Download Ans:- (i) Gluconic acid - Kendriya Vidyalaya No.2, Kribhco, Surat

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 of the 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-.a) one b) two c) four
Q-3.What type 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 and tetrahedral 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 both Schottky and 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 Fcenters
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)The cation 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)/(a3 x 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.
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 It occurs in compounds with low
high Co-ordination number. number.
3.
It occurs in compounds in
which cations and anions are
of similar size. Examples:
NaCl, KCl, KBr, CsCl.
It occurs in compounds in which
cations and anions differ in their size
to a large extent. Examples: ZnS,
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
and P 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 vapour pressure 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
molality of 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 of
vapour pressure of a solution?
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 (Hbonding) 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 in dil.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.15 Which 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 the particles.
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).
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.Molar conductance 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-7 How 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
=
106.4
nx2700C which is equal to 10800C,
n = +4
Q.10. Calculate e.m.f of the cell at 298K.
Cr Cr3+   Fe2+  Fe
0.1M
0.01M
Given: E Cr
3
Cr
= -0.75V, E Fe
Ans. E cell  E cell 
= -0.45V
2
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:
An aqueous solution of AgNO3 using silver electrode.
An aqueous solution of AgNO3 using platinum electrode.
A dilute solution of H2SO4 using platinum electrode.
An aqueous solution of CuCl2 using platinum electrode.
Ans
A-
At cathode
At Anode
+
Ag
Ag
At cathode
At Anode
B- Ag
O2
At cathode
At Anode
C- H2
O2
At cathode
At Anode
D-
Cu
Cl2
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 a reason.
E0Cu2+/Cu =0.34 ,
E0Ni 2+ /Ni = - 0.25
=0.34
Ans.
Cu2+
+
2e
Cu(s)
E0
Ni(s)
Ni 2+ + +2e
E0
=-
0.25
Cu2+ (aq.)
Ni(s)
Ni 2+ (aq.) +
Cu(s)
0
E = 0.59 V
0
Since the value of E cell 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)
At 25 0 C E0 Zn 2+ / 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
0
- ∆G
= 2 X 96500 X1.10
∆G 0
= -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 - E L = 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-16 Write 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)
Cu2+ + 2eCu (s) (Cathode)
2+
Zn (s) + Cu
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
2 H 2  4OH   4 H 2O  4e 
Anode
2H 2 ( g )  O2 ( g )  2H 2O(l )
Overall Reaction
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`s law ? calculate Λm∞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
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
2
Q1. If K = 10 (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
Q.11 If 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
equation in terms of rate of disappearance of Y2.
Ans.-d[Y2]/dt= k[X2] [Y2]2
2XY2 ,write the rate
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)
= 2.303/10-2 x 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.
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-19 The 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
8r= k [2A]n
dividing ii by I
8r/r = k [2A]n / k [A]n
8 =[2]n
n=3
order is 3
ii
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.
Ans a) 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
= 3, Ea = 30.146 kJ mol –1
k 275
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
INITIAL
con.[A]
con.[B]
RATE(mol/l/sec)
mol/l
Mol/l
1.
0.10
1.0
2.1 x 10-3
0
2.
0.20
1.0
8.4 x 10-3
3.
0.20
2.0
8.4x10-3
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)
p
q
R3= k[0.20] [2]
(3)
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
UNIT 5 SURFACE CHEMISTRY
Q1. What is adsorption?
Ans. The phenomenon of attracting and retaining the molecules of a substance
on the surface of a solid or a liquid is called adsorption.
Q2. 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.
Q3. What is desorption?
Ans. The reverse process i.e., removal of adsorbed substance from the surface
is called desorption.
Q4. What is occlusion?
Ans. The adsorption of gases on the surface of metals is called occlusion
Q5. How is the physically adsorbed mass of a gas on a unit mass of an
adsorbent related to pressure of the gas?
Ans. x/m = kp
Q6. Out of NH3 and N2 which gas will be adsorbed more readily on the surface
of charcoal and why?
Ans. NH3 as it is easily liquefiable gas.
Q7. 0.1 M AlCl3 solution is more effective than 0.1 M NaCl solution in
coagulating As2S3 sol while 0.1 M AlCl3 is less effective than0 .1 M Na3PO4
solutions in coagulating ferric hydroxide sol.
