Sure Shot Questions
UNIT 1 THE SOLID STATE
Q-1. What is the non- stoichiometry defect in the crystals?
Ans.These defects occur when the ratio ofthe cations and anions in the resulting compound is different
from that as indicated by the laws of the chemical combinations.
Q-2. How many atoms are there in a
simple or primitive unit cell
body centered cubic unit cell
face centered cubic unit cell
Ans-5.a) one b) two c) four
Q-3.Whattype of semiconductors are when Si doped with – a) Phosphorous and b) Gallium.
Ans. a) n-type and b) p-type
Q-4. What is the coordination no. of an octahedral void andtetrahedral void?
A-.Six(6) and Four(4)
Q-5. Which of the two will show the Schottky defect when added to the AgCl crystal
NaCl or CdCl2.
Ans.CdCl2 . This is because divalent Cd2+ ions occupy the Ag+ sites and thus produce cationic vacancies in
the crystal.
Q-6 What type of compounds show the Schottky defect?
Ans. Ionic compounds which have cations and anions of nearly the same size.
Q-7.Name the compound in which bothSchottkyand Frenkel defects are found together?
Ans-AgBr.
Q-8.What are F- centers? Why are the solids containing the F- centers are paramagnetic?
Ans.The free electrons trapped in the anion vacancies are termed as the F- centers
Q-9 A unit cell consists of a cube in which there are anions at each corner and one at the center of the
unit cell. The cations are the center of the each face .How many A) cations and B)anions make up the unit
cell? C)What is the simplest formula of the compound?
Ans.A)Thecation at the center of each face is shared by two unit cells.
Hence no. of cations= 6 X ½ = 3
B) The anion at each corner is shared by 8 unit cells .The anion at the center is not shared by any other unit
cell.
Hence no. of anions= 8 X 1/8 + 1 = 2
C) Since there are 3 cations and 2 anions the simplest formula of the compound is A3B2.
Q-10.Copper crystallises in face-centred cubic lattice and has a density of 8.930 g mol-3 at 293 K. Calculate
the edge length of unit cell. [At. mass of Cu = 63.5 a.m.u, Avogadro's constant NA = 6.02 x 1023].
Ans. Density = Mass of unit cell/Volume of unit cell
8.93 = (4 x 63.5)/(a3x 6.02 x 1023).
a3 = 47.24 x 10-24
a = 3.6 x 10-8 cm. = 360 pm.
Q.11 In corundum, oxide ions are arranged in hcp arrangement and the aluminium ions occupy 2/3
of the octahedral voids. What is the formula of corundum.
Ans
Number of oxide Ions = 8 x 1/8 = 1 per unit cell
Number of Al ions = 2/3
formula Al2/3 O or Al2 O3
OR
Rank of hcp = 6 = No. of octahedral voids.
No. of Al ions = 6 
2
=4
3
Formula Al4O6 or Al2O3.
1
Q12. What is the difference between Schottky and Frankel defect?
Ans.
Serial no.
Schottky defect
Frankel defect
1.
It decreases the density of the crystal.
It does not decrease the density of the crystal.
2.
It occurs in compounds with high Coordination number.
It occurs in compounds with low number.
3.
It occurs in compounds in which cations It occurs in compounds in which cations and anions
and anions are of similar size. Examples: differ in their size to a large extent. Examples: ZnS,
NaCl, KCl, KBr, CsCl.
AgCl, AgBr, Agl.
UNIT 2: SOLUTIONS
Q1.
0.1 mole of NaOH is present in 250 ml of solutions. Calculate the molarity(M).
Ans:
Molarity = No. of moles of solute. X
1000
Vol. Of Solution
= 0.1 X 1000 / 250 = 0.4 moles/litre.
Q2.
How many moles of urea are present in 5 litres a solution 0.02 M.
Ans:
Moles of Urea = Molarity X Vol. Of solution = 0.02 X 5 = 0.1 ml.
Q3.
What is effect of temperature on ?a) Molarity b) Molality.
Ans:
Molarity: It will decrease with the increase in temperature
Molality: It does not change with temperature.
Q4.
Why molality does not depend upon temperature?
Ans:
Molality is not related to the volume of solution or solvent but it is related to the mass of solvent
which does not change with temperature, hence the molality is independent of temperature.
Q5.
What is the relative lowering of the vapour pressure?
Ans:
Po A - P is known as relative lowering of the vapour pressure where Po A
Po A andP are the vapour pressure of pure solvent and solution respectively.
Q6.
What is the effect of the addition of non-volatile solution on the vapour pressure of pure liquid?
Ans:
It causes the lowering of vapourpressure i.e addition of non-volatile solute will decrease the vapour
pressure of the liquid
Q7.
Define the Raoult’s law.
Ans:
The vapour pressure of the solution containing the non-volatile solute is directly proportional to the
mole fraction of solvent.
Q8.
Define the molal elevation constant.
Ans:
Molal elevation constant is equal to the elevation of boiling points produce by one molal solution.
Q9.
What are the units of molal elevation constant?
Ans:
K kg mol–1 .
Q10. What is the relation between the elevation of boiling points and molalityof the solutions?
Ans:
Elevation of boiling point is directly proportional to the molality of solutions. Tb ∞ m.
Q11. What is the relation between osmotic pressure and relative lowering ofvapour pressure of a
solution?
2
Ans:
The relative lowering of vapour pressure is directly proportional to the osmotic pressure of the
solution.
P -Ps ∞ л
P
Q12. A and B liquids on mixing produced a warm solution. Which type of deviation is there and why?
Ans:
Negative type of deviation is present. In the negative deviation the solute-solution (A-A) interaction
and solvent-solvent(B-B) interaction will be weaker than solute-solvent(A-B ) interaction. Since the new
forces are stronger therefore heat is evolved and solution becomes warm.
Q13. Which type of deviation is shown by the solution formed by mixing cyclohexane and ethanol?
Ans:
Hydrogen bonds exist among the molecules of ethanol when cyclohexane is added to ethanol, the
cyclohexane molecule will come in between the ethanol molecules and disturb the hydrogen bonding. The
forces between ethanol and cyclohexane will be weaker than the previous forces (H-bonding) hence there
will be positive deviation.
Q.14 What mass of non - volatile solute (NH2CONH2 ) urea need to be dissolved in 50 gm of water in
order to decrease the vapour pressure of water by 25 %. What will be the molality of the solution.
Solution : Let the amt. Of urea dissolved in 50 g =x mol
Vap. Pressure of solution= 100-25=75 mm
Relative lowering indil.solutions =P0-Ps/P0=XB=Nb/Na
100-75/100=X/50/18
AMT. OF UREA ADDED TO 50 g. =25/36 X60 G/MOL =41.67 g.
Molality =no. of moles of solute/amt.of solvent in Kg =25/36/50/1000=13.9m
Q.15Which solution has higher concentration 1 Molar or 1 molar having same solute?
Ans.
(a) If density of solvent is 1 gm/ml then 1M solution is more concentrated then 1 molal.
(b) If density of solvent is less than 1 gm/ml then 1 molar solution is less concentrated than 1 molal.
(c) If density of solvent is more than 1 gm/ml then 1 molar solution is more concentrated than 1
molal.
Q16. Why osmotic pressure is a colligative property?
Ans-- Osmotic pressure is a colligative property because it depends on the number of theparticles.
Q17. What happens when red blood cells are placed in 0.1 per cent NaCl solution?
Ans—Water from NaCl solution passes into red blood cells (due to endosmosis), therefore they swell up and
burst.
Q18. Why molality is preferred over molarity?
Ans. Molality is unaffected by temperature as it does not involve volume.
Q19. Why solutions deviate from their ideal behavior?
Ans.Due to the difference in their intermolecular interactions.
Q20. What is value of van,t Hoff factor for (a) sodium sulphate (b) K4[Fe(CN)6]?
Ans. (a) 3 (b) 5
UNIT -3 ELECTROCHEMISTRY
Q1. What is the effect of temperature on the electrical conduction of
(i) Metallic conductor.
(ii) Electrolyte conductor?
Ans.With increase of temp.the electrical conduction of metals decreases whereas that of electrolyte
increases.
Q2. How is cell constant calculated from conductance values?
Ans. Cell constant = specific conductance/observed conductance.
Q3. What is the reference electrode in determining the standard electrode potential?
Ans. Normal hydrogen electrode (NHE).
3
Q-4. What are the units of specific conductance ?
Ans-.ohm-1 cm-1 or Scm-1 .
Q-5 What is the effect of the decreasing concentration on the molar conductivity of a weak electrolyte.?
A-ns.Molarconductance of a weak electrolyte increase with decrease of the concentration.
Q-6.Write an expression that relates the molar conductivity of a wek electrolyte to its degree of
dissociation.
Ans-. Degree of dissociation(αe )= Λem / Λωm
Q-7How does an electrochemical cell help in predicting the feasibility of a redox reaction ?
