinum, Noble Gases etc.
Combined States: The elements which are readily / easily attacked by moisture, Oxygen and Carbon
dioxide of air, occur in the combined state in form of their compounds are called minerals.
In the combined state, non-metals are found in the reduced form i.e. X  ( where x= F,Cl,Br,
I) while metals are found in the oxidized form i.e. is oxides eg. Fe2O3, Al2O3, SnO2, MnO2 etc.
Minerals :
The naturally occurring chemical substances in the form of which the metal occurs in
the earth along with impurities are called minerals.
Ores: The minerals from which the metal is conveniently and economically extracted is called an
Ores. Thus all ores are minerals but all minerals are not Ores.
Gangue or Matrix : The Earthy and silicious impurities with ores are known as matrix or Gangue.
Principal ores of some Important Metals :
Metal
Aluminium
Iron
Copper
Zinc
Ores
Composition
Bauxite
Kaolinite ( a form of Clay)
Hematite
Magnetite
AlOx(OH)3-2x
[ where 0 < x<1]
Fe2O3
Fe3O4
Iron Pyrite
Copper Pyrite
Malachite
Cuprite
Zinc Blend
FeS2
CuFeS2
CuCO3.Cu(OH)2
Cu2O
ZnS
Sintering :
It is a process of crushing ores to reasonable size before concentration of ores. It is
followed by heating of solid particles below its m.p. to change it into single mass.
Concentration :
The process of removal of unwanted materials like sand, clay , rock etc from
ore is known as concentration or benefication or enrichment or dressing.
Leaching or chemical separation : It is a process in which ore is treated with suitable reagent which
dissolves ores but not the impurities.
Beayer’s Process or purification of bauxite by leaching :
Bauxite ore is treated with caustic
soda (NaOH) , Al2O3 dissolves in concentrated solution leaving behind impurities.
68
Al2 O3 S   2 NaOH aq   3H 2 Ol   2 Na[ Al (OH ) 4 ]aq 
The aluminate in solution is neutralized by passing CO2 gas and hydrated Al2O3 is precipitated. At this
stage the solution is seeded with freshly prepared samples of hydrated Al 2O3, which induces the
precipitation.
2 Na[ Al (OH ) 4 ]aq  CO2  Al2 O3 .xH2 O  2 NaHCO3 aq
The precipitate of Al(OH)3 is filtered, dried and finally heated to about 1470 K to obtain pure
Al2O3.
K
Al2 O3 .xH 2 O 1470

 Al2 O3 s   xH 2 O g 
Concentration of Gold & Silver ores by leaching : In the metallurgy of Ag & Au the respective
metal is leached with a dilute solution NaCN or KCN in the presence of air for oxygen from which a
metal is obtained later by replacement.
4M s   8CN  aq   2 H 2 O  O2 g   4[M CN 2 ]

aq 
 4OH  aq 
Where M= Ag or Au
4[M CN 2 ]

aq 
 Zns   [Zn(CN ) 4 ]2 aq   2M s 
EXTRACTION OF CRUDE METAL FROM CONCENTRATED ORES :Conversion of ore into oxide i.e. de-electronation of ores : It is achieved by calcination and
roasting.

Calcination : It is the process of converting an ore into its oxide by heating it strongly below
its m.p. either in absence or limited supply of air.
This method is commonly used to convert metal carbonates and hydroxides to their
respective oxides.During the process of calcination, the following chemical changes occurs.
1)
Moisture is driven out.
2)
Volatile impurities of S, As & P are removed as their volatile oxides.
3)
Water is removed from hydrated oxides and hydroxides ores.
Al2 O3 .2 H 2 O Heat

 Al2 O3  2 H 2 O
Bauxite
Alumina

Fe2 O3 .3H 2 O 
 Fe2 O3  3H 2 O
Limonite
Ferric oxide
4) Carbonate ores are converted into their respective oxides by loss of carbon dioxides.

CaCO3 

CaO  CO2
Limestone
Calcium Oxide

CuCO3 .CuOH 2 

2CuO  H 2 O  CO2
Malachite
4) It makes the ore porous and hence easily workable in subsequent stages. Calcination is
usually carried out in a reverberatory furnace.
Roasting : It is the process of converting an ore into its metallic oxide by heating strongly at temp. is
sufficient to melt in excess of air. This process is commonly used for sulphide ores. The following
changes occurs during roasting.
1.Moisture is removed
2.Organic matter is destroyed
3.Non-Metallic impurities like that of sulphur, Phosphorous and Arsenic are oxidized
and are removed as volatile gases.
69
S 8  8O2  8SO2 
P4  5O2  P4 O10 
4 As  3O2  2 Al 2 O3 
4.
Ores are generally converted into metallic oxides.
2ZnS  3O2  2ZnO  2SO2 
2 PbS  3O2  2 PbO  2SO2 
2Cu 2 S  3O2  2Cu 2 O  2SO3
5 .It makes the ore porous & hence easily workable in subsequence stage.
1 MARK QUESTIONS
Q. 1. Why carbon reduction process is not applied for reducing
aluminium oxide to aluminium ?
Ans.
Because aluminium metal itself a very powerful agent and can easily
reduce CO formed during the reaction back to carbon.
Q. 2. Explain why thermit process is quite useful for repairing the
broken parts of a machine ?
Ans.
In thermit process, oxides of metals are reduced by aluminium in
allowed to fall between the broken parts of a machine.
heat
Fe2O3 (s) + 2 Al (s) ———
O3 + 2 Fe (l) + heat
2
molen
Q. 3.
f
G) of MgO (s) and CO (g) at 1273 K and
2273 K are given below :
f
G MgO (s) = – 941 KJ/mol at 1273 K
= – 344 KJ/mol at 2273 K
f
G CO (g)
= – 439 KJ/mol at 1273 K
= – 628 KJ/mol at 2273 K
On the basis of the above data, predict the temperature at which carbon
can be used as reducing agent for MgO (s).
70
Ans. The redox reaction is :
MgO (s) + C (s) ———
f
G°(Products) –
f
G°reactant
At 1273 K
– 439 – (– 941) = 502 KJ mol–1
At 2273 K
– 628 – (– 314) = – 314 KJ mol–1
The reaction is feasible at 2273 K.
Q. 4. Why is Zinc and not Copper used for the recovery of Silver from
the complex [Ag (CN)2] ?
Ans. Zinc is stronger reducing agent and more electropositive than Copper.
(E° = + 0.34V)
Q. 5. Graphite is used as anode and not diamond. Assign reason.
Ans. In
graphite
there
are
free
electrons
which
helps
in
electricalconductivity.
Q. 6.
How is granular zinc & zinc dust obtained ?
Ans.
Granular zinc is obtained by pouring molten zinc in cold water & zinc dust by
melting zinc & then atomising it with blast of air.
Q. 7.
How does NaCN act as a depressant in preventing ZnS from forming the
froth ?
Ans.
NaCN forms a layer of zinc complex, Na2 [Zn (CN)4] on the surface of ZnS
and thereby prevents it from the formation of froth.
Q. 8.
In the process of extraction of gold, Roasted gold ore :
O2
Roasted gold ore + CN– + H2O ———
–
[X] + Zn ———
Identify the complexes [X] & [Y].
Ans.
[X] = [Au (CN)2]–,
Q. 9.
Why is the reduction of a metal oxide easier if the metal formed is in
[Y] = [Zn (CN)4]2–
liquid state at the temp. of reduction ?
Ans.
The reduction of metal oxide is as :
M2O (s) + xM1 (s or l) ———
1
O (s)
x
71

The entropy of liquid metal is more than entropy of the metal in solid

–
Q. 10.
What is the role of collector in froth floatation process ?
Ans.
Collector enhances non-wettability of the mineral particles.
Q. 11.
At which temperature direct reduction of Fe2O3 by carbon is possible ?
Ans.
Above 1123 K, carbon can reduce Fe2O3.
Q. 12.
Why a very high cosmic abundance of iron is there ?
Ans.
A very high cosmic abundance of iron is due to its high nuclear binding
energy.
Q. 13.
Why refractory metals are used in construction of furnaces ?
Ans.
Refractory metals are used in construction of furnaces because they can
withstand high temperature.
Q. 14.
What is pyrometallurgy ?
Ans.
Extraction of metals using heat is called pyrometallurgy. It involves
concentration of ores, roasting calcination, smelting, reduction and refining of metals.
Sulphide, carbonate, oxide ores etc. are subjected to pyrometallurgy.
Q. 15.
How the most electropositive metals are isolated from their ores ?
Ans.
The most electropositive metals are isolated from their ores by electrolysis of
fused ionic salts.
SOME IMORTANT QUESTIONS
Q.1. In general which metals do you expect to occur in the nature in native state? Give examples.
Ans. : Metals such as Cu, Ag, Au, Pt etc. which lie below hudrogen in electrochemical series are not
readily attached by Oxygen, moisture and Carbon dioxide of the atmosphere and hence occur
in the native state in nature.
Q.2.
Copper and Silver lie below in the electrochemical series and yet they are found in the
combined state as sulphide in nature.Comment.
Ans. : Due to high polarizing power of Cu and Ag ions, their sulphides are even more stable than the
metals.
Q.3.
Which metals are generally extracted by electrolyte process? Which positions these metals
generally occupy in the periodic table?
Ans. : Electrolytic process is used for the extraction of active metals like Na, Ca, Mg, K, Al etc.
where all other methods fail. Except Al and few other metals, these metals belongs to S-block
elements of periodic table.
Q.4.
Although thermodynamically feasible, in practice, Magnesium metal is not used for the
reduction of Alumina in the metallurgy of aluminium.Why?
72
Ans. : Below the temperature(1623 K) correspoinding to the point of intersection of Al 2O3 and MgO
curves in Ellingham diagram, Magnesium can reduce Alumina because the ΔfGo value of
Al2O3 at temperatures below 1623 K, is less negative than that of MgO. Therefore below 1623
K Mg can reduce Al2O3 to Al. But Magnesium is much costier metal than Aluminium and
hence the process will be uneconomical.
Q.5.
Why is the reduction od a metal oxide easier if the metal formed is in the liquid state at the
temperature of reduction?
Ans. : Entropy is higher when a metal is in the liquid state than when it is in the solid state.
Therefore, the value of entropy change (Δs) of the reduction process is more on the +ve side
when the metal formed is in the liquid state and the metal oxide being reduced is in the solid
state. Since the value of T ΔS increase and that of ΔH remains constant, therefore the value of
ΔGo ( G 0  H 0  TS 0 _) becomes more on the –ve side and the reduction becomes
easier.
Q.6.
At a site, low grade copper ores are available and zinc and iron seraps are also available.
Which of the two seraps will be more suitable for reducing the leached copper ore and why ?
Ans. : The E0 value for the redox couple Zn2+/ Zn(-0.76V) is more negative than that of Fe2+/ Fe(0.44) redox couple. Therefore, Zinc is more reactive than iron and hence reduction will be
faster in case if Zinc seraps are used, But Zinc is a costier metal than iron so using iron scraps
would be more economical.
Q.7.
(A). Name the method used for refining of (i) Nickel (ii) Zirconium
(B). The Extarction of Au by leaching with NaCN envolves both Oxidation and Reduction.
Justify giving equations.
Ans. ; (A) [ i ] Mond Process : It is used to refine Nickel metal. When impure Nickel is heated in a
current of CO at 330-350 K, it forms volatile nickel tetracarbonyl complex leaving behind the
impurities. This complex again on heating at higher (450-470K) it undergoes thermal
decomposition giving pure Nickel.
350K
Ni  4CO 330

 NiCO 4
Impure
Nickel
Nickel tetracarbonyl
470K
NiCO 4 450

 Ni  4CO
Pure Nickel
[ ii ] Van Arkel method :
This method is very useful for preparing ultra pure metals by
removing all the Oxygen and Nitrogen present in the form of impurity in certain metals such as
Zirconium and titanium which are used in space technology.
In this method, crude Zirconium is heated in a evacuated vessels with iodine at 870 K, the
covalent volatile ZrI4 thus formed is separated. It is then decomposed by heating over a tungsten
filament at 2075 K to give pure Zirconium.
K
Zrs   2I 2 g  870
 ZrI 4 g  2075

 Zrs   2I 2 g 
Impure
Tungsten
Pure
filament
(B) During the leaching process, Au is first oxidized to Au+ by Oxygen (O2) of the air which then
combines with CN- (Cyanide) ion to form the soluble complex, Sodium dicyanoaurate(I).
4 Aus   8NaCN aq   2H 2 O  O2 g   4 NaAuCN 2 aq  4 NaOH aq
Impure
Soluble Complex
Gold is then extracted from this complex, by displacement method using a more electropositive Zinc
metal, In this reaction, Zn acts as a reducing agent. It reduces Au + Au while it itself gets oxidized to
Zn2+ which combines with CN- ions to form soluble complex, sodium tetracy anozincate(II)
73
2 NaAuCN 2 aq  Zns   2 Aus   Na2 ZnCN 4 aq
Thus extraction of Au by leaching with NaCN involved both oxidation and reduction.
Q.8.
Free energies of formation (ΔfG) of MgO(s) and CO(g) at 1273 K and 2273 K are given below :
ΔfG(MgO)(s) = -941 KJ/mol at 1273 K
ΔfG(MgO)(s) = -314 KJ/mol at 2273 K
ΔfG(CO(g)) = -439 KJ/mol at 1273 K
ΔfG(CO(g)) = -628 KJ/mol at 2273 K
On the basis of above data, predict the temperature at which carbon can be used as a reducing agent
for MgO(s).
Ans. : (a) At 1273 K
i) Mg  s  
1
O2  MgO  s  ;  f G  941Kj / mol
2
ii) C  s  
1
O2  CO  s  ;  f G  439 Kj / mol
2
The equation for reduction of MgO to Mg by C can be obtained by substracting equation (i) from
equation (ii). Thus
MgO s   C  s   Mg  s   CO g 
  r G  602KJ / mol
Since ΔrG of the above reduction reaction is +ve; therefore, reduction of MgO by C is not feasible at
1273 K.
(b)
At 2273 K
iii) Mg  s  
iv)
1
O2  MgO  s  ;  f G  314 KJ / mol
2
1
C  s   O2  CO  s  ;  f G  628KJ / mol
2
On substarcting Eq.(III) from Eq.(IV), we have
MgO s   C  s   Mg  s   CO g  ;  r G  314 KJ / mol
Since ΔrG for the above reduction reaction is –ve; therefore reduction of MgO by carbon at 2273 is
feasible.
Q.9. Metal sulphides occur mainly in rocks and metal halides in lakes and seas. Explain.
Ans.: Metal halides being soluble in water, get dissolved in rain water and are carried to lakes and
seas during weathering of rocks. On the otherhand, metal sulphides being insoluble are left behind in
the rocks as residue.
Q10. What are fluxes? How are they useful ?
Ans. : Flux is a substance that combines with gangue which may still be present in the roasted on the
calcined ore to form an easily fusible material called the slag.
Q.11. What is a slag?
Ans:
A slag is an easily fusible material which is formed when gangue still present in the roasted or
the calcined ore combines with the flux. For example: in the metallurgy of Iron, CaO(flux) combines
with silica(SiO2) gangue to form easily fusible calcium silicate (CaSiO3) slag.

CaCo3 

CaO  CO2
CaO  SiO 2  CaSiO3 Slag
Q.12. What is the Principle of Zone refining ?
74
Ans. : When the molten solution of an impure metal is allowed to cool, the pure metal crystallizes
out while the impurities remains in the solution.
Q.13. What is a depressant? Give one example.
Ans.
Compounds which prevent the formation of froth in froth flotation process are called
depressants. For example, NaCN can be used as a depressant for ZnS in the separation of ZnS from
PbS ore. It forms a layer of Zinc complex, Na2[Zn(CN)4] on the surface of ZnS and thus prevents it
from forming the froth.
Q.14. What is the role of a stabilizer in froth floatation process?
Ans.:
Chemical compounds like cresols and aniline which tend to stabilize the froth are called
stabilizers.
Q.15. What types of Ores are roasted?
Ans. : Sulphides ores are roasted to convert them into their oxides since Oxides are more easily
reduced to metal than sulphides.
Short Questions :01 Mark
Very Short answer questions :
FOR BELOW AVERAGE STUDENTS: 01 Mark
1.
How does minerals different from an Ore ?
2.
What do you mean by benefication process or concentration or enrichment or dressing ?
3.
What is a slag? Give one example of basic flux.
4.
Why do a few elements occurs in the native state while others do not ?
5.
Write constituents of bronze with their use.
6.
Why is Zinc not extracted from Zinc Oxide through reduction using CO?
7.
What is the role of cryolite in the metallurgy of Aluminium?
8.
Which type of metals are refined by distillation method?
FOR AVERAGE STUDENTS :01 Mark
1.
Name the chief form of occurrence of the following in the earth’s crust :
a) Aluminium,
2.
b) Iron,
c) Copper,
d) Zinc
An Ore sample of galena (PbS) is contaminated with Zinc blende(ZnS). Name ore
chemical which can be used to concentrated galena selectively by froth floatation process.
3.
How is cast iron different from pig iron?
4.
What is copper matte?
5.
Why pine oil is used in the froth floatation method?
For Bright Students:
1.
01 Mark
Copper and Silver lies below in the electrochemical series and yet they are found in the
combined state as sulphides in nature. Comment.
2.
Name the common element present in the anode mud in electrolytic refining of copper.
Why are they so present?
3.
What is meant by the term chromatography?
75
4.
What is the role of depressant in froth floatation process?
5.
What criterion is followed for selection of the stationary phase in chromatography?
6.
What is the role of collector & froth stabilizer in froth floatation process?
For Below Average Students:
02 Marks Questions
1.
Why copper metal is put in Silica lined converter ?
2.
Giving examples, differentiate between roasting and calcinations.
3.
Describe the method of refining of Nickel.
4.
Explain the Leaching of Silver or Gold.
5.
Write two important ores of Iron, Aluminium, coppere & Zinc with their chemical
formula.
6.
What is aluminothermy? Explain with an example.
FOR AVERAGE STUDENTS:
02 Marks Questions
1.
State the role of silica in the metallurgy of copper.
2.
Why is the extraction of copper from pyrite difficult than that from its oxide ore through
reduction?
3.
Write chemical reactions taking place in the extraction of Zinc from Zinc blend.
4.
Silver Ores and native gold have to be leached with metal cyanides. Suggest a reason for
this.
5.
What do you mean by vapour phase refining? Describe with suitable example.
6.
The extraction of Au by leaching with NaCN envolves both oxidation and reduction.
Justify giving equations?
7.
Copper can be extracted by hydrometallurgy but not Zinc explain.
FOR BRIGHT STUDENTS:
02 Marks Questions
1.
Out of C and CO, which is a better reducing agent for ZnO?
2.
What is the role of graphite in the electrometallurgy of Aluminium?
3.
What do you mean by leaching? Explain with an example.
4.
The Choice of a reducing agent in a particular case depends on thermodynamic factor,
how far do you agree with this statements? Support your opinion with two examples.
5.
Discuss some of the factors which need consideration before deciding on the method of
extraction of metal from its ore.
FOR BELOW AVERAGE STUDENTS:
03 Marks Questions
2.
What is the chief ore of iron? Write chemical reactions taking place in the
extraction of Iron from its Ores.
76
3.
3.
4.
5.
6.
Give the important uses of
a) Aluminium, b) Copper, c) Iron
Differentiate between
(i) Gangue and Flux
(ii) Minerals & Ores
What do you mean by following terms :
(i) Electro metallurgy
(ii) Hydrometallurgy (iii) Smelting
What is the significance of leading in the extraction of aluminium?
Explain the term :
(i) Distellation
(ii) Liquation
(iii) Poling in term of refining of metals.
FOR AVERAGE STUDENTS:
03 Marks Question
1.
2.
3.
Name the processes from which chlorine is obtained as a bye-product. What will
happen if an aqueous solution of NaCl is subjected to electrolysed.
Explain how the following metals are obtained from their oxides by the reduction
process.
a) Metal X which is low in the reactivity series.
b) Metal Y which is in the middle of the reactivity series.
c) Metal Z which is high up in the reactivity series
a) Describe a method for refining of Nickel.
b) The extraction of Ag by leaching with NaCN involves both oxidation &
reduction.
For Bright Students:
03 Marks
1.
Account for the following :
i)
The reduction of metal oxide is easier if the metal formed is in liquid state
at the temperature of reduction.
ii)
the reduction of Cr2O3 with Al is thermodynamically feasible yet it does
not occur at room temperature.
iii)
Is Carbon is satisfactory reducing agent for all metal Oxides ?
2.
Name the main ore of lead metal, write its formula. How is this metal obtained
from its oxides ?
3.
Discuss briefly the chromatographic method used for purification of elements.
4.
The value of ΔfG0 for formation of Cr2O3 is -540 KJ/mol and that of Al2O3 is -827
KJ/mol. Is the reduction of Cr2O3 possible with aluminium?
5.
How can you separate alumina from Silica in Bauxite ore associated with Silica?
Give equations if any.
77
UNIT
07
THE p-BLOCK
ELEMENTS
Nitrogen family (Group 15 Elements)
Members: Nitrogen, Phosphorous, Arsenic, Antimony, Bismuth
General electronic configuration:- ns2np3, valance electrons is five, the valency may be 3 and
5 both. Nitrogen is only trivalent but phosphorous forms pentavalent because of vacant dorbital. The oxidation states are mainly +3 & +5. Down the group +5 oxidation states
especially is unstable due to inert pair effect (i) the stability of higher oxidation state
decreases down the group. (ii) As stated in (i) is due to reluctance of ns2 electrons in bond
making process.]
General characteristics:
(i) Catenation: Increases down the group. Phosphorous has greater power of catenation than
nitrogen because in former the size is larger and availability of vacant d-orbitals, while in
later the size is smaller and stability is achieved through P∏ – P∏ bonding. Single N – N bond
is weaker than p – p bond.
(ii) Electro negativity trend: “N” because of higher electro negativity is able to make
effective H- bond. Therefore NH3 has higher B.pt than phosphine. Also this property causes
ammonia to be water soluble when ammonia makes H-bond with water.
(iii) Stability of Hydrides/ Basic characteristics: As size increases down the group the bond
length increases and hence stability of hydride decreases. Therefore acidic character increases
78
down the group. In terms of basic strength NH3 is more basic than PH3 because in “P” we
have larger size & vacant d- orbital. Therefore in spite of lone pair electrons in both cases
PH3 is less basic.
iv) Reactivity of Elements: Molecular Nitrogen has a triple bond and with very high bond
energy it is relatively inert. White “p” is more reactive than red “p” because in former we
have only P4 structure with lesser no of bonds while in later we have larger no. of bonds with
repeating P4 units.
(v) Anomalous character of “N”:
(a) Because of smaller size “N” is able to make P∏ – P∏ multiple bonds. This also causes N2 to
remain as discrete molecules and associated by weak Vander Waal’s forces. This is why N 2 is
possible but P2 is not. This also causes N2 to remain in gaseous form.
(b) “N” has no vacant d-orbital hence only valency of three is found while other in the
group like “P” is trivalent and pentavalent both.
(vi) Halides: “P” forms PCl3 & PCl5 both. PCl5 is unstable because of two longer axial
bonds. PCl3 fumes in moisture.
(vii) Oxo acids:
(a) N:- Two important oxo acids viz Nitric acid and Nitrous acid. Both are mono
protic.
The presence of nitrate ion can be detected by adding freshly prepare FeSO 4 solution
followed by adding 1-2 drop conc. H2SO4 by the side of test tube below tap water. This
makes a brown ring.
(b) P:- Three important oxo acids viz H3PO4, H3PO3, and H3PO2 . They are triprotic,
diprotic and monoprotic because of their structures.
(viii) Important compounds: Ammonia is prepared by Haber’s process.
Requirement for good yield of ammonia are high pressure, low tempt as per Le-chatelier’s
principle with iron oxide as catalyst and k2O, Al2O3 as promoters. Ammonia has sp3
hybridization. Lone pair electrons on “N” makes its shape pyramidal. Ammonia is a good
complexing agent because of lone pair electron on nitrogen atom.
(d) Phosphorous penta Chloride: “P” is sp3d- hybridized and structure is trigonal
bipyramidal.
OXYGEN FAMILY(Group 16 Elements)
MEMBERS:- Oxygen is the 1st element of the group and other members are sulphur,
Selenium, Tellurium and Polonium.
General Electronic configuration, Valency & oxidation state: The general electronic
configuration is ns2np4. The valence electrons are six and valency varies from 2 to 6. The
oxidation state varies from -2 to +6. The maximum oxidation state of oxygen is +2.
General Characteristics:
(i) Anomalous character of oxygen:
a) Because of smaller size it makes effective P∏ – P∏ bond. Therefore remains as discrete
molecule. But sulphur because of expansion of valency on account of vacant d- orbital exists
as solid.
79
b) Catenation: Oxygen has lesser ability of Catenation than sulphur because in later we have
vacant d-orbital. Therefore valency of sulphur goes from two to six. The single bond energy
in sulphur is larger than that of oxygen.
c) Paramagnetic behaviour: Because of two unpaired electrons in antibonding molecular
orbitals of oxygen it is paramagnetic. However sulphur is also paramagnetic in vapour-state.
d) Valency:- Only oxygen is divalent, others are tetravalent and hexavalent also.
e) Electronegativity: Only oxygen is able in making effective H- bonding. Hence H2O is a
liquid and H2S is gas.
ii) Stability of hydride/ Acidic character of hydride:- As size increases down the group,
the bond length between hydrogen and group member increases. This causes decreasing bond
energy down the group and acidic character of hydrides increases.
iii) Oxo acids: Sulphuric acid is the most important oxo- acid of sulphur, which is prepared
by contact processes:
iv) Allotropy:
a) Ozone is the allotrope of oxygen. High concn. Of ozone is dangerously explosive. Ozone is
thermodynamically unstable and converts into dioxygen. The free energy for this is highly
negative.
Because of this nascent oxygen so produced, Ozone acts as powerful oxidizing agent. Ozone
has two resonating structures.
b) Sulphur has two allotropes viz Rhombic (State at room temp) and Mono clinic (Prepared
by heating rhombic sulphur).
Halogen Family (Group 17 Elements)
Members:- F, Cl, Br , I and At .
General properties:
(1)
Electronic configuration - All these members have electrons (ns2np5) in
their valence shell.
(2)
Atomic and ionic radii - They have the smallest radii in their respective
periods. However size increases downward the group due to increase in
no. of shells.
(3)
Ionization enthalpy - They have very high ionization enthalpy. It
decreases down the group due to increase in size.
(4)
Electron gain enthalpy - They have maximum negative electron gain
enthalpy in their periods as they have one electron short to attain stable
electronic configuration. These values become less negative downwards
the group.
(5)
Electronegativity - They are highly electronegative in nature. It decreases
down the group.
(6)
Colour - All halogens are coloured . F2 – yellow, Cl2 – greenish yellow ,
Br2 – red , I2 – violet.
(7)
(8)
Oxidation state – All the members show -1 oxidation-state but Cl , Br and
I shows +1 , +3 , +5 and +7 in addition to -1 state.
Oxidising properties – The ready acceptance of electron by halogen make
them strong oxidizing agent. Fluorine is known to be the strongest
oxidizing agent among halogens.
80
(9)
Reactivity towards hydrogen :- With hydrogen they form hydride of the
type H- X. The acidic nature of these varies in the order HF < HCl < HBr
<HI .
Reactivity towards oxygen – With oxygen they form different oxides.
Fluorine forms two oxides OF2 and O2F2 but chlorine , bromine and iodine
form oxides using oxidation state of +1 to +7. The higher oxides are more
stable than lower oxides.
Reactivity towards other halogens – With other halogen they form
interhalogen compounds of the types XX’ , XX3’ , XX5’ , XX7’ (where X
is larger in size than X’)
Reactivity towards metal – They form metal halides . Their ionic nature
decreases as follows
MF > MCl > MBr >MI
Anomalous behaviour of fluorine
Fluorine is anomalous in many properties due to following reasons;
(1)
Small size and high charge density.
(2)
Absence of vacant d- orbital in the outermost shell.
(3)
The highest electronegativity.
(10)
(11)
(12)
INERT GASES (Group 18 Elements)
MEMBERS:- He , Ne, Ar, Kr,Xe & Rn
General Properties:1. All elements have octet configuration except He, which has duet. This makes them stable
in comparison to other elements.
2. Ionization enthalpy decreases down the group. Kr & Xe have low ionization energy &
hence they make compounds with highly electronegative elements like fluorine & oxygen.
3. Important compounds of xenon with its shape, hybridization & structure :-
sp3d
sp3d2
81
Sp3d3
sp3d2
sp3
OXO- ACIDS OF PHOSPHORUS
OXO- ACIDS OF SULPHUR
4. The compounds which led to discovery of inert gas compounds are O2+PtF6 – &
Xe PtF6 –
5. The xenon fluorides hydrolyse completely & partially leading to the formation of different
compounds like:XeF6 + H2O → XeO3 +HF (complete hydrolysis)
XeF6 + H2O → XeOF4 +HF (partial hydrolysis)
+
Binary Compounds
This section lists some binary compounds with halogens (known as halides), oxygen (known
as oxides), hydrogen (known as hydrides), and some other compounds of sulphur. For each
compound, a formal oxidation number for sulphur is given, but the usefulness of this number
is limited for p-block elements in particular. Based upon that oxidation number, an electronic
configuration is also given but note that for more exotic compounds you should view this as a
guide only. The term hydride is used in a generic sense to indicate compounds of the type
MxHy and not necessarily to indicate that any compounds listed behave chemically as
hydrides.
82
In compounds of sulphur (where known), the most common oxidation numbers of sulphur
are: 6, 4, 2, and -2.
Hydrides
The term hydride is used to indicate compounds of the type MxHy and not necessarily to
indicate that any compounds listed behave as hydrides chemically.
IMPORTANT QUESTIONS WITH MODEL ANSWER
(1 mark questions)
1. Why is phosphine weaker base than Ammonia?
Ans: P is bigger than N, therefore lone pair of electrons are less available on P than N.
2. Why is Ammonia a good complexing agent?
Ans: Because of lone pair of electrons on N.
3. Why Nitrogen is a gas whereas Phosphorous is a solid?
Ans: Nitrogen molecules have less Vander Waal’s forces of attraction whereas molecular
phosphorous has more Vander Waal’s forces of attraction.
4. Why is Nitrogen relatively inert?
Ans: Because of a triple bond with high bond energy.
5. Ammonia is water soluble. Why?
Ans: It forms hydrogen bond with water.
(2 marks questions)
1. Write shape of following molecules:
a) Ammonia
b) Phosphorous Pentachloride
Ans: a) Pyramidal
b) Trigonal bipyramidal
2. Ortho phosphoric acid is triprotic but phosphoric acid is diprotic. Why?
Ans: Because of their structures ortho phosphoric acid and phosphoric acid have three and
two ionisable hydrogen.
3. Give reasons:
a) Phosphorous pentachloride is unstable.
b) Phosphorous trichloride fumes in moisture.
Ans: a) Because of two longer axial bonds.
b) Because it reacts with water and forms hydrochloric acid.
(3 marks questions)
4. Nitrogen shows anomalous behaviour in the group. Why?
Ans: i) Small size
ii) High electro negativity
iii) Absence of vacant d-orbital.
5. How is ammonia prepared? Write chemical reaction involved in the Haber’s
process.
83
Ans: i) Ammonia is prepared by the combination of hydrogen and nitrogen in the ratio of 3:1.
ii) The used catalyst is iron oxide. The promotors are potassium and aluminium oxides.
iii) The required conditions are high pressure and moderate temperature.
N2 + 3H2 → 2NH3
06. Complete the following reactions :i) XeF6 + H2O → --- + ---- ( partial hydrolysis)
ii) XeF6 + H2O → --- + ---- ( Complete hydrolysis )
iii) Cl2 + H2O→
--- + ---Ans: i) XeOF4 + HF
ii) XeO3 + HF
iii) HOCl + HCl
07. Arrange the following in decreasing order :i)
NaF, NaI, NaCl,NaBr( Ionic nature)
ii)
PH3, NH3, ,AsH3,SbH3 ( Basic Strength)
iii)
HF,HBr,HCl,HI
( Acid Strength)
Ans i) NaF> NaCl > NaBr> NaI
ii) NH3> PH3> AsH3> SbH3
iii) HI> HBr> HCl> HI
08. Find hybridization and shape i) XeF2 ii) XeO3 iii) XeF4
Ans: i) sp3d, Linear
ii) sp3, Pyramidal
iii) sp3d2, Square planar
09. Give reasons
i) H2O is a liquid but H2S a gas.
ii) Ammonia has higher boiling point than phosphine
ii) Krypton & xenon make compounds.
Ans: i) Hydrogen bond is found in H2O but not in H2S.
ii) Hydrogen bond is found in NH3 but not in PH3.
iii) Because they have low ionization energy.
10. Write down the chemical reactions that take place in the manufacture of sulphuric acid by
contact’s process.
Ans: i) S + O2 → SO2
V2O5
ii) SO2 + O2
SO3
iii) SO3 + H2SO4 → H2S2O7
iv) H2S2O7 + H2O → 2H2SO4
SOME IMPORTANT QUESTIONS WITH ANSWERS
Q.1. Why does O3 act as powerful Oxidising agent ?
Ans: Due to ease with Which it liberates atoms of nascent oxygen (O3 O2 +0)
2Pbs (S) + 4O3 (g)  PbSO4(S) + 4O2 (g)
Q.2. Which forms of Sulphur show paramagnetic behaviour ?
Ans : In Vapour State sulphurPartly exists as S2 molecule which has two unpaired electrons
in the *
Orbitals hence exhibit paramagnetic
Q.3. Halogens have maximum negative electron gain enthalpy in the respective periods
in the P. T Why ?
Ans Due to smallest size and high effective nuclear charge , readily accept one electron
to acquire noble gas electronic configuration
84
Q4. Why is fluorine stronger Oxidising agent ?
Ans: Due to (a) Low enthalpy of dissociation
(b) high enthalpy of Hydration of FQ.5. How is the presence of SO2 detected ?
Ans : 1. It has pungent characteristic smell .
2. It decolourises KMn O4 solution
3. It turns acidified K2 Cr2 O7 green
Q.6. Are all the five bonds in PCl5 molecule equivalent ?
Ans – PCl5 has a trigonal bipyramidal structure and the three equatorial p-cl bonds are
equivalent . while trhe
two axial bonds are different and longer than equatorial bonds.
Q.7. Why H3 PO3 is dibasic and H3PO4 is tribasic ?
Ans In H3PO3 only two replacable hydrogen but inH3PO4 , three hydrogens are replacable .
Q.8. Why is helium used in diving apparatus ?
Ans: Because heliun is very low solubility in blood
Q.9. Give reason for bleaching action of Cl2
Ans Due to oxidation
Cl2+H2O 2HCl+ O
Coloured substance +O colourless substance
Q10 Write the formula of tear gas .
Ans : CCl3 NO2
Q11. Which aerosol deplete ozone layer ?
Ans C.F.C
FOR BELOW AVERAGE
[01 Mark Question:]
A.
Give reasons
01. N2 is relatively inert in terms of reactivity.
02. NH3 is a good complexing agent.
03. O3 acts as a powerful oxidizing agent.
04. NH3 forms hydrogen bond but PH3 does not.
05. Bond angle of NH4+ is higher than NH3.
06. H2S is less acidic than H2Te.
07. H2O is a liquid but H2S is a gas.
08. NH3 acts as a lewis base.
09. Halogens act as strong oxidizing agent.
10. Dioxygen is a gas but sulphur is a solid at room temperature.
11. Nitrogen does not form NCl5 whereas phosphorous does.
85
12. N2 is a gas but phosphorous is a solid.
13. N2 is possible but P2 is not,Why?
14. NH3 is more basic than PH33
15. F2 and Cl2 are gases but I2 is a solid
16. NH3 is soluble in water but PH3 is not.
17. PH3 has lower boiling point than NH3.
18. PCl5 is unstable.
19. Catenation in nitrogen is weaker than phosphorous.
20. Stability of hydride of group 15 decreases down the group.
21. The +5 oxidation state of Bi is less stable than +3.
22. The basic character of hydride of group 15 decreases down the group.
23. NH3 is pyramidal.
24. One of the noble gases only xenon is known to form real chemical compounds.
FOR AVERAGE
01. Fluorine is the strongest oxidizing agent among halogens.
02. H3PO3 is a diprotic acid.
03. Bond angle in PH3 is less than NH3
04. Noble gases form compound with fluorine and oxygen only.
05. White P is more reactive than red P.
06. PCl5 is more covalent than PCl3.
07. SO2 is air pollutant.
08. Basic character among the hydrides of group 15 elements decreases with increasing
atomic nos.
09. H3PO4 is triprotic but H3PO2 is monoprotic
10. BiH3 is a reducing agent.
11. Electron affinity of fluorine is less than that of chlorine.
12. Sulphur vapours show paramagnetic behaviour.
13. Halogens are coloured.
FOR ABOVE AVERAGE
01. H3PO3 and H3PO2 show reducing character.
02. Catenation in oxygen is lesser than sulphur.
03. Group 16 members are called chalcogens.
04. PKa value of HOCl is higher than that of HOClO.
05. Nitrogen shows Maximum covalency of four.
06. NH3 has higher dipole moment than NF3.
86
07. Trimethylamine is pyramidal but trisilylamine is planar.
08. High concentration of ozone can be dangerously explosive.
09. Identify the neutral molecule which is isoelectronic with ClO-. Is that molecule a
lewis base?
10. PKa(I) for H2SO4 is larger than PKa(II).
FOR BELOW AVERAGE
B.
Complete the following reactions:01. XeF2 + H2O
02. XeF2 +PF5
03. XeF4 +H2O
04. XeF6 +H2O
05. XeF4 +SbF5
06. I2+H2O+Cl2