Ans. AlCl3 – contains Al+3 ion and
NaCl – contains Na+ ions
Greater the charge on the opposite ion greater will be coagulation.(Hardy
Schulze Rule) Similarly PO4-3 ion is more effective than Cl- ion in coagulating
Fe(OH)3 solution.
Q8. Why delta is formed at the point where river enters the sea?
Ans Due to coagulation of colloidal clay.
Q9. Preparation of colloid such as protein can be made more stable if the
colloid is dialysed? Why
Ans. This is because dialysis helps in removing undesirable ions from colloidal
preparation in order to stabilize the colloid.
Q10. Give reasons for the development of electrical charge
on colloidal particles?
Ans : Due to preferential adsorption of common ion
Q-11. Write Freundlich adsorption isotherm equation.
Ans. Log x/m =log k+ 1/n log p.
Q-12. What is Kraft` temperature?
Ans. Micelles form ionic surfactants only above a certain temperature called as
Kraft`s temperature.
Q-13.What is the difference between a sol and a gel?
Ans. Both are colloidal solutions. A sol has a solid dispersed phase in liquid as
dispersion medium. A gel has liquid as dispersed phase and solid as dispersion
medium.
Q14. 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 can be ascertained.
Q15. 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
Q16. 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-17. What are macromolecular colloids?
Ans. When certain substances having big size molecules, called
macromolecules having large molecular masses are dissolved in a suitable
liquid, they form a solution in which the molecules of the substance. i.e., the
dispersed phase particles have size in the colloidal range. Such substances are
called macromolecular colloids. For example, starch, cellulose, proteins and
gelatin.
Q-18. 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-19. How the following sols are produced?
Sulphur sol 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.
Q20- 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
2) More viscous low conductivity.
Q21 Name the dispersion medium in hydrosol, alcosol, benzosols and
aerosol.
Ans- WATER, ALCOHOL, BENZENE AND GASES.
Q-22. Describe Bredig`s arc method for the preparation of colloidal solutions.
Ans. Electric arc is struck between the electrodes of the metal immersed in the
dispersion medium. The intense heat produced vaporizes some of the metal,
which then condenses to form particles of the colloidal size.
Q-23. Give some importance of adsorption in daily life.
Ans. 1) Gas masks 2) decolorizers 3) volumetric analysis 4) removes hardness of
water. 5) Chromatography 6) heterogeneous catalysis. 7) De-humidizers.
Q24. 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.
Q25. What is a colloidal state? Give differences between a true solution,
colloidal solution and a suspension.
Ans. (a) Colloids:
Property
True solution
Colloidal state
Suspension
(i) Nature
Homogenous
Heterogeneous
Heterogeneous
Less than 1 nm =
Greater than 100
(ii) Particle size
1 nm to 100 nm
-9
10 m
nm.
(iii) Diffusion
Diffuse rapidly
Diffuse slowly
Do not diffuse
Completely
Invisible to naked
Visible to naked
(iv) Visibility
invisible and do not eye but scatter
eye
scatter light
light
(v) Tyndall effect Do not exhibit
Exhibit
Do not exhibit
These can be
These can also be
precipitated by
coagulated by
These can or
(vi) Coagulation
adding suitable
adding suitable
cannot coagulated
electrolyte
electrolyte
Q-26. What are lyophilic and lyophobic colloids? Give differences between
them.
Ans.
Lyophilic Colloids
(i) These are prepared by shaking or
warming the dispersed phase in
dispersion medium and do not require
any electrolyte for their Stabilizations
(ii) These are reversible
Lyophobic Colloids
These are difficult to prepare. Special
methods are used. Addition of
stabilizer is a must for their
stabilization.
(ii) These are irreversible
(iii) These are generally unstable and
(iii) These are very stable and are not
are coagulated by the addition of an
easily coagulated
electrolyte
(iv) Particles carry no or very little
(iv) Colloidal particles carry +ive or -ive
charge depending on PH of the solutioncharge
(v) Viscosity is much higher than
(v) Viscosity is same as that of the
medium
medium.
(vi) Surface tension is usually less than (vi) Surface tension is usually the same
medium
as that of medium
(vii) Colloidal particles migrate towards
(vii) Migration in electric field : Particles
cathode or anode under the influence
may or may not migrate in an electric
of an electric field depending upon
field
charge on them
(viii) Visibility : Particles are seen under (viii) Particles are invisible, but can be
an ultra-microscope
detected by ultra-microscope
(ix) Tyndall effect : Less distinct
(ix) More distinct
(x) Cataphoresis : Measurement is
(x) Cataphoresis can be measured
difficult
easily.