Ans-14.If E0 of the cell is +ve it will yield –ve ∆G0 Value which indicates the reaction is spontaneous.
∆G0 = -n F E0
Q.8 Why  m for acetic acid cannot be determined experimentally?
Ans. Molar conductivity of weak electrolytes keeps on increasing with dilution and does not become
constant even at very large dilutions.
Q-9 The electrolysis of a metal salt solution was carried out by passing a current of 4 amp for 45 minutes.
It resulted in deposition of 2.977 g of a metal. If atomic mass of the metal is 106.4g/mol, calculate the
charge on the metal cation.
Ans. Quantity of electricity passed = 4 x 45 x 60 C= 10800C.
Let charge on metal ion be n+
Quantity of electricity required to deposit 106.4g of metal = n x96500C
Quantity of electricity required to deposit 2.977g of metal=
n  96500  2.977
= nx2700C which is equal to
106.4
10800C,
n = +4
Q.10. Calculatee.m.f of the cell at 298K.
CrCr3+ Fe2+ Fe
Given: E Cr 3
Cr
= -0.75V, E Fe 2

Ans. E cell  E cell 
0.1M
= -0.45V
Fe

0.0591
Cr 3
log
6
Fe 2



2
3
= +0.2606V
Q.11. Predict the products of electrolysis of each of the following:
a) An aqueous solution of AgNO3 using silver electrode.
b) An aqueous solution of AgNO3 using platinum electrode.
c) A dilute solution of H2SO4 using platinum electrode.
d) An aqueous solution of CuCl2 using platinum electrode.
AnsAt cathode
A- Ag
At cathode
B- Ag
At cathode
C- H2
At cathode
D- Cu
At Anode
Ag+
At Anode
O2
At Anode
O2
At Anode
Cl2
4
0.01M
Q-12. Describe the characteristics of variation of molar conductivity with dilution for (a) weak and (b)
strong electrolytes.
Ans.
Strong
Mol.con.
weak
Con 1/2
with increasing dilution the molar conductance increases rapidly because of greater extent of ionization at
greater dilution.
These are dissociated more or less completely at low dilutions. Their molar conductances therefore show
slight increase with increasing dilution. At low conc, the interionic attractions are more and when dilution
increases these attractions become weak and the conductance increases gradually.
Q-13.Can nickel spatula to be used to stir a copper sulphate solution? Support your answer with areason.
E0Cu2+/Cu =0.34 ,
E0Ni 2+ /Ni = - 0.25
2+
0
=0.34
Ans. Cu
+ 2e
Cu(s)
E
Ni(s)
Ni 2+ +
+2e
E0 =-0.25
2+
2+
Cu (aq.)
+
Ni(s)
Ni (aq.) +
Cu(s)
E0 = 0.59 V
Since the value of E0cell is positive , nickel spatula will displace copper from copper sulphate solution,
hence nickel spatula cannot be used for stirring.
Q-14. Calculate the maximum possible electrical work that can be obtained from the following cell under
standard conditions at 25 0 c.
Zn(s) / Zn 2+ (aq.)// Cu2+ (aq) /Cu(s)
0
0
2+
At 25 CE Zn / Zn= -0.76 V , E0 Cu2+ (aq) /Cu(s) = +0.34 V.
Ans. Maximum electrical work = + ∆G 0
- ∆G 0
= n F E0
Cu2+ (aq.)
+ Zn(s)
Zn 2+ (aq.) +
Cu(s)
0
E = E R –E L
= 0.34 –(- 0.76) =1.10
- ∆G 0
= 2 X 96500 X1.10
0
∆G
= -212.3 Kj
Q-15. Calculate the equilibrium constant for the reaction at 25 0 c
Cu2+ (aq.)
+ Ni(s)
Ni 2+ (aq.) +
Cu(s)
0
2+
0
2+
E Cu /Cu =0.34 ,E Ni /Ni = - 0.25 R = 8.314 J/K/mol F= 96500 C.
Ans. E CELL
E R- EL = 0.34 – (-0.25) = 0.59 V
- ∆G 0 =2 X 96500X 0.59
∆G 0 = -113.87 k J = -113870 j
∆G 0 = - 2.303 RT log K
-113870 = -2.303 X8.314 X298 log K
log K= 113870 / 2.303 X 8.314 X 298 = 19.9597
K= 9.051 X 1019.
Q-16Write the reaction taking place at cathode and anode during the operation of Daniel cell.
Ans. A Zn - Cu cell is known as Daniel cell. The reactions taking place are described as under :
Zn (s) Zn2+ + 2e- (Anode)
5
Cu2+ + 2e- Cu (s) (Cathode)
Zn (s) + Cu2+ Zn2+ + Cu (s) (Overall reaction)
Q-17. A conductivity cell whose cell constant is 3.0/cm is filled with 0.1 M acetic acid solution , its
resistance is 4000 ohms.
find molar conductance of 0.1 M acetic acid.
Degree of dissociation of acetic acid.
Molar conductance of acetic acid at infinite dilution is 400 ohm –1 cm2 mol-1.
Ans. Λm = 1000/M X 1/R X l/a =1000/.1 X 1/4000 X 3 = 7.5 S cm2 mol-1
α= (Λm)c/ (Λm)ω = 7.5 /400 =0.01875
Q- 18. What is the basic principle of fuel cell ? give two advantages of fuel cell.
Ans. FUEL CELL is used to convert the chemical energy of fuel into electrical energy
Chemical reactions: Reaction at
O2 ( g )  2 H 2O(l )  4e   4OH  (aq)
Cathode
Anode
2 H 2  4OH   4 H 2O  4e 
Overall Reaction
2 H 2 ( g )  O2 ( g )  2 H 2O(l )
Advantages: 1-it does not create pollution
2- it has high efficiency.
Q19. Describe normal hydrogen electrode and its application.
Ans. The construction of normal hydrogen electrode is explained below :
Application :
(i) The potential of normal hydrogen electrode has been arbitrarily taken as zero. Thus in combination with
other electrodes, cell can be constructed and the potentials of electrodes can be determined.
(ii) Normal hydrogen electrode can be used to determine the pH of a solution.
Q-20.What is Kohlrausch`slaw ?calculateΛm∞acetic acid
Given Λm∞HCl=426 S cm2 mol-1
Λm∞NaCl= 126 S cm2 mol-1
Λm∞ CH3COONa =91 S cm2 mol-1
Ans. : According to this law, molar conductivity of an electrolyte, at infinite dilution can be expressed as the
sum of contributions from its individual ions
Λm∞ (CH3COOH) = Λm∞ CH3COO- + Λm∞ H+
= Λm∞ CH3COO-+ Λm∞ Na+ - [Λm∞ Na+ +Λm∞Cl-] + [+Λm∞ H+ +Λm∞Cl-]
∞
=Λm (CH3COONa) –Λm∞ (NaCl)+Λm∞ (HCl)
=[91-126+426] Scm2/mol
=391 S cm2 mol-1
Q-21. State and explain Faraday`s laws of electrolysis?
What is the relationship between the chemical equivalent and electrochemical equivalent?
Ans. Faraday – First law of electrolysis : The amount of substance deposited during electrolysis is directly
proportional to quantity of electricity passed.
W  zit
Faraday – Second Law : The amount of different substances librated by the same quantity of electricity
passing through the electrolytic solution is proportional to their chemical equivalent weights
6
Z = EQUIVALENT WT. IN GM./ 96500
Q.22 Silver is electrodeposited on a metallic vessel of surface area 800cm2 by passing a current of 0.2
ampere for 3 hours. Calculate the thickness of the silver deposited. Given the density of silver as 10.47
gcm-3 (atomic mass of Ag = 107.92 U).
Ans The quantity of electricity = 0.2amp x3 x60 x60sec= 2160C.
2160C of electricity deposit Ag=
107.95  2160
= 2.42g
96500
Mass= area x thickness x density
2.42g= 800cm2 x thickness x 10.47gcm-3
Hence thickness = 2.89 x10-4 cm.
UNIT
4 CHEMICAL KINETICS
Q1. If K = 102 (rate constant), will the reaction go to completion?
Ans. Yes.
Q2.What is the order of a reaction when ethyl Acetate is hydrolysed?
Ans2. CH3COOC2H5 + H2
3COOH + C2H5OH
It is a Pseudo-unimolecular reaction.
Q3.Why is a finely divided substance more efficient catalyst?
Ans3.Because the surface area of a finely divided substance is very large.
Q4.What are the units of rate constant for a zero order reaction?
Ans4.Time-1 mole+1 litre-1.
Q5.What are the units of rate constant for a second order reaction?
Ans5.Litre+1 mole-1 sec-1.
Q6. The rate of a reaction at different times (- dx/dt) is found as
Time (minutes)
Rate (in mol s-1)
0
2.80 x 10-2
10
2.78 x 10-2
20
2.81 x 10-2
What is the order of a reaction?