FOR AVERAGE
01.
02.
03.
04.
05.
06.
07.
CaF2+H2SO4 
NaOH +Cl2 
F2 + H2O 
PCl3 + H2O 
PCl5 + H2O 
C + H2SO4 
Ca(OCl)Cl+HCl 
FOR ABOVE AVERAGE
01. P4 + KOH +H2O
02. XeF4 + O2F2
03. Ca3P2 + H2O
04. POCl3 + H2O
05. PH3 +HNO3
06. CuO +NH3
07. BrO3- +F2 +OH-
C.







Deduce structural formula of the following;
For below average:-
NH3, PCl5, XeF2, XeF4, XeF6, SF6, IF7
For average:-
H3PO4, H3PO2, H3PO3, SF4, XeO3
For above average:-
Cl2O7, P4O10, N2O5, XeOF4, XeOF2, H2S2O8.
87
FOR BELOW AVERAGE
D.
What happens when;
1.
Conc. H2SO4 is added to CaF2.
2.
SO3 is passed into water.
3.
Chlorine is passed through a hot and conc. Solution of alkali, like Ba(OH)2.
4.
XeF4 undergoes hydrolysis.
5.
XeF6 undergoes hydrolysis.
6.
PCl5 is heated.
7.
NaCl is heated with H2SO4 in presence of MnO2.
8.
Cl2 is passed through a solution of NaI in water.
9.
SO2 gas is passed through a solution of Fe(III) salt.
10
White P is heated with NaOH solution in an inert atmosphere of CO2.
FOR BELOW AVERAGE & AVERAGE
E. Arrange the following as mentioned:
1.
M-F, M-Cl, M-Br, M-I.
(Decreasing order of ionic nature)
2.
F2, Cl2, Br2, I2.
(Increasing order of bond energy)
3.
H2O, H2S, H2Se, H2Te.
4.
NH3, PH3, AsH3, SbH3. (Increasing order of basic nature)
5.
HI, HBr, HCl, HF.
6.
HI, HBr, HCl, HF. (Increasing order of reducing power)
7.
HOCl, HOClO, HOClO2, HOClO3. (Increasing order of acid strength)
8.
H2O, H2S, H2Se, H2Te.
9.
HClO4, HBrO4, HIO4.
(Decreasing order of boiling points)
(Increasing order of thermal stability)
(Increasing order of acid strength)
(Decreasing order of acid strength)
F. Some more additional questions; (3 & 5 marks each)
1. Explain the process and necessary conditions involved in the preparation of NH3,
H2SO4 or HNO3.
2. What are interhalogen compounds? How are they classified?
3. Describe the chemistry of ring test.
4. Why do noble gases have comparatively larger atomic size? Give the reasons which
prompted Bartlett to prepare first noble gas compound.
5. Explain the chemistry of group 15 elements with respect to catenation, stability of
hydrides, and relative stability of oxidation states.
6. Describe the chief uses of fluorine, chlorine and their compounds.
How is bleaching powder prepared? Mention two of its uses .
88
7. Write the three steps involved in the manufacture of sulphuric acid by the contact
process
8. What are the condition to maximize the yield of H2SO4 in contaqct process ?
9. How can H2SO4 reacts with cu metals ?
10. How ammonia is manufactured by Haber’s process on large scale ?
11. Mention the Conditions to maximize the yield of ammonia .
12. How does amminia acts as a Lewis base ,show it with the help of a chemical .
13. Illustrate how copper metal can give diff. products on reaction with HNO3 .
14. How HNO3 is prepared by ostwalds process ? Write reaction only at diff . stages
15. Balance the following equation :(i) I2 + HNO3 HIO3+NO2+H2O (ii) C+ HNO3 CO2 +NO2 +H2O
(iii) S + HNO3(conc.) H2SO4 +NO2+ H2O (iv) P4+ HNO3 (conc.) H3PO4+
NO2+H2O
(V) HNO3 HNO3+ NO+ H2O
89
UNIT
08
THE d- BLOCK AND
f-BLOCK ELEMENTS
STUDY MATERIAL
Transition elements:-The elements lying in the middle of periodic table belonging to gr.3 to
12 are known as transition elements.
Electronic configuration :-
(n-1)d 1-10 ns 1-2 .
Lanthanoids:-The 14 elements immediately following lanthanum, i.e .Cerium(58) to
Lutetium(71) are called lanthanoids.
Actinoids:-The 14 elements immediately following Actinium(89),with atomic no.90 to 103
are called actinoids.
Inner transition elements: Elments in which last electron inter in f-sub-shell of
antipenultimate energy level.
Solvation energy or energy of hydration:-The amount of energy change when 1 mole of
solid substance is changed into vapour state.
CHARACTERISTICS OF TRANSITION ELEMENTS:Properties of transition metals
i. metallic in nature
ii. show variables oxidation states.
iii. complex compounds.
iv. coloured compounds.
v. good catalyst
vi. paramagnetic compounds.
vii. Interstitial compounds
Reason due to
High enthalpy of atomization
Presence of vacant d- orbitals
Small size, High charge density
d-d Transition
Suitable surface area in d-orbital
Unpaired electron in d-orbital
Similar atomic size & Vacant space in lattice
Lanthanoid Contraction:-There is steadly decrease in the atomic and ionic radii as the
atomic no, increases. It is because ,for every additional proton in the nucleus ,the
corresponding electron goes into the 4f shell which is too diffused to screen the nucleus as
effectively as more localized for the outermost electron increases steadily with the atomic no.
of lanthaoids. This contraction in size is called lanthanoid contraction.
TRANSITION ELEMENTS FORM :Basic Oxide :- Sc2O3,TiO, Ti2O3, VO,V2O3, MnO
,FeO, CuO, CoO, NiO, Cu2O
Acidic Oxide:-V2O5,Mn2O7,CrO3, OsO4
AmphotericOxide:-CuO,TiO2,ZnO,CrO3, Cr2O3,
Mn2O3,MnO2,Mn3O4Fe3O4
Spinel:- ZnFe2O4,
Non stoichiometric compounds:Fe0.91O or Fe0.95O
Ionic Halides;All flouoro i.e. CuF2,AgF
90
Due to lowest oxidation states
Due to highest oxidation states
Due to intermediate oxidation states
Tri & divalent occupies octa & tetrahedral hole
Due to defects in lattice crystal
Due to lower oxidation states
Covalent Halides:-CuI2,CuCl2,CuBr2,AgCl,AgBr
Due to higher oxidation states
Application of d- & f- block elements:1. Iron and steel used in construction,as catalyst in Haber’s process
2. CrO2 is used in magnetic tapes
3. Ni is in hydrogenation
4. W (Tunguston) in making electrical filament
5. AgBr in photography
6. Cu , Ag, Au are coinage metals
7. Pt is catalyst in manufacture of HNO3
8. V2O5 is catalyst in contact process
9. Titanium is in nuclear reactions.
10. PdCl2 is in Wacker’s process.
Differences between Lanthanoids and Actinoids:-
Lanthanoids
Actinoids
Common oxidation state is +3, besides this
+2 ,+4
Have smaller tendency of complex forming
Don’t form oxo- ions
Non-Radioactive Except Promethium
Compounds are less basic
Common oxidation state is +3, besides this
+4,+5,+6, &+7 also.+2 quite rare
Strong tendency of complex forming
Do form oxo-ions i.e. UO2+, PuO2,UO+
All are Radio-active
Compounds are more basic
Important compounds of transition metals:1.Potassium dichromate K2Cr2O7
Ore:- Chromate FeCr2O4
Preparation:4FeCr2O4+ 8Na2CO3 + 7O2 8Na2CrO4+ 2Fe2O3 + 8CO2
2Na2CrO4+ H2SO4
 Na2Cr2O7 + Na2SO4 + H2O
Na2Cr2O7 + 2KCl

K2Cr2O7 + 2NaCl
Structure:-
91
Properties:1. Oxidising property:
(a)
Iodide to iodine:
Cr2O7 2- + 14H+ + 6e-  2Cr 3+ + 7H2O
6I 3I2 + 6e--------------------------------------------------------------Cr2O7 2- + 14H+ + 6I-  2Cr3+ + 3I2 + 7H2O
(b)Ferrous to ferric:
Cr2O7 2- + 14H+ + 6e-  2Cr 3+ + 7H2O
6Fe2+
 6Fe3+ + 6eCr2O7 2- + 6Fe2+ + 14 H+  2Cr3+ + 6Fe3+ + 7H2O
2.Chromyl Chloride test:K2Cr2O7 + 4NaCl + 6H2SO4 2KHSO4 + 4NaHSO4 + 2CrO2Cl2 + 3H2O
Uses:
 It is used as an oxidizing agent
 For tanning of leather
 As disinfectant
 In volumetric analysis
 In chromyl chloride test.
2. Potassium permanaganate KMnO4
Ore:- Pyrolusite - MnO2
Preparation:2MnO2+ 4KOH + O2 2K2MnO4 + 2H2O
4H+ + 3MnO42- 2MnO4- + MnO2 +2H2O
Structure:
Below
Properties:1. In acidic solutions:Iodine is liberated from potassium iodide
10 I - + 2MnO4- + 16 H+  2Mn 2+ + 8H2O + 5 I 2
Fe 2+ ion(green) is converted to Fe 3+ (yellow)
Uses:
92
5Fe 2+ +MnO4 - + 8 H+  Mn 2+ + 4H2O + 5 Fe3+
2. In neutral or faintly alkaline solution:Oxidation of iodide to iodine
2MnO4-+ H2O + I-  8MnO2 + 6SO4 2- + 2OHUses:



As disinfectant
As oxidizing agent
For qualitative detection of halides, sulphites, tartarate
In dry cell
QUESTION BANK
Below Average : 01 Marks
1.
What are d Block Elements ?
2.
3.
4.
5.
Ans: Those elements in which d-orbital are progressively filled up are known as
d-orbital.
What is general electronic configuration of d – Block elements ?
Ans: (n-i) d1-10 ns1-2
Why are Zn , Cd , Hg are not regarded as transition metals ?
Ans: It is because neither they nor their ions have incompletely filled d- orbitals .
What is the highest oxidation state shown by Cr. (24) , Mn(25) ?
Ans: Cr. Shows (+6) and Mn Shows (+7) Oxidation state .
Why do transition metals have high enthalpy of atomization ?
Ans : it is due to strong metallic bonds and additional covalent bonds
02 Marks
1.
What are lanthanides ? Write electronic configuration of lanthanides .
Ans The 14 elements after lanthanum (57) are called lanthanides . There outer
electronic
configuration is (n-2) f1-14 (n-1)d 0-1 ns2
2.
For Some of first row of transition elements the E o value are
V
-1.18
v
Cr
-0.91v
Mn
-1.18 v
Fe
-0.44 v
Co
-0.28 v
Ni
-0.25 v
Give suitable explanation for the irregular trend in their values .
Ans – it is due to irregular variation of sum of sublimation and ionization
energy in these elements
.
3.
Why is orange solution of K2Cr2O7 turns yellow on adding Na OH ?
Ans : It changes to Cr O4-2 ions which are yellow in colour .
Cr2O7-2+2OH-1  2CrO4-2 yellow + H2O
4.
Co (II) is stable in aqueous solution but in the presence of strong ligands
and in air get oxidized
Co+3, Why ?
Ans:- Strong ligands and air Oxidized Co+2 to Co+3 by providing energy released
due to force of attraction between them and metal ions .
5.
Why Ce+3 Can be easily Oxidised to Ce+4 ?
93
Cu
+0.34V
Ans : It is because Ce+4 has stable electronic configuration .
03 Marks
1. Give balanced equation for the preparation of K2Cr2O7 from chromate ores . What
happens when acidified dichromate soln. react with (i) FeSo4 (ii) H2s gas ?
Ans : 4FeCr2O4 + 16 NaOH + 7O2 8 Na2CrO4 + Na2SO4+ 2Fe2O3
2Na CrO4 + 2H2SO4 + (Conc) Na2Cr2O7 + H2O + Na2SO4
Na2Cr2O7 + 2KCl  K2 Cr2 O7 + 2 NaCl
(i) 6Fe +2 + Cr2 O7-2 + 14 H+  6Fe+3 + 2Cr+3 +7H2O
green
(ii) 3H2S + Cr2 O7-2 + 8H+3S+ 2Cr+3 + 7H2O
2. Describe the preparation of KMnO4 . How does the acidified permanganate Soln.
reacts with (i) Iron (ii) Sulphurdioxide (iii)Oxalic acid ?Write the ionic eqn for the
reactions
Ans : 2MnO2 + 4KOH + O2K2MnO4 + 2H2O
MnO4-2  MnO4-1 +e(a) [ Fe +2 Fe +3 + e-]x 5
5e-+8H+ MnO4-1 Mn+2 + 4H2O
5 Fe +2 + 8H+ MnO4- 5Fe+3 + Mn+2+ 4H2O
(b) 2KMnO4 + 2H20 +5SO2 K2SO4 + 2MnSO4 + 2H2SO4
COO
│2CO2 + 2e x 5
COO
{5e- + 8H+ + MnO4-1 Mn +2 + 4 H2O) }x 2
16 H+ + 2 MnO4 -1 + 5C2O4-2 5CO2)2 + 2Mn+2 + 8H20
3. What is an lanthanoid contraction ? What is its cause ? Give its consequences .
Ans The decrease in atomic and ionic size with increase in atomic no. in lanthanoid series
is called lanthanoid contraction . The cause of lanthanoid contraction is poor shielding
effect of d and f electron , due to which effective nuclear charge increases .
. consequences :(i) The ion ic size of lanthanoid is similar to each other , they occur , together and their
separation becomes difficult .
(ii) The ionization energy of 5 d – transition series are higher than 4 d – transition series .
(iii)There is close resemblance between elements of 4 d and 5 d transition series . due to
similar ionic
Average 01 Mark
(i)
Why does V2O5 acts as catalyst ?
Ans : It is because V shows variable O.S .
(ii)
Why are transition metals and there compounds generally coloured ?
Ans It is due to present of unpaired electrons in transition metals and their ions ,
that is
why they absorbed light from visible region and undergoes d-d
transition and radiate complementary colour .
(iii) Why are Cr+ , Ag+ & Sc+3 Colour less ?
Ans It is because they do not have unpaired electrons ,therefore can not under goes d-d
transition .
94
(iv)
(v)
Why do Zr and Hf show similar properties ?
Ans It is due to similar ionic size which due to lanthanide contraction .
Why is Cu+1 diamagnetic and Cu +2 paramagnetic ?
It is because Cu+ does not have unpaired electrons where as Cu+2 has unpaired
electrons in d-sub cell .
02 Marks
(i)
Why are transition metal fluorides are ionic in nature where as chlorides and
bromides are covalent in nature .
Ans - It is because of fluorine is most electronegative where as Cl and Br are
less electro negative .
Greater the difference in electro negativity more will be ionic character .
(ii)
Why Mn+2 is more stable than Mn+3 but Fe +3 is more stable than Fe+2 ?
Ans Mn+2 has half filled d- orbital ,therefore it is more stable than Mn+3
Fe+3 has half filled d- orbital, therefore it is more stable than Fe +2 .
(iii)
Ans
light
Why are lanthanoid colored and paramagnetic L
It is due to presence of unpaired electrons, they are paramagnetic and they absorb
From visible region and undergoes f-f transition and radiate complementary
colour .
03 Marks
(i)
What is meant by disproportionation ? Give two examples of above reaction
in aqn. Solution .
Ans Disproportionation means the same substance get oxidized and
reduced, examples
Cu+  Cu+2 + Cu
3CrO4-3 + 8H+ 2CrO4-2 + Cr+3 + 4H20
(ii)
Explain the following , giving reason .
(a)
Transition metal and many of their compounds show paramagnetic behavior
(b) The enthalpy of atomisation of the transition metal is high .
(c)
The Transition metal generally form coloured compounds .
Ans(a) It is due to unpaired electrons .
(b) It is due to strong metallic bond and additional Covalent bonds due to presence of
unpaired electron in d- orbitals
(c) It is due to presence of unpaired electrons ,they undergo d-d transition .
Above Average : 01 mark
(i)Why is K2Cr2O7 generally preferred over Na2Cr2O7 in a volumetric
analysis .Give reason .
Ans : It is because Na2Cr2O7 is hygroscopic in nature that is why it absorbs moisture from
atmosphere , therefore its standard solution can not be prepared .
(ii) Why do transition metals have higher I.E than s –block elements ?
Ans: It is due to smaller atomic size than s-block elements , effective nuclear charge is more
, terefore I.E is high.
(iii) Why are Ti+2 and other transition metals and their ions paramagnetic ?
95
Ans: These ions have unpaired electrons in d- subshell.
(iv) Complete the following eqn.
TiCl4+H2O -----------+---------Ans : TiO2+4HCl
(v) Which trivalent cation is largest in lanthanoid series ?
Ans La+3
Marks 02
(1) Write an equation in ionic form to represent oxidizing action of Cr2 O72 ion in acidic
Medium and draw its structure . Draw structure of CrO4 -2 also .
Ans- ionic equation 6 e- + 14 H+ + Cr207-22Cr+3 +7H2O
(2) Co(ii) is stable in aquous soln. but in presence of strong ligands and in air get
Oxidised to CO +3 , Why ?
Ans Stronger ligands and air Oxidised Co +2 To Co +3 by providing energy released due to
force of attraction
Between them and metal ions .
(3)What happen when acidified KMnO4 is added to ferrous sulphate soln.?
Ans MnO4-1 + 5 Fe +2 + 8H+  Mn+2 +5Fe+3 +4H2O
The colour of KMnO4 becomes coloursless and Fe+2 is oxidized to Fe+3
Above Average 03 Marks
(1) Compare the chemistry of actinoids with that of lanthanoids with reference to (i)
electronic
configuration
(ii) Oxidation state (iii) Chemical reactivity .
Ans : Electronic config. – In lanthanoids 4f orbitals are progressively filled whereas in
actinoids
5f orbitals are progressively filled up .
(ii)
Oxidation state lanthanoids Show +3 oxidation state & same elements of this series
show
+2 & +4 oxidation state also .
Actinoids show +3 , +4, +5,+7 O.S. Although +3 , +4are the most common .
(iii)
Chemical reactivity :- Actinoids are more reactive then lanthanoids due to bigger
atomic
size and lower I.E
************************
96
UNIT -09
CO-ORDINATION
COMPOUND
STUDY MATERIAL
1 MARK & 2 MARKS QUESTIONS
Q. 1. A cationic complex has two isomers A & B. Each has one Co3+, five NH3, one Br
and one SO42. A gives a white precipitate with BaCl2 solution while B gives a
yellow precipitate with AgNO3 solution.
Ans.
(a)
What are the possible structures of the complexes A and B ?
(a)
Will the two complexes have same colour ?
(a)
[CO (NH3)5 Br] SO4 and [CO (NH3)5 SO4] Br
(b)
NO
Q. 2. FeSO4 solution mixed with (NH4)2SO4 solution in 1 : 1 molar ratio gives the test of
Fe2+ ion but CuSO4 solution mixed with aqueous ammonia in 1 : 4 molar ratio
does not give the test of Cu2+ ion. Explain why ?
Ans.
When FeSO4 and (NH4)2SO4 solution are mixed in 1 : 1 molar ratio, a double salt is
formed. It has the formula FeSO4 (NH4)2SO4 . 6 H2O. In aqueous solution, the salt
dissociates.
When CuSO4 and NH3 are mixed in the molar ratio of 1 : 4 in solution, a complex
[Cu (NH3)4] SO4 is formed.
Q. 3. If to an aqueous solution of CuSO4 in two tubes, we add ammonia solution in one
tube and HCl (aq) to the other tube, how the colour of the solutions will change ?
Explain with the help of reaction.
Ans.
In first case, colour will change from blue to deep blue.
[Cu (H2O)4]2+ + 4 NH3 ———
) ]2+ + 4 H2O
3 4
deep blue
While in second case, its colour will change to yellow.
[Cu (H2O)4]2+ + 4 Cl– ———
]2+ + 4 H2O
4
yellow
Q. 4. A, B and C are three complexes of Chromioum with the empirical formula
H12O6Cl3Cr. All the three complexes have Cl and H2O molecules as the ligands.
Complex A does not react with conc. H2SO4. Complexes B and C lose 6.75% and
97
13.5% of their original weight respectively on heating with conc. H 2SO4. Identify
A, B and C.
Ans.
Data suggests that the complexes are hydrate isomers.
As comples A does not lose any molecule of H2O on heating which shows that no
water molecule of H2O is outside the co-ordination sphere.
2
O]0] Cl3
= × 266.5 = 18 au
B = [Cr (H2O)5 Cl] Cl2 . H2O
isomer exists as a dihydrate :
[Cr (H2O)4 Cl2] Cl . 2 H2O
SCN–
F–
Q. 5. Fe3+ ———
———
excess
excess
What are (A) and (B) ? Give IUPAC name of (A). Find the spin only magnetic
moment of (B).
Ans.
A = Fe (SCN)3, B = [FeF6]3–
IUPAC name of A = trithiocyanato iron (III)
E. C. of Fe (III) = d5, unpaired e–s = 5
Spin only magnetic moment =
5(5  2) B. M.
= 5.916 B. M.
Q. 6. A complex is prepared by mixing COCl3 and NH3 in the molar ratio of 1 : 4, 0.1
M solution of this complex was found to freeze at – 0.372 °C. What is the formula
of the complex ? Given that molal depression constant of water Kf = 1.86 °C/m.
Ans.
The oretical
f
= Kf . m
= 1.86 × 0.1
= 0.186°
98
= 0.372°
f
f
is double of the theoretical value this shows that each molecule of
[CO (NH3)4 Cl2] Cl
Q. 7. How t2g and eg orbitals are formed in an octahedral complex ?
Ans.
In an octahedral complex, positive metal ion is considered to be present at the centre
and negative ligands at the corners. As lobes of dx² – y² and dz² lie along the axes, i.
e. along the ligands repulsions are more and so is the energy. The lobes of the
remaining three d-orbitals lie between the axes i. e. between the ligands, the
repulsions between them are less and so is the energy.
Q. 8. Dimethyl glyoxime is added to alcoholic solution of NiCl2. When ammonium
hydroxide is slowly added to it, a rosy red precipitate of a complex appears.
Ans.
(a)
Give the str. of the complex showing hydrogen bond.
(b)
Give oxidation state and hybridisation of central metal ion.
(c)
(a)
Identify whether it is paramagnetic or diamagnetic.
(b)
O. S. = + 2
bybridisation = dsp²
(c)
diamagnetic as no unpaired electron.
Q. 9. Explain the reason behind a colour of some gem stone with the help of example.
Ans.
The colours of many gem stones are due to the presence of transition metal ions &
colour are produced due to d-d transition. For example the mineral corundum Al2O3 is
colourless when pure but when various M3+ transition metal ions are present in trace
amounts various gem stones are formed. Ruby is Al2O3 containing about 0.5 – 1%
Cr3+.
99
Q. 10. How many EDTA (lethylendiamine tetra acetic acid) molecules are required to
make an octahedral complex with a Ca2+ ion.
Ans.
EDTA is a hexadentate ligand therefore only one EDTA molecule is required to form
octahedral complex.
Q. 11. What is the hybridisation of central metal ion and shape of Wilkinson’s catalyst
?
Wilkinson’s catalyst is (PH3P)3 RhCl. In this Rh has dsp² hybridisation and square
Ans.
planar shape.
Q. 12. Which vitamin is a complex compound of cobalt ?
Ans.
Vitamin B12 is a complex compound in which central metal atom is cobalt.
Q. 13. Write the IUPAC name of [CO (NH3)4 B12]2 [ZnCl4].
Ans.
Tetraamminedibromocobalt (III) tetrachlorozincate (II)
Q. 14. What is the value of x in the complex HxCO (CO)4 on the basis of EAN rule. (At.
No. Co` = 27)
x = 36 – (27 + 4 × 2)
Ans.
= 36 – 35
=1
Q. 15. Why is the silver plating of copper, K [Ag (CN)2] is used instead of AgNO3 ?
Ans.
This is because if AgNO3 is used Cu will displace Ag+ from AgNO3. The deposit so
obtained is black, soft, non-adhering. To get a good shining deposit, [Ag (CN)2]– are
used as it is a stable complex, the conc. of Ag+ is very small in the solution. As such
no displacement of Ag+ ions with Cu is possible.
Short answer type question
Q. 1
Write the formula of the following compound.
(a) Tetraamineaquachlorido Cobalt (III) chloride .
(b) Potassium tetrahydrooxozincate (II)
(c)Potassium trioxalatoaluminate (II)
(d) Pentaaminecarbonato cobalt (III) chloride
Ans- (a) Co (NH3)4 (H2O) a} Cl2 (b) K2 [Zn (OH) 4]
(c) K3 (Al (C2O4)3) (d) [Co (NH3)3 (CO) 3] Cl
Q2
Write the IUPAC name of the following Co-Ordination compound.
(a)[ Pt (NH3)2 Cl (NO2) Cl(NO2)] (b) K3 [Cr(C2O4)3]
(c)[Co(NH3)5CO3 ] Cl (d) [ Co Cl2 (en)2] Cl
Ansa)
b)
c)
d)
Dammie chloridonitrito –N-Platinum(II)
Pottassium trioxalato chromate (III)
Pentaamine carbonato cobalt (III) chloride
Dichloroidobis (ethane -1, 2 –diamine) cobalt (III) chloride
100
Q.3
[NiCl4]2 is paramagnetic while [Ni (CO) 4] is diamagnetic through both one
tetrahedral Why?
(i) In [ Ni (CO) 4 ] Ni is in Zero Oxidation state but [NiCl4]2 it is in +2 oxidation state
(ii) In the presence of Co ligand , the unpaired d electronics of Ni pair up but Cl being a
weak legand is
unable to pair up the unpaired electron
Q4
Explain [Co (NH3) 6]3+ is an inner oriental complex when [Ni (NH3)6]2+ is an outer
oriental
Complex.
Ans : In [Co (NH3) 3+ Complex d2 SP3 hybridisation involved forming inner d orbital
Where as In Ni (NH3) 6 complex involved in SP3d2
Hybridiasation forming outer d –orbitals
Q5
What is meant by unidentate , didentate and ambidentate ligands . give two
example of each
Ans:
When a legend is bound to a metal ion through a single diner atom, is said to be
unidentate for ex- Cl-, H2O & NH3
When a legande can bound through two donor atoms as in H2N CH2 CH2NH2, or
C2 O4 (oxalate) is said to be didentate legend. When several donor atoms
are present in a single ligand, the ligand is said to be polydentate ligand which can
ligate through two different atoms is called ambidenate ligand Ex-NO2, SCN
Q6.
Ans:
What are difference between double salt and a complex salt.
(1) both double salts as well as complex are formed by the combination of two or
more stable compounds in stiochometric ratio
(2) They differ in the fact that double salts such as carnallite –KCl. MgCl2. 6H20
, Mohr’s salt FeSO4 (NH4) 2 SO4 . 6 H20 & Potash alum K2SO4 Al2(SO4) 3
.12H2O etc dissociatic into simple ions when dissolve in water .
However when complex ions such as [Fe (CN) 6]4 of
K4 Fe (CN) 6 do not dissociate into Fe2+ and CN- ions
Q. 7.
Ans:
Discuss the nature of bonding in metal carbonyls
The homoleptic carbonyls are formed by most of the transition metal. These
Carbonyls. Show Simple tetrahedral , trigonal bypyramidal and octahydral
Structure .
It is Shown on poge no. 255 of unit 9 NCERT Vol –I
Q. 8.
Ans –
Mention any four characteristics of Co- Ordination Compound
(1) Hard ness of water is estimated by simple titration with Na2 EDTA.
(2) Extraction of metals like Ag and Au make use of complex formation .
(3) Purification of metals by the formation of Co- Ordination compounds
(4) Biological application like pigment formation photosynthesis, Chloropyll.
Haemoglobin pigment of blood .
List various types of isomerism possible for Co- ordination compounds .
given an example .
Q9.
Ans
A)Stereoisomerism
(a) Geometrical isomerism – Cis & trans of [ Pt (NH3)2cl2]
(b) Optical isomerism – d & l of [Co (en)3]3+
(B) Structural isomerism
(a) Linkage isomerism [ Co (NH3) 5 (NO2)]Cl2
(b) Co –ordination isomerism –[ Co(NH3)6] [Cr(CN)6]
(c) Ionisation [Co(NH3)5 SO4] Br & [Co(NH3) 5 Br ] SO4
(d)Solvate isomerism – [Cr (H20) 6 ] Cl3 .
101
Q10 (Fe ( H2O)6) 3 + is strongly paramagnetic whereas [ Fe(CN )6 ]3 is weakly
paramagnetic . Explain .
In [Fe (CN) 6 ]3 only one electron is unpaired & is strong ligand forms d2 sp3
hbridisation shows weekly paramagnetic where as in presence of H2O in
(Fe(H20)6)3+ weak ligand and five electron are unpaired and forms SP3 d2
hybridization shows strongly panamagnetic .
Q11
Calculate the ovelall complex dissociation equilibrium constant for cu
(NH3)4 2+ ion , given that β4 for this complex is 2.1x10 13
Ans : The over all dissociation constant is the reciprocal of over all stability
constant i.e
1 β/4 = 4.7X 10 -14.
01 Marks Question (For Below Average )
1. Define ligands
2. What do yoy mean by complex compounds
3. Explain Co-ordination number.
4. Calculate Co-ordination no. of [Co(NH3)4 (Cl) (NO2) ] cl complex
5. What do you mean by oxidation no/state of co-ordination compounds .?
6. Write the example of linkage ammonium
7. Write two use / application of Co-ordination compounds .
8. Write example of chelating lingads
9. What do you mean by chelate effect?
10. Write the IUPAC name of [Cu (NH3)5Cl] Cl2 complex
Marks Question (For Below Averege )
1. How will you distinwich Homoleptic & hetroleptic complex
2. What are the limitation of VBT
3. What are the merits of CFT
4. Explain t2g and eg in the field of CFT Splitting.
5. What is meant by stability at a Co-ordination Compound in solution? State the
factor on which stability Of the complex depends.
6. What is spectrochemical serious? Explain the difference between strong field ligand
weak field ligand.
7. Give reason why FeSO4 Soln mixed with (NH4) 2SO4 soln in 1:1 molar ratio given
the test of Fe2+ ion but cuso4 soln mixed with aqueas ammonia in 1:4 molar ratio
does not given the test of Cu2+ ion
8. out of given two Co-ordination compound which is Chiral & why explain
102
(a) cis – [CrCl2 (ox)2]3- (b) [Cr Cl2(ox)2]39. Give evidence that
[Cu (NH3)5Cl SO4 and [Co (NH3)5 SO4]Cl are ionization isomerism .
AVERAGE & ABOVE AVERAGE STUDENTS
01 Marks question
1. [NiCl4]2- paramagnetic which [ Ni(C04)] is diamaqgnetic why .
2. Predict the no of unfaired electron in [Pt (CN)4]23. Predict the oxidation no of cobalt in K [ Co (CO)4]
4. Explain magnetic moment of the complex
5. Calculate Co- ordination no. of the complex
cis –[ Cr (en )2Cl2] Cl
6. Discuss role of Co-ordination compound in the field of medicinal chemistry
7 Discuss nature of bonding in (Ni(CO))4
8. How many Co-ordination numbers are there in ethylenediamine ?
9. Name two factors that fever a metal ions forming complex .
10. What is meant by the denticity of a ligand .
02 Marks (Bright achiever)
1. Explain giving one examples role of Co-ordination Compound in biological system
& Analytical system
2. (A) How many ions are present / formed from the complex [CuNH3)6] Cl2
(b) Amongst the following which one has the highest magnetic moment value?
3. [Fe (CN6] 4 – and [Fe (H20) 6] 2 are of different colour in dilute solution. why?
(b) Draw figure to show the splitting at d-orbital in an octahedral crystal field.
4. Explain Aqueous copper Sulphate Solution gives
(a) A green ppt with aqueous potassium fluoride.
(b) A bright green solution with aqueous potassium chloride
5. A Co- ordination compound has the formula CoCl3 4NH3. it does not liberate ammonia
but precipitate chloride ion as ACL. Give the IUPAC name of the complex and write
its structural formula.
6. Explain why chelating complex is more stable than unchealated complex
7. Name two complexes which are used in medicines.
103
***********
UNIT
-10
HALO ALKANE AND HALO
ARENE
FOR AVERAGE
01 mark questions
Q.1. Write down the IUPAC name of the following organic compounds: (a) CH3CHCl2
(b) CH3CH2CH2CH(C(CH3)3)CH(I)CH2CH3
(c)
H5C2
Cl
Ans: - (a) 1,1- Dichloroethane
(b) 3-Iodo – 4 – (1,1 – dimethyl ethyl ) heptane
(c) 1- Chloro – 4 – ethyl cyclo hexane
Q.2.
Ans:-
Write down the structures of the following organic compounds
(a) 1- Bromo – 4 – sec. butyl – 2 – methyl benzene
(b) 2 – Chloro – 3 – methyl pentane
(c ) Vinyl chloride
(a)
C2H5
CH
Br
CH3
CH3
(b) CH3CH2CH(CH3)CH(Cl)CH3
( c) CH2= CHCl
Q.3.Write down the structures of the following organic compounds: (a) Allyl Chloride
(b) Teflon
( c) D.D.T.
(d) P.V.C.
Ans: - (a) CH2=CH-CH2Cl
( b) [-CF2-CF2-]n
Cl
(c ) CCl3CH
Cl
(d) [-CH2 – CH - ]n
104
Cl
Q.4.Which one of the following has the highest dipole moment, and
why?
(a) CH2Cl2
(b) CHCl3
( c) CCl4
Ans:- CH2Cl2 has the highest dipole moment since both the Cl- atoms are
present on one side (on the head) of c – atom and therefore cause a maximum
dipole moment. In CHCl3 and CCl4, two Cl – atoms and four Cl – atoms cancel
out their dipole moments.
Q.5. What happens when
a) Methyl Chloride is treated with KCN
b) ChloroBenzene is subjected to hydrolysis
c) Propene is treated with Cl2 in the presence of U.V. light OR is heated.
d) Chlorobenzene is treated with acetyl chloride in presence
e) of anhyd. AlCl3
f) Chloroform is slowly oxidized by air in presence of light.
Ans:a) CH3 – Cl + KCN
CH3 – CN + Kcl
Methyl cyanide
b)
Cl + HOH H+
OH + HCL
Phenol
U.V. Light or
( c) CH3 – CH = CH2 + Cl2
CH2 – CH= CH2 + HCL
Heat
Cl
Allyl Chloride
O
Cl
Cl
(d)
anhyd.
COCH3
Cl + CH3-C-Cl
+
AlCl3
COCH3
Air
(e) CHCl3 +
1
O2
2
COCl2 + HCl
FOR AVERAGE
2 marks questions
105
Q.1. Which are the possible mono chloro structural isomers expected to be
formed
on free radical mono chlorination of (CH3)2CHCH2CH3.
Ans:- There are four possible product as follows
CH2(Cl) CH – CH2 – CH3,
CH3
(CH3)2 C – CH2 – CH3,
CH3
(CH3)2 CH – CH – CH3 and
Cl
(CH3)2 CH – CH2 – CH2Cl
Q.2. Explain the following reactions :
(a) Sandmeyer’s reaction
(b) Elimination reaction
Q.3. Arrange the compounds in increasing order of their boiling pts.
(a) CH3CH2CH2CH2Br, CH3CH2CHBrCH3 , (CH3)3C Br
(b) CH3Br, CH2Br2, CHBr3
Ans:
(a) (CH3)3C-Br< CH3CH2CHBrCH3< CH3CH2CH2CH2Br
Boiling point increases.
Boiling point decreasing on increasing the branching
(b)
CH3Br< CH2Br2< CHBr3
Boiling point increases
Boiling point increases due increasing molecular mass.
Q.4. Write the mechanism of the following reaction:
Frideal Craft Acylation ( in Chlorobenzene)
Ans:Reaction
Cl
Cl
106
Cl
Anhy.
COCH3
+
+ CH3COCl
AlCl3
COCH3
Reaction Mechanism:
O
Cl
CH3C+ +
CH3-CO—Cl + AlCl3
-AlCl4Cl
Cl
+
CH3CO
-
H
CH3CO
+AlCl4
- AlCl3
+
HCl
Q.5. Explain why: (a) H2SO4 cannot be used along with KI in the conversion of an
alcohol to an alkyl halide.
(b) Alkyl halide though polar are immiscible with water.
Ans: (a)
(b)
H2SO4 converts KI to corresponding HI and then oxidise it into
iodine.
When halo alkane interacts with water molecule , less amount of
energy is released which is not sufficient to break the original Hbond between water molecule and to form new H-bond with halo
alkane and water.
For Average
3 marks questions
Q.1. Identify A,B,C,D,E and R in the following chemical reaction .
Organic Peroxide Aq. KOH
KMnO4
a) CH3CH=CH2 + HBr
b)
CH3
A
B
C
CH3
Na/Dry ether
CH3
CH3
CH3
Alk.
CH3
107
Mg
RX
D
H2O
E
Ans:(a) A
B
C
CH3CH2CH2Br
CH3CH2CH2OH
CH3CH2COOH
(b) R
(CH3)3CD
(CH3)3CMgX
E
(CH3)3CH
Q.2. Write down the IUPAC name of the following organic compounds:
(a)
Br
Br
b) O- Br- C6H4CH(CH3)CH2CH3
c) (CH3)3CCH=CHC6H4I-p
Ans:a) 4,4-Dibromo pentene
b) 2-methyl- 2-(o-bromophenyl) butane
c) 3,3-Dimethyl-1-(p-iodophenyl) butane
Q.3. How can the following interconversions are carried out
(a) Ethanol to but-1-yne
(b) Benzene to 4-bromo nitro benzene
( c) Toluene to benzyl alcohol
Ans:-
Conc. H2SO4
1. Br2
a) CH3CH2OH
CH2=CH2
0
160-170 C
2.Alc.KOH
CH= CH
NaN
H2
C2H5Cl
CH3CH2C= CH
CH=
CNa
NO2
b)
Conc.HNO3
NO2
Br2/ CCl4
Br
CH3
c)
CH2Cl
U.V.Light
108
Q.4.Identify the following pair of structures :
a)
CH3
HO
H
H
CH3
HO
H
Cl
H
Cl
C6H5
b)
C6H5
Br
H
CH3
H
Br
CH3 CH3
CH3
Ans:a) Dia stereomers
b) Structural isomers
Q.5. What is the formula of tartaric acid? Does it show optical isomerism or not.
Draw the structures of its optical isomers.
Ans: Tartaric acid exists in three optical isomeric form.
COOH
COOH
HO
COOH
H
H OH
H
OH
H
OH HO
H H
OH
COOH
COOH
COOH
l-T.A
Meso-T.A.
d-T.A.
FOR ABOVE AVERAGE
MARKS QUESTION:109
Q1. Write down the IUPAC name of the following organic compound.
(a)CH3 CH=C (Cl) CH2 CH (CH3)2 (b)CH3CH Br CH Cl CH3
(c) CH3 C (Cl) (C2H5) CH2 CH3 (d) CH3 C(P-Cl-C6H4)2 CH (Br)CH3
Ans:- (a) 3-chloro-5-methyle hex-2-ene
(b) 2-Choloro-2-ethyl butane.
(C) 2-Bromo-3-chloro butane.
(d) 3-Bromo-2,2-bis(p-chloro phenyl) butane
Q2. Write down the structure of the following organic compounds.
(a) 1-chloro-3-ethyl cyclohexane.
(b) P-bromo chloro benzene.
(c) Ethylene di chloride.
(d) Ethylidene di chloride.
(e) Benzoyl chloride.
Ans:- (a)
(b)
CH2-Cl
CH2-Cl
Cl
C2H5
Cl
(c)Br
Cl
(d) CH3-CH
Cl
COCl
(e)
FOR ABOVE AVERAGE
02 Marks
Q1. Write the structure of the major organic product in each of the following
reaction.
(a)
Br
NaSH