(xi) A Large amount of electrolyte is
(xi) A small amount of electrolyte is
needed for their coagulation.
needed for their coagulation.
(xii) Examples : Mostly organic
(xii) Generally in organic substances
substances e.g. starch gums, gelatin
e.g. metal sols, sulphides and oxides
etc.
sols.
(xiii) Conductivity can rarely be
(xiii) These show high conductivity
measured over a considerable range of
which can be measured
conc.
UNIT 6 General Principal and isolation of elements
Q.1- What is the role of cryolite in electrometallurgy of aluminium?
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.
Ans:
Name one each of
a) Silica b) Lime
a) Acidic flux
b) Basic flux
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.
Ans:
Q.8.
I.
II.
III.
What is the significance of Leaching in the extraction of Aluminium?
Leaching is significant as it helps like SiO2, Fe2O3 etc from the Bauxite ore.
State the role of
Depressant in froth flotation process.
Silica in the metallurgy of copper.
Graphite rod in the electrolytic reduction of alumina.
Ans.
i.
ii.
iii.
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.
It acts as flux.
Graphite rod acts as anode in the extraction of aluminium.
UNIT 7
p-Block Elements
1. N2 is less reactive at room temperature? because of high bond
dissociation enthalpy
2. PCl5 more covalent than PCl3? because of pentavalent cation 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.
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.NO2 dimerise Because 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 I2 Because 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 vander waal force
of attraction and larger size.
34. Draw the structure of XeOF4, XeO3, XeF4, XeF6, BrF3, H3PO3, H2SO5, H2S2O7
UNIT 8 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.
Q.2. Give
Ans.
comparison between lanthanides and actinides.
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.
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.
UNIT 9
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.
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 a difference in the nature of ligands. Cl− is a weak field ligand and it does not cause the pairing of
unpaired 3d 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
4s electrons to shift to the 3d orbital, thereby giving rise to sp3 hybridization. Since no unpaired electrons
are present 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 If
theNH3 causes the pairing, then only one 3d
pairing. Therefore, Ni can undergo d2orbital
sp3
is empty. Thus, it cannot undergo d2sp3
hybridization.
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−
Answer
(i) [Cr(C2O4)3]3−
(ii) [PtCl2(en)2]2+
(ii) [PtCl2(en)2]2+
(iii) [Cr(NH3)2Cl2(en)]+
(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+
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 field ligand.
Answer. A spectrochemical series is the arrangement of common ligands in the increasing order
of their crystal-field splitting energy (CFSE) values. strong field ligands cause higher splitting in the
d orbitals than weak field ligands.
I− < Br−< S2−< SCN−<Cl−< N3< F−< OH−< C2O42−∼ H2O < NCS−∼ H−< CN− < NH3< en ∼ SO32−<
NO2−<phen< CO
CHEMISTRY MIND MAP
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.
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 + NaX
X=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 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
Aldehydes and ketones are reduced to alkanes on reaction with hydrazine
followed by heating with potassium hydroxide in ethylene glycol.
CH3 –CO - CH3

13. Aldol condensation
CH3 –CH2 - CH3
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.
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.
S
NO
NAME OF
THE
COMPOUND
CHEMICAL
REAGENT
OBSERVATION
PAIRS OF COMPOUNDS THAT CAN BE
DISTINGUISHED
1
R-X
Aqueous NaOH &
AgNO3 solution
Precipitate
1) Ethyl chloride & chlorobenzene
AgCl – White
Ethyl chloride gives white ppt
AgBr – Pale
yellow
Chlorobenzene does not give ppt.
AgI – Dark
yellow
2) Cyclohexylbromide & bromobenzene
Cyclohexylbromiide gives pale yellow ppt
Bromobenzene does not give the test.
3) Benzyl chloride & chlorobenzene
Benzyl chloride gives white ppt
Chlorobenzene 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 – conc
HCl & anh ZnCl2
10 – Does not
react.