Ans6. Since rate of reaction is constant, it is a zero order reaction.
Q7.A sample of a radio-active substance takes 80 days to lose half of its radio activity. What time will be
required to lose 1/8 of its radio activity.
Ans7.
80 + 80 + 80 = 240 days.
Q8.If half life period of a first order reaction is 80 days, what is the value of rate constant?
Ans8. K = 0.693/80 d-1
= 0.0693/8 = 0.00866 d-1.
Q9.What is the order of a reaction?
If rate = K|A|1/2.|B|-3/2.|C|2
Ans9. n = 1/2 - 3/2 + 2/1 = 2/2 = 1.
Q.10 A gas decomposition of AB follows the rate law; rate = K AB 3/4 .Write units of K.
Ans Mol-1/4 L ¼ time-1
7
Q.11If half-life of a reaction is inversely proportional to initial concentration of the reactant, what is the
order of the reaction?
Ans.Second order
Q.12 What is the order of the reaction represented in the following graph?
Rate
conc
Ans. First order
Q-13 What is the order of the reaction represented in the following graph?
RATE
C
Ans ZERO ORDER.
Q-14. For the assumed reaction : X2 + 2Y2
disappearance of Y2.
Ans.-d[Y2]/dt= k[X2] [Y2]2
2XY2 ,write the rate equation in terms of rate of
Q15. A first order reaction takes 69.3 minutes for 50% completion. How much time will be needed for 80%
completion?
Ans. t1/2 = 69.3 minutes
k = 0.693/69.3 = 10-2 min-I
Now t = 2.303/k log a/(a - x)
= 2.303/10-2x 0.6990 = 160.9 minutes.
Q16. In a certain first order reaction, half the reaction was decomposed in 500 seconds. How long will it
take for 90% completion?
Ans. K = 0.693/(t1/2) = 0.693/500
= 0.001386 sec-1
Now, t = 2.303/0.001386 x log (100)/10
= 1661 sec.
Q17. Find the value of rate constant for the reaction
A + B AB
if the rate of the reaction is 5 x 10-5 (mole/l)m-1 and |A| and |B| are 0.05 and 0.01 mol respectively.
8
Ans. A + B
AB
R = K[A][B]
K = (5 x 10-5)/(0.05 x 0.01)
= 10-1
= 0.1 (mol/l)-1min-1
Q.18 Following reaction takes place in one step
2NO (g) + O2 (g)  2 NO2 (g)
How will the rate of above reaction change if the volume of the reaction vessel is diminished to one third
of its original volume? Will there be any change in order of reaction with the reduced volume?
Ans.Rate of the reaction will become 27 times, order will remain unchanged
Q-19The rate of decomposition of a substance A becomes 8 times when its concentration is doubled.
What is the order of the reaction?
Ans. Rate= k [A]n
i
r = k [A]n
ii
8r= k [2A]n
dividing ii by I
8r/r = k [2A]n/k [A]n
8 =[2]n
n=3
order is 3
Q.20 The kinetics of the reaction: A + 2B  Products; obeys the rate equation Rate = k AX BY. For it,
find
Order of the reaction
Apparent molecularity of reaction
Order of reaction when B is in large excess.
Ansa) X+Y
b) 3
c) X
Q.21 In the start of summer, a given sample of milk turns sour at room temperature (300K) in 48 hours. In
a refrigerator at 275K, milk can be stored three hours longer before it sours. Calculate activation energy of
the souring of milk.
Ans.
k 300
–1
k 275 = 3,Ea = 30.146 kJ mol
Q.22. Decomposition of SOCl2 at 600K in gas phase is a first order reaction with rate constant of 2.3  10-5
sec –1 .Calculate the % of SOCl2 that would remain undecomposed after 200 minutes of reaction time.
Ans.log
A   =0.12
A 
A   =1.318
A 
 % of SOCl2 that remains undecomposed is
100
= 75.8%
1.318
Q-23.Consider the following data for the reaction
A + B
Products
Determine the order of the reaction w.r.t. A and B and the overall order of reaction.
s.n.
Initial
Initial con.[B]
INITIAL
con.[A] mol/l Mol/l
RATE(mol/l/sec)
9
1.
0.10
1.0
2.1 x 10-3
2.
0.20
1.0
8.4 x 10-3
0.20
2.0
8.4x10-3
0
3.
A-6) The rate law may be written as
Rate = k[A]p [B]q
R1 = k[0.10]p [1]q
(i)
2.1 x 10-3= k[0.10]p [1]q
R2 = k[0.20]p [1]q
8.4 x 10-3 = k[0.20]p [1]q
(2)
R3= k[0.20]p [2]q
8.4x10-3 = k[0.20]p [2]q
dividing eq. (2) by (1)
8.4 x 10-3 /2.1 x 10-3= k[0.20]p [1]q /[0.10]p [1]q
or 4 = 2 p
p =2 i.e. order w.r.t A =2
SIMILARLY by dividing (3) by(2) we get
Q=0 i.e. w.r.t B =0
Overall order = 2+0= 2
(3)
UNIT 5 SURFACE CHEMISTRY
Q1.What is adsorbate and adsorbent?
Ans. The substance adsorbed on the surface is called adsorbate and the substance on which it is adsorbed is
called adsorbent.
Q2.What is desorption?
Ans. The reverse process i.e., removal of adsorbed substance from the surface is called desorption.
Q3.What is occlusion?
Ans. The adsorption of gases on the surface of metals is called occlusion
Q4. Give reasons for the development of electrical charge
on colloidal particles?
Ans: Due to preferential adsorption of common ion
Q-5. Write Freundlichadsorption isotherm equation.
Ans. Log x/m =log k+ 1/n log p.
Q-6. What is Kraft` temperature?
Ans. Micelles form ionic surfactants only above a certain temperature called as Kraft`s temperature.
Q7. Explain electrophoresis.
Ans. Movement of colloidal particles under the influence of electric fields is termed as electrophoresis.
When electric current is passed through a colloidal solution, colloidal particles are discharged at an
electrode depending upon the type of charge on them. In this way nature of the charge on colloidal particle
10
can be ascertained.
Q8.What is Brownian movement? What is its advantage?
Ans. The movement of colloidal particles in Zig-zag way, when viewed through ultra-microscope is called
Brownian movement. This is due to unequal bombardment of colloidal particles by the particles of
dispersion medium.
As a result of Brownian movement colloidal solution is stable and colloidal particles do not settle at the
bottom of vessel or it is not coagulated
Q9.What is peptization? How a colloidal solution can be prepared with its help?
Ans. Process of conversion of precipitate to colloid is termed as peptization.
By adding Fe+++ in a suspension of ferric hydroxide colloidal solution of hydrated ferric oxide can be
prepared.
+ Fe+++
Q-10. What is positive and negative adsorption?
Ans.Positive adsorption: When there is a more concentration of adsorbate at the surface of the adsorbent
than that in the bulk, it is called positive adsorption.
Negative adsorption: When there is a less concentration of a solute (Adsorbate) on the surface layer than in
the bulk part of the solution it is called negative adsorption.
Q-11. How the following sols are produced?
Sulphursol collodion
Ans. By bubbling H2S gas through an oxidizing agent like bromine water.
H2S + Br2
2HBr + S
cellulose nitrate colloid can be prepared by dispersing it in a mixture of ethanol and ether. This is
commercially called as collodion.
Q12- Differentiate between oil in water and water in oil type of emulsions.
Ans-OIL IN WATER- 1) when diluted with water no separate layer is formed.
2) Less viscous and more conductive.
WATER IN OIL – 1)When diluted with water show separate layer
Q13.Explain Tyndall effect. What is its significance?
Ans. (1) If a beam of light is passed through a sol, the beam is visible due to scattering of light by the
colloidal particles. It is called Tyndall effect and the illuminated path as Tyndall cone or beam. The intensity
of scattered light depends upon the difference of the refractive index of the dispersed phase and that of
dispersion medium.
Lyophobic sols have appreciable difference hence show the Tyndall effect, while in the lyophilic sols this
difference is very small. So the Tyndall effect is very small.
General Principal and isolation of elements
Q.1- What is the role of cryolite in electrometallurgy of aluminium?
11
Ans- alumina cannot be fused easily because of high melting point. Dissolving of alumina in cryolite furnishes
Al3+ ions, which can be electrolyzed easily.
Q.2 Describe the method of refining of nickel.
Ans- Mond Process, Ni is heated in a stream of CO forming a volatile complex, which then decomposes at
higher temperature to give Ni.
At 330-350K: - Ni + 4CO → Ni (CO) 4
At 450-470K Ni (CO)4 → Ni + 4 CO
Q.3- What is Zone Refining? Explain with example.
Ans- It is based on the principal that impurities are more soluble in molten state of metal than solidified
state.