CH2Br
110
Cl
Heat
 

+ C2H5ONa ethanol
(b)
O2N
NO2
Ans:-
Br
OC2H5
(b)
(a)
CH2SH
NO2
O2N
Q2. Give the Preparation and uses of chloropicrin chlorotone.
Q3. Explain the formation of two products in the following reaction.
CH3Ch2Cl + H2O
CH3CH=CH CH2OH + CH3 Ch(OH)CH=CH2
Q4. (a) Write all Possible isomers of C7H7Cl
(b) Write the Structural formula of optical isomer.
Of the component having molecular formula C4H9Br.
Ans:- (a) 4 isomers
(b)
H
CH3-C-C2H5
Br
Q5.What mass of propene is obtained from 34.0 gm of 1-iosopropane on
treating with ethanolic KOH if yield is 36%
Ans:-3.02 gm.
Q6. Arrange the following compound according to reactivity towards
nucleophilic substitution reaction with reagent mentioned.
4 – nitro chlorobenzene, 2, 4 – dinitro chloro benzene, and 2, 4, 6 – trinitro
chloro benzene with Ch3ONa
For Above Average
111
03 Marks
Q1. An Organic compound ‘A’ having molecular formula C4H8 on treatment
with dil H2SO4 gives ‘B’ – ‘B’ on treatment with ione Hcl and anhydrous
Zncl2 gives ‘C’ and on treatment with sodium ethoxide gives back ‘A’ . Identify
the compound ‘A’ , ‘B’ and ‘C’ and write equation involved.
Ans:
A
CH3 − C = CH2
|
CH3
B
CH3 − C(OH) − CH3
|
CH3
C
CH3 − C(Cl) − CH3
|
CH3
Q2. Although chlorine is an electron – withdrawing group, yet it is ortho, paradirecting in
electrophilic aromatic substitution reactions why ?
Ans: Chlorin because of its –I- effect withdraws electron from the benzene ring
and hence tends
to destabilize the intermediate carbocation formed during the electrophilic
substitution.
Q3.Predict the order of reactivity of the following compound in SN1 and SN2
reaction.
(a) The four isomeric bromobutone
(b) C6H5CH2Br, C6H5CH(C6H5)Br, C6H5CH(CH3)Br, C6H5C(CH3)C6H5Br
Ans:-
CH3CH2CH2CH2Br<(CH3)2CH CH2Br<CH3 CH2 CH(Br)CH3<(CH3)3C-Br
Reactivity towards SN1 Reaction
CH3CH2CH2CH2Br>(CH3)2CH CH2Br>CH3 CH2 CH(Br)CH3>(CH3)3C-Br
Reactivity towards SN2 Reaction
112
(b) Reactivity towards SN1 Reaction
C6H5C(CH3)(C6H5)Br>C6H5(CH(C6H5)Br>C6H5CH(CH3)Br>C6H5CH2Br
Reactivity towards SN2 Reaction
C6H5C(CH3)(C6H5)Br<C6H5(CH(C6H5)Br<C6H5CH(CH3)Br<C6H5CH2Br
Q4. Explain the following .
(a) Allyl chloride is hydrolysed more readily than n-propyl chloride.
(b) Vinyl chloride is hyolrolysed more slowly than ethyl chloride
Ans:- (a) Allyl chloride readily undergoes ionization tpo produce sesonance
stabilized allye
carbocation. Science carbocation are rective species, therefore
allyle cotion reading combines with OH ions to form allyl alcohol.
In contrast n-propyl chloride does not
undergo ionization to
produce n- propyl chloride.
a. Vinyl chloride get reacting stabilization Carbon-chlorine bond
acquires some double bond character. In contrast in ethyl chloride,
the carbon-chlorine bond is a pure single bond. This Vinyl chloride
under goes hydrolysis more slowly than ethyl chloride.
UNIT 11
ALCOHOLS, PHENOLS
AND ETHERS
)
113
1 MARK QUESTIONS
Q. 1. What is the main product obtained when vapours of t-butyl alcohol are passed
over copper at 300° ?
Ans.
Isobutylene (2-Methyl propene)
Q. 2. What is usually added to ethyl alcohol to make it unfit for drinking purposes ?
Ans.
Methanol and Pyridine.
Q. 3. Phenol has smaller dipole moment than methanol.
(OR)
Why are dipole moments of phenols smaller than dipole moments of alcohols ?
Ans.
Due to electron-withdrawing effect of the benzene ring, the C — O bond in phenol is
less polar but in case of methanol due to electron-donating effect of — CH3 group, C
— O bond is more polar.
Q. 4. Name the products obtained when anisole is treated with HI.
Ans.
Phenol and methyl iodide.
Q. 5. Why are Grignard reagents soluble in ether but not in benzene ?
Ans.
Grignard reagents from co-ordination complexes with ether but not benzene since the
former has lone pair of electrons but the later does not.
Q. 6. Alcohols are easily protonated than phenols. Justify.
Ans.
In phenols lone pair of electrons on the oxygen atom are delocalised over the benzene
ring due to resonance and hence are not easily available for protonation. In contrast in
alcohols, the lone pairs of electrons on the oxygen atom are localized due to absence
of resonance and hence are easily available for protonation.
Q. 7. Di-tert butyl ether cannot be made by Williamson’s synthesis. Explain why ?
Ans.
To prepare di tert-butyl ether by Williamson’s synthesis, we need tert-butyl bromide
and Sodium tertiary butoxide. Since tert-butyl bromide being 3°-alkyl halide prefers
to undergo elimination rather than substitution, therefore the product obtained is
isobutylene rather than ditertiary butyl ether.
Q. 8. While separating a mixture of ortho and para-nitro phenols by steam distillation,
name the isomer which will be steam volatile. Give reasons.
114
Ans.
In o-nitrophenol, there is intramolecular hydrogen bonding as follows :
In p-nitrophenol, there is intramolecular hydrogen bonding as follows :
Due to intermolecular H bonding in p-nitrophenol, its b. p. is much higher than that of
o-nitrophenol. Hence o-nitraphenol due to its lower b. p. is steam volatile while pnitrophenol is not.
Q. 9. How an — OH group attached to carbon in the benzene ring activates benzene
towards electrophilic substitution ?
Ans.
The lone pair of electrons present on oxygen atom enter into resonance with the
benzene ring. As a result, the electron density becomes higher at o- and p- position
and due to higher electron density, the ring gets activated towards electrophilic
substitution.
Q. 10. Ethers are cleaved by acids not by based. Why ?
115
Ans.
The C — O — C bond in ethers like the C — OH bond in alcohols is quite strong. In
order to weaken it, the oxygen atom must be protonated. A subsequent nucleophile
attack by a strong nucleophile such as Br– as I– ion on the less hindered carbon atom
of the protonated :
+
C—O—C
|
H
bond brings about the cleavage of ethers to form an alcohol and an alkyl halide. The
acids only can provide the H+ ion required for protonation of O atom of ether and
therefore only acids can bring about the cleavage of ethers and not bases.
Q. 11. Phenols do not undergo substitution of the — OH group like alcohols. Explain.
Ans.
The C — O bond in phenols has some double bond character due to resonance and
hence cannot be easily cleaved by a nucleophile. In contrast, the C — O bond in
alcohols is a pure single bond and hence can be easily cleaved by a nucleo phile.
Q. 12. Alcohols acts as weak bases. Explain.
Ans.
The oxygen atom of the hydroxyl group has two lone pairs of electrons. Therefore
alcohols accept a proton from strong mineral acid to form oxonium ions. Hence act as
weak bases.
Q. 13. Write the mechanism of hydration of ethene to yield ethanol.
Ans. H2O + H+ ———
O+
3
Step (i) : — Protonation of alkene to form carbocation by electrophilic attack :
H
—C=C
+
H
H — O+ — H ——— — C — C+
+ H2O
Step (ii) : — Nucleophilic attack of water on carbocation :
H
H
— C — C+
+
H2O ——— — C — C — O+ — H
H
Step (iii) : — Deprotonation to form an alcohol :
H
H
—C—C—O —H
H
+
+
H2 O
OH
——— — C — C —
+
H3O+
Q.1 Why is it that secondary alcohols can only undergo a single oxidation step in contrast to primary
alcohols? (2 marks)
1. Once the oxidation has reached the ketone stage, (primary alcohols on oxidation yield aldehyde where as
secondary alcohols on oxidation yield a ketone), it is impossible to put more oxygen atoms on the relevant
116
carbon atom without rupturing the backbone of the molecule.
Q1. Why phenol is acidic ?
Ans: In phenol dissociation takes place as follows.
OH
O−
+H+
(Phenol)
( Phenoxideion )
The conjugate base of phenol ie. Phenoxide ion is resonance stabilized. This is why
the negative charge on oxygen atom is delocalized through out the ring. So, the oxygen
present in phenoxide ion has less tendency to form undissociated phenol molecule and
equilibrium lies towards right direction.
Q2. While separating a mixture of ortho and para nitrophenols by steam distillation, name
the
isomes which will be steam volatile. Give reason.
117
Ans: O- nitro phenol will be steam volatile because there is intramolecular. H-bonding in its
molecule.
O
OH
N
Intramolecular hydrogen bonding
\\
O
But in case of p- nitro phenol there is association of molecules of pnitro phenol due to intermolecular hydrogen bonding.
O
O
O
O
\\
//
N
OH
N
|
|
|
|
OH
|
N
|
OH
\\
O
O
Q.1 Give the mechanism of formation of alkene from alwhol in preserve of acid catalyst.
H H
| |
H-C-C-O-H
| |
HH
H H
| |
H- C – C - OH2
| |
H H
Ethyl alcohol
(Protonated alcohol)
-H2O
HH
| |
H-C=C-H
-H+
H-C-C-+
| |
| |
H H
H H
(Ethene)
(Carbocation)
Q2. Discuss the mechanism of fornsation of ether from alcohol in presence of aid catalyst.
H H
| |
H-C-C-O-H + H+
H3CCH2 - +OH2
| |
HH
Ethylacohol
(Pronated alcocol)
H H
| |
H-C-C-O-H + H3CCH2-O+H2
| |
HH
HH
| |
| |
+
H-C-C-O -C-C-H
| | | | |
118
HH
H HH
HH
-H+
HH
HH
| |
| |
H-C-C-O-C-C-H
| | | | |
H HH HH
Q3.
Discuss Williamson systhecis.
Ans:- In this method symmetrical and unsymmehical ethers are formed. In this synthesis an
alkyle is allowed to react with sodium alkoxide.
R – X + R’ – O – Na
R – O – R’ + NaX
Ethers contains substitutes alkylgsoups can be prepared by this
method. The reaction involveds SN2 attack of an alkoxide ion on primary alkyl halide.
CH3
|
H3 – C – O-Na+ + CH3 – Br
|
CH3
CH3
|
H3C – O – C-CH3 + NaBr
|
CH3
In case of secondary and tertiany holides elimination completes over substitution. It a
3 alkyl holide is used an alkene is the only reaction product and no ether is formaed.
CH3
|
H3C-C-Br + Na+OCH3
|
CH3
H3C-C=CH2+NaBr+CH3OH
|
CH3
( 2- methyl propene )
Q4. Ortho and paranitrophenols are more acidic than phanel. Draw the resonance structures
of the corresponding phonoxide ion.
Ans:- Electronwithdrawal by nitro group makes the plenoxide ion more resonance stabilized
and so the strength of phenel increases.
SOME MORE IMPORTANT QUESTIONS WITH ANSWERS
BELOW AVERAGE
1 mark questions:Q.1. Give the IUPAC name of CH3O – CH – CH3
CH3
Ans:- 2- Methoxypropane
119
Q.2. Give the IUPAC name of C6H5--O – CH2 – CH2 – CH – CH3
1
2
3
4
CH3
Ans:- 3- Methyl butoxy benzene
1 CH2OH
H3C -- CH – CH2 – CH-- CH - CH3
Q.3. Give the IUPAC name of
6
5
4
3
CH3
2
OH
Ans : - 2,5 – Dimethyl hexane – 1,3 - diol
Q.4 .
Give the IUPAC name of
CH3
CH3 -- C -- OH
CH3
Ans : - 2-Methyl propanol
Q.5. Give the IUPAC name of
OH
Ans:- Benzene – 1,4- diol
OH
Q.6. Write the structures of 1- Ethoxypropane
H
H
H
Ans: -
C2H5O - C - C – C - H
H H
H
Q.7. Write the structure of 2- Ethoxy – 3 – methyl pentane
H
Ans: -
H
CH3
H- C- C– C 120
OC2H5 H
C --
C -H
H H
H
H
H
Q.8. Illustrate Riemer Tiemann reaction with one example
Ans :OH
OH
CHO
KOH, CHCl3
(Phenol)
H+
(2- hydroxy benzaldehyde)
Q.9. Why phenol is acidic ?
Ans:- Due to resonance stabilization of its conjugate base phenoxide ion.
Q.10. Convert anisole to phenol?
OCH3 + HI
OH
Ans:+ CH3 I
(Anisole)
(Phenol)
2 Marks Questions:Q.1. Give the equations of reaction for the preparation of phenol from cumene.
Q.2. Why phenol is more acidic than etanol.
Q.3. Explain the following with an example: (i)
Kolbe’s reaction
(ii) Williamson’s ether synthesis
Q.4. Show how will you synthesise pentan- 1- ol using a suitable alkyl halide?
Q.5. Write chemical reaction for the preparation of phenol from cumene?
Q.6. Write the equation of hydration of ethane to yield ethanol.
Q.7. Convert the following :(i) Ethyl alcohol and Acetic acid
(ii) Propan – 2 – ol to propene
Q.8. Explain esterification with example.
121
Q.9. Convert the following : (i)
Aniline to Phenol
(ii) Phenol to picric acid
Q.10. Alcohols are comparatively more soluble in water than hydrocarbons of
comparable molecular masses. Explain this fact.
3 marks questions
Q.1. The following is not an appropriate reaction for the preparation of t-butyl
ether.
C H3
C H3
C2H5 ONa +
CH3 - C - CH3
CH3 -- C --
OC2H5
(i)
(ii)
CH3
CH3
What would be the major product of this reaction?
Write a suitable reaction for the preparation of t- butyl ether.
Q.2.
(i) Explain why is ortho - nitro phenol more acidic than methoxy
phenol ?
(ii) Why o-nitro phenol is steam volatile while p-nitro phenol has
higher boiling point.
(iii) Give reason for the higher boiling point of ethanol in comparison
to Methoxy methane.
Q.3.
(i) What happens when phenol is treated with FeCl3?
(ii) Distinguish between phenol and benzyl alcohol.
(iii) Explain the coupling reaction with one example.
Q.4. Account for the following : (i) Alcohols act as weak bases.
(ii) Phenol has smaller dipole moment than methanol.
(iii)Phenols do not give protonation reaction easily.
Q.5.
(i) Draw the structures of all isomeric alcohols of molecular formula
C5H12O.
(iii) Classify the isomers of alcohols as Primary, Secondary and
Tertiary alcohols.
122
Q.6. While separating a mixture of ortho and para nitro phenols by
steam
distillation, name the isomers which will be steam volatile. Give
reason.
Q.7. Convert the following: (i) Acetaldehyde to Isopropylalcohol
(ii) Acetone to t - butyl alcohol.
Q.8. Arrange the following set of compounds in order of their increasing
boiling points.
(a) Pentan – 1 – ol, Butan – 1- ol, Butan – 2- ol, ethanol, Propan
–
1- ol,
(b) Pentan -1-ol, n- butane, pentanal, ethoxyethane and
methanol.
Q.9.
(i) What is the Lucas test?
(ii) Distinguish Primary, Secondary and tertiary alcohols with
the
help of Lucas
test.
Q.10.
(i) Why the presence of –OH group attracted to benzene ring
activates the ring towards electrophilic substitution?
AVERAGE
01 mark questions : Q.1. Write down the IUPAC name of
H3C
CH3
OC2H5
Q.2. Write down the IUPAC name of
OH
CH3
CH3
Q. 3. Write down the structure of the product of the following reaction: H3C-CH= CH2
H2O/ H+
………………….
Q.4. Distinguish between
123
OH
COOH
and
Q.5. Distinguish between
O
COCH3
C-C6H5
and
Q6. Write the structure of the product of the following reaction: O
NaBH4
H3C – CH2 - CH- C -H
………………….
CH3
Q.7. Give the mechanism of dehydration of alcohols to alkenes.
Q.8. Explain Williiamson’s synthesis with one example.
Q.9. Alcohols are comparatively more soluble in water than hydrocarbons of
comparable
molecular masses.
2 marks questions
Q.1. How is 1-propoxypropane synthesized form propan – 1- ol. Write
mechanism of this
reaction.
Q.2. Write the equation of the following reactions : (i) Friedel Craft reaction
(ii) Nitration of anisole
Q.3. Explain the following with an example :
(i)
Kolbe reaction
(ii) Reimer Tiemann’s reaction
Q. 4. Convert the following: (i)
Phenol to bezene
(ii) Phenol to benzoquinone
Q.5. Distinguish between the following pair of compounds :
124
CH2OH
OH
(i)
and
(ii)
O
C
and
O
H
C
C- H
H
Q.6. Write the reactions of Williamson’s synthesis of 2- ethoxy- 3methylpentane. Starting from ethanol and 3- methylpentan-2-ol.
Q.7. You are given benzene , Conc.H2SO4 and NaOH. Write the equations for
the
preparation of Phenol using these reagents.
Q.8. Name the reagents used in the following reactions:
(i)
(ii)
Benzyl alcohol and Benzoic Acid
Butan- 2- one to Butan – 2- ol
Q.9. Explain the acidity of phenol in the light of resonance structures.
Q.10. Distinguish primary, secondary and tertiary alcohols using Copper as
catalyst.
3 marks questions
Q.1. (a) When 3- methyl butan – 2 – ol is treated with HBr, then following
reaction takes place
Br
H3C – CH – CH – CH3
HBr
H3C – CH – CH2 –
CH3
H3C OH
H3C
(b) Explain the nitration of anisole
125
Q.2. Write the equations of the following reactions: (i) Bromination of anisole in ethanoic acid medium.
(ii) Friedel-Craft’s acetylation of anisole.
Q.3.
(i)
(ii)
Explain why is ortho - nitro phenol more acidic than o- methoxy
phenol?
Write the mechanism of the reaction of HI with methoxy
methane.
Q.4. How are the following conversions carried out?
(i)
(ii)
(iii)
Q.5.
(i)
or
Ethyl magnesium chloride
Benzyl chloride
Propane
Propan – 1- ol
Benzyl alcohol
Propan – 2- ol
Explain the preparation of ethers by acid dehydration of secondary
tertiary Alcohols is not a suitable method.
(iii)
Explain why propanol has higher boiling point than that of the
hydrocarbon , butane?
ABOVE AVERAGE
01 mark questions
Q.1. What is meant by hydrocarbonation-oxidation reaction?
Illustrate with an example.
Q.2. Ortho and para nitrophenols are more acidic than phenol. Draw the
resonance structures of the corresponding phenoxide ions.
Q.3. Distinguish between anisole and phenol?
Q.4. Preparation of ethers by acid dehydration of secondary or tertiary alcohol
is
not a suitable method. Why?
Q.5. Convert Salicylic acid to Aspirin.
Q.6. Predict the product of the following reaction.
126
OH
CCl4/ OH -H+
Phenols
Q.7. Predict the major product of acidic catalysed dehydration of butan-1-ol.
Q. 8.
CH3
CH3 -- C – O -- CH3 + HI
……………
CH3
Predict the product in the above
reaction
(t-butyl methyl ether )
2 marks questions: -
Q.1. Give the major products that are formed by heating each of the
following ethers unit HI.
(i)
CH2-O
(ii)
CH3
H3C -- CH2 – CH2 – O-- CH – CH2—CH3
CH3
Q.2. Why anisole undergoes bromination with bromine in ethanoic
acid
even in absence of iron (III) bromide catalyst.
Q.3. During dehydration of alcohols to ethers, the method is suitable
for
the preparation of ethers having primary alkyl groups only,
why?
127
Q.4. Explain the fact that in aryl alkyl ethers
(i) The alkoxy group activates the benzene ring towards the
electrophilic substitution.
(iii) It directs the incoming substituents to ortho and para
positions
in benzene ring.
Q.5. Show how could you synthesise the following alcohols from
appropriate alkenes?
CH3
OH
(i)
OH
(ii)
Q.6. Write Structures of the products of the following reactions:
O
(i)
NaBH4
CH2-C-OCH3
O
(ii)
NaBH4
H3C – CH2 - CH-- C HO
………………….
CH3
3 marks questions: -
Q.1.
(a) Show how the following alcohols prepared by the reaction of
a
suitable are Grignard reagent on methanol?
(i) H3C – CH- CH2OH
………………….
128
CH3
(ii)
CH2OH
(c) Give the mechanism of conversion of alkene into alcohol in
presence of acid catalyst.
Q.2.
(a) Which of the following is an appropriate set of reactants for
the preparation of 1- methoxy-4 – nitrobenzene and why?
(i)
(ii)
Br
ONa
+ H3COONa
+ CH3Br
NO2
NO2
c) The bond angle in alcohols is slightly less than the
tetrahedral angle.
Q.3.
(i) Arrange the following compounds in the increasing order of
their acidic strength :
Propan – 1 – ol ; 2,4,6 trinitrophenol; 3 – nitrophenol; 3,5
dinitro-phenol; Phenol; 4-methyl phenol
(ii) Predict the product of the reaction given below:CH3
CH3 -- C -- Br + Na+ŌCH3
………………..
CH3
129
Q.4.
(i)
What is the cause of large difference of B.P.s between ethyl
alcohol, Ethylene glycol and glycerol?
(ii) Of benzene and phenol, which is more easily nitrated and
why?
Q.5. Give the structures and IUPAC names of the products expected
from the following reactions:
(a) Catalytic reduction of butanal.
(b) Hydration of propene in the presence of dilute sulphuric
acid.
(c) Reaction of propanone with methylmagnesium bromide followed
by hydrolysis.
130
UNIT 12TH
ALDEHYDE, KETONS AND
CARBOXYLIC ACIDS
ALDEHYDES, KETONES AND CARBOXYLIC ACIDS
BASIC CONCEPTS:
1. Aldehydes and ketones :These are carbonyl compounds and contain carbonyl (
>C=O) functional group.They have general formula RCHO and RCOR’ respectively.
The carbon atom of the carbonyl group is SP3 Hybridised. The >C=O bond is polar
due to the different electronegativities of the constituent atoms. Aldehydes
(RCHO), ketones(R2C=O) and carboxylic acids (RCOOH) are most important and
widely used compounds in organic chemistry. They play an important role in
biochemical processes of life.These compounds and their derivatives are used in
many food products, pharmaceuticals, for artificial flavouring, as solvents and for
preparing a number of materials.
The structure of carbonyl group is given below
2. PREPARATION OF ALDEHYDES:
(a) By oxidation of alcohols
(b) By Dehydrogenation of alcohols
(c) By ozonolysis of alkenes
131
(d) By hydration of alkynes
(e) Rosenmunds reaction
(f) Treatment of acid chloride with dialkylcadmium gives ketones
(g)From nitriles and esters
This rection is called Stephens reaction.
Nitriles are selectively reduced by isobutylaluminium hydride (DIBAL-H) to imines
followed by hydrolysis to give aldehydes.
Esters are also reduced to aldeydes with DIBAL-H.
Nitriles react with Grignard reagent which on hydrolysis gives ketones.
132
(h) Aromatic aldeydes are prepared from aromatic hydrocarbons by the
following methods:
( a) Etard’s reaction: Toluene reacts with chromyl chloride (CrO 2Cl2) in Cs2 or CCl4 to
form benzaldehyde.
(b) Gatterman- Koch reaction: When a mixture of CO and HCl gas is passed through
benzene in presence of anhydrous AlCl3 + CuCl, benzaldehyde is formed.
(c) Friedel- Craft’s acylation:
133
Physical Properties of Aldehydes and Ketones
(a)
Physical State : Most of the aldehydes ( except formaldehyde which is gas )
are liquid at room temperature. The lower ketones are colourless liquids and
have a pleasant smell. The higher members are colourless solids. Aromatic
ketones are usually solids with a pleasant smell.
(b)
Boiling Points : Aldehydes and ketones have relatively high boiling points as
compared to hydrocarbons of comparable moleculer masses. It is due to the
reason that aldehydes and ketones contain polar carbonyl group and
therefore they have stronger interactions dipole dipole interactions between
the opposite ends of C=O dipoles.
These dipole dipole interactions are however , weaker than intermolecular
H-bonding in alcohols.Consequently boiling points of aldehydes and ketones
are relatively lower than the alcohols of comparable molecular masses.
Dipole- Dipole interaction between two carbonyl groups
(C) Solubility :
The lower members of aldehydes and ketones (upto four carbon atoms) are
soluble in water . It is due to their capability of forming hydrogen bonds with
water molecules . The solubility of these compounds in water decreases with
the increase in the size of alkyl group. It is because of the increase in the
magnitude of non polar parts in the molecule. However higher homologous
are soluble in organic solvents.
134
CHEMICAL PROPERTIES OF ALDEHYDES AND KETONES:
Aldehydes and ketones are highly reactive compounds. Both Aldehydes and ketones
undergo nucleophilic addition reactions.
Explanation: The reactive nature of aldehydes and ketones is because of the presence of a
polar carbonyl group. As the oxygen atom is more electronegative, therefore, it pulls the
electron density around itself acquiring a partial negative charge whereas a partial
positive is developed on the carbon atom. The positively charged carbon atom of carbonyl
is then readily attacked by the nucleophilic species for initiation of the raction. This leads
to the formayion of an intermediate anion which further undergoes the attack of H + ion
or other positively charged species to form the final product. The nucloephilic recation
may be catalysed by acids or bases. The reaction in general , may be represented as:
Aldehydes are generally more reactive than ketones due to the following reasons:
i) Presence of two alkyt groups cause more steric hindrance in the approach of
nucleophile to carbonyl carbon. In aldehydes only one alkyl group is present.
ii) In ketones +I effect of two alkyl groups reduce the positive charge of carbonyl
carbon and reduces its electrophilicity.
SOME IMPORTANT NUCELOPHILIC ADDITION REACTIONS:
a) Addition of hydrogen cyanide( HCN)
Presence of base catalyses the reaction
b) Addition of sodium hydrogensulphite( NaHSO3)
135
c)Addition of alcohols:
d)Addition of ammonia and its derivatives:
136
e) Reduction reactions:
i) Using LiAlH4 or NaBH4 to alcohols
iii)
Reductio
n to Hydrocarbons:
Clemmensen reduction
137
Wolf- Kishner reduction
f) Oxidation :Aldehydes are easily oxidized to carboxylic acids with common oxidizing
agents like KMnO4, K2Cr2O 7, Nitric acid and even with mild oxidizing agents like Tollen’s
reagent and Fehling solution also.
Ketones are generally oxidized under vigorous conditions with strong oxidizing agents.
Oxidation of ketones involve C-C bond cleavage and as a result a mixture of carboxylic
acids are formed.
Tollen’s Test: The test is used to distinguish between aldehydes and ketones. Aldehydes
form silver mirror with ammonical silver nitrate( Tollen’s reagent). Ketones do not react
with it.
Fehling’s test: Fehling reagent= fehling solution ‘A’ (aq CuSO 4) + Fehling solution B( Alk.
Sodium potassium tartarate)
Haloform reaction: Aldehydes and ketones Having atleast one –CH3 Group linked to the
carbonyl carbon give this reaction.
138
Iodoform test: ( using the reagent NaOH+I2)
is used to identify the presence of
group or
in a compound.
g) Condensation reactions:
i) Aldol condensation: This recation is given by those aldehydes and ketones which have
α-hydrogen atom. α-hydrogen of carbonyl compounds is acidic due to strong electron
withdrawing nature of carbonyl group in presence of a base.
Aldols have both -OH group and carbonyl group.
ii) Cross aldol condensation:
139
iv)
Cannizar
o’s reaction:
This is a disproportionation reaction in which aldehyde undergoes self reduction and
oxidation.
h) Electrophilic substitution reaction: When carbonyl group is attached to benzene ring, it
deactivates the ring and it is meta- directing.
Uses of aldehydes and ketones:
i)
Used as
solvents and flavouring agents.
ii)
Formaldehyde is used as formalin( 40% solution) to preserve biological
specimens.
140
iii)
Used as
starting material for the preparation of a number of organic compounds, Dyes,
polymers etc.
CARBOXYLIC ACIDS
(i) PREPARATION:
a)
primary alcohols and aldehydes
From
b)
From
alkylbenzene :
The alkyl chain in alkyl substituted benzene is completely oxidized to –COOH
irrespective of the length of the chain. Primary and secondary alkyl groups also get
oxidized, only tertiary group remains unaffected.
c)
From
nitriles and amides :
d)
From
Grignard reagent:
e)
From
acyl halides and anhydrides :
141
f)
From
esters :
(ii) PHYSICAL PROPERTIES:
a)
Lower
members of carboxylic acids (upto 9-C) are liquids at room temperature, and have
foul odour.
b)
Higher
members are solid and odourless.
c)
Carboxyli
c acids remain associated (dimer) form in vapour state and in aprotic solvents(like
organic solvents).
d)
In protic
solvents like H2O carboxylic acids remain in dissociate form.
e)
The
solubility of carboxylic acids decreases with increasing number of C-atoms as the
non-polar hydrophobic alkyl part gets bigger.
f)
Carboxyli
c acids are soluble organic solvents like benzene, ether, chloroform, etc.
CHEMICAL REACTIONS OF CARBOXYLIC ACIDS:
142
Acidic character of carboxylic acids:- carboxylic acids are weaker than mineral acids but
stronger than alcohols and simple phenols. Carboxylic acids are stronger acids than
phenols because carboxylate ion is more resonance stabilised than phenoxide ion. As in
carboxylate ion resonance structure –ve charge is on electronegative O-atoms while it is
on lesser electronegative C-atoms in phenoxide ion resonance structure.
Alcohols are even less acidic as alkoxide ion shows no resonance.
Carboxylic acids dissociates in water to give carboxylate ion and hydronium ion.
Higher the Pka value less is the acidic strength.
Effect of Substituents on the Acidic Strength of Carboxylic Acids :- Electron withdrawing
group (EWG) increases the acidic strength as it increases the polarity (-I effect) of –O-H
bond facilitating the release of H+ ion.
Electron withdrawing group stabilizes the carboxylate ion by dispersing –ve charge.
Presence of electron donating group(EDG) decreases the acidic strength as it decreases
the polarity of –O-H bond due to its +I effect. Also EDG destablishes the carboxylate ion
by further increase e-density towards the carboxylate ion. The effect of some EWG s is as
follows:
Closer the presence of EWG to the carboxylic group more is the acidic strength.
143
When carboxylic group is attached to vinyl or phenyl group its acidic strength is due to
resonance.
The presence of EWG increases the acidic strength in benzene ring while presence of EDG
decreases acidic strength.
Reactions
(i)
Formatio
n of anhydrides
(ii)
Esterifica
tion
(iii)
Formatio
n of acyl chlorides
Thionyl chloride (SOCl2) method is preferred because both the by-products are gaseous
as a result pure product is obtained.
(iv)
Reaction
with Ammonia
144
(V)
Reduction
(v)
Sodalime
decarboxylation
(vi)
Kolbe’s
electrolysis
(vii)
Hell
Volhard Zelinsky (HVZ) reaction: By this method α- substituted carboxylic acids
can be prepared.
This reaction is given by only those carboxylic acids which have α- H. ‘X’ in α- position of
carboxylic acid can be replaced by any group in its reaction with aq KOH, alc.KCN, etc.
145
(viii)
Aromatic
carboxylic acids undergo electrophilic substitution reaction. The –COOH group
attached to benzene ring is a deactivating group and is m-directing.
Recap: Summary

Aldehydes, ketones and carboxylic acids have carboxyl group and are highly polar
compounds.

Carboxylic acids although contain group but do not give the reactions given by
aldehydes and ketones.

Aldehydes and ketones have higher boiling points than the corresponding
hydrocarbons and ethers due to higher polarity (stronger dipole-dipole interactions)

Carboxylic acids have also higher boiling points due to extensive H-bonding.

Carboxylic acids are stronger acids than phenols and alcohols but weaker than
mineral acids.

Aromatic acids are in general more acidic than aliphatic acids due to the presence
of electron withdrawing benzene ring.









Higher the pKa value of an acid weaker is the acid, i.e., it is poor proton donor.
Ketones are more polar and have higher boiling point than aldehydes of
comparable molecular mass.
Aldehydes are more reactive towards nucleophilic substitution reactions than
ketones.
Aldehydes oxidise to give carboxylic acids.
Aldehydes on reduction give 1º alcohols.
Ketones on reduction give 2º alcohols.
Ketones are oxidised only under drastic conditions and breaking of C—C bond
takes place at the carboxyl group.
Aldol condensation is given by only those aldehydes and ketones which have
Cannizaro’s reaction is given by only those aldehydes which do not have
146




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

Aldehydes and ketones can be converted to corresponding hydrocarbons (having
same number of C-atoms) by Clemenson’s reduction and Wolf-Kishner reduction.
Carboxylic acids on reduction give 1º alcohols.
Sodalime decarboxylation method can be used to reduce a C-atom in carboxylic
acids.
Hell Volhard Zelinsky reaction is used to introduce a substitute like CN, —X, —OH,
etc., at in a carboxylic acid.
Carbonyl group and carboxylic group both are ring deactivating and m-directing
groups when attached to benzene ring.
Aldehydes can be oxidised with mild oxidising agents like Tollen’s reagent and
Fehling’s solution but not ketones.
Aromatic aldehydes are oxidised by Tollen’s reagent but not with Fehling’s
solution.
·Formic acid is the only acid which gives the tests given by aldehyde groups.
1 MARK QUESTIONS
Q. 1. Name the reaction and the reagent used for the conversion of acid chlorides to
the corresponding aldehydes.
Ans.
Name : Rosenmund’s reaction
Reagent : H2 in the presence of Pd (supported over BaSO4) and partially poisoned by
addition of Sulphur or quinoline.
O
||
O
Pd/BaSO4
||
R — C — Cl + H2 ————————
— C — H + HCl
+ S or quinoline
Q. 2. Suggest a reason for the large difference in the boiling points of butanol and
butanal, although they have same solubility in water.
Ans.
The b. pt. of butanol is higher than that of butanal because butanol has strong
intermolecular H-bonding while butanal has weak dipole-dipole interaction. However
both of them form H-bonds with water and hence are soluble.
Q. 3. What type of aldehydes undergo Cannizaro reaction ?
Ans.
Aromatic and aliphatic aldehydes which do not contain
Q. 4. Out of acetophenone and benzophenone, which gives iodoform test ? Write the
reaction involved.
(The compound should have CH3CO-group to show the iodoform test.)
147
Ans.
Acetophenone (C6H5COCH3) contains the grouping (CH3CO attached to carbon) and
hence given iodoform test while benzophenone does not contain this group and hence
does not give iodoform test.
C6H5COCH3 + 3 I2 + 4 NaOH ———
Acetophenane
3
+ C6H5COONa + 3 NaI + 3 H2O
Iodoform
I2/NaOH
C6H5COC6H5 ———
Q. 5. Give Fehling solution test for identification of aldehyde gp (only equations).
Name the aldehyde which does not give Fehling’s soln. test.
Ans.
R — CHO — 2 Cu2+ + 50 H– ———
–
+ Cu2
O
2
Benzaldehyde does not give Fehling soln. test.
(Aromatic aldehydes do not give this test.)
Q. 6. What makes acetic acid a stronger acid than phenol ?
Ans.
Greater resonance stabilization of acetate ion over phenoxide ion.
Q. 7. Why HCOOH does not give HVZ (Hell Volhand Zelinsky) reaction but
CH3COOH does ?
Ans.
CH3COOH contains
-hydrogen and hence does not give HVZ reaction.
Q. 8. During preparation of esters from a carboxylic acid and an alcohol in the
presence of an acid catalyst, water or the ester formed should be removed as
soon as it is formed.
Ans.
The formation of esters from a carboxylic acid and an alcohol in the presence of acid
catalyst in a reversible reaction.
H2SO4
RCOOH + R’OH
Carboxylic acid
RCOOR’ + H2O
alcohol
Ester
To shift the equilibrium in the forward direction, the water or ester formed should be
removed as fast as it is formed.
Q. 9. Arrange the following compounds in increasing order of their acid strength.
Benzoic acid, 4-Nitrobenzoic acid, 3, 4-dinitrobenzoic acid, 4-methoxy benzoic
acid.
Ans.
Since electron donating gps. decrease the acid strength therefore 4-methoxybenzoic
acid is a weaker acid because methoxy sp. is E. D. G. than benzoic acid. Further since
electron withdrawing gps. increase the acid strength, therefore both 4 nitrobenzoic
148
acid and 3, 4-dinitrobenzoic acids are stronger acid than benzoic acid. Further due to
the presence of additional NO2 gp at m-position wrt COOH gt, 3, 4-dinitrobenzoic
acid is a lattice stronger acid than 4-nitrobenzoic acid. Thus the increasing order of
acid strength will be :
4-methoxybenzoic acid < benzoic acid < 4-nitrobenzoic acid < 3, 4, dinitrobenzoic
acid.
Q. 10. How is tert-butyl alcohol obtained from acetone ?
Ans.
Q. 11. Give IUPAC name of the following compound :
Ans.
2-methylcyclopent-3-ene-1-oic acid
Q. 12. How will you distinguish between methanol and ethanol ?
Ans.
By Iodoform test :
Ethanol having
-methyl gp will give yellow ppt. of iodoform whereas methanol
-methyl gp will not give ppt. of iodoform.
Q. 13. Distinguish between :
Ans.
(i)
Acetaldehyde and acetone
(ii)
Methanoic acid and Ethanoic acid.
(i)
Acetaldehyde will give positive tests with Tollen’s reagent and Fehling Solns.
whereas acetone will not give these test.
(ii)
Methanoic acid gives Tollen’s reagent test whereas ethanoic acid does not due
to difference in their boiling points.
Q. 14. Why are aldehydes more reactive than ketones ?
Ans.
It is because of 2 reasons :
The carboxyl compounds (both aldehydes & ketones) undergo nucleophilic addition
reaction.
149
(i)
+ I effect : The alkyl group in ketones due to their e– releasing character
decrease the electrophilicity / + ve charge on c-atom and thus reduce its
reactivity.
(ii)
Steric hindrance : Due to steric hindrance in ketones, they are less reactive.
Q. 15. Give the composition of Fehling A and Fehling B ?
Ans.
Fehling A = aq. CuSO4
Fehling B = alkaline sodium potassium tartarate
(Rochelle Salt)
Q. 16. Name one reagent which can distiguish between 2-pentanone and 3-pentanone ?
Ans.
2-pentanone has a CH3CO-group, hence gives positive iodoform test.
3-pentanone does not have a CH3CO-group, hence does not give positive iodoform
test.
Iodoform test ——— 2/NaOH
O
||
I2
CH3CH2CH2 — C — CH3 + NaOH ———
3
(yellow ppt.)
O
||
CH3 — CH2 — C — CH2 — CH3 ———
Q. 17. Why pcc cannot oxidise methanol to methane and while KMnO4 can ?
Ans.
This is because pcc is a mild oxidising agent and can oxide methanol to methanal
only.
While KMnO4 being strong oxidising agent oxidises it to methanoic acid.
Q. 18. Would you expect benzaldehyde to be more reactive or less reactive in
nucleophlic addition reaction than propanal ? Explain.
Ans.
C-atom of carbonyl group of benzaldehyde is less electrophilic than C-atom of
carbonyl group in propanal. Polarity of carbonyl group is in benzaldehyde reduced
due to resonance making it less reactive in nucleophilic addition reactions.
O
||
C
O
H
–
H
+
There is no such resonance effect in propanal and so the polarity of carboxyl group in
it is more than in benzaldehyde. This makes propanal more reactive than
benzaldehyde.
150
Q. 19. What are Hemiacetal and acetal ?
Ans.
Hemiacetal and acetals are formed by addition of alcohols on carboxyl compounds.
Q. 20. Why does methanal not give aldol condensation while ethanol gives ?
Ans.
This is because only those compounds which have
-
Q. 21. Why does methanal undergoes Cannizaro’s rxn ?
Ans.
Because it
Q. 22. Arrange the following in order of increasing boiling points :
CH3CH2CH2OH, CH3CH2CH2CH3, CH3CH2 — OCH2CH3, CH3CH2CH2CHO
Ans.
CH3CH2CH2CH3 < C2H5OC2H5 < CH3CH2CH2CHO < CH3 (CH2)2 OH
(hydrogen)
(ether)
(aldehyde)
——————————————————
increase in bond polarity.
(alcohol)
Q. 23. Why does solubility decreases with increasing molecular mass in carboxylic
acid ?
Ans.
Because of increase in alkyl chain length which is hydrophobic in nature.
Q. 24. Although phenoxide ion has more no. of resonating structures than carboxylate
ion, carboxylic acid is a stronger acid. Why ?
Ans.
Conjugate base of phenol —
non equivalent resonance structures
in which –ve charge is at less electronegative C-atom and +ve charge is at more
electronegative O-atom.
O
–
+
O
+
O
–
+
O
–
–
151
–
O
In carboxylate ion, – ve charge is delocalised on two electronegative O-atoms hence
resonance is more effective.
O
O
R—C
R—C
O
O
]
O–