2 – Forms
turbidity after
few min
0
30 - Forms
turbidity
immediately
Yellow ppt of
iodoform (CHI3)
Iodoform test –
NaOH &I2
Note: When both
the alcohols given
are primary / sec
apply this test
To distinguish primary , sec & ter alcohols
Primary – Methanol, Ethanol, n – alkyl
alcohols / alkan – 1-ol, benzyl alcohol
Secondary – Isopropyl alcohol, alkan – 2 –
ol
Tertiary – ter- buytl alcohol (2- methyl –
propan – 2- ol)
Alcohols with CH3 – CH – group give 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
Violet colour
1) Phenol & ethyl alcohol
Phenol gives violet colour
Ethyl alcohol does not give this
4
Aldehydes
Tollens test –
ammoniacal
AgNO3 , warm
Silver mirror
test
Aldehydes give positive test.
1. Acetaldehyde (Propanal) & Acetone
(Propanone)
Acetaldehyde (Propanal) gives Ag mirror
Acetone (Propanone) does not give this
test.
Iodoform test –
NaOH &I2
Note: When both
the compounds
given are
aldehydes/
ketones apply this
test
Ald. & ket. with CH3 – C – group give
this test
Yellow ppt of
iodoform (CHI3)
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
Formic acid
( Methanoic
acid)
Tollens test
Effervescence
due to evolution
of CO2 gas
Effervescence
due to evolution
of CO2 gas
All carboxylic acids give this test
HCOOH is the only carboxylic acid that
gives this test
HCOOH and CH3COOH
HCOOH gives this test . CH3COOH does not
give this test
6
Amines
Primary
amines
Hinsberg test –
10 –Soluble
Benzene sulphonyl
chloride + NaOH
20 – Insoluble
Carbylamine test –
Alcoholic KOH +
CHCl3
30 – does not
react with
Hinsberg
reagent
Very unpleasant
smelling
isocyanide gas
evolves
To distinguish primary , sec & ter
amines
Alkyl amine – 10
Dialkylamine -20 (N- alkyl alkanamine)
Trialkylamine( N,N – dialkyl alkanamine - 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.
Reddish orange
dye
Azodye test –
Aniline
NaNO2 + HCl +
phenol ( 00 to 5
C)
0
Diethylamine does not give this test
Aniline & Benzylamine
Aniline gives this test
Benzylamine does not give this test.
MECHANISM FOR PREPARATION OF ALCOHOL FROM ALKENE
MECHANISM FOR PREPARATION OF ETHER FROM ALCOHOL
MECHANISM FOR DEHYDRATION OF ALCOHOL TO ALKENE
ACIDITY OF PHENOL OR
o-p DIRECTING NATURE
STABILITY OF PHENOXIDE ION
UNIT 10 HALOALKANES & HALOARENES
1.What are ambident nucleophiles ? Give two eg.
Nucleophiles that possess two nucleophilic centres. Eg CN- & NO2
-
2. Why is chloroform stored in closed dark colour bottle ?
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?
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.
(a) Due to resonance in benzene ring there is partial double bond in C-X bond
which is difficult to break.
(b) 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 Alc. 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 ;
H3C
CH3
H
H
CH3
(b)Write Wurtz fittig reaction .
Ans. (a)
Br
4-Bromo-3-methyl –pent-2-ene.
(b)
2Na + X-R
X
R
+ 2NaX
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
HBr/peroxide
b)CH3CH=CH2
KOH
CH3CH2CH2Br
alc. KOH
c) CH3CH2CH2Br
CH3CH2CN
CH3CH2CH2OH
HBr
CH3CH=CH2
13.(a)Convert Nitrobenzene to m-Bromoiodobenzene.
CH3CHBrCH3
(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
CH3CH2OH
KCN
CH3CH2Cl
HBr/peroxide
b)CH3CH=CH2
KOH
CH3CH2CH2Br
alc KOH
c) CH3CH2CH2Br
CH3CH2CN
CH3CH2CH2OH
HBr
CH3CH=CH2
CH3CHBrCH3
UNIT 11 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.
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 R CH2OH > R2 CHOH
> 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
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.
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
Ans. Propan-1-ol < 4-methylphenol < phenol < 3-nitrophenol < 2,4, 6trinitrophenol.
14.
I)
Butanol to Butanoic acid
ii).
Ethanol to propanone
iii).
Phenol to salicylic acid
Ans.
IV).
Methanol to Ethanol
V.
Ethanol to propanol
vi.
Phenol to Benzyl Alcohol
Ans.
VII.