In this method, a rod of impure metal is moved slowly over circular heater. The portion of the metal being
heated melts & forms the molten zone. As this portion of the rod moves out of heater, it solidified while the
impurities pass into molten zone. The process is repeated to obtain ultrapure metal and end of rod
containing impure metal cutoff.
Q.4- Describe the method of refining of Zirconium and Titanium.
Ans- Van Arkel process – it is used for obtaining ultrapure metal. The impure metal is converted into volatile
compound, which then decomposes electrically to get pure metal.
At 850K: - Zr impure) + 2 I2 → ZrI4
At 2075K:- ZrI4 → Zr (pure) + 2 I2
Q.5.
Name one each of
a) Acidic flux
b) Basic flux
Ans:
a) Silica b) Lime
Q.6.
What is basic difference between Calcination and Roasting?
Ans:
Calcination the concentrated ore is heated either in the absence or in the limited supply of air on the
othar hand roasting is carred out in excess of air.
Q.7.
What is the significance of Leaching in the extraction of Aluminium?
Ans:
Leaching is significant as it helps like SiO2, Fe2O3etc from the Bauxite ore.
Q.8.
State the role of
I. Depressant in froth flotation process.
II. Silica in the metallurgy of copper.
III. Graphite rod in the electrolytic reduction of alumina.
Ans.
i.
It prevent certain sulphides like ZnS to enter the froth in presence of Pbs, Therefore, Helps in their
separation. Sodium Cyanide is used as depressant in the separation of ZnS and PbS.
ii. It acts as flux.
iii. Graphite rod acts as anode in the extraction of aluminium.
p-Block Elements
1. N2 is less reactive at room temperature? because of high bond dissociation enthalpy
2. PCl5 more covalent than PCl3? because ofpentavalentcation has more polarizing power than PCl3
3. BiH3 strogest reducing agent ammomgest all hydrides of grp 15 element because of lowest bond
dissociation enthalpy due to larger size of Bi
4. Bond angle in PH4+ is higher than that in PH3. Because PH4+ has sp3 hybridization with no lone pair,
whereas PH3 has pyramidal shape with one lone pair.
5. NH3 form H – bond but PH3 does not. because more electronegativity and small size of N
6. PCl5 is ionic in nature in the solid state. Because it exist as [PCl4]+ [PCl
7. P4 is more reactive than N2 due to larger size of phosphorous than nitrogen.
12
8. All the P-Cl bonds in PCl5 are not equivalent.Due to BP – BP electron repulsion.
9. Bi (v) strong oxidising agent than Sb(v) Due to inert pair effect
10. BiH3 strongest reducing agent amongst all the hydride of group 15 elements
Because of its
lowest bond dissociation enthalpy
11. R3P= O exist but R3N= O does not. Because absence of d – orbital
12. Nitrogen show catenation properties less than phosphorus because of N – N bond is weaker than P – P
bond due to e – e repulsion due to small size.
13. H3PO3 is dibasic and H3PO4 is tribasic. Because of presence of ionisable two P – OH bond in H3PO3
whereas presence of ionisable three P – OH bond in H3PO4
14. Nitrogen does not form pentahelides. Because of absence of d – orbital and small size
15. NO2dimeriseBecause of presence of unpaired or odd electron on N
16. Stability of +5 oxidation State decreases and +3 oxidation state stability increases down the group of
15 group elements. Because of inert pair effect.
17. Which neutral molecule would be isoelectronic with ClO- ? OF2 and ClF
18. H2O liquid but H2S gas because of presence of H – Bonding in water whereas H2S is not
19. O3 act as powerful oxidizing agent because of libration of nascent oxygen.
20. H2S is less acidic than H2Te. Because of S atomic size less than Te. Therefore S cannot release H+ ion
easily.
21. SF6 is kinetically an inert. Because of steric effect
22. H2S is more acidic than H2O.
Because of S atomic size more than O. Therefore S can release H+
ion easily.
23. All the bonds in SF4 are not equivalent Because of all bonds are in different planes.
24. Interhalogen compound are more reactive than halogens Because of interhalogen bond are weaker or
low bond dissociation enthalpy than halogen.
25. ICl more reactive than I2Because weaker bond in ICl than I2
26. Halogens are strong oxidizing agents Because high electronegativity, therefore can gain electrons
easily.
27. F2 is most reactive of all the other halogens. Because inter electronic repulsion due to small size
28. F2 is stronger oxidizing agent than Chlorine. Because inter electronic repulsion due to small size of
fluorine
29. F2 has low bond enthalpy than Cl2. Because inter electronic repulsion due to small size of fluorine
30. Negative value of electron gain enthalpy of Fluorine less than Chlorine. because inter electronic
repulsion due to small size of fluorine
31. Noble gas Xenon forms known Compounds. Because of larger atomic size and high polarizing power of
Xenon.
32. Noble gases have comparatively larger atomic size. Because inter electronic repulsion and low
ionization enthalpy
33. Noble gas has very low boiling points. Because of presence of weak vanderwaal force of attraction and
larger size.
d- and f- Block elements
Q.1. Why d-block elements or their compounds are good catalytic agents?
Ans.Since a good catalyst must provide a large surface to adsorb or to form activated complex with
reactants. d-block elements or their compounds have this unique property and are thus good catalytic
agents.
13
Q.2. Give comparison between lanthanides and actinides.
Ans.
Lanthanides
Actinides
1. These show lanthanide contraction.
1. These show actinide contraction.
2. These are separated by ion-exchange behaviour.
2. These also show ion exchange behaviour.
3. Most of lanthanides are colourless.
3. Most of the ions of actinides are coloured.
4. These have lesser tendency to form complexes.
4. These have greater tendency to form complexes.
5. Their magnetic behaviour can be easily explained.
5. Their magnetic behaviour cannot be easily
explained.
6. These show mostly +3 oxidation state.
6. These show even higher oxidation state +3 to +7.
Q.3. (a) Why transition elements exhibit:(i) variable oxidation states (ii) form complexes (iii) form coloured compounds (iv) and show catalytic
properties.
Ans. (a) (i)incomplete d-sub shell and participation of ns and (n - 1)d electrons in bond formation
(ii)small size, high nuclear charge, presence of valence vacant d-orbitals, d-block elements or their ions
accommodate lone pairs of electrons from electron pair donor species and thus complexes are formed for
example
(iii)The Transition metal ions have unpaired d-electrons and d-d transition
(iv)the presence of unpaired electrons in d-orbitals.
Q.4.Most of the transition elements are paramagnetic?
Ans. due to presence of one or more unpaired electrons in atomic orbitals.
14
Q5.What is the lanthanoid contraction? What are its causes and consequences?
Ans. Lanthanoid contractions – The cumulative effect of the regular decrease in size
or
radii of Lanthanoid with increase in atomic number is called Lanthanoid
contraction.
Causes - The shape of f orbitals is diffused. They have poor shielding effect due to which the effective
nuclear charge increase with increase in atomic number. This causes a decrease in atomic radii
Consequences – Due to Lanthanoid contraction1. Radii of the members of the third transition series is similar to those of second
transition series.
2. It becomes difficult to separate Lanthanoids.
Q.6. Assign reasons for the following:
(i) The enthalpies of atomisation of transition elements are high.
(ii) The transition metals and many of their compounds act as good catalysts.
(iii) Scandium (Z = 21) does not exhibit variable oxidation states and yet it is regarded as a transition
element.
Ans. (i) This is because transition metals have strong metallic bonds as they have a large number of unpaired
electrons.
(ii) The catalytic activity of transition metals is attributed to the following reasons:
(a) Because of their variable oxidation states transition metals form unstable intermediate compounds and
provide a new path with lower activation energy for the reaction.
(b) In some cases, the transition metal provides a suitable large surface area with free valencies on which
reactants are adsorbed.
(iii)This is because scandium has partially filled d orbitals in the ground state
(3d1 4s 2).
Q.7.Zinc, cadmium and mercury are not considered as transition metals. Why?
Ans. Zinc, cadmium and mercury have fully filled d10 configuration. Therefore they are not considered as
transition metal.
CO ORDINATION COMPOUNDS
Q.1.Why Coordination compounds are coloured?
Ans. Due to d-d transition.
Q.2What is Homoleptic complexes?
The complexes in which a metal is bound to only one kind of donor groups. E.g., [Co(NH3)6]3+
Q.3.Write IUPAC name of the following [CO(NH3)6Cl3
Ans. Hexaaminocobalt(III)chloride
Q.4. Draw figure to show splitting of d- orbitals in an octahedral crystal field.
Q.5. Explain structure of[ [CO(NH3)63+ ] on the basis of valence bond theory.Structure on the basis of
valence bond theory
In the diamagnetic octahedral complex,[Co(NH3)6]3+ , the cobalt ion is in +3 oxidation state and has the
electronic configuration 3d6. The hybridisation scheme is as shown in diagram.