R—C
O–
Q. 25. There are two — NH2 group in semicarbazide. However, only one is involved in
formation of semicarbazones. Why ?
Ans.
Although semicarbazide has two — NH2 groups but one of them is involved in
resonance.
O
||
H2N — C — NH2NH2
—
O–
+
|
N = C — NH — NH2
2
—
O–
..
| +
N — C = NH — NH2
2
As a result, e– density on one of the — NH2 group is reduced and hence it does not act
as nucleophile.
— NH2 group is not involved in resonance and is available for
nucleophilic attack.
2 MARKS QUESTIONS
Q. 1. Arrange the following carboxyl compounds in increasing order of their reactivity
in nucleophilic addition reactions. Explain with proper reasoning :
Benzaldehyde. p-tolualdeyde, p-nitrobenzaldehyde, Acetophenone.
Ans.
Acetophenone is a ketone while all others are aldehydes, therefore it is least reactive.
In p-tolualdehyde, there is methyl group (CH3) at para position w.r.t. to the carboxyl
gp, which increases electron density on the carbon of the carboxyl gp by
hyperconjugation effect thereby making it less reactive than benzaldehyde.
H
H—C—
H
—C—H
H—C=
H
H
=C
O–
H
On the other hand, in p-nitrobenzaldehyde, the NO2 gp is a powerfuil electronwithdrawing gp. It withdraws electrons both by inductive and resonance effect
thereby decreasing the electron density on the carbon atom of carboxyl gp. This
facilitates the attack of the nucleophile and hence makes it more reactive than
benzaldehyde.
152
etc.
Therefore, the overall order of increasing reactivity :
acetophenone < p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde.
Q. 2. Arrange the following compounds in increasing order of their boiling points.
Explain by giving reasons.
CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3.
Ans.
The molecular masses of all these compounds are comparable :
CH3CHO (44), CH3CH2OH (46), CH3COCH3 (46), CH3CH2CH3 (44).
CH3CH2OH exists as associated molecule due to extensive intermolecular hydrogen
bonding and hence its boiling point is the highest (351 K). Since dipole-dipole
interaction are stronger in CH3CHO than in CH3OCH3, hence boiling point of
CH3CHO (293 K) is much higher than that of CH3OCH3 (249 K). Further, molecules
of CH3CH2CH3 have only weak Vander Waals forces while the molecules of
CH3OCH3 have little stronger dipole-dipole interactions and hence the boiling point of
CH3OCH3 is higher (249 K) than that of CH3CH2CH3 (231 K). Thus the over all
increasing order of boiling points is :
CH3CH2CH3 < CH3OCH3 < CH3CHO < CH3CH2OH
Q. 3. Which acid of each pair shown here would you expect to be stronger ?
CH3CO2H or FCH2CO2H
Ans.
increases the electron
the electron density in the
lizes the carboxylate
O — H bond thereby
ion by intensifying
making the release of a
the – ve charge.
the
carboxyate
density in the O — H
ion by
dispersing
bond thereby making
charge.
the relase of a proton
proton easier.
difficult.
153
the – ve
Thus due to lesser electron density in the O — H bond and greater stability of
FCH2COO– ion over CH3COO– ion FCH2COOH is a stronger acid than CH3COOH.
Q. 4. Which acid is stronger and why ?
Ans.
F3C —
— COOH
F3C —
—C
or
H3
— COOH
CH3
—C
CF3 has a strong – I effect.
CH3 has a weak + I effect.
It stabilises the carboxylate ion
It stabilises the carboxylate ion
by dispersing the – ve charge.
by intensifying the – ve charge.
Therefore due to greater stability of F3C — C6H4 — COO– (p) ion over CH3 —
C6H4COO– (p) ion, F3C — C6H4 — COOH is a much stronger acid than CH3 — C6H4
— COOH.
Q. 5. Arrange the following compounds in increasing order of their reactivity towards
HCN. Explain it with proper reasoning.
Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone.
Ans.
Addition of HCN to the carboxyl compounds is a nucleophilic addition reaction.
The reactivity towards HCN addition decreases as the + I effect of the alkyl gp/s
increases and/or the steric hindrance to the nucleophilic attack by CN– at the carboxyl
carbon increases. Thus the reactivity decreases in the order.
——————— + I effect increases———————
——————— Steric hindrance increases———————
——————— Reactivity towards HCN addition decreases ———————
In other words, reactivity increases in the reverse order, i. e.,
Ditert-butyl Ketone < tert-Butyl methyl Ketone < Acetone < Acetaldehyde
Q. 6. Write structural formulae and names of four possible aldol condensation
products from propanal and butanal. In each case, indicate which aldehyde acts
as nucleophile and which as electrophile.
154
Ans.
(i) Propanal as nucleophile as well as elecrophile.
OH
CH3
5 4
3|
2|
1
CH3CH2CHO + CH3CH2CHO ——
CH2 — CH — CH — CHO
3
Propanal
Propanal
3-hydroxy-2-methyl pentanal
(ii) Propanal as electrophile and butanal as nucleophile.
OH CH2 — CH3
5 4 3| 2| 1
CH3CH2CHO + CH3CH2CH2CHO ——
CH2CH — CH — CHO
3
Propanal
Butanal
2-ethyl-3-hydroxy pentanal
(Electrophile)
(Nucleophile)
(iii) Butanal as electrophile and propanal as nucleophile.
OH CH3
6 5 4
3|
2|
1
CH3CH2CH2CHO + CH3CH2CHO + ——
CH2CH2 — CH — CH — CHO
3
Butanal
Propanal
3-hydroxy-2-methyl pentanal
(Electrophile)
(Nucleophile)
(iv) Butanal both as nucleophile as well as an elecrophile.
OH CH2CH3
6 5 4
3|
2|
1
CH3CH2CH2CHO + CH3CH2CH2CHO + ——
CH2CH2 — CH — CH — CHO
3
Butanal
Butanal
2-ethyl-3-hydroxy hexanal
(Electrophile)
(Nucleophile)
Q. 7. An organic compound with the molecular formula C9H10O forms 2, 4-DNP
derivative, reduces Tollen’s reagent and undergoes Cannizaro reaction. On
vigorous oxidation, it gives 1, 2-benzenedicarboxylic acid. Identify the
compound.
Ans.
(i)
Since the given compound with M. F. C9H10O forms a 2, 4-DNP derivative and
reduces Tollen’s reagent, it must be an aldehyde.
(ii)
Since it undergoes Cannizaro reaction, therefore CHO gp. is directly attached
to the benzene ring.
(iii)
Since on vigorous oxidation, it gives 1, 2-benzene dicarboxylic acid, therefore
it must be an ortho substituted benzaldehyde. The only o-substituted aromatic
aldehyde having M. F. C9H10O is 2-ethyl benzaldehyde. All the reactions can
now be explained on the basis of this structure.
155
COO–
C2H5
[Ag (NH3)2]+ OH–
———————
Tollen’s reagent
Silver
mirror 2-ethyl benzoate
[O]
————
CHO
C2 H 5
COOH
COOH
2-ethyl benzaldehyde 1, 2-benzene dicarboxylic acid
(M. F. C9H10O)
2, 4-dinitrophenyl hydrozene
NO2
CH = NNH —
— NO2 + H2O
C2H5
2, 4-DNP derivative
Q. 8. Explain why o-hydroxybenzaldehyde is a liquid at room temperature while
p-hydroxybenzaldehyde is a high melting solid.
Ans.
Due to interamolecular H-bonding ortho-hydroxy benzaldehyde exists as discrete
molecule whereas due to intermolecular H-bonding, p-hydroxybenzaldehyde exists as
associated molecules. To break these intermolecular H-bonds, a large amount of
energy is needed. Consequently, p-hydroxybenzaldehyde has a much higher m. pt.
and b. pt. than that of o-hydroxy benzaldehyde. As a result, o-hydroxy benzaldehyde
is a liquid at room temperature while p-hydroxy benzaldehyde is a high melting solid.
Q. 9. Identify A, B and C and give their structures :
O
CH3
COCH3
Br2
H+
———
———
NaOh
I
156
H12O)
17
Ans.
The given compound (I) contains CH3CO gp and hence in the presence of Br2/NaOH
undergoes haloform reaction to give sodium salt of carboxylic acid (A) and
bromoform CHBR3 (B). (A) on protonation gives the corresponding acid (II). (II)
-ketoacid readily undergoes decarboxylation to give 2-methylcylohexanane
(C).
O
CH3
COCH3
Br2/NaOH
——————
Haloform reacn
O
3
CH3
+
H+
———
COO–
(B)
I
(A)
— CO2)
———————
Dexcarboxylation
H+
——————
-keto acid)
2-methyl
cyclo
hexanone
(C) M. F. = C7H12O
1
Write the structure of the following compounds
(i)
4-OXO pentanal
(ii)
2, 4 Dimethyl pent 3-one.
(iii)
3-Methylbutanal
(iv)
4-chloropentane -2-one
(v)
3-brome-4-Phenyl pentanoic acid
(vi)
Para-methyl Benzaldehyde
(vii) 4-Methyl pent-3-ene-2-one
Ans:- (i)
HOHH O
| || | | ||
H-C-C-C-C-C-H
|
| |
H H H
(ii)
CH3 O CH3
|
|| |
H3C-C ----C-C-CH3
|
|
H
H
(iii) CH3-CH-CH2-CHO
157
Ch3
(iv) CH3-CH-CH2-C-CH3
Cl
O
(v) CH3 - CH - CH - CH2 - COOH
C6H5 - Br
(vi) CH3 -
-CHO
(vii) CH3-C-CH=C-CH3
||
|
O
CH3
2
(i)
Write IUPAC Names of the followings:CH3
CHO
(ii)
O
CH3
(iii) (CH3 )2 CHCOCH((CH3 )2
(iv) CH 3 – CH (OCH3) CHO
(v) CH3-CH-CH2CH2-CHO
|
CH3
(vi) PHCOPH
(vii) CH3-CH2-CH-CH2-CH-CHO
|
|
Br
CH3
(viii) C6H5CH=CH-CHO
(ix) (CH3)3-CCH2COOH
Ans:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
3.
Arrange the following compound in increasing order of their boiling points.
(a)
(b)
(c)
(d)
(e)
3- Methyl cyclohexane carbaldehyde
2- Methyl Cyclo hexanone
2,4 Dimethyl Pantan -3- One
2- Methoxy Propanal
4-Methyl pertanal
Diphenylmethanone
4-Bromo-2-Methyl Hexanal.
3-Dhehyl Propernal
3,3-dimethyl Bytasnance acid
CH3-CHO
CH3-CH2-OH
CH3-O-CH3
CH3-CH3
CH4
158
ANS:- CH4 < CH3-CH3<CH3-O-CH3<CH3-CHO<CH3-CH2-OH
4.
Arrange the following compound in the increasing order of their properties as
indicated
(i)
(ii)
Acetaldehyde, Acetone, Di-tert-Butyl Ketone(Reaction towards HCN)
Benzoic acid, 4 –Nitro benzoic acid, 3, 4 Dinitro benzoic acid, 4-methoxy benzoic
acid (acid strength)
Ans:- (1) Di-tert-Butyl < Ketone < Acetone< Acetaldehyde
(2) 4-methoxy Benzoic Acid< Benzoic Acid<4 –Nitro benzoic acid<3, 4
Dinitro benzoic acid
5
Give simple chemical test to distinguish between the following pair of compounds:(i)
Propanal & propanone
(ii)
Benzaldehyde and Acetophenone
(iii) Ethanal & Propanal
(iv)
Acetophenone & Benzophenone
Ans:(i) Propanal & propanone
(ii) Benzaldehyde and Acetophenone
Tollen’s reagent Test
(i) Ethanal & Propanal
(ii) Acetophenone & Benzophenone
By Iodoform Test.
6. How will you distinguish?
(i)
(ii)
Phenol & Benzoic Acid
Benzoic Acid & Ethyl benzoate.
Ans:- By Sodiumbicarbonate test, Benzoic acid gives effervescence.
COOH
COONa
|
+ NaHCO3
+ CO2
+ H2O
Phenol and ethyl benzoate do not give this Test.
7. How will you distinguish the following pairs:(i) Pentan- 2-one and Pentan- 3-one
(ii) Propanol & Propanal
(iii) Methanal & Ethanal
O
O
||
||
Ans:- CH3-CH2-CH2-C-CH3,
CH3-CH2- C-CH2-CH2
Will Give +ve Iodoform test
(iii)
Do not give +ve Iodoform test
Propanol will give sodium metal test.
159
Propanol will give +ve Fehling’s Solution Test
8. Arrange them in the increasing order of reactivity in esterification reaction
i.
ii.
CH3OH, (CH3)3COH, (CH3) 2-CH-OH, CH3CH2OH
(CH3) 3CCOOH, CH3COOH, (CH3) 2-CHCOOH,HCOOH
Ans:- (i) (CH3)3COH<(CH3) 2-CH-OH< CH3CH2OH< CH3OH
(ii) (CH3) 3CCOOH< (CH3) 2-CHCOOH< CH3COOH< HCOOH
9 Arrange the following acid derivatives towards increasing order of nucleophelic
Substituion reaction
(a) RCONH2, RCOOCOR, RCOCI & RCOOR
(b) Acid derivative unlike aldehyde and ketones show mucleophic substitution
Ans:- (a) RCONH2 < RCOOR
AMIDE
ESTER
<R.CO.O.COR
ACID ANHYDRIDE
<RCOCl
ACID CHLORIDE
(b) Acid derivatives although contain >C=O group, yet do not under go the
usual properties of carbonyl groups due to the presence of resonance.
9. What happens:01. When primary alcohol vapours are passed over Cu metal at 573 k
Cu,573K
Ans:- RCH2-OH
RCHO
Corresponding Aldehyde is formed.
02. When secondary alcohols treated with chromic anhydride (CrO3)
CrO3
Ans:-R-CH-R’
R-C-R’
|
[O]
OH
O
Corresponding Ketone is formed.
03. Ethyne treated with H2O in the presence of H2SO4 & HgSO4
H2O
Ans: CH = CH
CH3-CHO
H2SO4-HgSO4
Corresponding Acetaldehyde is formed.
04. Toulene is treated with chromyl chloride (CrO2Cl2) followed by hydrolysis.
Ans:
CH3
CHO
|
(i) CrO2Cl2
|
(ii) H2O
Corresponding Benzaldeiyde is formed.
05. Propanone is treated with HCN
160
Ans:-
CN
|
CH3-C-CH3
|
OH
CH3-C-CH3+HCN
||
O
Corresponding Cyanohydrim of problem is formed.
06. Ethanal is treated with NaHSO3.
Ans:CH3CHO+NaHSO3
CH3-CH-SO3H
|
ONa
CH3-CH-SO3Na
|
OH
Corresponding Sodiumbisulphite of ethanol is formed.
07. Proponal is treated with Methyl magnesium bromide.
Ans:H
|
[CH3CH2-C-OMgBr]
|
CH3
H2O
CH3-CH2CH-OH
|
CH3
CH3-CH2CHO+CH3MgBr
Corresponding Butane 2–0% is formed.
08. 3- Pentanone is treated with KMnO4 at high pressure
Ans:O
||
KMnO4
CH3-CH2CHO-C-CH2CH3
CH3-CH2-COOH+CH3-COOH
High Press
Corresponding A mixene of ethonoic acid and propanoic and propanoic and is formed
09. Ethanal is treated with dil NaOH.
Ans:dilNaOH
2CH3CHO
CH3-CH-CH2-CHO
CH3-CH=CH-CHO
|
-H2O
OH
Aldol:Condensation taken-place giving rise to a mixume of salt of Bemoic and
Benzylalcohal.
10. Benzaldehyde is treated with hot Alkali
Ans:2
CHO
COONa
161
CH2OH
+
Disproportionation rection takes place giving rise to a mixture of salt of
acid and Benzylalcohol.
10. Acetic acid is treated with PCl5.
Ans:PCl5
CH3COOH
CH3COCl
Acetylchloride is formed.
11. Benzoic acid is treated with NH3
Ans:COOH
(i) NH3
(ii)
CONH2
Benzamide is formed.
11. COMPLETE THE FOLLOWINGS
(i)
O
||
C-Cl
H2
CHO
Pd- BaSO4
1. AlH(i-Bu)2
(ii)CH3-CN
CH3CHO
2 H2O
NH2
NC
(iii)
CHCl3 + alc. KOH
(iv)
O
||
COC2H5
+
AlCl3
C2H5-C-Cl
CS2
(v) CH3-C = C-H
Hg2+,H2SO4
H2O
CH3COCH3
(vi)
H
Conc NaOH
2
HCOONa + CH3OH
C =0
H
162
Benzoic
O
||
(vii)
NO2
|
NO2
|
=N-NH--
+ H2N -NH
-NO2
NO2
(viii)
CH2-CH2-CH3
COH
1. KMnO4
KOH
2. H2O
I LiAlH4
(ix) CH3-COOH
CH3CH2OH
2. H3O+
12. Transformation:
i. Ethanol to butane 1,3 diol;
Ans:[O]
C2H5OH
CH3CHO
H
dil NaOH
|
CH3-C-CH2-CHO
|
OH
[H]
CH3-CH-CH2-CH2-OH
|
OH
Butane 1,3,-diol
ii.
Ans:-
3 nitro brome benzene to 3 - nitro benzoic acid
NO2
NO2
NO2
|
|
|
Mg/dryether
CO2
Br
MgBr
NO2
|
C-O-MgBr
COOH
||
O
3- Nitro
Benzoic acid
(iii) Butanal to butanoic acid
Ans:[O]
CH3-CH2-CH2-CHO
CH3-CH2-CH2-COOH
KMnO4
(iv) Acetic acid to Ethan amine
163
Ans:-
NH3
CH3COOH
(i) LiAlH4
CH3CONH2
CH3CH2NH2
(ii) H2O
(v) Bromo benzene to benzoic acid
Ans:Br
Mg
Dry ether
O
MgBr
CO2
(vi) Benzene to m-nitro acetophenone
Ans:NO2
Conc
conc
HNO3-H2SO4
||
COOH
C-OMgBr
H2 O
|
NO2
CH3COCl
AlCl3(Anhy)
COCH3
(vii) Benzoic acid to benzaldehyde.
Ans:-
O
||
COOH
C-Cl
PCl5
CHO
H2
Pd-BaSO4
QUESTION BANK FOR BRIGHT STUDENTS
Q.1
Complete each synthesis by giving missing starting material, reagent or products
(i)
(ii)
(iii)
164
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
Ans: Try yourself
165
Q.2
Predict the products of the following reactions:
(i)
(ii)
(iii)
(iv)
ANS: (i)
(ii)
166
(iii)
(iv)
Q.3
Write the structures of products of the following reactions;
(i)
(ii)
(iii)
(iv)
ANS: (i)
167
(ii)
(iii)
(iv)
168
Q.4
Write structural formulas and names of four possible aldol condensation products from
propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which
as electrophile.
ANS:(i) Taking two molecules of propanal, one which acts as a nucleophile and the other
as an electrophile.
(ii) Taking two molecules of butanal, one which acts as a nucleophile and the other as an
electrophile.
(iii) Taking one molecule each of propanal and butanal in which propanal acts as a
nucleophile and butanal acts as an electrophile.
(iv) Taking one molecule each of propanal and butanal in which propanal acts as an
electrophile and butanal acts as a nucleophile.
Question 5:
An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The
molecular mass of the compound is 86. It does not reduce Tollens’ reagent but forms an
addition compound with sodium hydrogensulphite and give positive iodoform test. On
vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of
the compound.
ANS: % of carbon = 69.77 %
169
% of hydrogen = 11.63 %
% of oxygen = {100 − (69.77 + 11.63)}%
= 18.6 %
Thus, the ratio of the number of carbon, hydrogen, and oxygen atoms in the organic
compound can be given as:
Therefore, the empirical formula of the compound is C 5H10O. Now, the empirical formula
mass of the compound can be given as:
5 × 12 + 10 ×1 + 1 × 16
= 86
Molecular mass of the compound = 86
Therefore, the molecular formula of the compound is given by C5H10O.
Since the given compound does not reduce Tollen’s reagent, it is not an aldehyde. Again,
the compound forms sodium hydrogen sulphate addition products and gives a positive
iodoform test. Since the compound is not an aldehyde, it must be a methyl ketone.
The given compound also gives a mixture of ethanoic acid and propanoic acid.
Hence, the given compound is pentan−2−ol.
The given reactions can be explained by the following equations:
170
Q.5
Arrange the following compounds in increasing order of their property as indicated:
(i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity
towards HCN)
(ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid
strength)
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid
(acid strength)
ANS:(i) When HCN reacts with a compound, the attacking species is a nucleophile, CN−.
Therefore, as the negative charge on the compound increases, its reactivity with HCN
decreases. In the given compounds, the +I effect increases as shown below. It can be
observed that steric hindrance also increases in the same
171
Hence, the given compounds can be arranged according to their increasing reactivities
toward HCN as:
Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde
(ii) After losing a proton, carboxylic acids gain a negative charge as shown:
Now, any group that will help stabilise the negative charge will increase the stability of the
carboxyl ion and as a result, will increase the strength of the acid. Thus, groups having +I
effect will decrease the strength of the acids and groups having −I effect will increase the
strength of the acids. In the given compounds, −CH3 group has +I effect and Br− group has
−I effect. Thus, acids containing Br− are stronger.
Now, the +I effect of isopropyl group is more than that of n-propyl group. Hence,
(CH3)2CHCOOH is a weaker acid than CH3CH2CH2COOH.
Also, the −I effect grows weaker as distance increases. Hence, CH 3CH(Br)CH2COOH is a
weaker acid than CH3CH2CH(Br)COOH.
Hence, the strengths of the given acids increase as:
(CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH
(iii) As we have seen in the previous case, electron-donating groups decrease the
strengths of acids, while electron-withdrawing groups increase the strengths of acids. As
methoxy group is an electron-donating group, 4-methoxybenzoic acid is a weaker acid
than benzoic acid. Nitro group is an electron-withdrawing group and will increase the
strengths of acids. As 3,4-dinitrobenzoic acid contains two nitro groups, it is a slightly
stronger acid than 4-nitrobenzoic acid. Hence, the strengths of the given acids increase as:
4-Methoxybenzoic
acid
< 3,4-Dinitrobenzoic acid
<
Benzoic
acid
<
4-Nitrobenzoic
acid
Q.6
Predict the products formed when cyclohexanecarbaldehyde reacts with following
reagents.
(i) PhMgBr and then H3O+
(ii)Tollens’ reagent
(iii) Semicarbazide and weak acid
(iv)Excess ethanol and acid
(v) Zinc amalgam and dilute hydrochloric acid
172
ANS:(i)
(ii)
(iii)
(iv)
(v)
173
Q.7
How will you prepare the following compounds from benzene? You may use any inorganic
reagent and any organic reagent having not more than one carbon atom
(i) Methyl benzoate (ii) m-Nitrobenzoic acid
(iii) p-Nitrobenzoic acid (iv) Phenylacetic acid
(v) p-Nitrobenzaldehyde.
ANS:(i)
(ii)
174
(iii)
(iv)
(v)
175
Q. 8
Name the reaction and the reagent used for the conversion of acid chlorides to the
corresponding aldehydes.
Ans. Name : Rosenmund’s reaction
Reagent : H2 in the presence of Pd (supported over BaSO4) and partially poisoned
byaddition of Sulphur or quinoline.
Q.9 How is tert-butyl alcohol obtained from acetone?
Ans.
Q. 10. Why pcc cannot oxidise methanol to methane and while KMnO 4 can ?
Ans. This is because pcc is a mild oxidising agent and can oxide methanol to methanal only.
While KMnO4 being strong oxidising agent oxidises it to methanoic acid.
Q. 11. What are Hemiacetal and acetal ?
Ans. Hemiacetal and acetals are formed by addition of alcohols on carboxyl compounds.
Q. 12. Which acid is stronger and why ?
F3C — C6H4— COOH or H3C — C6H4— COOH
Ans.
CF3 has a strong – I effect. CH3 has a weak + I effect.
It stabilises the carboxylate ion It stabilises the carboxylate ion
by dispersing the – ve charge. by intensifying the – ve charge.
Therefore due to greater stability of F3C — C6H4 — COO– (p) ion over CH3 —C6H4COO– (p) ion,
F3C — C6H4 — COOH is a much stronger acid than CH3— C6H4 —COOH.
176
Q. 13. An organic compound with the molecular formula C9H10O forms 2, 4-DNP
derivative, reduces Tollen’s reagent and undergoes Cannizaro reaction. On vigorous
oxidation, it gives 1, 2-benzenedicarboxylic acid. Identify the compound.
Ans. (i) Since the given compound with M. F. C9H10O forms a 2, 4-DNP derivative and
reduces Tollen’s reagent, it must be an aldehyde.
(ii) Since it undergoes Cannizaro reaction, therefore CHO gp. is directly
attached to the benzene ring.
(iii) Since on vigorous oxidation, it gives 1, 2-benzene dicarboxylic acid, therefore it must be
an ortho substituted benzaldehyde. The only o-substituted aromatic aldehyde having M. F.
C9H10O is 2-ethyl benzaldehyde. All the reactions can now be explained on the basis of this
structure.
[Ag (NH3)2]+ OH[O]
Ag + C2H5—C6H4—COO —————-CHO—C6H4—C2H5 ——> COOH—C6H4
Tollen’s reagent 2-ethyl benzaldehyde
|
Silver mirror 2-ethyl benzoate
COOH
1, 2-benzene
acid
FOR BELOW AVERAGE
01 Marks QUESTIONS
1.
Write the structure of the following compounds
(viii)
(ix)
(x)
(iv)
(v)
(vi)
(vii)
2.
4-OXO pentanal
2,4 Dimethyl pent 3-one.
3-Methylbutanal
4-chloropentane -2-one
3-brome-4-Phenyl pentanoic acid
Para-methyl Benzaldehyde
4-Methyl pent-3-ene-2-on
Write IUPAC Names of the followings:O
(i) CH3
CHO
(ii)
CH3
(iii) (CH3 )2 CHCOCH((CH3 )2
(v) CH3-CH-CH2CH2-CHO
|
CH3
(vi) PHCOPH
(vii) CH3-CH2-CH-CH2-CH-CHO
|
|
(iv) CH 3 – CH (OCH3) CHO
177
dicarboxylic
Br
CH3
(viii) C6H5CH=CH-CHO
(ix) (CH3)3-CCH2COOH
FOR AVERAGE
02 Marks QUESTIONS
3.
Arrange the following compound in increasing order of their boiling points.
(f) CH3-CHO
(g) CH3-CH2-OH
(h) CH3-O-CH3
(i) CH3-CH3
(j) CH4
4.
Arrange the following compound in the increasing order of their properties as indicated
(iii)
(iv)
Acetaldehyde, Acetone, and Di-tert-Butyl Ketone (Reaction towards HCN)
Benzoic Acid, 4–Nitro benzoic acid, 3, 4 Dinitro benzoic acid, and
4-methoxy
benzoic acid (acid strength)
FOR ABOVE AVERAGE
02 Marks QUESTIONS
5. Give simple chemical test to distinguish between the following pair of compounds:(v)
Propanal & propanone
(vi)
Benzaldehyde and Acetophenone
(vii) Ethanal & Propanal
(viii) Acetophenone & Benzophenone
6.
7.
How will you distinguish?
(iv) Phenol & Benzoic Acid
(v) Benzoic Acid & Ethyl benzoate.
How will you distinguish the following pairs:
(i)
Pentan 2-one and Pentan 3-one
(ii) Propanol & Propanal
(iii) Methanal & Ethanal
FOR AVERAGE
02 Marks QUESTIONS
8. Arrange them in the increasing order of reactivity in esterification reaction.
iii.
iv.
CH3OH, (CH3)3COH, (CH3) 2-CH-OH, CH3CH2OH
(CH3) 3CCOOH, CH3COOH, (CH3) 2-CHCOOH,HCOOH
12. Arrange the following acid derivatives towards increasing order of nucleophelic
Substituion reaction
(a) RCONH2, RCOOCOR, RCOCI & RCOOR
(b) Acid derivative unlike aldehyde and ketones show mucleophic substitution
178
FOR ABOVE AVERAGE
02 Marks QUESTIONS
10.
Write mechanism for the followings:(a)
(b)
(c)
11.
Nucleophilic addition in aldehydes and Ketones
Aldol condensation
Esterification of carboxylic acids
What happens:1. When primary alcohol vapours are passed over Cu metal at 573 k
2. When secondary alcohols is treated with chromic anhydride (CrO3)
3. Ethyne is treated with H2O in the presence of H2SO4 & HgSO4
4. Toulene is treated with chromyl chloride (CrO2Cl2) followed by hydrolysis.
5. Propanone is treated with HCN
6. Ethanal is treated with NaHSO3.
7. Proponal is treated with Methyl magnesium bromide.
8. 3- Pentanone is treated with KMnO4 at high pressure
9. Ethanal is treated with dil NaOH.
10.
Benzaldehyde is treated with hot Alkali.
11. Acetic acid is treated with PCl5.
12. Benzoic acid is treated with NH3.
12.
(i)
COMPLETE THE FOLLOWING.
O
C-Cl
H2
?
Pd- BaSO4
1. AlH(t-Bu)2
(ii)
CH3-CN
?
2 H2O
NH2
|
CHCl3 + Alc KOH
(iii)
?
O
(iv)
AlCl3
+
C2H5-C-Cl
179
?
CS2
Hg2+,H2SO4
(v) CH3-C = C-H
?
(vi)
H
Conc NaOH
?
2 C =0
H
O
||
(vii)
NO2
|
+ H2N -NH
(viii)
NH2
CH2-CH2-CH3
1. KMnO4
KOH
2. H2O
I LiAlH4
(ix)
CH3-COOH
2. H3O+
13. Road Map problems
O
||
(i) R2 C
Identity A & B.
NH3
H2| Ni
A
B
Cu
(ii) CH3-CH2-OH
di NaOH
A
B
573 K
O
||
1. CH3MgBr
2,4,DNP
180
(iii) CH3-C –H
A
B
2. H2O
O
||
(iv) CH3-C-CH3
1. CH3MgBr
Cu
A
B.
2. H2O
753 K
HgSO4
CH3MgX
[o]
(v) HC = CH
A
B
C
H2SO4
FOR ABOVE AVERAGE
02 Marks QUESTIONS
14. Transformation
(i) Ethanol to butane 1, 3 diol;
(ii) 3- nitro bromobenzene to 3- nitro benzoic acid
(iii) Butanal to butanoic acid
(iv) Acetic acid to Ethan amine
(v) Bromo benzene to benzoic acid
(vi) Benzene to m-nitro acetophenone
(vii) Benzoic acid to benzaldehyde
15.
IDENTIFICATION TYPE QUESTIONS
(a)
An organic compound ‘A’ molecular formula C3H6O is resistant to oxidation but
forms a compound ‘B’ (C3H8O) on reaction. ‘B’ reacts with HBr to form bromide ‘C’ with
an treated with alcoholic KOH form an alkene D C3H6 deduce the structure A,B,C,D.
(b)
An organic compound ‘A’ with molecular formula C8H8O forms an orange red
precipitate with 2-4 DNP reagent & with yellow precipitate with on heating with iodine in
the presence of sodium hydroxide. It neither reduce Tollens or fehling reagent nor does it
decolorize bromine water. On drastic oxidation with chromic acid, it gives a carboxylic acid
(B) having molecular formula n C7H6O2. Identify the compound “A” and “B” and explain the
reaction involved.
(c)
An organic compound “A” C8H6 on treatment with dilute H2SO4 containing HgSO4
gives compound “B” which can also be obtained from a reaction of benzene with acid
chloride in presence of AlCl3? “B” on treated with I2 in aq KOH gives C and yellow
compound ‘D’ identify A,B,C,and D . Give the chemical reaction involved
FOR BELOW AVERAGE
01 Marks QUESTIONS
16. NAME REACTIONS: (Write Notes on)
(i) Rosen mund reduction
(ii) Cannizaro reaction
(iii) Cross aldol condensation
181
(iv)
(v)
(vi)
Aldol Condensation
Clemensen’s reduction
Wolff – Kishner reaction
17. DESCRIBE THE FOLLOWING TERMS:
(i) Acetylation
(ii) Decarboxylation
(iii) Silver mirror test
FOR ABOVE AVERAGE
01 Marks QUESTIONS
18. GIVE THE REASON FOR THE FOLLOWINGS:
(i)
The boiling points of aldehyde and ketones are lower than the corresponding
alcohols and carboxylic acid .
(ii)
The boiling point of carboxylic acid is higher than the Corresponding esters
(iii) Aldehydes and ketones under goes a number of nucleophlic addition reaction
(iv)
Aldehydes under go nucleophlic addition reaction more readily than ketones
(v)
Formaldehyde gives Cannizaro reaction where as acetaldehyde gives aldol
condensation
(vi)
Di- tert. butyl ketone does not give NaHSO3 adduct but acetone gives
(vii) Floroacetic acid is a stronger acid than chloroacetic acid.
(viii) The PKa value of chloroacetic acid is lower than PKa value of acetic acid
(ix)
>C = O group behaves differently in aldehyde and acid .
(x)
The melting point of an aliphatic carboxylic acid containing an even number of
carbon atoms is higher than next lower and next higher homologues containing
Odd number of carbon atoms.
(xi)
Electrophilic substitution on benzoic acid takes place at meta position.
**********************************************
182
UNIT -13
ORGANIC COMPOUNDS
CONTAINING NITROGEN
AMINIES
Alkyl amines are more basic than ammonia and aryl amines are less basic than alkyl amines
due to presence of electron density on N-atom.
This is as follows:
..
..
..
NH3 < R-NH2 > Ar-NH2
In alkyl amines magnitude of electron density on N-atom is increased
By presence of electron releasing group ie alkyl groups whereas the magnitude of electron
density is decreaed on N –atom in aryl amines
Due to electron withdrawing group ie aryl group .
The less basicity of aryl amine is due to resonatig structures.
+
NH2
183
BASICITY AMONG PRIMARY, SECONDARY & TERTIARY AMINES:
Basicity among alkyl amines is found as secondary amines > pr.amines> tert. Amines
. The irregular trends among amines is explained on the basis of electron doner alkyl group
attached to an atom. Pr. Amines have less no. of alkyl group,hence less basic in secondary
amines presence of two alkyl groups make more basic than pr. Amines but in tert. Amines
presence of three alkyl groups creats steric hindrance resulting in repulsion between elctrons
makes low electron density on N-atom.
..
..
..
R---N---R > R---NH2 > R2---N----R
The above basicity of amines can be explained on the basis of Kb &
pKb. higher the Kb values or lower the pKb value shows higher basic characters .
Sec. amines > tert. Amines < pr. Amines
pKb
3.00
3.29
3.25
PREPARATION OF AMINES:I.
(a)
REDUCTION OF NITRO COMPOUNDS—
H2/Pd
ArNO2 ------------------------ ArNH2
Ethanol
Sn/ HCl
(b) ArNO2 --------------------ArNH2
II.
Ammonolysis of alkyl halides---
An alkyl or aryl halides react with ethanolic solution of ammonia undergo nucleophilic
substitution in which halogen atom is replaced by amino gr.
RX
RX
RNH2----------- R2NH ------------ R3N
PR Amine.
SEC.
TERT.
DISADVANTAGE—By this method a mixture of amines are prepared and not separated
easily.
III. Reduction of nitriles -H2/Ni
RCN------------------ RCH2 NH2
a.
Reduction of amides----LAH
RCONH2 --------------------- RCH2NH2
WATER
V.
Hoffmann bromamide degradation reaction—
Pr. amines are prepared by this reaction in which amide combines with bromine in aq. Or
ethanolic solution of sodium hydroxide.
RCONH2 + Br2 +4 NaOH ------ RNH2 +Byproducts
184
Important chemical properties--(i) Basic properties-- Presence of electron density on N atom is the reason of basic
property. Higher the electron density on N atom , higher the basic character .
(ii) Solubility in water--- Lower alkyl amines are soluble in water due to H- bonding .
Hihger akyl amines are insoluble in water.
v.
Acylation reaction --- Amines combines with acyl halides to form N-alkyl ethanamine.
C2H5NH2 + CH3COCl ------ C2 H5 NHCOC H3
Carbylamine reaction ---Primary amine combines with chloroform in ethanolic
potassium hydroxide to form isocyanide with pungent smell.
Heat
RNH2 +CHCl3 + 3 KOH -------- RNC + 3 KCl + 3H2O
vii.
Reaction with nitrous acid -----NaNO2+HCl
RNH2 + HNO2 -------------ROH + N2 +HCl
HOH
vi.
DIAZOTIZATION REACTION & DIAZONIUM SALTS :The conversion of primary aromatic amines into diazonium salt is known as diazotization
reaction.The diazotization reaction is characterize in the joining two N-atoms (diazo)through
double bonds.
PREPARATION:- The commonly known diazonium saltie benzene diazonium chloride is
repared by reacting aniline with sodium nitrite and hydrochloric acid.
273—278 K
ArNH2 + NaNO2 +HCl ---------------------ArN2 +Cl- +NaCl +2H2O
PROPERTIES OF DIAZONIUM SALTS:-It is a colourless crystalline solid. It is in soluble in water .It is easily decomposable in dry
state.
IMPORTANT REACTIONS:-(I)
Sandmeyer’s reaction :- The reaction of diazonium salt with halo acid in
presence of copper halide to have halobenzene is called sandmeyer’s reaction.
Ar-N2Cl + CuCl + HCl ------- Ar-Cl + N2
(II)
Gatterman reaction :--- The reaction of diazonium salt with halo acid in
presence of copper powder is called Gatterman reaction.
Ar-N2Cl + Cu + HCl ---------- Ar-Cl + N2 + CuCl
(III)
Hydrolysis reaction ----- Diazonium salt on hydrolysis upon 283 K gives
phenol.
HOH, 283K
Ar-N2Cl -------------------------- Ar-OH + N2 + HCl
(IV)
Reduction of diazonium salt ----- Diazonium salts on reduction with mild
reducing agent like phosphorous acid or ethanol reduced to arene .
Ar-N2Cl + H3PO2 + HOH --------- ArH + N2 + H3PO3 + HCl
(V)
Nitration reaction ------
Diazonium salt on treatment with
185
HBF4 followed by sodium nitrite & copper at high temperature gives nitroarene.
+
- NaNO2
Ar-N2Cl + HBF4 -- Ar-N2BF4 --------- ArNO2 + N2 + NaBF4
(V)
Coupling reaction ----The reaction between diazonium salt with aryl
compounds through –N=N- is called coupling reaction.
Ex.—Benzene diazonium chloride reacts with phenol in basic medium , phenol coupled
with diazonium salt to the para position of phenol to produce p-hydroxy azobenzene( orange
dye).
OH–
Ar-N2Cl + ArOH -------------- Ar-N=N-ArOH + Cl- + HOH
1 MARK QUESTIONS
Q. 1. Why the presence of a base is essential in the ammonolysis of alkyl halides ?
Ans.
During ammonolysis of alkyl halides, the acid liberated during the reaction combines
with the amine formed to form amine salt. To liberate free amine from the amine salt,
a base is needed.
Q. 2. Although — NH2 gp is an ortho and para directing gp, nitration of aniline gives
along with ortho and para, meta derivatives also.
Ans.
Nitration is carried out with a mixture of Conc. NO3 + Conc. H2SO4 (nitrating mix). In
the presence of these acids, most of aniline gets protonated to form anilinium ion.
Therefore, in the presence of acids, the reaction mixture consists of aniline and
anilinium ion. Now – NH2 gp in aniline is O, p-directing and activating while –N+H3
gp in anilinium ion is m-directing and deactivating hence a mixture of all three–ortho,
para and meta derivatives is formed.
Q. 3. Pkb of aniline is more than that of methyl amine.
Ans.
In aniline, the lone pair of electrons on the N-atom are delocalized over the benzene
ring. As a result electron density on the nitrogen decreases. In contrast in CH2NH2, + I
effect of CH3 increase the electron density on the N-atom. Therefore, aniline is a
weaker base than methylamine and hence its Pkb value is higher than that of
methylamine.
Q. 4. Aniline gets coloured on standing in air for a long time. Why ?
Ans.
Due to strong electron-donating effect (+ R effect) of NH2 gp, the electron density on
the benzene ring increases. As a result, aniline is easily oxidised on standing in air for
a long time to form coloured products.
186
Q. 5. CH3CONH2 is a weaker base than CH3CH2NH2.
Ans.
Due to resonance, the lone pair of electrons on the nitrogen atom in CH3CONH2 is
delocalised over the keto gp. There is no such effect in CH3CH2NH2. Due to reduction
in electron density on N of CH3CONH2, it is a weaker base than CH3CH2NH2.
Q. 6. Aromatic primary aminies can’t be prepared by Gabriel phthalimide synthesis.
Ans.
The preparation of aromatic primary amines (Aniline) by Gabriel phthalimide reaction
requires the treatment of pot. phthalimide with C6H5Cl or C6H5Br, which is a
nucleophilic substitution Rxn. Since aryl halides do not undergo nucleophilic
substitution under ordinary laboratory conditions, therefore C6H5Cl or C6H5Br does
not react with pot. phthalimide to give N-phenyl. Phthalimide and hence aromatic
primary amines can’t be prepared by this method.
Q. 7. Accomplish the following conversions :
(i) Nitrobenzene to benzoic acid
Ans.
(ii) Benzyl Chloride to 2-phenylethanamine
NH2 NaNO2 + HCl
(i)
NO2
(i) Fe/HCl
273 – 278 K
—————
—————
O
O
(ii) NaOH
Diazotization
N+NCl–
CuCN/HCN
—————
O
Benzene
Diazanium
Aniline
Chloride
Benzonitrile
COOH
H3+O
—————
Hydrolysis
O
Benzoic acid
(ii)
O
2 1
CH2CH2NH2
CH2CN
CH2Cl2
—————
– KCl
Benzyl Chloride
phenylethanamine
O
LaAlH4
—————
Redue
Phenyl ethanenitrite
O
2-
Q. 8. Give the structures of A, B and C in the following compounds :
NaCN
OH–
NaOH + Br2
(i) CH3CH2I ———————
——————
——————
Partial Hydrolysis
NH3
(ii) CH3COOH ———————
D
NaOBr
——————
187
NaNO2/HCl
——————
CN
O
Ans.
O
||
—
C
—
NH2
3
–
NaCN
(i) CH3CH2I ——————
OH
CH
CN
——————
3
2
(Partial Hydrolysis)
Propanenitrile (A)
Ethanamide (B)
NaOH + Br2
———————————
Hofmann bromamide reacn
NH3
(ii) CH3COOH ———
CH3OH
D
Ethanoic acid
— NH2
Methanamine (C)
NaOBr
CONH2 ———————
3
3
NaNO2/HCl
NH2 ——————
3
(Hofmann bromamide
Ethanoimide (A)
reac.)Methanamine (B)
Methanol
(C)
Q. 9. Why is it difficult to prepare pure amines by ammonolysis of alkyl halides ?
Ans.
By ammonolysis of alkyl halides, a mixture of primary, secondary and tertiary amines
is formed. So it is difficult to separate it.
Rx
Rx
Rx
NH3 ———
— NH2 ———
—
NH
———
2
1° amine
2° amine
Rx
—
N
———
3
3° amine
4
N+ X–
Quaternary
Salt
Q. 10. Can tertiary amines undergo acetylation reactions ? Explain.
Ans.
For an amine to undergo acetylation, it should have a replaceable hydrogen atom.
Tertiary amines cannot undergo acetylation reactions because these do not have
replaceable hydrogen atom.
Q. 11. Sulphanilic acid has acidic as well as basic group; but it is soluble in alkali but
insoluble in mineral acids. Explain.
Ans.
Sulphanilic acid exists as Zwitter ion as :
NH 3+
O
SO 3–
In the presence of dil. NaOH the weakly acidic — NH3+ group transfers its H+ to OH–
to form soluble salt P — NH2C6H4SO3–Na+. On the other hand, — SO3+ group is a very
weak base and does not accept H+ from dil HCl to form NH3C6H4SO3H and therefore,
it does not dissolve in dil HCl.
Q. 12. Why are aliphatic amines more basic than aromatic amines ?
Ans.
In Aromatic amines, due to resonance, NN-atom is less available.
188
+
NH2
NH2
+
NH2
+
NH2
–
NH2
–
–
In aliphatic amines, due to e– releasing nature of alkyl groups lone pair of e– on Natom is more available.
Q. 13. Explain why :
(i)
Ethylamine is soluble in water but aniline is not.
(ii)
Aniline does not undergo Friedel Crafts reaction.
(iii)
Diazonium salts of aromatic amines are more stable than that of aliphatic
amines.
Ans.
(i)
Ethyl amine dissolves in water due to formation of H-bonding with water
molecules.
H
H
|
|
H—N------------H—O------- H—N--------H—O----|
|
C2H5
H
|
|
C2H5
H
H—N—H
|
C2H5
However in Aniline, due to larger hydrophobic alkyl part extent of H-bonding
decreases considerably
ater.
(ii)
Aniline being a Lewis base reacts with Lewis acid AlCl3 or FeCl3 to form a
salt.
C6H5NH2
Lewis base
+
AlCl3
———
H5NH2+AlCl3–
6
Lewis acid
As a result, N-atom acquires + ve charge and hence acts
deactivating group for electrophilic reaction.
189
as a strong
(iii)
Diazonium salts of aromatic amines are more stable due to dispersal of +ve
charge on benzene ring due to resonance.
+
NN
+ –
N=N
N+=N –
+
N+=N
–
+ –
NN
+
+
In aliphatic amines, there is no such dispersal of + ve charge due to absence of
resonance.
Q. 14. Why 1° amines have got higher boiling point than 3° amines ?
Ans.
Due to presence of two H-atoms, in 1° amines they undergo extensive hydrogen
bonding while due to absence of H-atom in 3° amines do not undergo H-bonding.
-------- H
R
H-------|
|
|
R—N—H-------- N—H--------N—R
|
|
H
H--------
--------H—N—R
|
H
Due to extensive hydrogen bonding in 1° amines, they have higher boiling point – 7
than 3° amines.
Q. 15. How can you distinguish between 1° and 2° amine ?
Ans.
(i)
Carbylamine test :
R — NH2 + CHCl3 + 3 KOH ———
— NC + 3 KCl + H2O
(pungent smelling)
2° amines do not give this test.
(ii)
Aryl sulphonyl chloride test :
|
H
190
Q. 16. How can you distinguish between aromatic and aliphatic amines ?
Ans.
The diaronium salts of Aromatic amines undergo coupling reactions with phneol or
andine to form coloured compounds used as dyes.
No such reactions are given by aliphatic amines.
Q. 17. How can tri substitution of Bromine be prevented in aniline ?
OR
How can we prepare mono bromo aniline ?
Ans.
Because of high reactivity of aromatic amines substitution tends to occur at o- and ppositions. Monosubstituted aniline can be prepared by protecting — NH2 group by
acetylation with acetic anhydride, then carrying out substitution followed by
hydrolysis of substituted amide.
2 MARKS QUESTIONS
Q. 1. Give increasing order of reactivity towarads electrophilic substitution reaction of
the following compounds :
Ans.
CH3
N
O
O
CH3
CH3
CH3
CH3 — N+ — CH3
CH2 — N+ (CH3)3
O
O
Higher the electron density in the benzene ring, more reactive is the aromatic
compound towards electrophilic substitution reaction. Now due to the presence of a
lone pair of electrons on the N-atom which it can directly donate to the benzene ring.
N (CH3)2 (due to two alkyl groups on N) is a much stronger electron donating gp than
191
CH3 gp. (only one alkyl group on N). The remaining two gps contain a positive charge
on the N-atom and hence act as electron withdrawing gps. But in (CH3)3 N+ — gp, the
+vely charged N is directly attached to the benzene ring, therefore, its electron
withdrawing ability is much stronger than — CH2N+ (CH3)2. From the above, it
follows that the electron density in the benzene ring increases in the order :
Therefore, their reactivity towards elecrophilic substitution reactions also increases in
the same order.
Q. 2. Which one is more acidic ? Explain :
Ans.
Due to powerful – I effect to the F-atom, it withdraws electrons from N+H2 gp. As a
result, electron density in the N — H bond of p-fluoroanilinium ion decreases and
hence release of a proton from p-fluoroanilinium ion is much more easier than from
anilinium ion. Therefore, p-fluoroanilinium ion is more acidic than anilinium ion.
Q. 3. Explain the order of basicity of the following compounds in (i) Gaseous phase
and (ii) inaqueous soln. :
(CH3)3N, (CH3)2NH, CH3NH2, NH3
Ans.
Due to + I effect of alkyl gps, the electron density on the N-atom of 1°, 2° and 3°
amines is higher than that on the N-atom in NH3. Therefore, all amines are more basic
than NH3.
(i)
In gaseous phase, solvation effects are absent and hence the relative basicity of
amines depends only on + I effect of the alkyl gps. Now since + I effect
increases in going from 1° to 2° to 3° amine, so the basicity of amines
decreases in the order :
192
3° amine > 2° amine > 1° amine
(CH3)3N > (CH3)2NH > CH3NH2 > NH3
(ii)
In aq. soln, the basicity depends upon two factors :
(a)
+ I effect of CH3 gp and
(b)
Solvation effect.
Stabilization of the conjugate acid (formed addition of a proton to amine) by
H-bonding explained above on the basis of + I effect, the order will be :
(CH3)3N > (CH3)2NH > CH3NH2
On the basis of Stabilisation of conjugate acids by H-bonding alone as explained
below :
The order will be :
CH3NH2 > (CH3)2NH > (CH3)3N
The combined effect of these two opposing factors is that (CH3)2 NH is the strongest
base. In case of CH3NH2 and (CH3)3 NH, the stability due to H-bonding predeminates
over stability due to + I effect of CH3 gp, thereby making CH3NH2 stronger than
(CH3)3 NH. So the overall order in aq. soln will be :
(CH3)2 NH > CH3NH2 > (CH3)3N > NH3
Conceptual Questions
Q 1. Give reasons:--a. Primary amines are more basic than ammonia.
b . Sec. amines are more basic than pr.amines.
Ans— a. In pr. Amines, alkyl group increases electron density on the N-atom making
it more basic than ammo
b. In sec. amine, there are two alkyl groups increase electron density on N-atom more
than N-atom in pr. amines in which there is one alkyl group.
193
Q.2.
Why are amines always less acidic than comparable alcohols?
Ans.--- Amines are less acidic than alcohols because of oxygen higher electronegative
& smaller size. Therefore O-H breakes easily than N-H bond.
Q.3.
Why is amide more acidic than an amine ?
Ans.—In amide there is C=O group attached NH2 , electron withdrawing & increases
acidic nature of amide whereas in amines there is alkyl group which is electron
releasing and makes it more basic & hence less acidic.
Q.4.
How is aniline distinguished from N-methyl aniline ?
Ans.—On adding CHCl3 & KOH , aniline gives offensive smell forming isocyanide
compound whereas N-methyle aniline not.
Q.5.
Why cyanide ion act as ambident nucleophile ?
Ans.---- Cyanide ion can link through either carbon or nitrogen, therefore it is an
ambident nuclophile.
Q6.
How would you remove nonpolar impurities from an amine ?
Ans.--- Adding HCl , amine will form salt whereas nonpolar impurities will not react .
Amines can be regenerated by adding NaOH .
Q7.
Why tert. Amines donot undergo acylation reaction ?
Ans.-- Tert. Amines donot undergo acylation reaction because they donot have Nattached to H- atom.
Q8.
How will you convert the following.
(a) Aniline to acetanilide
(b) Benzene to aniline
Ans:- (a) C6H5NH2+CH3COCl
(b) C6H6+HNO3 Conc
NaOH
ConeH So4
2
333K
C6H5NHCOCH3 +HCL
3H
2
C6H5NO2
C6H5NH2
Sn/Hcl
Q9.
Why does the boiling point of CH3NH2 less than HCOOH?
Ans:- HCOOH is associated with intermolecular H-bonding in greater extent then
CH3NH2 and therefore HCOOH has higher boiling point than CH3NH2 .
Q10. What is Zwitter ion? Write the formula of Zwitter ion of alanine.
Ans:- The ion of a molecule having +ve change at one end and –ve charge at other
is called zwitter ion.
CH3 –CH-COOH
CH3-CH-COO-
CH2
NH2
Q11. How will you prepare aniline from nitrobenzene industrially?
Ans:- Aniline is prepared industrially by reduction of nitrobenzene in presence o of
Fe/Hcl.
NO2
NH2
+ 6[H]
Fe/Hcl
+ 2H2O
194
H2
and
Q12.
Aromatic primary amines are not prepared by gabrial phthalimide
synthesis. Why?
Ans:- Because aryl halides donot undergo nucleophilic substitution reaction.
Q13.
Convert the following(i)
Acetamide to ethanamine
(ii)
Benzene to autophenone
(iii)
Anitine to 4- bromo aniline
(iv)
Methanamine to ethanamine
(v)
Phenol to aniline
(vi)
Ethyl alcohol ( ethanol) to ethanamine
Na/C H OH
2 5
Ans:- (i) CH2CONH2
CH3CH2NH2
(ii) C6H5 + CH3COCl
Alcl3
(iii) C6H5NH2 + CH3cocl
Br C6H4NHCOCH3
H2O/H+
C6H5COCH3
C6H5NHCOCH3
BrC6H4NH2
H O/H
2
CH COOH
3
NH2
OR
Br
(iv) CH3NH2
CH3CH2NH2
(v)
OH
or
CH3OH
C6H6OH
(vi) C6H5OH + NH3
14.
NH
3
Zncl
2
Al O
2 3
Heat
CH3Cl
CH3CN
C6H5NH2
C6H5NH2
Account for the following(i) Nitration of toluene is easier than benzene
(ii) Ethylanmine is soluble in water where as aniline is not.
(iii) pKh of aniline is more than that of methylamine.
(iv) Diazonium salts of aromatic amines are more stable than these of
aliphatic amines.
(v)Methylamice in water reacts with ferric chloride to precipitate hydrated
ferric oxide.
(vi) Primary amires have higher boiling point then tertiary amines.
(vii) Silver chloride dissolves in aqeous methylamine solution.
(viii) Aniline readily reacts with bromine to give 2,4,6 tribromo-aniline.
Ans:- (i) The methyl group of toluene is electron releasing group and releases
eleases electron density to the benzene ring and hence makes the joining of
electrophile NO2+ more easier and faster.
(ii) Ethylamine has more electron density on the N-atam with electron releasing nature of
ethyl group while aniline has electron gainer benzene ring & deereaus electron density on
195
the N-atom & hemee formation of H-bond with water becomes difficult and hence not
soluble.
(iii) Aniline is less basic than methylamine due to electron releasing benzene ring and
hence pKb value is more & Kb value is less.
(iv) Diazonium salt of aromatic amines have dipolar N2-cl group which shows more
intermolecular attraction than those of aliphatic amines and hemee more stable.
(v) Methylamine is basic in nature and have attracts to the cl group of Fecl3 in presence of
H of water and hence precipitate the Ferric oxide hydrated.
(vi) Primary amires have less surface and less spherical shape and hence expericnce
greate varder waal’s force of attraction and hence greater boiling point
(vii) Silver chloride has tendency to form complex compound with aqeous methyl amine
and hence dissolves.
(vii) Aniline has electron releasing amine group, NH2 and increases electron density on
the ortho- and para- positions of the benzene ring and hence form 2,4,6 tribromo-aniline.
Amines are basic even more basic than ammonia among amines, alkyl amines are more basic
than aryl amines.
Explanation:- The basic nature of amines is because of lone pair of electrons on the N-atom
which can be donated to other.
..
R – NH2
..
NH2
Alkyl amines have electron donor alkyl group ( R ) that increases the electron density on the Natem and makes the alkyl amines more basic while aryl amines the alkyl amines more basic
while aryl amines have electron gainer benzene ring. Benzene is rich in electrons but having
tendency to gain electrons instead of donating to the N-atom & hence the electon density on the
N- atom becomes less and hence less basic. The low basicity of aryl amines may be explained
broadly by ellustrating resonating structures of anyl amines ie aniline.
The low basicity of anylamires is with resonance stabilization of the compound i.e
..
NH2
+
NH2
+
NH2
+
NH2
I
II
III
IV
..
NH2
V
Out of I--V resonating structures the figures II ,III and IV are involved in the
charge separation. The fig. II electrons are always rotating between 6 carbon
and not available permanently to the N-atom for strong basic nature.
196
atoms
In comparision to the aryl amires the alkyl amines have no such type of resonance and
hence the movement of electrons is always alkyl group to the N-atom of the functional
group amino group (NH2).
R – NH2
This phenomena makes the permanent displacement of the electrons towards N-atom
makes it more basic.
FOR BELOW AVERAGE
01 Mark Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Write IUPAC Name - (CH3)2CHNH2
Write IUPAC Name - CH3NHCH(CH3)2
Write IUPAC Name - CH3CH2N( CH3)2
Classify the given amines into primary, secondary& tertiary.
(CH3 )2CHNH2 ,( CH3 )2 CHNH2, CH3CH2N(CH3)2
Arrange the following in decreasing order of PKb values—
(CH3 )2 CHNH2 ,( CH3 )2 CH NHCH3, CH3CH2N(CH3)2
Arrange the following in increasing order of basic strength
Aniline, Dimethylaniline, Diethyl amine &Methyl amine
Which of the compounds is most basic –
Aniline, p-Nitroaniline & p-Toludine
Which of the following compounds is more soluble in waterAniline, Diethyl aniline & Ethyl amine
Write the hybridized state of amines.
What is the geometry of amines?
What is the structural formula of Zwitter ion of Sulphanilic acid
Complete the following –
CuCN
ArN2+X- +CN- ------------- --------+N2 + X –
Convert –
Aniline to benzonitrile
Convert –
Aniline to sulphanilic acid.
What do you mean by ambident Nucleophiles ? Give an example.
02 Marks Questions
1.
2.
3.
4.
5.
Give reasons --a. Amines are basic in nature.
b. Aryl amines are less basic than ethyl amines.
How would you distinguish between Primary & secondary amines?
Out of ethyl alcohol & ethyl amine, which has higher boiling point and why?
Illustrate each of the following with an example –
a Sandmayer reaction
b. Diazotization reaction.
How will you convert --a. Nitrobenzene to phenol
b. Aniline to chlorobenzene
197
6.
7.
8.
9.
Account for the following--(i) Primary amines are higher boilng points than comparable tertiary amines.
(ii)
Conc. Sulphuric acid used in nitration of Benzene.
Write the chemical equations for one example of each the following--(i)
Coupling reaction
(ii)
Hoffmann bromamide reaction.
How will you convert--(i) Methanamine to ethanamine
(ii) Ethanamine to methanamine.
Complete the following –
NaCN
OHNaOH/ Br2
(i)
CH3CH2I--------- A------------B--------------------- C
(ii)
NH3
NaOBr
NaNO2/ HCl
CH3COOH------------ A------------- B ------------------ C
heat
10.
11.
12.
13.
14.
15.
How will you convert --(i) Ethanoic acid into methanamine
(ii) Methanol to ethanoic acid
Write short notes on the following(i) Carbylamine reaction
(ii) Acetylation reaction
Why cannot aromatic primary amine be prepared by Gabrial phthalimide synthesis?
Explain, Why it difficult to prepare pure amines by ammonolysis of alkyl halides?
Silver chloride dissolve in methylamine solution, Give reason.
Write the structure of the following compounds –(i) TNT (ii) picric acid
03. Marks Questions
1.
Give reasonsa.
Aniline is weaker base than cyclohexaamine.
b.
Alkylamines have less PKb value than arylamines.
c.
Methylamine in water reacts with ferric chloride to precipitate
ferrichydroxide.
2.
What happens when
a. Aliphatic/aromatic amines react with chloroform in alcoholic potash.
b. Aniline combines with bromine water at room temperature.
c. Arene diazonium chloride combines with fluoroboric acid.
3.
Predict the products—
(i)
Arene diazonium chlorides react with phenol in presence of alkali.
(ii)
4-Nitrotoluene combines with bromine & further reduced by tin in
Presence of acid.
(iii) Aniline is treated with Acetylchloride in presence of pyridine.
4.
Give one chemical test to distinguish between them(i)
Ethylamine and aniline
(ii)
Ethylamine and dimethylamine
(iii) Aniline and benzylamine.
198
5.
Identify the compound A, B, C in the following reactions(i)
Br2/NaOH
HNO2
A----------------- B ---------------C
Sn/ HCl
NaNO2/ HCl
(ii) A---------------- B ----------------------- C
(iii)
Br2/ KOH
A------------------ B -------------------- C
Heat
6.
Comment on the followinga. Hoffmann’s bromamide reaction.
b. Carbylamine reaction
c. Diazotization reaction
7.
Write the chemical reactions a. Hinsberg’s test for all amines.
b. Test to distinguish aliphatic amines & aromatic amines.
8.
Write the reactions & mention the product with IUPAC name—
(i) Aniline combines with nitric acid in presence of sulphuric acid.
(ii) Aniline combines with sulphuric acid at 453—473K.
(iii) Arene diazonium chloride combines with ethanol.
Give reasons(i) Silver chloride dissolves in aqeous methyl amine solution.
(ii) Tertiary amines do not undergo acylation reaction.
(iii) Aniline readily reacts with bromine to give 2,4,6-tribromoaniline.
What is the Gabrial phthalamide synthesis ?
Condensation of aniline and benzaldehyde gives compound A that is
hydrogenated to give compound B .Identify A and B .
What happens when
(i) Nitropropane reduced with LiAlH4
(ii) Ethyl isocyanide undergoes hydrolysis
(iii) Benzene diazonium chloride reacts with phenol in basic medium.
9.
10.
11.
12.
FOR AVERAGE
01. Mark Questions
1
2.
3.
Why do amines react as Nucleophiles?
How is m-nitroaniline obtained from nitroarene?
Write IUPAC name of the compound—
H2 N
4.
5.
OCH3
Give chemical test to distinguish between benzyl amine & aniline .
Complete the following reactions—
Conc. H2SO4
199
6.
7.
8.
9.
ArNH2 --------------------- ------------------How would convert, aniline to benzonitrile?
What is Mendius reaction?
Why does ammonolysis of alkyl halides not give pure Amines?
How will you convert aniline to phenol?
02
Marks Questions
1.
2.
3.
4.
5.
Accompalish the following conversions –
a. Chlorobenzene to p-chloroaniline
b. Nitromethane to dimethyl amine
Give reasonsa.
Aliphatic amines are stronger base than aromatic amines.
b.
Diazonium salts of aromatic amines are more stable than aliphatic
amines.
Write the chemical reaction for the following with one example.
a. Coupling reaction
b. Sandmeyer reaction.
How will you carry the following conversions—
a.
Aniline to benzonitrile
b.
Aniline to 2, 4, 6-tribromo aniline
Describe the test of primary amines & secondary amines with chemical
equations.
03. Marks Questions
1.
2.
3.
4.
5.
6.
Starting the necessary chemical reactions, conditions write the chemical Equations to obtained
chlorobenzene from aniline.
Identify X & Y.
NH3
H2/Ni
RCOR -------------- X -------------- Y
Write short notes on the following--a. Carbylamine reaction.
b. Diazotisation reaction.
c. Hoffmann Bromamide reaction.
Describe a chemical test to identify pr. Amines, sec, amines and tert. Amines’
An aromatic compound A on treatment with aq. Ammonia & heating forms
compound B which on further heating with bromine and KOH forms another
compound, Formula C6H7N. Identify the compounds A, B and C.
Account for the following--a.
pKb value of aniline is more than that of methyl amine
b.
Ethylamine is soluble in water but aniline is not.
a.
Aqueous methyl amine reacts with Ferric chloride to give precipitate
hydrated ferric oxide.
FOR ABOVE AVERAGE
01. Mark Questions.
1.
2.
3.
4.
5.
An organic compound X having molecular formula C2H7N On treatment nitrous acid
gives an oily yellow substance. Identify X’
Name the product obtained by action of aniline with Bromine.
What happens when ethyl amine dissolves in liq. Ammonia?
How would you prepare Orange Dye?
Give the reagents used in Gatterman reaction.
200
02 Marks Questions.
1.
2.
3.
4.
5.
Give reasons—
a. Tert. Amines do not undergo acylation reaction.
b. Aniline cannot be prepared by Gabrial synthesis’
Accomplish the conversions—
b. Benzoic acid to aniline
c. Propanoic acid to ethanoic acid
What happens whenb. Aromatic amines react with nitrous acid.
c. Aryl diazonium chloride reacts with phenolic in basic medium.
Distinguish between the pair of the compounds—
a. Ethanamine & diethylamine
b. Phenol & aniline.
Write the product for the following reactionsa. Nitroethane is treated with lithium Aluminium Hydride.
b. Ethyl-isonitrile is hydrolysed in the presence of acid’
03. Marks Questions.
1.
2.
3.
4.
Write short on the following—
a. Gattermann reaction.
b. Hoffman bromamide reaction
c. Gabrial phthalamide reaction.
An organic compound A with molecular formula C2H5NO2 , reacts with nitrous acid
to give C2H4O3N, B. On reduction A Gives another compound C with molecular
formula C2H7N On treatment with nitrous acid gives D which shows iodoform test,
Identify A.
Distinguish between.
(A)
Aliphatic amines & aromatic amines.
(B)
Aniline & phenol
(C)
Diethylamine & triethylamine
Accomplish the conversions—
a.
b.
c.
Nitrobenzene to benzoic acid.
Benzene diazonium chloride to p-hydroxy azobenzene.
Chlorobenzene to p-chloroaniline.
5.
An aliphatic compound X, molecular formula C3H7NO, reacts with bromine in
presence of KOH to produce another compound Y . The compound Y further reacts
with nitrous acid to form Ethanol & nitrogen gas. Identify X & Y and write chemical
reactions involved.
6.
Give suitable reasons—
(A)
(B)
(C)
Ethylamine dissolves in liquid ammonia.
Aniline does not undergo Friedel Crafts reaction
Gabriel phthalamide synthesis is preferred for synthesizing
Primary amines.
************
201
UNIT -14
BIOMOLECULES
A biomolecule is a molecule that naturally occurs in living organisms. Biomolecules
consist primarily of carbon and hydrogen, along with nitrogen, oxygen, phosphorus and
sulfur. Other elements sometimes are incorporated but are much less common.
A diverse range of biomolecules exist, including:



Small molecules:
o Lipid, Phospholipid, Glycolipid, Sterol
o Vitamin
o Hormone, Neurotransmitter
o Carbohydrate, Sugar
o Disaccharide
Monomers:
o Amino acid
o Nucleotide
o Phosphate
o Monosaccharide
Polymers:
o Peptide, Oligopeptide, Polypeptide, Protein
o Nucleic acid, i.e. DNA, RNA
o Oligosaccharide, Polysaccharide
Saccharides
Monosaccharides are carbohydrates in the form of simple sugars. Examples of
monosaccharides are the hexoses glucose, fructose, and galactose and pentoses, ribose, and
deoxyribose
Disaccharides are formed from two monosaccharides joined together. Examples of
disaccharides include sucrose, maltose, and lactose
Monosaccharides and disaccharides are sweet, water soluble, and crystalline.
Polysaccharides are polymerized monosaccharides, complex, unsweet carbohydrates.
Examples are starch, cellulose, and glycogen. They are generally large and often have a
complex, branched, connectivity. They are insoluble in water and do not form crystals.
Shorter polysaccharides, with 2-15 monomers, are sometimes known as oligosaccharides.
Amino acids
Amino acids are molecules that contain both amino and carboxylic acid functional groups. (In
biochemistry, the term amino acid is used when referring to those amino acids in which the
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amino and carboxylate functionalities are attached to the same carbon, plus proline which is
not actually an amino acid).
Amino acids are the building blocks of long polymer chains. With 2-10 amino acids such
chains are called peptides, with 10-100 they are often called polypeptides, and longer chains
are known as proteins. These protein structures have many structural and functional roles in
organisms.
There are twenty amino acids that are encoded by the standard genetic code, but there are
more than 500 natural amino acids. When amino acids other than the set of twenty are
observed in proteins, this is usually the result of modification after translation (protein
synthesis). Only two amino acids other than the standard twenty are known to be
incorporated into proteins during translation, in certain organisms:


Selenocysteine is incorporated into some proteins at a UGA codon, which is normally
a stop codon.
Pyrrolysine is incorporated into some proteins at a UAG codon. For instance, in some
methanogens in enzymes that are used to produce methane.
Besides those used in protein synthesis, other biologically important amino acids include
carnitine (used in lipid transport within a cell), ornithine, GABA and taurine.
Protein structure
The particular series of amino acids that form a protein is known as that protein's primary
structure. Proteins have several, well-classified, elements of local structure and these are
termed secondary structure. The overall 3D structure of a protein is termed its tertiary
structure. Proteins often aggregate into macromolecular structures, or quaternary structure.
Metalloproteins
A metalloprotein is a molecule that contains a metal cofactor. The metal attached to the
protein may be an isolated ion or may be a complex organometallic compound or organic
compound, such as the porphyrin group found in hemoproteins. In some cases, the metal is
coordinated with both a side chain of the protein and an inorganic nonmetallic ion. This type
of protein-metal-nonmetal structure is found in iron-sulfur clusters.
Vitamins
A vitamin is a compound that cannot be synthesized by a given organism but is nonetheless
vital to its survival or health (for example coenzymes). These compounds must be absorbed,
or eaten, but typically only in trace quantities. When originally discovered by a Polish doctor,
he believed them to all be basic. He therefore named them vital amines. The l was dropped to
form the word vitamines.
1. Carbohydrates – Carbohydrate is a class of compounds that include polyhydroxy
aldehyde, polyhydroxy ketone and large polymeric molecules that can be broken
down to polyhydroxy aldehyde and ketones. Example: sugar, glucose, starch, gums
etc. They are derived mainly from plants.
2. Monosaccharides – These include non-hydrolysable carbohydrates. These are
soluble in water e.g. glucose, fructose, etc.
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3. Aldose – The monosaccharides contaning a aldehydic group are called aldose.
Example: glucose.
4. Ketose - The monosaccharides contaning a ketonic group are called ketose. Example:
fructose.
5. Disaccharides – These carbohydrates, which on hydrolyses yield two molecules of
monosaccharides are called disacchaides. They are crystalline solid. Soluble in water
and sweet in taste. Example: cane sugar, maltose, lactose.
6. Oligosaccharides – Those carbohydrates which yields 2 to 10 monosaccharides
molecules on hydrolysis are called oligosaccharides. Example: Raffinose. On
hydrolysis it gives glucose, fructose and galactose.
7. Polysaccharides - Those carbohydrates, which produce large no, of monosaccharides
units are called polysaccharides. They are formed by linking together a large no. of
monosaccharide units through glycosidic linkage. e.g – starch, amylose, glycogen,
cellulose, etc.
8. Sugars - In general, monosaccharides and oligosaccharides are crystalline solids,
soluble in water and sweet in taste. These are collectively called sugars. e.g glucose,
sucrose.
9. Non-sugars – The polysaccharides are amorphous, insoluble in water and tasteless
are known as non-sugars e.gss starch and cellulose.
10. Reducing sugars – Those carbohydrates which contains free aldehydic or ketonic
group and reduces Fehling’s solution and Tollen’s reagent are called reducing sugars.
e.g. all monosaccharides, maltose and lactose.
11. Non-reducing sugars – Those sugars which do not have free aldehydic or ketonic
group and do not reduce Fehling’s solution and Tollen’s reagent are called nonreducing sugars. e.g. sucrose.
12. Glucose – Glucose has one aldehydic group, one primary alcoholic ( - CH2OH ) and
four secondary alcoholic group ( - CHOH ) and gives reaction with
(i) HCN – Glucose reacts with hydrogen cyanide to give glucose cyanohydrin
CH2OH – (CHOH)4 – CHO + HCN  CH2OH - (CHOH)4 – CH(OH)(CN)
(ii) Tollen’s reagent – Glucose reduces Tollen’s reagent to metallic silver.
CH2OH – (CHOH)4 – CHO + Ag2O  CH2OH - (CHOH)4 – COOH + 2Ag
(iii) Glucose on prolong heating with HI forms n-hexane. Suggesting all the
six carbon atoms in glucose are linked linearly.
HI
CH2OH – (CHOH)4 – CHO
CH3- (CH2)4 – CH3
Red P
n-hexane
13. Proteins – Proteins are complex ployamides formed from amino acids. They are
essential for proper growth and maintenance of body. They are long polymers of
amino acids linked by peptide bonds.
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14. Myoglobin – It is a protein which stores oxygen in muscle tissue. It consists of only
single polypeptide chain associated with heme unit.
15. Amino acids – The compounds, which contain carboxylic acid group and amino
group, are called amino acid. It forms proteins.
16. Essential amino acids – Those amino acid which are not synthesized by our body
called essential amino acid. Example: valine, lysine, etc.
17. Non- Essential amino acids – Those amino acid which are synthesized by our body
called non-essential amino acid. It is also called dispersible amino acids. Example:
glycine, alanine, etc.
18. Enzymes – They are essential biological catalysts, which are needed to catalyse
biochemical reaction. Example: maltase, lactase. Almost all enzymes are globular
proteins. They are highly specific for a particular reaction and for a particular
substrate.
1 MARK QUESTIONS
Q. 1. Which carbohydrate is called grape-sugar ? Give its condensed structural
formula.
Ans.
Glucose is called grape-sugar ? Its condensed structural formula is :
CHO
|
(CHOH)4
|
CH2OH
Q. 2. Which of the following is
-amino acid ?
Histidine, trypsin, cysteine, proline
Ans.
Trypsin, it is a protein made from amino acids.
Q. 3. Write the structure of Zwitter ion formed from Alanine.
Ans.
COO–
|
CH3 — C — H
|
NH3+
Q. 4. Wreite the name and structure of the simplest amino acid which can show
optical activity.
Ans
COOH
|
CH3 — C — H
|
NH2
Alanine
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Q. 5. How many hydrogen bonds are present between (i) A and T (ii) C and G in a
double helix structure ?
Ans.
(i)
between A and T there exist two hydrogen bonds :
A----------T
----------
(ii)
between C and G there exist three hydrogen bonds :
---------C---------- G
----------
Q. 6. When RNA in hydrolysed there is no relationship among the quantities of four
bases obtained like DNA. What does this fact indicate about structure of RNA ?
Ans.
This indicate that RNA has a single strand structure.
Q. 7. Where does the water present in the egg go after boiling the egg ?
Ans.
After boiling, the water soluble globular protein of egg while get denatured and it
cogulates into hard and rubbery insoluble mass.
Q. 8. Which vitamin is helpful in healing wound and cuts ? What is the chemical name
of this vitamin ?
Ans.
Vitamin-C is helpful in healing wounds and cuts, its chemical name is Ascorbic acid.
2 MARKS QUESTIONS
Q. 1.
Ans.
-helix structure of proteins ?
-helix structure, polypeptide chain of amino acids coils as a right handed screw
because of the formation of all possible Hydrogen bonds between — NH group at
each amino residue and > C = 0 group of adjacent turn of helix.
Q. 2. Name the water insoluble fraction of Starch. Name the monomer of this.
Ans.
-glucose.
Q. 3. What are the products of hydrolysis of (i) lactose (ii) sucrose. Also name the
Ans.
enzyme used for reaction.
Lactase
(i) C12H22O11 + H2O ————
Lactose
Invertase
(ii) C12H22O11 + H2O ————
Sucrose
H12O6 + C6H12O6
6
glucose
glactose
H12O6 + C6H12O6
6
glucose
fructose
Q. 4. How will you prove that all the carbon atoms of glucose are in straight chain ?
206
Ans.
The reaction of glucose with HI gives n-hexane and it proves that all sin-carbon-atom
are in straight chain.
HI
Ho CH2 — (CHOH)4 — CHO ———
— (CH2)4 — CH3
3
Heat
n-hexane
Q. 5. Enumerate two reactions of glucose which cannot be explained by its open chain
structure.
Ans.
(i) Glucose does not give Schiff‘s Test although it contains aldehyde group.
(ii) Glucose does not form crystaline product with NaHSO3.
Q. 6. B-complex is an often prescribed Vitamin. What is complex about it ? What is its
usefulness ?
Ans.
It is a type of Vitamin which contains B1, B2, B6 and B12. It required to release energy
from food and to promote healthy skin and muscles. Its deficiency causes beri-beri
(Vitamin B1) and Anaemia (Vitamin B12).
Q. 7. What are anomers ? Give two points of difference between two anomer of
glucose.
Ans.
The pair of optical Isomers which differ in the orientation of H and OH gp only at C 1
Carbon atom are called anomers.
Difference between two anomers of glucose :
-D (+) glucose
-D (+) glucose
(1) The specific rotation is + 111°.
(1) The specific rotation is +
19.2°.
(2) The – OH gp at C1 is below the plane.
(2) The – OH gp at C–1 is above the
plane.
Q1. What monosaccharides could be obtained from the hydrolysis of each of the following:
(a) Sucrose (b) Cellulose (c) Starch (d) Glycogen (2 marks)
Q2. Why is ATP an energy rich molecules? (2 marks)
Q3. Shown below is a molecule of ATP. ATP on hydrolysis form ADP. Indicate the point of
cleavage in ATP molecule when it forms ADP on hydrolysis. (2 marks)
207
Q4. Why is sucrose a non-reducing sugar? (2 marks)
Q5. Why are maltose and lactose reducing sugars? (2 marks)
Q.1 What monosaccharides could be obtained from the hydrolysis of each of the following:
(a) Sucrose (b) Cellulose
(c) Starch (d) Glycogen (2 marks)
1. (a) Glucose and fructose
(b) Glucose
(c) Glucose
(d) Glucose
Q.2 Why is ATP an energy rich molecules? (2 marks)
2. ATP is energy rich molecule because of the presence of four negatively charged oxygen
atom very close
to each other. These four negatively charged oxygen atoms experience very high repulsive
energy. When
ATP is hydrolysed, the hydrolysis of bonds results in decrease in repulsive forces and
consequently a
large amount of energy is released.
Q.3 Shown below is a molecule of ATP. ATP on hydrolysis form ADP. Indicate the point of
cleavage in ATP
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molecule when it forms ADP on hydrolysis.
(2 marks)
3. During hydrolysis, the bond between second and third phosphate can be easily broken to
form ADP the
point of cleavage is shown in diagram below. Point of cleavageto form ADP.
Q.4 Why is sucrose a non-reducing sugar? (2 marks)
4. Sucrose is a non-reducing sugar because anomeric
carbon of both monosaccharides are
involved in
glycoside or acetal formation.
Q.5 Why are maltose and lactose reducing sugars? (2 marks)
5. Maltose and lactose are reducing sugars because in one of the monosaccharide units there
is a homiacetal
group that can be opened to give free –CHO group.
FOR BELOW AVERAGE
Questions of one mark
Q
1. What are biomolecules ?
2. Give two examples of monosaccharies.
3. Give name the monomers which constitutes lactose .
4. Give the name of linkage which found between the monomer units of sucrose
5. Name the two essential amino acids.
6. Name the source of vit. E.
7. Who proposed the double helical structure of DNA ?
FOR AVERAGE
1. What are monosaccharides ?
209
2. Give an example of branched polysaccharide .
3. What are non-reducing sugars ?
4. What happen when glucose is treated with HI ?
5. Draw the pyranose structure of alpha-D-glucose
6. Name the components of starch .
7. Where glycogen is present in human body ?
8. Name the primary constituents of proteins.
9. What are the expected products of hydrolysis of maltoses10.Which type of proteins constitute enzymes ?
FOR ABOVE AVERAGE
1. Why is maltose a reducing sugar?
2. Give the name of most abundant carbohydrate present on the earth.
3. Write down the reaction between acetic anhydride and glucose.
4. Give the formula of glycine .
5. What is the function of glycogen in human body.
6. Draw the structure of zwitter ion.
7. What do you mean by denaturation of proteins ?
8. What is the activation energy for acidic hydrolysis of sucrose ?
FOR BELOW AVERAGE
Questions of two marks
1. What are main functions of hexose sugars?
2. Give two applications of carbohydrates in plants.
3. Give the functions of cellulose.
4. How we classify the amino acids according to the need in the body ?
5. What happen when denaturation of proteins takes place? Explain with examples
6. Give importance and source of vit. B.
7. Explain peptide linkage
FOR AVERAGE
1. Draw the furanose and pyranose structure of hexose sugar.
2. Explain acidic and basic amino acids with examples
3. Differentiate between fibrous and globular proteins.
4. What is the difference between nucleoside and nucleotide.
5. How can we classify the vitamins .Give source of vit A & C.
210
FOR ABOVE AVERAGE
1. Draw the hawarth projection formula for maltose.
2. Differentiate between tertiary and quaternary structure of
proteins
3. Differentiate amylase and amylopectin
4. What are functional difference between DNA and RNA.
5. Name the factors affecting enzymatic activity.
FOR BELOW AVERAGE
Questions of three marks
1. How carbohydrates are classified according to monomer units present in them ?
2. Give two chemical reaction for preparation of glucose.
3 .How starch is different from cellulose ?
4. What do you mean by secondary structure of proteins ? Explain.
5. Give main source of and function of vit. B & vit. D
FOR AVERAGE
1.
Give the reaction of glucose with (i) Br2 (ii) NH2OH (iii) HNO3.
2.
How can you show that glucose have cyclic structure?
3.
Explain the alpha –helical and beta pleated structure of proteins.
4.
Explain the mechanism of enzyme catalysis.
5. Give difference between DNA & RNA. Give applications of RNA
FOR ABOVE AVERAGE
1.
What do you mean by DNA finger-printing ? Enlist their uses.
2.
How do you explain the amphoteric nature of amino acids ?
3.
Give biological application of DNA.
4.
Two strands of DNA are not identical but complimentary. Explain.
5.
Supply of vit. C needs continue for human body. Why ?
Some Important Questions with Answers
1.
Give one example of branched polysaccharide.
Ans. - Starch
2.
Why sucrose is known as invert sugar ?
Ans- Because after hydrolysis optical rotation is changed.
211
3.
What D &L denotes ?
Ans- Relative configuration of particular hydrocarbon with respect to glyceraldehydes.
4.Amino acids have relatively high melting point as compared to corresponding
haloacids. Explain.
Ans: Amino acids have relatively higher m.p as compared to corresponding haloacids.
We know that amino acids have amino (basic) group and carboxylic (acid) group. A
proton which comes from carboxylic group is accepted by amino group. Thus amino
acid exists in the form of dipolar ion i.e. zwitter ion. Due to this ionic form amino
acids become crystalline solids and due to this salt like structure show higher melting
point.
5
Ans:
Amino acids are amphoteric in nature/behavior. Explain.
Due to presence of dipolar ion i.e. zwitter ion, amino acid can reacts with acid and
bases.
6
Why can’t vitamin C be stored in our body? Or
Why except B12 or vitamin B and C can’t be stored in our body?
Vitamin B except B12 & C are water soluble vitamins. Therefore these vitamins are
regularly supplied from outside because they are readily excreted in urine and can’t be
store in our body.
Ans:
7
Ans:
What is glycogen? Why glycogen is also known as animal starch?
Glycogen is polysaccharide of glucose. Carbohydrates are stored in animal body as
glycogen. It is found in liver, brain. It is also known as animal starch because its
structure is similar to amylopectin and is highly branded.
8.
Ans:
What is DNA fingerprinting? Enlist their uses.
It is well known fact that every individual has unique fingerprints. These occur at the
tips of the fingers and have been used for identification for a long time. But these can
be altered by surgery. A sequence of base on DNA is also unique for a person and
information regarding this is called DNA fingerprinting. Following are the uses of
DNA printing.
in forensic laboratories for identification of criminals.
To determine paternity of an individual.
To identify the dead bodies in any accident by comparing the DNA’s of parent or
children.
To identify racial groups to rewrite biological evolution.
a)
b)
c)
d)
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UNIT – 15
POLYMER
Polymer: - Polymer compounds are very high molecular masses formed by the
combination of Simple molecules by covalent bonds.
Ex. nCH2=CH2 —→ (— CH2— CH2—)n
Ethene
Polythene
( Monomer)
( Polymer)
Homopolymer :- A polymer from one type of monomers is called homopolymer
Ex. Polythene.
Copolymer :-
A polymer from two or more type of monomers is called copolymer
Ex. Terylene or decron, Nylon 66.
Classification of polymers :- Polymer are classified in number of ways described as
follows
(A) Natural polymers :- The polymers obtained from nature (plants & animals ) are
called natural polymers. Eg. Strarch, Cellulose, Natural
rubber and Proteins etc.
(B) Synthetic Polymers :- The polymers which are prepared in laboratory are called
Synthetic or man made polymers. Eg. Polythene, PVC,
Nylon, Teflon.
Classification of polymers on the basis of structure
(a) Linear polymer :- These are polymers in which monomeric units are linked together
to form linear chains. Eg. Polythene, PVC, Teflon
(b) Cross linked polymers:- These are polymers in which the monomers unites are crosslinked together to form a three- dimensional network. Ex. Bakelite, Synthetic rubber.
(c) Branch chain polymers:- The polymers in which monomers are joined to form long
chains with side chains. Ex. Low density polythene.
Classification of polymers on the basis of molecular forces.
(a) Elastomers:- The polymers that have elastic character like rubber are called
elastomers.
213
Ex. Natural rubber
(b) Fibers:- They have strong intermolecular forces between the chains. These forces are
either H-bonding or dipole interactions.
Ex. Nylon-66, dacron, silk etc.
(c) Thermoplastics :- These are the polymers which can be easily softened repeatedly
heated and hardened when cooled with little change in their properties.
Ex. Polythene, PVC.
(d) Thermositting Polymers :- These are the polymers which undergo permanent
change on heating. They become hard and infusible on heating. Ex. Bakelite. Etc.
Classification of polymers on the basis of mode of synthesis.
(a) Addition polymers:- A polymers formed by direct addition of repeated
monomers without the elimination of by product molecule is called addition
polymers. Ex. Polythene, PVC,PTFE, etc
(b) Condensation polymers :- A polymer formed by the condensation of two or
more than two monomers with the elimination of simple molecules like water,
ammonia, hydrogen sulphide, alcohol, etc. is called condensation polymer. In
this each case monomer generally contains two functional groups. Ex. Nylon
66 and Nylon 6, Bakelite, Daecron, etc.
High density polythene (Zeiglar Natta catalyst)
It is prepared by heating of about 333- 343K under a pressure of 6-7 atm in the
presence of Zeiglar Natta catalyst ( Tri ethyel aluminium Titenium tetra chloride )
333-343K, 6-7 atam
n CH2 = CH2 ——————————→
(— CH2— CH2)n
Zeiglar Natta catalyst
Uses:- It is used in manufacture of containers ( buckets , Dust bin, etc)
It is used in manufacture of different house wares pipes etc.
214
Polymer (origion Greek) poly means many, mers means part
Defination: Very large molecules having high molecular mass (103-107 u)
Macromolecule: other name for polymer
Polymerization: the process of formation of polymer
Classification of polymer:
1. Based on sources:
a. Natural: from plants and animal sources for example starch
cellulose rubber etc.
b. Semi synthetic: polymer with modification of natural polymer
for example rayon.
215
c. Synthetic prepared in laboratory for example nylon, ethylene
etc.
2. Based on structure of polymer:
a. Linear: long and straight change for example PVC
b. Branched: linear polymer with branches for example low
density plastic
c. Cross linked or network: contains strong covalent bond
between various linear polymer for example Bakelite, melamine
.
3. Based on mode of polymerization:
a. Addition polymerization: repeated unit of monomers having
unsaturation in its structure
i.
Homo-polymer: when monomer is of same kind for
e.g. polyether, PVC
ii.
Copolymer polymer having two different monomers
for e.g. Buna-S, Buna- N
216
b. Condensation polymerization: repeated condensation reaction
between mostly two different bi-functional monomer unit with
the elimination of water alcohol etc. for e.g. nylon 66, Dacron
4. Based on molecular forces
a. Elastomers: rubber like solid with stretchable property up to
some extant for e.g. Buna-S, Buna N.
The force between chains is weak Vanderwall force.
b. Fiber: thread like solid with high tensile strength
Force between chain is hydrogen bonding for e.g. polyester,
nylon 6, 6
c. Thermo plastic: linear or slightly branched long chain molecule,
soften on heating. For example polyethylene PVC. Etc.
d. Thermosetting: cross linked molecule once setup can’t not be
soften on heating. for example Bakelite, melamine
Decreasing order of force of attraction between the chains
Thermosetting > Fiber> Thermoplastic> Elastomers
Difference between thermoplastic and thermosetting
Type of polymerization reaction:
Addition polymerization or chain growth:

Due to increase in the length of chain

Due to addition of similar monomer it is addition polymerization e.g. polythene
Steps of addition polymerization with free radical mechanism:
Chain initiation :by molecules like Benzoyl peroxide(C6H5OCOOCOC6H5)
Chain propagation : through free radical
Chain terminalisation :combination of free radicals.
Q .which catalyst is used in HDP like polythene?
A. Zeigler-Natta catalyst.
Q. what is Zeigler-Natta catalyst?
A. Triethylaluminium (C2H5)3Al and titanium tetrachloride(TiCl4)
Condensation polymerization or step growth:
Since each step produces a functionalized species and is independent of each other,this is
called step groth.
As,monomers with different functional groups are condensed,so, it is Condensation
polymerization.
e.g. Condensation between Hexamethylene-di-amine[ NH2-(CH2)6-NH2] and Adipic
acid[HOOC-(CH2)4-COOH in Nylon-6,6.
Q.why nylon is named as such?
A.It is Acroname of New-York and London.
Q .why is it called 6,6?
A. As the monomers contain six carbon each i.e. Hexamethylene- di - amine[ NH2-(CH2)6NH2] and Adipic acid [HOOC-(CH2)4-COOH.
Q. what is the monomer of nylon-6?
217
A. Caprolactum [NH2-(CH2)5-COOH].
Q. Which properties of polyester make it more usable fibre ?
A. Crease resistant, Dip dry, Blending with other natural fibre like cotton, wool etc.
Q. what are different kinds of blended polyester?
A. Tere-Cot (Terelene with Cotton)
Tere-Wool (Terelene with Wool)
Tere-Silk (Terelene with silk)
Q.A Tere-Cot shirt is marked as 60-40.What does it mean?
A. It contains 60% terelene and 40% cotton.
Q. What is the source of natural rubber?
A. Milky latex from rubber tree.
Q. Which countries are the greater producer of rubber ?
A. Sri –Lanka, Malaysia, Indonesia, South America.
Q. What is the monomer of natural rubber?
A. Isoprene (2-Methylbut-1,3-diene).
Q. What is a vulcanized rubber?
A. When natural rubber is heated with sulphur, it forms cross-linkage of sulphur between
linear polymers. Such rubber is called vulcanized rubber.
Q. How is vulcanized rubber than natural rubber?
A. vulcanized rubber
Natural rubber
1. Non sticky
1.Sticky
2. Linkage of sulphur
2.Linear chain
3. Less absorbent of water
3. Absorbs water
4. Easily oxidized
4. Less oxidized
5. Working temp. -40 to 120 0C
5. Working temp. 40 to 120 0C
6. Tough and hard
6. Soft
Q. What is Neoprene rubber?
A.2-Chlorobut -1,3-diene
Q. Why is Neoprene rubber non –inflammable ?
A. Due to presence of chloro group .
Q. What is Bio-degradable polymer?
A. The polymers which can be degraded by microbes. e.g. PHBV and Nylon-2 ,Nylon-6.
Q. Give full form of PVC, PTFE,PHBV?
A. PVC: Poly Vinyl Chloride
PTFE : Polytetrafluoroethene
PHBV : poly-β-hydroxybutyrate-co-β-hydroxy valerate.
Name of polymer
Monomer
structure
Uses
Addition polymer
Polythene
Ethene
CH2=CH2
Carry bags,toys,buckets
PTFE(Teflon)
Tetraflouroetene CF2=CF2 Non -stick cookware,Oilseals,gasket
Polyacrylonitrile
Acronitrile
CH2=CHCN Substitute for wool(mink)
Polypropene
Propene
CH2=CHCH3 ropes,pipes
Polystyrene
Styrene
CH2=CHC6H5 Insulator,Wrapping material
PVC
Vinyl Chloride
CH2=CHCl Raincoats,flooring,water pipes
(Thermocol)
Condensation polymer
Nylon 6,6
Adpic Acid
COOH-(CH2)4COOH
Parachute Rope, Socks,
Stockings
Hexamethylene- di – amine [ NH2-(CH2)6-NH2]
Nylon 6
Caprolactum
[NH2-(CH2)5-COOH] Parachute Rope, Socks,
Stockings
Dacron
Ethylerne Glycol
OH-(CH2)2OH
Terephthalic Acid
COOH-C6H4-COOH Blended Fibers, Socks, Stockings
218
Bakalite
Formaldehyde
HCHO
Electric switches, Handel of
pressure cooker
Phenol
C6H5OH
Melamine
Formaldehyde
HCHO
Unbreakable Crockery
Melamine
C3N6H6
1 MARK QUESTIONS
Q. 1. Write an equation for the Chemistry involved when a drop of hydrochloric acid
make a hole in nylon stockings.
Ans.
The (— CO — NH —) amide bond in nylon gets hydrolysed.
Q. 2. Fibres are of crystalline structure. Why ?
Ans.
Fibres have strong intermolecular forces of attraction which leads to close packing of
their chains and impart crystalline structure.
Q. 3. Which artificial polymer is present in bubble gum or chewing gum ?
Ans.
Bubble gum or chewing gum contains synthetic Styrene-butadiene rubber.
Q. 4. Name the polymer used for making medicinal Capsule.
Ans.
PHBVUC Polyhydroxy butyrate-CO- -hydroxy valerated.
Q. 5. Which polymer is used in making electrical goods and why ?
Ans.
Bakelite because of its electrical insulator property.
Q. 6. Is (CH2 — CH — C6H5)n a homo polymer or a copolymer.
Ans.
It is a homo polymer and the monomer from which it is obtained in styrene C6H5 CH
= CH2.
Q. 7. Which colligative property is used to determine the molecular masses of the
polymers ?
Ans.
Osmotic pressure is the colligative property used to determine the molecular masses
of polymer.
2 MARKS QUESTIONS
Q. 8. Write the names of monomers and structure used for getting the following
polymers ?
PmmA, PVC, Teflon
Ans.
(i)
PVC — monomer — Vinylchloride, CH2 = CH — Cl
219
(ii)
Teflen — monomer — Tetrafluoro ethylene, F2C = CF2
(iii)
PMMA — monomer — Methyl methacrylate,
CH3
|
CH2 = C — COOCH3
Q. 9. Identify the monomer in the following polymer structures :
Ans.
(i)
HOCH2 — CH2OH
(ii)
ethane-1, 2-dial
Melamine
N
H2N
N
and
HOOC —
O
— COOH
Benzene-1, 4-dicarboxylic acid
Formaldehyde
NH2
N
and
HCHO
NH2
Q. 10. How do double bonds in the rubber molecule influence their structure and
reactivity ?
Ans.
Due to presence of double bond the rubber molecule show cis-configuration because
which the polymer chains can not come close to each other and get boiled but as they
have their hindrance problem. This causes them to have weak Vander Waals
intraction between the molecules consequently they get elastic property, solubility in
organic solvents etc.
Q. 11. Why are the No. of 6, 6 and 6 put in the names of nylon-6, 6 and nylon-6 ?
Ans.
The no. of 6, 6 in nylon-6, 6 implies that both the monomers of nylon-6, 6 namely
hexamethylene diamine and adipic acid contain six carbon atom each. The no. of 6 in
nylon-6 contains six carbon atom each. The no. of 6 in nylon-6 indicates that its
monomer Caprolectum has six carbon atom in its molecule.
Q. 12. Could a copolymer be formed in both addition and condensation polymerisation
or not ? Explain.
Ans.
Yes, though copolymers are mostly addition polymers like styrcue butadiene rubber
and butyl rubber, the term can be used for condensation polymers also eg. for
example.
220
(i)
Addition polymerisation of styrene and butadiene form copolymer styrene
butadiene rubber.
n CH2 = CH — CH = CH2 + n C6H5CH = CH2 ———>
C6H5
(— CH2 — CH = CH — CH2)n — CH2 — CH —)n
(ii)
Condensation polymer became thylene diamine and adipic acid form nylon-6,
6.
n HOOC — (CH2)4 — COOH + n H2N — (CH6) — NH2 ———
adipic acid
hexamethylene
— n H2O
diamien
(— NH (CH2)6 — NH — CO — (CH2)4 — CO —)n
Q. 13. Arrange the following polymer in increasing order of their molecular forces :
Ans.
(a)
nylon-6, 6, Buna-S, polythene.
(b)
nylon-6, Neoprene, polyvinyl chloride.
(a)
Buna S < Polythene < Nylon-6, 6.
(b)
Neoprene < Polyvinyl Chloride < Nylon-6.
Q. 14. Explain the difference between Buna— S and Buna — N.
Ans.
Buna — N is a copolymer of 1, 3-butadiene and acrylo nitrile, Buna — S is a
copolymer of 1, 3-butadiene and styrene.
Q. 15. Why should we always use purest monomer in free radical mechanism ?
Ans.
Monomer has to be as pure as possible because the presence of any other molecule
during free radical polymerisation can act chain initiator or inhibitor which will
interfere with the normal polymerisation reaction.
BELOW AVERAGE
01 Marks Questions
Q1: Write the monomer of Polyvinyl chloride.
Ans: Vinyl chloride.
Q2: Write the polymer of phenol and formaldehyde that is used for making combs.
Ans: Bakelite, (a) Phenol (b) Formaldehyde
Q3: Write the monomer of Glyptal polymer.
Ans: (a) Ethylene glycol (b) Phthalic acid
Q4: What is Synthetic polymer ? Give one example.
Ans: Man made polymers are called Synthetic polymers. Polyethene, PVC.
Q5: What is approximate molecular mass of Polymer ?
Ans: It is approximate molecular mass 103-107u.
Q6: What are monomers of Nylon 6.6
221
Ans: (a) Hexamethylene diamine and adaipic acid.
Q7: Write the Mnomers of Buna-s.
Ans. 1, 3 butadine and styrene
Q8: What are carben fibers ? Give two examples.
Ans: A long thread solid posses high tensile strength. Eg. Terrylen and Nylon.
Q9: What is addition polymerization ?
Ans: A chemical reaction in which monomer unit is repeated to produce a molecule without
losing of molecule like water or ammonia. Ex. Polythene, Polyvinyl chloride
Q10: Give one example of each for low density polythene and high density polthene.
Ans: LDP. Squeze bottle and toys
HDP. Buckets, dustbin, bottles.
02 Marks Questions
Q1. Differentiate between thermo plastic and thermo setting Plastic.
Ans:
Thermo Plastic Polymer
Thermo setting Polymer
1. Liner or slightly branched long
1. Cross linked or heavy branched
change molecules
molecule
2. Posses intermediate inter
2. This can not be reused
molecular force of attraction.
3. Ex. Bakelite, Urea-Formal dehyde
3. Ex. Polythene, Polystyrene
resins
Q2: Give two Polymerisation reactionds for each.
(a) Addition polymers and
(b) Condensation Polymers
Ans: Addition Polymers :nCH2=C
(
CH2
CH2)n
Condensation Polymers
n H2N(CH2)6 NH2 + n HOOC(CH2) COOH [NH(CH2)2NHCO(CH2)4CO)]n+nH2O
Q3: What are elastomers ? Give one example.
Ans: The rubber like material which are elastic in nature. Eg. Natural rubber.
Q4: Write the chemical structure of Glyptal
Ans: Ethylene Glycol-OHCH2CH2OH
Phthalic acid
COOH
COOH
222
Q5 : Define Copolymers.Give two examples.
Ans: The Polymer in which repeating units are combined with two or more differents types of
monomers
Eg. Polyster and terrylene(Dacron)
Q6: What is Neoprene? Mention any two uses.
Ans: A synthetic rubber made by chloroprene. It is a synthetic rubbers.
Uses: Conveyor belts in coal mines & gaskets.
Q7: Define Synthetic rubbers. Whether it is homopolymer or co-polymer?
Ans: Any Vuclanised rubbers like polymers. Which is getting stretched to about twice lengths
and returns to the original shape and size as the external force is released.
Eg. Homopolymers.
Q8: Explain the difference between Buna-N and Buna-S.
Ans: Buna-N:1. It is obtained by the copolymerization of 1,3 – butadiene and acrylonitrile in the
presence of a Peroxide Catalyst.
2. It is resistant to action of Petrol.
3. It is used in making in oil seals tank lining etc
Buna-S:1. It is obtained by the Copolymerisation of buta diene and styrene.
2. It is used for automobile tyres.
Q9: What do you understand by Vulcanised Rubbers?
Ans: When natural rubber is heated with sulpher at 373K to 415K. The vulcanized rubber
has excellent elasticity with low water absorption tendency. The probable structure of
vulcanized rubber molecule as follows;
CH3
CH2
CH2
C
CH
S
S
C
CH
CH2
CH2
CH3
Q10: Explain the terms:(a) PDI
223
(b) PMMA
Ans: PDI :- Polydispersity Index: The ratio of the mass average molecular mass is called
PDI
PMMA:- It is polymer of ethyl methylacrylate.
AVERAGE STUDENT
Very Short Answer 01 Marks
Q1: Write the monomer of Polystyrene?
Ans: Styrene.
Q2: Write the structure of monomer of Polystyrene?
Ans: Ar
CH = CH2
Q3: Write the monomer of Neoprene?
Ans: Chloroprene.
Q4. Write the name of monomers of Melamine-Formaldehyde.
Ans: Melamine and Formaldehyde.
Q5: What is the monomer of Polyacrylonitrile?
Ans: Acrylonitrile
Q6. Write the polymer of Tetrafluoro ethane.
Ans: Teflon.
Q7: Define Copolymers?
Ans: The polymer is obtained by two or more different monomers.
Q8: Is ( NH
CHR
CO )n is a homopolymer or a copolymer?
Ans: It is a Copolymer.
Q9: Why should one always use purest monomer in free radical polymerization?
Ans: The impurities present in monomer may combine with free radical that slow the rate
of polymerization.
Q10: What is meant by PTEE ? Give its popular name.
Ans: Polytetra fluoroethylene. Teflon.
02 Marks Questions
Q1: Differntiate between addition polymer and condensation polymer.
Ans:
Addition Polymer
Condensation Polymer
1. Large no. of unsaturated monomers
1. Large no. of monomers having different
combine together to form polymers Eg.
functional group with release of small
Polythene and polystyrene
molecule like water combine to form
polymers. Eg. Nylon and Teflon
224
Q2: Distinguish between thermoplastic and thermo setting polymers.
Thermo plastic polymer
Thermo setting plastic
1. The inter moliculer forces are
1. The inter molecular force are
intermediate.
maximum.
2. there are no cross linkage bond
2. They are cross linkaged.
3. Eg. Polythene and polystyrene
3. Eg. Bakelite
Q3: Distinguish between Homopolymers and Copolymers.
Homopolymers
Copolymers
1. The polymers which are form by only
1. The polymers which are form by two or
one type of monomers Eg. Teflon & PVC
more types of monomers .Eg. Terrylene &
Buna-S
Q5: Depict a free radical mechanism of addition polymerization of isoprene.
.
R* + CH2 = C—CH = CH2
—→
R — CH2— C—CH = CH2
|
|
CH3
CH3
↕
.
R — CH2— C=CH— CH2
|
CH3
↓
CH3
CH2 ----and
C=C
R —— CH2
H
225
Trans 1, 4 structure
CH3
H
C=C
R —— CH2
CH2 ---
Cis 1, 4 structure
Q6: Why are the numbers 6, 6 and 6 put in the name of Nylon 66 and Nylon 6?
Ans: Nylon 66 means that it is formed by two monomers each containing six carbon
atoms. While in Nylon 6 means that it is formed by a monomer containing six carbon
atoms.
Q7: Explain the structural difference between polyacrylats and polyesters
Ans: PolyacrylateCH2 = C—R
|
COOR
And Polyster –
— C—O—
||
O
Q8: What is PHBV ?
Ans: It is a poly hydroxyl butyrate. Co-β- hydroxyl Valerate.
It is copolymer of 3- hydroxyl butanoic acid and 3- hydroxyl pentanoic acid in which
monomer units linked by ester linkage.
Q9: Explain the difference between Buna-N and Buna-S.
Ans:
1.
Buna-N:It is optained by the copolymerization of 1,3 – butadiene and acrylonitrile in the
presence of a Peroxide Cxtalyst.
2.
It is resistant to action of Petrol.
3. It is used in making in oil seals tank lining etc
226
Buna-S:1. It is obtained by the Copolymerisation of buta diene and styrene.
2. It is used for automobile tyres.
Q10. Arrange the following Polymers in increasing order of inter molecular forces.
(a) Nylon 66, Buna-S and Polythene
(b) Nylon 6, Neoprene and PVC
Ans: (a) Buna-S, Polythene and Nylon 66.
(b) Neoprene, PVC and Nylon 6.
BRIGHT STUDENT
Very Short Answer Question: 01 Mark
Q1. What does stand for PDI
Ans: Polydispersity Index.
Q2. Write the name of neoprene polymer?
Ans: Chloroprene
Q3: Write the name of reagent use for initating a free radical chain reaction.
Ans: Tertiary Butyle peroxaide.
Q4: What are carbon fiber?
Ans: A long thread solid posses high tensile strength. Eg. Terrylen and Nylon
Q5: A polymer is use for making combs, name the polymer.
Ans : Baklite
Q6: A polymer is used for making paints and lacquers, name it.
Ans: Glyptal.
Q7:
Name the monomer of polyvinylchloride.
Ans: Vinylcloride.
Q8: Arrange in increasing order of inter molecular forces.
Nylon 66, Buna-S and polyethene.
Ans: Nylon 66 > polyethene> Buna-S
Q10. Arrange the following Polymers in increasing order of inter molecular forces
Nylon 6, Neoprene and PVC
Ans : Nylon 6, PVC and. Neop
SHORT ANSWER
02 Marks
Q1: Write the structure of monomer of polystyrene.
Ans: Styrene. Ar— CH=CH2
Q2: Clasify the polymers as addition and condenstation polymers.
Nylon 66, Buna-S, Polythene.
Ans: Nylon 66 condensation polymer.
227
Buna-S and polythene—Addition polymer.
Q3: Classify the polymers as addition and condensation polymer—
Nylon 6, Neoprene and PVC.
Ans: Nylon 6 —Condensation polymer
Neoprene and PVC— Addition polymer.
Q4: What is Biodegradble polymer?
Ans: The polymers which are Biodegrable and not causes any environmental population is
called Biodegrable . Eg. PHBV and Nylon 2 Nylon 6.
Q5. Who do double bonds in rubber molecules influence there structure and reactivity.
Ans: Greater the number of double bound the rubber will be less reactive because of
greater inter molecular forces of attraction among the monomers.
Q6: What is PHBV? Give its one use.
Ans: It is a poly β hydroxybutyrate—co β hydroxyl valerate .
It is obtain by the copolirisation of 3— hydroxyl butanoic acid and 3— hydroxy
pentanoic acid.It is used in speciality packaging , orthopaedic devices.
228
UNIT -16
CHEMISTRY IN
EVERYDAY LIFE
STUDY MATERIAL
1.
2.
3.
4.
5.
6.
7.
Tranquilizers- they are chemicals which are used for treatment of mental diseases
.they act on higher centers of central nervous system. Example Equanil , Seconal
,luminal or Barbituric acid etc.
Antibiotics- Chemical substances which are produced by micro-organisms(such as
moulds and bacteria) and are capable of destroying other micro organism are called
antibiotics .Example penicillin ,Ampiciline.
Broad spectrum Antibiotics- Antibiotics which are effective against several different
types of harmful micro-organisms and thus, capable of curing several infections are
called broad spectrum antibiotics Example chloromycetin, Tetracycline .
Antihistamins :- Antihistamins are amines which are used as drugs to control the
allergy effects produced by histamines. Example
diphenyl hydrazine
or
promethazine .
Antipyritics :- Chemical substances which are used to bring down the body
temperature in high fevers are called antipyretic. Example Aspirin, paracetamole or
Analgin etc.
Disinfectants :- Chemical substances which kill micro –organism or stop their growth
but are harmful to human tissue are called disinfectants .example 1.% solution of
phenol (2)SO2 in very low concentration.
Analgesics:- chemical substance used for relieving pain are called analgesics . Example
novalgin, profanes etc.
8.
Antacids:- These substances react with hydrochloric acid present in the stomach
neutralize it partially. Example magnesium hydroxide , Aluminum hydroxide etc . it
can be used as tablet or in aqueous suspension .
9.
Antifertility drug Those drug which control the birth of the child are called antifertility
drugs. Steroids are the active ingredients of the pill functioning as an antifertility
agents. some commonly used pills contains a combination of ethynlestradiol and
norethindrone.
10.
Anticancer agents Those drugs which are used in treatment of cancer are used as
anticancer agents e.g cis-platin.
11.
Antimicrobials Disease in man and animal which may be caused by bacteria, virus
and other micro-organisms microbes. Any organism that causes diseases is called a
pathogen.
Many body secretions either kill the microbes or inhibit their growth. Examples are
lysozyme in tears, nasal secretion and saliva, fatty acids and lactic acid in sweat and
sebaceous secretions and hydrochloric acid in stomach.
The control of microbial diseases can achieved in three ways by
(1)A drug which kills the organisms in the body(bacteria) e.g penicillin
(Bactricidal).
(2) A drug which inhibits or arrests the growth of the organism (bacteriostatic)
(3) Increasing immunity and resistance to infection of the body.
Antibiotics are the class of drugs used as antimicrobials.
229
12.
Chemicals in food :Many chemicals are added to food for their preservation and
enhancing their appeal. These include flavouring , sweetness, dyes, antioxidants, etc.
13.
Preservatives—Those chemicals which prevent undesirable changes in flavour,
colour, texture, appetitic appeal during a storage are called preservative. They delay
these changes and prevent spoilage of food due to microvial growth. The most
common preservative is sodium benzoate(C6H5COONa) .
Class 1 preservatives-includes table salts, sugar , vegetable oil.
Class 2 preservative—sodium benzoate ,salts of sorbic acids and propanoic acid.
14.
Antioxidants--- Antioxidants are the important class of compounds which prevent
oxidation of food materials.
These compounds retard the action of oxygen on the food and thereby help in
preservation. These act as sacrificial materials i.e. they are more reactive towards
oxygen than the materials they are protecting. Example
(1) Butylated hydroxyl anisole (BHA)
(2) Butylated hydroxyl toluene (BHT)
15.
Artificial sweetness
sweetness are another type of food additives e.g ortho
sulphobenzimide (saccharin) is marketed as soluble sodium or calcium salt. It is 300
times sweeter than cane sugar. It is lifesaver for diabetic patients and is to control
intake of calories
16.
Soap—They are sodium or potassium salts of higher fatty acids like steric acids, oleic
acid and palmitic acid. Soaps are formed by heating fat with aqueous sodium
hydroxide solution the reaction is called saponification.
O
║
CH2-O-C-C17H35
O
CH2-OH
║
|
CH-O - C-C17H35
+3NaOH =→ CH-OH +3C17H35COONa
O
|
║
CH2-OH
Sodium stearate
CH2-O-C-C17H35
Glyceryl stearate
Glycerol
17.
Detergents: They are sodium or potassium salts of sulphonic acids e.g sodium alkyl
benzene sulphonate.
Types of Detergents:(i)
Anionic detergents:- When detergents bears an anion at the soluble end of the
chain, it is called anionic detergents.ex. It is used in toothpastes and house
hold work.
CH3(CH2)16CH2OH+H2SO4
CH3(CH2)16CH2OSO3H
NaOH(aq)
- +
CH3(CH2)16CH2OSO3 Na
- H2O
Lauryl alcohol
Lauryl hydrogen sulphate
230
Sodium lauryl sulphate
(Anionic detergents)
18.
Advantage of detergents:
(a)
They work well even with hard water.
(b)
They are more effective than soaps.
(c)
They can work well even with acidic water.
1.
(d) They can work well with woolen garments.
Chemotherapy : Treatment of diseases with the help of suitable drugs.
2.
Antipyretic : Reduce body temperature. Eg: Aspirin, Analgin.
3.
Analgesic : Pain Killer. Eg : Aspirin, Morphin, Heroin, Analgin.
4.
Tranquiliser : Decreases anxiety. Eg: Luminal, Seconal.
5.
Antibiotic
:
Cure
Chloroamphenicol.
6.
Antimalarial : Eg: Quinine, Chloroquinine.
7.
Antifertility Drug : Birth control. Eg: A pill containing estrogen and
progesterone.
8.
Preservatives : Chemicals used for preventing spoilage of food materials.
Eg: Sodium Benzoate.
9.
Artificial sweetening agent : Eg: Sacharin, Aspartame (methyl ester).
infections.
Eg:
Penicilin,
Tetracyclin,
10. Cationic Detergents : Eg: Trimethylstearylammoniumchloride.
11. Anionic Detergents : Eg: Sodiumdodecylsulphate.
12. Non-ionic Detergents : Eg: Pentaacrythritolmonostearate.
231
13. Biodegradable Detergents : Detergents having straight hydrocarbon chain and easily
decomposed by microorganisms. Eg: Sodiumlaurylsulphate, sodiumdodecyl
benzenesulphonate.
Important Questions :
16. What is understood by Chemotharepy.
17. Explain the terms with an example :
a) Antibiotic.
b) Tranquillizers.
c) Antihistamin.
d) Antipiratics.
e) Disinfactant.
18. Describe the following substances with one suitable examples :
a) Food Preservatives.
b) Sweetning agent.
c) Enzymes.
d) Antioxident.
19. Explain the following terms with one examples :
a) Biodegradable detergents.
b) Non-biodegradable detergents.
c) Anionic detergents.
d) Non- ionic detergents.
20. What are analgesic drugs. How are they classified and when are they usually
recommended for use.
21. Aspirin drugs help in preventing of heart attack, explain.
22. Diabetic patient are advised to take artificial sweeteners instead of natural sweetener.
Why?
23. Which type of detergent produce less foam.
24. Which type of detergent would you use to formulate toilet cleaner and Why?
25. Name the antiseptic you will add to soap to make it useful for control of pimples.
26. Discuss two ways in which drugs prevent attachment of natural substrate on active
site of enzymes.
232
27. Write one use of followings :
a) Zentac.
b) PCM.
c) Ranitidine.
d) Equanil.
e) Novestrol.
f) Morphin.
g) ChloroamPhenicol.
h) Bithional.
Hints :
5. Analgesics – Pain Killer
Classification – Narcotic – Morphin (to relief post operative pain)
Non Narcotic – Aspirin, PCM (to reduce fever , to prevent blood platelets coagulation)
6. Aspirin act as blood thiner and causes no coagulation.
7. The artificial sweetners are either non metabolized by body or do not produce
carbohydrate like glucose when metabolized.
8. Nonionic detergents
9. Cationic detergents
10. Bithional
12. Zentac – Antacid
PCM – Antipyretics, Analgesics
Ranitidin – Antacids
Equanil – Tranquillizer
NonVestrol – Antifertility drugs
Morphine – Analgesics
ChloroamPhenicol – Antibiotics
Bithional – Anticeptics
1 MARK QUESTIONS
Q. 1. Give an example of a chemical substance which can act both as an antiseptic and
disinfectant.
Ans.
Phenol.
Q. 2. What is the chemical name of antiseptic chloroxylenol ?
Ans.
4-Chloro-3, 5-dimethyl phenol.
Q. 3. Which alkaloid is used to control hypertension ?
Ans.
Reserpine.
233
Q. 4. Which alkaloid is used to treat malaria ?
Ans.
Quinine.
Q. 5. Identify the following :
HO —
Ans.
— NHCOCH3
Paracetamol.
Q. 6. What is the use of the following compound :
OCOCH3
COOH
Ans.
Analgesic.
Q. 7. Which artificial sweetner has the lowest sweetness value ?
Ans.
Aspartame.
Q. 8. Which artificial sweetner has the highest sweetness value ?
Ans.
Alitame.
Q. 9. What kind of a drug is Chlorpheniramine maleate ?
Ans.
Anti histamine drug.
Q. 10. What is the chemical name of aspirin ?
Ans.
2-Acetoxy benzoic acid.
Q. 11. Name the antibacterial present in toothpaste.
Ans.
Triclosan.
Q. 12. Identify the drug :
Ans.
Penicillin G.
Q. 13. Which antibiotic contains NO2 group attached to aromatic nucleus in its
Ans.
structure ?
Chloramphenicol.
234
Q. 14. What is the use of barbituric acid ?
Ans. Tranquilizer.
Q. 15. Name one analgesic drug which is also used to prevent heart attacks.
Ans. Aspirin.
Q. 16. What is the application of the following molecule in everyday life ?
CH3 (CH2)11 CH2 —
Ans.
O
— SO3–Na+
It is a detergent.
Q. 17. Give an example of a drug that is antipyretic as well as analgesic.
Ans. Paracetamol.
Q. 18. Out of morphine, N-Acetyl-para-aminophenol, Diazepam and tetrahydrocatenol,
Ans.
which can be used as analgesic without causing addiction ?
N-Acetyl-para-aminophenol.
Q. 19. Name a sugar substitute which is 160 times as sweet as surcrose.
Ans. Aspartame.
2 MARKS QUESTIONS
Q. 1. When a mixture of salicylic acid, acetic anhydride and acetic acid is refluxed,
Ans.
what is the product obtained and what is its use in everyday life ?
Aspirin used as analgesic.
Q. 2. Distinguish between a narrow spectrum and broad spectrum antibiotic.
Ans. A narrow spectrum antibiotic works against a limited range of microbes whereas a
broad spectrum antibiotic works against a large variety of microbes.
Q. 3. What is Salvarson ? To which class of drugs does it belong ? For what distance is
Ans.
it used ?
Antimicrobial agent.
Used for the treatment of STD called Syphilis.
Q. 4. How does apirin act as an analgesic ?
Ans. It inhibits the synthesis of prostaglandins which stimulates inflammation of the tissue
and cause pain.
Q. 5. What are barbiturates ? To which class of drugs do they belong ?
Ans. 5, 5-derivatives of barbituric acid are called barbiturates. They belong to the class of
tranquilizers. They also act as sleep producing agents.
Eg. Luminal, Veronal.
Q. 6. What is tincture iodine ? What is its use ?
Ans.
Alcoholic Solution of I2.
Used as an antiseptic.
235
Q. 7. To what class of medicines does chloramphericol belong ? For what disease can
it be used ?
Ans.
Broad spectrum antibiotic
Used to cure typhoid, dysentry, acute fever.
Q. 8. Pick the odd one amongst the following on the basis of their medicinal properties
mentioning the reason.
Ans.
(i)
Luminal, seconal, phenacetin, equanil
(ii)
Chloroxylenol, phenol, chlorampherical, bithional
(i)
Phenacetin is an antipyretic while all the rest are tranquilisers.
(ii)
Chlorampherical is an antibiotic while all the remaining are antiseptics.
Q. 9. Why is bithional added to soap ?
Ans.
Acts as an antiseptic and reduces the odours produced by bacterial decomposition of
organic matter on the skin.
Q. 10. Why are cimetidine or ranitidine better antacids than sodium bicarbonate or
magnesium or aluminium hydroxide ?
Ans.
If excess of NaHCO3 or Mg (OH)2 or Al (OH)3 is used, it makes the stomach alkaline
and thus triggers the release of even more HCl. In contrast, cimetidine or ranitidine
prevent the interaction of histamine with the receptor cells in the stomach wall and
thus release lesser amount of HCl.
Q. 11. Why is the use of aspartame limited to cold foods and drinks ?
Ans.
It decomposes at baking or cooking temperatures and hence can be used only in cold
foods and drinks.
Q. 12. What problem arises in using alitane as artificial sweetener ?
Ans.
It is a high potency artificial sweetener. Therefore it is difficult to control the
sweetness of the food to which it is added.
Q. 13. What is the purpose of adding antioxidants to food ? Give two examples of
antioxidants.
Ans.
They are added to prevent oxidation of fats and oils present in food thus preventing
food from becoming rancid.
Eg. BHA (Butylated hydroxy anisole)
BHT (Butylated hydroxy toluene)
FOR BELOW AVERAGE
01 Mark Questions
1.
2.
3.
4.
Define the term chemotherapy
Define the term Tranquilizers and give one example.
Define the term Anti Oxidants and give one example.
Name a food preservative which is most commonly used by food products.
236
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
List two major classes of antibiotics and give one example of each class.
Why is bithional added to the toilet soap?
Give an example of a narcotic which is used as an analgesic.
Define antiseptic and give one example.
Name one medical compound each that is used to treat (1) hypertension (2) general
body pain.
What is function of wide spectrum antibiotics and give one example.
How are antiseptic different from disinfectant? Give one example of each of them.
Soap is a weak antiseptic itself. What may be added to soap to improve its antiseptic
action?
What type of drug is chloramphenicol?
Why ethanol is added to soap?
Name the medicine which can act as analgesic as well as antipyretic. Give its
chemical name.
Give one example of bactericidal antibiotic and bacteriostatic antibiotic.
Why is use of aspartame limited to cold foods and drinks
What type of drug is Phenacetin?
How are synthetic detergents better than soaps?
What is tincture of Iodine?
FOR BELOW AVERAGE STUDENTS
A1.
The branch of science which deals with the treatment of diseases
using suitable chemicals is known as chemotherapy.
A12.
They are chemicals which are used for treatment of mental diseases. Example –
Equanil.
Antioxidants are the compounds which prevent oxidation of food materials.
Examples, BHA (Butylated hydroxy anisol) and BHT( Butylated hydroxy
toluene.
Class I- preservatives – Table salt, sugar and vegetable oil.
Class II – preservatives – Sodium benzoate
(a) Simple antibiotics- Penciline, Ampiciline
(b) Broad spectrum antibiotics – Chloromycetin.
Bithional is added to soaps to impart them antiseptic properties. Such soaps are used
to reduce odor due to bacterial action on skin surface.
Morphine or heroin.
They are chemicals which kill or prevent the growth of microorganisms. They are
applied to living tissues. Example- KMnO4 Solution, 0.2 % solution of phenol etc.
(i) Equanil
(ii) Novalgin
Antibiotics which are effective against several different types of harmful micro
organisms and capable of curing several infections are called broad spectrum
antibiotics. Example- Chloromycetin, tetracycline.
Antiseptic is used in living tissue where as disinfectants are used in nonliving things.
Example of Anticeptic- Tincture of iodine ( 2-3 % iodine in alcohol – water solution)
and iodoform act as antiseptic. Example of disinfectants – 2% solution of phenols.
Bithional.
A13.
Broad spectrum antibiotics.
A14.
To make transparent the soap.
A15.
Aspirin.
A2.
A3.
A4.
A5.
A6.
A7.
A8.
A9.
A10.
A11.
A16. Ofloxacin and chloramphenicol respectively.
A17.
It is unstable at cooking temperature.
A18.
Antipyretics.
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A19.
Broad spectrum antibiotics.
A20.
2-3% Iodine in ethanol is called tincture of iodne.
FOR AVERAGE
01 Mark Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
What are main constituents of Dettol?
What are food preservatives?
Name the sweetening agent used in the preparation of sweet for a diabetic patient.
Why do soaps not work in hard water?
If water contains dissolve calcium hydrogen carbonate, out of soaps and synthetic
detergents which one will you use for cleansing cloths?
Why should not medicine be taken without consulting doctors?
Name two narcotics which are used as analgesics?
What is an antipyretics ? Give an example.
What are antagonists and agonists ?
Pick out the odd amongst the following compounds on the basis of there
medicinal properties.
Luminal, Seconal, Phenacetin, Equanil.
FOR ABOVE AVERAGE
01 Mark Questions
1.
2.
3.
4.
5.
6.
Write the chemical formula of
(i) Sodium stearate (ii) Sodium palmitate
Write chemical equation of soaponification.
What is biodegradable detergents?
Write chemical formula of (i) Sodium dodecyl benzene sulphonate and (ii) Sodium
lauryl sulphate.
What are pathogens ?
What is the chemical formula of cataionic detergents ?
FOR BELOW AVERAGE
02 Mark Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
Mention one important use of each of the following
(i) Equanil
(ii) sucralose
Name a broad spectrum antibiotics and state two diseases for which it is prescribed.
State the function along with one example each of (i) Antihistamines (ii)
Antioxidents.
Describe the following with an example.
(i) Antimicrobials
(ii) Analgesics
Name the medicines used for the treatment of the following diseases (i) Tuberculosis
(ii) Typhoid
Describe the following with example.
(i) Preservatives
(ii) Biodegradable detergents
Describe the following with giving examples
(i) Edible colours
(ii) Antifertility drugs
Define the followings and give one examples.
(i) Antipyrites
(ii) Antibiotics
Give one important use of each of the following :
(i) Bithional
(ii) Chloramphenicol (iii) Streptomycine
(iv) Paracetamol
238
FOR AVERAGE
02 Mark Questions
1.
Why do we require artificial sweetening agents?
2.
Which process is involved in holding the drugs to the active site of enzymes?
3.
What are different types of synthetic detergents ? Explain with sutable
Examples.
4.
How are synthetic detergents better than soap?
5.
Describe the cleansing action of detergents.
FOR ABOVE AVERAGE
02 Mark Questions
1.
What are biodegradable and non biodegradable detergents?
2.
Explain the following terms with suitable examples.
(a) Cationic detergents
(b) anionic detergents.
3.
Why are detergents usually preferred to soap for washing cloths?
4.
Why biodegradable detergents are more prefer now a days.
5.
Define the following terms with suitable examples.
(i)
Antacids
(ii) Antihistamines (iii) Antibiotics
(iv) Antifertility drugs
(v) Foodpreservatives.
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