Ethanal to propan -2- ol
Ans.
UNIT 12 ALDEHYDES KETONES & CARBOXYLIC ACIDS
1.
Arrange the following set of compounds in order of
increasing boiling points.
CH3CH2CH2CHO, CH3CH2CH2CH2OH, CH3CH2 OCH2 CH3
,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 towards
nucleophilic 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.
Acetaldehyde undergoes aldol condensation. Why ?
Ans. Because it contains alpha hydrogen.
7.
Formaldehyde & benzaldehyde do not undergo aldol
condensation. Why ?
Ans. Because they do not contain alpha hydrogen.
8.
Formaldehyde & benzaldehyde show Cannizaro reaction.
Why ?
Ans. Because they do not contain alpha hydrogen
9.
Carboxylic acids are higher boiling liquids than alcohols. Why
?
Ans. 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.
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
UNIT 13 AMINES
1.
Ammonolysis is not preferred for preparation for primary
amines.Why ?
Ans. Because it gives mixture of amines.
2.
Alkyl amines have higher b.pts than alkanes of comparable
molecular mass. Why?
Ans. Because alkyl amines form intermolecular hydrogen bond.
3.
Alcohols have higher b.pts than alkyl amines of comparable
molecular mass. Why?
Ans. 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.
(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.
Methylamine in water reacts with ferric chloride to
precipitate hydrated ferric oxide. Why ?
Ans. Methylamine is more basic than water. So it forms OH- with water.
The OH- ions react with ferric chloride and gives ppt.
8.
Aniline does not undergo Friedel- Crafts reaction. Why ?
Ans. In Friedel- Crafts reaction AlCl3 which is used a catalyst is a Lewis
acid. It reacts with aniline and forms salt.
9.
Nitration of aniline gives a substantial amount of mnitroaniline besides ortho and para derivatives. Why ?
Ans. During nitration, in presence of strong acidic medium aniline is
protonated to form the anilinium ion which is meta directing.
10.
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
CΞN
NaNO2+HCl
0 – 5 0C
Aniline
KCN
CuCN
Benzonitrile
12. Account for the following :(i) Pkb of aniline is morethan that of methylamine.
(ii) Ethylamine is soluble in water whreas aniline is not.
(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.
UNIT 14 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 nonreducing reagents.
(Q.2) Define glycosidic linkage?
(1
Mark)
(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.
(2
Marks)
(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?
(2
Marks)
(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?
(2
Marks)
(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?
(3
Marks)
(Ans) Amino acids are usually colourless, crystalline solids. These are water soluble ,
high melting solids and behave like salts rather than simple amines or 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 proteis eg 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
RNA
uracil
Ribose sugar
Protein synthesis
Single helical
What happens when glucose reacts with
(i)
(ii)
(iii)
Br2 water
HI
Conc. HNO3
Ans:- (i) Gluconic acid
(ii)
(iii)
n-hexane
Saccharic acid
UNIT 15 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.
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
1. They
have
three
without cross-linking.
dimensional network of
covalent bonds with cross
links.
2. Upon
heating
these
2.Upon heating they retain
polymers become soft and
their strength and does
becomes hard again on
not become soft.
cooling.
3. These plastics can be 3.These plastics cannot be recycled.
recycled. They can be
melted and reused.
4.Example: PVC, Nylon, 4.Example: Phenol formaldehyde resin,
urea formaldehyde resin.
Polystrene.
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
Buna-S
Dacron
Neoprene
CH2=CH-CH=CH2 (Butadiene), CH2=CH-C6H5 (Styrene)
HOCH2-CH2OH ( Ethylene Glycol), HOOC-C6H4-COOH (Terepthalic acid)
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-βhydroxybutyrate-co-β-hydroxyvalerate (PHBV), Nylon-2-nylon-6.
10.
Give the differences
vulcanised rubber.
between
natural
rubber
and
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.
UNIT 16 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?
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
micro-organism 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
examples.
artificial
sweetening
agents?
Give
two
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
sodium-dodecylbenzene sulphonate.
Non-biodegradable detergents are not easily degraded by micoorganisms. 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 ( Butylated Hydroxy 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) Asprin drug helps in preventing the heart attack.
(ii) Detergents are non-biodegradable while soaps are biodegradable.
Ans. (i) Asprin is found to help in prevention of heart attack because of its antiblood 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.
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