15
Six pairs of electrons, one from each NH3 molecule, occupy the six hybrid orbitals. Thus, the complex has
octahedral geometry and is diamagnetic because of the absence of unpaired electron. In the formation of
this complex, since the inner d orbital (3d) is used in hybridisation, the complex, [Co(NH3)6]3+ is called an
inner orbital or low spin or spin paired complexes.
Q.6. [NiCl4]2− is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why?
Answer Though both [NiCl4]2− and [Ni(CO)4] are tetrahedral, their magnetic characters are different. This is due to
difference in the nature of ligands. Cl− is a weak field ligand and it does not cause the pairing of unpaired 3
electrons. Hence, [NiCl4]2− is paramagnetic.
In Ni(CO)4, Ni is in the zero oxidation state i.e., it has a configuration of 3d8 4s2.
But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also, it causes the 4
electrons to shift to the 3d orbital, thereby giving rise to sp3 hybridization. Since no unpaired electrons are presen
in this case, [Ni(CO)4] is diamagnetic.
Q.7.Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex.
Ans.
[Co(NH3)6]3+
Oxidation state of cobalt = +3
Electronic configuration of cobalt = d6
[Ni(NH3)6]2+
Oxidation state of Ni = +2
Electronic configuration of nickel = d8
NH3 being a strong field ligand causes the pairing.
If NH3 causes the pairing, then only one 3d orbital is
Therefore, Ni can undergo d2sp3 hybridization. empty. Thus, it cannot undergo d2sp3 hybridization.
Therefore, it undergoes sp3d2 hybridization.
Hence, it is an inner orbital complex.
Hence, it forms an outer orbital complex.
Q.8. Draw the structures of optical isomers of:
(i) [Cr(C2O4)3]3−
(ii) [PtCl2(en)2]2+
(iii) [Cr(NH3)2Cl2(en)]+
Answer
16
(i) [Cr(C2O4)3]3−
(ii) [PtCl2(en)2]2+
(iii) [Cr(NH3)2Cl2(en)]+
Q.9. Draw all the isomers (geometrical and optical) of:
(i) [CoCl2(en)2]+
(ii) [Co(NH3)Cl(en)2]2+ (iii) [Co(NH3)2Cl2(en)]+
Answer(i) [CoCl2(en)2]+
In total, three isomers are possible.
(ii) [Co(NH3)Cl(en)2]2+
17
Trans-isomers are optically inactive.
Cis-isomers are optically active.
(iii) [Co(NH3)2Cl2(en)]+
Q.10. What is spectrochemical series? Explain the difference between a weak field ligand and a strong fie
ligand.
Answer. A spectrochemical series is the arrangement of common ligands in the increasing order of their crysta
field splitting energy (CFSE) values. strong field ligands cause higher splitting in the d orbitals than weak fie
ligands.
I− < Br−< S2−< SCN−<Cl−< N3< F−< OH−< C2O42−∼ H2O < NCS−∼ H−< CN− < NH3< en ∼ SO32−< NO2−<phen< CO
ORGANIC CHEMISTRY
NAME REACTIONS
1. Sandmeyer’s reaction
Benzene diazonium salt with cuprous chloride or cuprous bromide results in the
replacement of the diazonium group by –Cl or –Br.
18
2. Wurtz reaction
Alkyl halides react with sodium in dry ether to give hydrocarbons containing double
the number of carbon atoms present in the halide.
2 RX+
2 Na
R-R +2 NaX
Wurtz-Fittig reaction
A mixture of an alkyl halide and aryl halide gives an alkylarene when treated with
sodium in dry ether.
3. Fittig reaction
Aryl halides when treated with sodium in dry ether give
diphenyl.
4. Finkelstein reaction
R-X + NaI→ R-I + NaXX=Cl, Br
5. Swarts reaction
H3C-Br +AgF→ H3C-F + AgBr
6. Kolbe’s reaction
When phenol reacts with sodium hydroxide, phenoxide ion generated undergoes
19
electrophilic substitution with carbon dioxide and forms salicylic acid.
8.
Reimer-Tiemann reaction
On treating phenol with chloroform in the presence of sodium hydroxide, a –CHO
group is introduced at ortho position of benzene ring.
9. Williamson synthesis
An alkyl halide is treated with sodium alkoxide to form ether.
CH3Br + C2H5ONa

CH3 – O - C2H5
10. Rosenmund reduction.
Acyl chloride (acid chloride) is hydrogenated over catalyst, palladium on barium
sulphate to give aldehyde..
11.Clemmensen reduction
Aldehydes and ketones are reduced to alkanes on treatment with zinc- amalgam and
concentrated hydrochloric acid.
CH3 –CHO  CH3 - CH3
12. Wolff-Kishner reduction
20
Aldehydes and ketones are reduced to alkanes on reaction with hydrazine followed by
heating with potassium hydroxide in ethylene glycol.
CH3 –CO - CH3
CH3–CH2 - CH3
13. Aldol condensation
Aldehydes and ketones having at least one α-hydrogen undergo a reaction in the
presence of dilute alkali form β-hydroxy aldehydes (aldol) or β-hydroxy ketones
(ketol),
respectively.
14. Hell-Volhard-Zelinsky reaction
Carboxylic acids having an α-hydrogen are halogenated at the α-position on
treatment with chlorine or bromine in the presence of small amount of red phosphorus
to give α-halocarboxylic acids.
15. Hoffmann bromamide degradation reaction
Primary amines can be prepared by treating an amide with bromine in an aqueous or
ethanolic solution of sodium hydroxide.The amine so formed contains one carbon less
than that present in the amide.
21
Important reactions
i) Chloroform is slowly oxidised by air in the presence of light to an extremely
poisonous gas, carbonyl chloride, also known as phosgene. It is therefore stored in
closed dark coloured bottles completely filled so that air is kept out.
ii) Phenol is manufactured from the hydrocarbon, cumene (isopropylbenzene)
iii) Carbylamine reaction
Aliphatic and aromatic primary amines on heating with chloroform and ethanolic
potassium hydroxide form isocyanides or carbylamines which are foul smelling
substances. Secondary and tertiary amines do not show this reaction.
iv) Coupling reaction
Benzene diazonium chloride reacts with phenol or aniline to form azo dyes.
22
S
NO
NAME OF
CHEMICAL
THE
REAGENT
COMPOUND
OBSERVATION
PAIRS OF COMPOUNDS THAT CAN BE
DISTINGUISHED
1
R-X
Precipitate
1) Ethyl chloride &chlorobenzene
Aqueous NaOH&
AgNO3 solution
AgCl – White
AgBr – Pale
yellow
AgI – Dark
yellow
Ethyl chloride gives white ppt
Chlorobenzene does not give ppt.
2) Cyclohexylbromide&bromobenzene
Cyclohexylbromiide gives pale yellow pp
Bromobenzene does not give the test.
3) Benzyl chloride &chlorobenzene
Benzyl chloride gives white
pptChlorobenzene does not give ppt.
4) Vinyl iodide &Allyl iodide
Vinyl iodide gives pale yellow ppt.
Allyl iodide does not give the test.
2
R - OH
Lucas test –
concHCl&anh
ZnCl2
10 – Does not
react.
To distinguish primary , sec &ter
alcohols
20 – Forms
turbidity after
few min
Primary – Methanol, Ethanol, n –
alkyl alcohols / alkan – 1-ol, benzyl
alcohol
30 - Forms
turbidity
immediately
Secondary – Isopropyl alcohol, alkan –
2 – ol
Yellow ppt of
iodoform
(CHI3)
23
Tertiary – ter- buytl alcohol (2methyl – propan – 2- ol)
Iodoform test
– NaOH&I2
Alcohols with CH3 – CH – group give
this test
Note: When
both the alcohols
given are
primary / sec
apply this test
OH
1) Ethyl alcohol & methyl alcohol (both
are 10)
Ethyl alcohol gives yellow ppt
Methyl alcohol does not give this
test.
2) Pentan – 2- ol&Pentan – 3- ol (both
are 20)
Pentan – 2- ol gives yellow ppt
Pentan – 3- ol does not give this test.
DISTINCTION TESTS
3
Phenol
Neutral ferric
chloride (FeCl3)
Violet colour
1) Phenol & ethyl alcohol
Phenol gives violet colour
Ethyl alcohol does not give this
test
4
Aldehydes
Tollens test –
ammoniacal
AgNO3 , warm
Silver mirror
Aldehydes give positive test.
1. Acetaldehyde (Propanal) & Acetone
(Propanone)
Acetaldehyde (Propanal) gives Ag
mirror
Acetone (Propanone) does not give this
test.
24
Iodoform test
– NaOH&I2
Note: When
both the
compounds
given are
aldehydes/
ketones apply
this test
Yellow ppt of
iodoform
(CHI3)
Ald. &ket. with CH3 – C – group
give this test
O
1) Acetaldehyde &benzaldehyde
Acetaldehyde gives yellow ppt.
Benzaldehyde does not give this test.
1) Acetaldehyde & formaldehyde
Acetaldehyde gives yellow ppt.
Formaldehyde does not give this
test.
3) 2 – Pentanone& 3 – Pentanone
2 – Pentanone gives yellow ppt.
3 – Pentanone does not give this
test.
4) Acetophenone&benzophenone
Acetophenone gives yellow ppt.
Benzophenone does not give this test.
5.
Carboxylic
acids
Sodium
bicarbonate test
Effervescence
due to
evolution of
CO2 gas
All carboxylic acids give this test
Formic acid
( Methanoic
acid)
Tollens test
Effervescence
due to
evolution of
CO2 gas
HCOOH is the only carboxylic acid
that gives this test
HCOOH and CH3COOH
HCOOH gives this test . CH3COOH does
not give this test
25
6
Amines
Primary
amines
Hinsberg test –
10 –Soluble
Benzene
sulphonyl
chloride + NaOH
20 – Insoluble
Carbylamine test
– Alcoholic KOH
+ CHCl3
3 – does not
react with
Hinsberg
reagent
0
Very
unpleasant
smelling
isocyanide gas
evolves
To distinguish primary , sec &ter
amines
Alkyl amine – 10
Dialkylamine -20 (N- alkyl alkanamine)
Trialkylamine( N,N – dialkylalkanamine 30
To distinguish primary amines from
other amines
1)Ethylamine
&Diethylamine
Ethylamine gives this test.
Diethylamine does not give this test
2) Aniline &Dimethylamine
Aniline gives this test.
Aniline
Reddish
orange dye
Azodye test –
NaNO2 + HCl +
phenol ( 00 to 5
C)
Diethylamine does not give this test
Aniline &Benzylamine
0
Aniline gives this test
Benzylamine does not give this test.
MECHANISM FOR PREPARATION OF ALCOHOL FROM ALKENE
26
MECHANISM FOR PREPARATION OF ETHER FROM ALCOHOL
MECHANISM FOR DEHYDRATION OF ALCOHOL TO ALKENE
27
ACIDITY OF PHENOL OR
o-&p- DIRECTING NATURE
STABILITY OF PHENOXIDE ION
28
HALOALKANES & HALOARENES
1.What are ambident nucleophiles ? Give two eg.
Nucleophiles that possess two nucleophiliccentres.Eg CN- & NO22. Why is chloroform stored in closed dark colourbottle ?
In the presence light & air it forms poisonous gas , phosgene.
2CHCl3 +O22COCl2 + 2HCl
3. Alkly halides though polar are immiscible with water. Why ?
Because they do not form intermolecular H bond with water.
4. Arrange the following set of compounds in order of increasing boiling
points.
1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane
NOTE : SN1 : 10< 20< 30 Tertiary cabocation is more stable
SN2 : 30 <20 <10 Steric hindrance is less in primary
5. Which one in the following pairs would undergo SN2 reaction faster?
Because iodine is a better leaving group.
6. Which one in the following pairs would undergo SN1 reaction faster?
29
7. An alkyl halide having molecular formula C4H9Cl is optically active .What is
its name?
Ans.= 2-chlorobutane
8. Explane why haloarenes are less reactive towards nucleophilic substation
reactions than Haloalkanes.
Due to resonance in benzene ring there is partial double bond in C-X bond which is
difficult to break.
(a) It is difficult for electron rich nucleophiles to approach electron rich benzene ring
in haloarenes.
9. How do the products differ with ethyl bromide react separately with ;
(a) Alc. KCN and Alc. AgCN. B) Aq. KOH and Aic. KOH.
Ans.
(a)
Alc.KCN
C2 H5 CN+
KBr
C 2H5Br
Alc.AgCN
C2 H5 NC+ AgBr
(b)
Aq.KOH
C2H5OH
+
KBr
C2H5 Br
Alc.KOH
CH2=CH2 +KBr +H2O
10(a) Give the IUPAC name of the following ;
30
H3C
CH3
H
H
CH3
Br
(b)Write Wurtzfittigreaction .
Ans. (a)
4-Bromo-3-methyl –pent-2-ene.
(b)
R
X
+ 2NaX
2Na + X-R
Dry ether
11. Arrange the following compounds in order of reactivity towards SN2
displacement:
2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
ANS:2-Bromo-2-methylbutane (ter) < 2-Bromopentane(sec) < 1Bromopentane (pri)
Because of less steric hindrance in 1-Bromopentane
12. Convert the following:(a) ethanol to propanenitrile
(b) propene to 1-propanol
(c) 1-Bromopropane to 2-Bromopropane
Ans a)
PCl5
CH3CH2OH
KCN
CH3CH2Cl
CH3CH2CN
HBr/peroxide
KOH
31
b)CH3CH=CH2
CH3CH2CH2Br
alc. KOH
c) CH3CH2CH2Br
CH3CH2CH2OH
HBr
CH3CH=CH2
CH3CHBrCH3
13.(a)Convert Nitrobenzene to m-Bromoiodobenzene.
(b)Draw the structure of the product of the Bromination of following
compounds in presence of FeBr3.
(b)Because –NO2 deactivates the ring but OCH3 activates.
14. Convert the following:(a) ethanol to propanenitrile
(b) propene to 1-propanol
(c) 1-Bromopropane to 2-Bromopropane
Ans a)
PCl5
KCN
CH3CH2OH
HBr/peroxide
CH3CH2Cl
CH3CH2CN
KOH
32
b)CH3CH=CH2
CH3CH2CH2Br
alc KOH
c) CH3CH2CH2Br
CH3CH2CH2OH
HBr
CH3CH=CH2
CH3CHBrCH3
ALCOHOLS PHENOLS & ETHERS
1.
Phenol is acidic in nature. Why?
Ans.
Phenol is acidic in nature because:
a) phenol , due to resonance, the positive charge rests on oxygen making the
shared pair of electrons more towards oxygen and hydrogen as H+
b) The carbon attached to OH is SP2 hybridize and is more electronegative, this
decreases the electron density on oxygen, increasing the polarity of O-H bond
and ionization of phenol.
c) The phenoxide ion formed by loss of H+ is more resonance stabilized than
phenol itself.
2.
o- nitrophenol has lower boiling point ( is more volatile ) than p –
nitrophenol.
Ans.
P- nitrophenol has intermolecular hydrogen bonding which increases the boiling
point while in o- nitro phenol due to presence of intra molecular hydrogen
bonding, there is a decrease in boiling point and increase in volatility.
33
3.
Methanol is miscible with water while iodomethane is not.
Ans.
Methanol can form intermolecular hydrogen bonding with water but there
is no hydrogen bonding in iodomethane and water. Therefore methanol in miscible
in water.
4 Alcohols have higher boiling points than isomeric ethers.
Ans.
Alcohols can form intermolecular hydrogen bonds due to their high polarity
whereas, ether cannot. Therefore alcohols have higher boiling points than
isomeric ethers.
5.
Ethers are soluble in water alkanes are not.
Ans.
Ethers can form H- bonding with water molecule whereas alkenes cannot.
Therefore ethers are soluble in water and alkanes are not.
6.
The order of acidic strength in alcohols is RCH2OH>R2CHOH>R3 COH
Ans.
In alcohols, the acidic strength is due to polar nature of O-H bond. An electron
releasing group e.g., alkyl groups, increases electron density on oxygen tending
to decrease the polarity of O-H bond. This decreases the acid strength.
Therefore the order of acid strength is
34
7. During preparation of ester from alcohol and acid, water has to be
removed as soon as it is formed.
Ans.
8.
Ethers cannot be prepared by dehydration of secondary or tertiary
alcohols.
Ans.
For secondary and tertiary alcohols, elimination competes over substitution and
alkenes are formed on acidic dehydration as the reaction follows Sn1
mechanism. Therefore the acidic dehydration of secondary or tertiary alcohols
does not give ethers.
9.
Reaction of anisole with HI gives methyl iodide and phenol.
Ans.
10.
o-nitrophenol is more acidic than o-methoxyphenol. Why ?
Ans.
Because of electron withdrawing nitro group in o-nitrophenol there is
effective delocalisation negative charge.
35
11.
Phenol is more acidic than alcohols. Why?
Ans.
Phenoxide is more stabilised by resonance than alkoxide ion.
12.
What is denaturation of alcohols ?
Ans.
Commercial alcohol is made unfit for drinking by adding copper sulphate/
pyridine.
13.
Arrange the following compounds in increasing order of their acid
strength:
Propan-1-ol, 2,4,6-trinitrophenol, 3-nitrophenol, phenol, 4methylphenol
Propan-1-ol < 4-methylphenol < phenol < 3-nitrophenol < 2,4, 6-trinitrophenol.
Ans.
14.
I)
Butanol to Butanoic acid
ii).
Ethanol to propanone
iii).
Phenol to salicylic acid
Ans.
36
IV).
Methanol to Ethanol
V.
Ethanol to propanol
vi.
Phenol to Benzyl Alcohol
Ans.
VII.
Ethanal to propan -2- ol
Ans.
ALDEHYDES KETONES & CARBOXYLIC ACIDS
37
1.
Arrange the following set of compounds in order of increasing boiling
points.
CH3CH2CH2CHO, CH3CH2CH2CH2OH, CH3CH2 OCH2CH3 ,CH3CH2CH2 CH3
Ans. CH3 ,CH3CH2CH2 CH3 <CH3CH2 OCH2 CH3 < CH3CH2CH2CHO < CH3CH2CH2CH2OH
2.
Ketones are less reactive than aldehydes towards nuclephilic addition
reaction. Why?
Ans. Two electron releasing alkyl groups in ketones reduce the electrophilicity of
carbonyl. Also due to steric hindrance nucleophilic attack is not easier.
3.
Benzaldehyde is less reactive than propanal towardsnucleophilic
addition reaction. Why ?
Ans. Resonance in benzaldehyde reduces the electrophilicity of carbonyl carbon
4.
(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6trimethylcyvlohexanone does not.
Ans. The three methyl groups reduce the elecrophilicity of carbonyl and also offer
steric hindrance to the nucleophilic attack
(ii) In semicarbazide ,the NH2 group directly attached to the -C = O group is not
involved in the formation of semicarbazones.Give reason.
The lone pair of electrons of the – NH2 group directly attached to –C = O group
is involved in resonance. Hence it does not act nucleophile
5.
Arrange the following compounds in increasing order of their reactivity
towards nuclephilic addition reaction.
Methanal, propanone, ethanol
Ans. Propanone<Ethanal<Methanal
6.
Ans.
Acetaldehyde undergoes aldol condensation. Why ?
Because it contains alpha hydrogen.
7.
?
Ans.
8.
Ans.
9.
Ans.
Formaldehyde &benzaldehyde do not undergoaldol condensation. Why
Because they do not contain alpha hydrogen.
Formaldehyde &benzaldehyde show Cannizaro reaction. Why ?
Because they do not contain alpha hydrogen
Carboxylic acids are higher boiling liquids than alcohols. Why ?
Carboxylic acids form more extensive hydrogen bond than alcohols and exisi as
dimer.
10. Carboxylic acids are more acidic than phenol. Why ?
OR
Ka of carboxylic acid is more than that of phenol.Why ?
OR
pKa of carboxylic acid (acetic acid) is more than that of phenol. Why?
Ans. Carboxylate ion is more stabilised than phenoxide ion due to resonance with
negative charge on more electronegative oxygen atom.
11. Carboxylic acids do not show the reactions of carbonyl group. Why ?
Ans. Due to resonance the double bond character is reduced in carboxylic
acids.
12. Compare the acid strength of the following acidsHCOOH , CH3COOH , C6H5COOH
Ans. HCOOH > C6H5COOH > CH3COOH
13.
Why chloroacetic acid stronger than acetic acid.
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Ans- Cl is electron withdrawing group thus chloroacetate ion is more stable than
acetate ion.
14.
Carry out the following conversion—
a. Acetylene to Acetaldehyde
b. Toluene to Benzaldehyde
Ans—a.CHΞCH + H2O
H2SO4/HgSO4
CH2=CHOH
CH3CHO
CH3
CHO
b.
[O]
CrO2Cl2
+ H2O
AMINES
1.
Ans.
Ammonolysis is not preferred for preparation for primary amines.Why ?
Because it gives mixture of amines.
2.
Alkyl amines have higher b.pts than alkanes of comparable molecular
mass. Why?
Because alkyl amines form intermolecular hydrogen bond.
Ans.
3.
Ans.
Alcohols have higher b.pts than alkyl amines of comparable molecular
mass. Why?
Because in alcohols the strength of H – bond is more than that in amines since
O is more electronegative than N.
4.
Arrange the following set of compounds in order of increasing boiling
points.
n-Butylamine(10), Diethylamine(20),Ethyl dimethylamine(30) [isomeric
amines]
Ans. Ethyl dimethylamine(30) <Diethylamine(20) < n-Butylamine(10)
Reason - 30 amine – No H bond
In primary amine 2 hydrogen atoms are available for H- bond. In sec amine only
one H atom available for H- bond.
5.
(i) Ethylamine (aliphatic amines) is a stronger base than aniline
(aromatic amine). Why ?
(ii) pKb of aniline is more than that of methyl amine. Why?
Ans. (i)
In ethyl amine the electron releasing alkyl group increases the electron
density on nitrogen. In aniline the electrons on N is involved in resonance.
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(ii)
In methyl amine the electron releasing alkyl group increases the electron
density on nitrogen. In aniline the electrons on N is involved in resonance.
6.
Why cannot primary amines be prepared by Gabriel phthalimide ?
Ans. Aryl halides do not undergo nucleophilic substitution with the anion formed by
phthalimide.
7.
Ans.
8.
Ans.
9.
Ans.
10.
Methylamine in water reacts with ferric chloride to precipitate hydrated
ferric oxide. Why ?
Methylamine is more basic than water. So it forms OH-with water. The OH- ions
react with ferric chloride and gives ppt.
Aniline does not undergo Friedel- Crafts reaction. Why ?
In Friedel- Crafts reaction AlCl3 which is used a catalyst is a Lewis acid. It reacts
with aniline and forms salt.
Nitration of aniline gives a substantial amount of m-nitroaniline besides
ortho and para derivatives. Why ?
During nitration, in presence of strong acidic medium aniline is protonated to
form the anilinium ion which is meta directing.
Arrange the following In decreasing order of basic strength:
C6H5NH2,C6H5(CH3)2,(C2H5)2NH and CH3NH2
Ans- (C2H5)3N >CH3NH2 > C6H5NH2 > C6H5N(CH3)2 .
11.
How will you convert Aniline to Benzonitrile?
AnsN2+Cl-
NH2
NaNO2+HCl
0 – 5 0C
CΞN
KCN
CuCN
Aniline
Benzonitrile
12. Account for the following :(i) Pkbof aniline is morethan that of methylamine.
(ii) Ethylamine is soluble in water whreas aniline is not.
40
(iii)Why has aniline a weak basic nature than aliphatic amines.
Ans- (i) aniline has an electron withdrawing phenyl group so it is the weaker base
than ammonia. Methyl group in methylamine is electron donating group so it is
stronger base than ammonia.
(ii) ethyl group in ethylamine is comparatively a small group and causes no
hindrance in formation of hydrogen bonding and hence it is soluble in water.
(i)
Due to resonance in aniline.
BIOMOLECULES
(Q.1) What is difference between Reducing and non-reducing sugars or
carbohydrates?(1 Mark)
(Ans) All those carbohydrates which contain aldehydic and ketonic group in the
hemiacetal or hemiketal form and reduce Tollen’s reagent or Fehling’s solution are
called reducing carbohydrates while others which do not reduce these reagents are
called non-reducing reagents.
(Q.2) Define glycosidic linkage?
(Ans) The two monosaccharide units are joined together through an ethereal or
oxide linkage formed by the loss of a molecule of H2O. Such a linkage between two
monosaccharide units through oxygen atoms is called glycosidic linkage.
(Q.3) What is Milk sugar? Give its characteristics.
(Ans) Lactose occurs in milk so, it is called milk sugar. Lactose on hydrolysis
with dilute acids yields an equimolar mixture of D-glucose and D-galactose. It is a
reducing sugar since it forms an osazone. It undergoes mutarotation and also reduces
Tollen’s or Fehling’s solution.
(Q.4) What do you understand by denaturation of proteins?
(Ans) When a protein in its native form, is subjected to physical change like in
temperature or chemical change like change in pH, the hydrogen bonds are disturbed.
Due to this, globules unfold and helix get uncoiled and protein loses its biological
activity. This is called denaturation of protein.
(Q.5) Give a broad classification of vitamins?
(Ans) Vitamins are complex organic molecules.They can be broadly classified as:
(i) Water soluble vitamins: These include vitamin B-complex and vitamin C.
(ii) Fat soluble vitamins: These are oily substances that are not readily soluble in
water. However, they are soluble in fat. These include vitamins A,D,E and K. Nucleic
acids are bipolar (i.e. polymers present in the living system). They are also called
polynucleotides since the repeating structural unit of nucleic acids is a nucleotide
(Q.6) Give a short note on Zwitter ion?
(Ans) Amino acids are usually colourless, crystalline solids. These are water
soluble , high melting solids and behave like salts rather than simple amines or
41
carboxylic acids. This behaviour is due to the presence of both acidic (carboxylic
group) and basic (amino group) groups in the same molecule. In aqueous solution, the
carboxyl group can lose a proton and amino group can accept a proton, giving rise to a
dipolar ion known as zwitter ion.
Q7.
What is the primary structure of proteins?
Ans:- The sequence of amino acids in proteins.
Q. 8Define globular & Fibrous proteins
Ans: The protein with spherical shape and soluble in water are globular proteiseg Albumin and Insulin
The protein having thread like structure and insoluble in water are fibrous proteins eg keratin and myosin.
Q.9
Differentiate between DNA & RNA
DNA
thymine
Deoxy ribose sugar
Genetic information
Double helical
Q.10 What happens when glucose reacts with
RNA
uracil
Ribose sugar
Protein synthesis
Single helical
(i) Br2 water
(ii) HI
(iii) Conc. HNO3
Ans:- (i) Gluconic acid
(ii)
(iii)
n-hexane
Saccharic acid
POLYMERS
1. Define the term homopolymerisation giving an example.
Ans. Polymerisation process involving only single monomeric species is known as
homopolymerisation. For example formation of polyethene from ethane.
2. Write the name and structure of the monomer of polymer: PVC.
Ans. The monomer of PVC is vinyl chloride.
42
CH2=CH-Cl
3. Give an example of elastomers.
Ans. Natural rubber of Buna-S.
4. What are the monomer units of Bakelite?
Ans. Phenol and Formaldehyde.
5. Write the name and structure of monomer of natural rubber.
Ans. Isoprene or 2-Methylbuta-1,3-diene.
6. What are the monomer units of terylene?
Ans. Ethylene glycol and terepthalic acid.
7. Differentiate between thermoplastic and thermosetting polymers. Give
one example of each.
Ans.
Thermoplastic polymer
Thermosetting polymer
1. They are linear polymer without
1. They have three dimensional
cross-linking.
network of covalent bonds with
cross links.
2. Upon heating these polymers
2.Upon heating they retain their
become soft and becomes hard
strength and does not become
again on cooling.
soft.
3. These plastics can be recycled. 3.These plastics cannot be recycled.
They can be melted and reused.
4.Example:
PVC,
Nylon, 4.Example: Phenol formaldehyde resin,
Polystrene.
urea formaldehyde resin.
8. Explain each of the following giving suitable example of each:
(i) Elastomers
(ii) Condensation polymers
(iii)Addition polymers
Ans. (i) Elastomers: Polymers which are rubber like with elastic properties are
called Elastomers.
(ii) Condensation polymers: The polymer formed by the repeated condensation
reaction between two same or different bifunctional monomeric units with the
elimination of simple molecules like water, ammonia etc., are called as condensation
polymers.
(iii)Addition polymers: The polymers formed by the repeated addition of monomer
molecules possessing double or triple bond are called addition polymers.
8. Write the name and structures of the monomers of the following
polymers:
Buna-S, Dacron and Neoprene
Ans.
Polymer
Monomer
43
Buna-S
Dacron
CH2=CH-CH=CH2 (Butadiene), CH2=CH-C6H5 (Styrene)
HOCH2-CH2OH
(
Ethylene
Glycol),
HOOC-C6H4-COOH
(Terepthalic acid)
Neoprene
CH2=C(Cl)-CH-CH3 (Chloroprene)
9. What are biodegradable polymers? Give two examples.
Ans. Polymers which can be easily degraded by micro-organisms present in nature
are called biodegradable polymers. Common examples are poly-β-hydroxybutyrateco-β-hydroxyvalerate (PHBV), Nylon-2-nylon-6.
10.
Give the differences between natural rubber and vulcanised
rubber.
OR
How does vulcanisation change the character of the natural rubber?
Ans.
Natural rubber
It becomes soft at temperature above
335K and brittle at temperature below
283K
It shows high water absorption
capacity.
It is highly soluble in non-polar
solvents.
It is readily attacked by oxidising
agents.
Vulcanised rubber
It has a much wider useful
temperature range.
It absorbs water to a much lesser
extent.
It is less soluble in non-polar solvents.
It is not readily attacked by oxidising
agents.
CHEMISTRY IN EVERYDAY LIFE
1. What is meant by narrow spectrum antibiotics?
Ans. Antibiotics which are mainly effective against gram positive of gram
negative bacteria are called as narrow spectrum antibiotics. For example
Penicillin G is a narrow spectrum antibiotic.
2. What is meant by broad spectrum antibiotics?
Ans. These are medicines which are effective against several different types of
micro-organisms. For example tetracycline, chloramphenicol etc.
3. What do you understand by the term ‘Chemotherapy’?
Ans. Use of chemicals for therapeutic effect is called as Chemotherapy.
4. Name a drug used in case of mental depression?
Ans. Equanil, barbituric acid derivatives such as seconal, neuronal are used as
antidepressents.
5. Give an example of Narcotic which is used as analgesic.
Ans. Morphine is a narcotic which is used as an analgesic i.e. to reduce pain.
6. What is the role of Bithional in toilet soaps?
44
Ans. Bithional acts as antiseptic when added to soap. Thus, it increases the
antiseptic properties of the soap.
7. Define antipyretic. Give two examples.
Ans. These are the chemicals which are used to bring down the body temperature
during high temperature. For example: Asprin, Paracetamol.
8. What is tincture of iodine? For what purpose it is used?
Ans. Alcohol water solution containing 2-3% iodine is known as tincture of iodine. It
is used as antiseptic.
9. Give an example of non-ionic detergent.
Ans.Lauryl alcohol ethoxylate.
10.
Name the chemical responsible for antiseptic properties of Dettol.
Ans. Chloroxylenol and terpineol in a suitable solvent.
11.
To which class of drugs cimetidine and ranitidine belongs?
Ans.Anti-histamines.
12.
Name a food preservative which is most commonly used by food
producers.
Ans. The most common food preservative used by food producers is sodium
benzoate.
13.
Describe antiseptics giving suitable examples.
Ans.Antideptics are the chemical substances which prevent the growth of microorganism and may even kill them. Boric acid, hydrogen peroxide, bithional, is
common antiseptics.
14.
How are transparent soaps manufactured?
Ans. Transparent soaps are manufactured by dissolving soap in ethanol and
evaporating the excess solvent.
15.
What are artificial sweetening agents? Give two examples.
Ans. These are the chemical substances which are sweet in taste but do not add to
calorie intake. Some common examples are aspartame saccharin, alitame etc.
16.
Why the use of aspartame is is limited to cold food and drinks?
Ans. This is because aspartame decomposes at baking and cooking temperatures.
Hence aspartame can be used only with cold drinks and foods.
17 What are biodegradable and non-biodegradable detergents? Give one
example of each.
Ans. Detergents which are easily degraded by micro-organisms present in the
rivers, ponds etc., are called as biodegradable polymers. For example –PHBV, sodiumdodecylbenzenesulphonate.
45
Non-biodegradable detergents are not easily degraded by mico-organisms.
For example sodium-1,1-dimethyldecylbenzene sulphonate.
18. Explain the following giving one example of each type:
a. Antacids
b. Disinfectants
Ans. (i) Antacids: The chemicals which control the release of acid in gastric juices or
neutralises the excess acid in gastric juices are called antacids.
(ii) Disinfectants: The chemical substances which are used to kill microorganisms but
they cannot be applied on living tissues are called as disinfectants. For example 1%
solution of phenol in water is a disinfectant.
19. Explain the following giving one example of each type:
a. Tranquilizers
Ans. a. Tranquilizers: These are chemicals used for the treatment of stress and mild
or even severe mental diseases. For example, veronal, serotonin, derivatives of
barbituric acid etc.
20. Describe the following giving one example:
a. Antifertility drugs
b. Antioxidants
Ans. Antifertility drugs: These drugs are used to prevent unwanted pregnancies. For
example Norethindrone
Antioxidants: Chemicals which prevent the oxidation of food stuff during
storage etc., are called as antioxidants. Vitamin C and E are natural oxidants. BHA (
ButylatedHydroxy Anisole) is a synthetic antioxidant.
21. Why are Cimitidine and ranitidine better antacid than sodium hydrogen
carbonate?
Ans. Excessive intake of sodium hydrogen carbonate can make stomach alkaline and
trigger the production of even more acid. Also these treatments controls only the
symptoms of hyperacidity and not the cause. Cimetidine, ranitidine are better antacids
because they control the cause of hyperacidity. It is observed that hormone Histamine
stimulates the production of pepsin and hydrochloric acid in the stomach. Drugs like
cimetidine and ranitidine prevent the interaction of histamine with the receptors
present in the stomach wall. This results in production of lesser amount of acids and
controlling hyperacidity.
22. Account for the following:
(i) Aspirin drug helps in preventing the heart attack.
(ii) Detergents are non-biodegradable while soaps are biodegradable.
46
Ans. (i) Aspirin is found to help in prevention of heart attack because of its anti-blood
clotting action. The blood clots if formed are mainly responsible for heart attack.
(ii) Common detergents which contain branched chain hydrocarbon part in their
molecules are non-biodegradable. This is because micro-organisms present in the
waste water cannot metabolise branched chain hydrocarbons. On the other hand soaps
contain linear chain hydrocarbons, which can be biodegraded more easily and water
pollution is prevented.
47