higher …...also slow down the effect of ageing. example : vitamin c, vitamin e and b-carotene....
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HIGHER SECONDARY EXAMINATION - MARCH 2016 CHEMISTRY ANSWER KEY (Scheme of valuation )
PART – I Q.No C h o i c e BOOKLET SERIES - A Choice BOOKLET SERIES - B
1 d. AgBr d. +6
2 d. en c. T - Shape
3 c. amino acid b. Cataphoresis
4 c. Nitronium ion a. TiO2
5 b. 3 c. 16
6 c. 16 a. 4.2 105 sec-1
7 b. Monazite a. Peroxide
8 a. Zn2+ b. Monazite
9 b. MnSO4 d. AgBr
10 a. Silver sol b. MnSO4
11 d. +6 a. Entropy increases
12 a. Peroxide b. 2:1
13 a. Sucrose a. Sucrose
14 b. 2:1 d. en
15 b. Decreases c. amino acid
16 a. Ethylmethylamine d. KiloJoulemol-1
17 a. Entropy increases b. 6
18 b. 6 a. Ethylmethylamine
19 c. 1.44t b. 2 – pentanone
20 c. p – dimethylaminobenzene a. Zn2+
21 d. R─CH3 c. p – dimethylaminobenzene
22 d. KiloJoulemol-1 b. q = 0
23 b. 2 – pentanone d. Calcium oxalate
24 c. T - Shape b.
25 b.
b. 3
26 a. TiO2 d. R─CH3
27 b. q = 0 a. Silver sol
28 a. 4.2 105 sec-1 b. Decreases
29 b. Cataphoresis c. Nitronium ion
30 d. Calcium oxalate c. 1.44t
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Q.No PART ─ II Marks
31. What is bond order ? Bond order may be defined as half the difference between the number of electrons in bonding molecular orbitals (Nb) and the
number of electrons in antibonding molecular orbitals (Na)
3
(or)
3
3
32.
Z = 17 Z* = Z – S
Z* = 17 – [(0.35 × No. of other electrons in nth shell) + (0.85 × No. of electrons in (n –1)th shell) + (1.00 × total number of electrons in the inner shells)]
Z* = 17 – [(0.35 × 6) + (0.85 × 8) + (1 × 2)] Z* = 17 – 10.9 = 6.1
1+1
3
33. What is inert pair effect ?
The tendency of being less availability for ns electron in bonding. The inert pair effect increases down the group with the increase in atomic number.
3
34.
3
35. Give the percentage composistion of nichrome and its uses . Nichrome: Cr = 15%, Ni = 60% Fe = 25%
It is used in resistance wires for electrical heating
2 1
36. What is the action of heat on K2Cr2O7 ?
3
37. What is Q – value of a nuclear reaction ?
The amount of energy absorbed or released during nuclear reaction is called Q-value of nuclear reaction.
Qvalue = (mp-mr) 931 MeV where mr - Sum of the masses of reactants
mp - Sum of the masses of products
2
38. Sketch a) Simple Cube b) Face – Centred cube
c) Body centered cube a) b) c)
31 3
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Q.No PART ─ II (Contd.) Marks
39. For a chemical reaction the values of ΔH and ΔS at 400 K are –10 k cal mol–1 and 20 cal. deg–1 mol–1 respectively. Calculate the value of ΔG of the reaction?
ΔG = ΔH – TΔS
ΔG At 400 K; ΔG = –10,000 – (20 × 400) ΔG = –18,000 cals. mole–1
1 1 1
3
40. Define reaction quotient . Reaction quotient „Q‟ is defined as the ratio of product of initial
concentrations of products to the product of initial concentrations of reactants under nonequilibrium conditions.
3
41. Write any three characteristics of simple reaction. (Any Three characteristics)
1. Occurs in single step.
2. Overall order values are small. Total and pseudo order values lie between 0,1,2 and 3.
3. No side reactions 4. Products are formed directly from the reactants 5. Experimental rate constant values agree with the calculated
values. Theories of reaction rates apply well on simple reactions.
31 3
42. What is threshold energy ? Reactant molecules come into contact through collisions for
transformation into product molecules. Since, not all collisions are successful in producing the product molecules, all colliding
molecules must possess certain minimum energy called as the threshold energy which is needed to make the collisions effective and successful.
(or) Threshold energy = Activation energy + Energy of colliding the molecules.
3
43. Why colloidal system in gas in gas does not exist?
A colloidal solution of gas in gas is not possible, as gases are completely miscible and always form true solutions.
3
44. State Kohlrausch’s law . „„At infinite dilution wherein the ionisation of all electrolytes is
complete, each ion migrates independently and contributes a definite value to the total equivalent conductance of the
electrolyte‟‟. (OR) At infinite dilution, according to Kohlrausch‟s law, the total equivalent conductance of the electrolyte,
where are the cationic and anionic equivalent conductances at infinite dilutions
and n+ and m– correspond the valency of cations and anions furnished from each molecule of the electrolyte.
3
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Q.No PART ─ II (Contd.) Marks
45. Draw the structure of S─cis and S─trans form of 1,3 ─ butadiene.
(OR)
1
+
1
3
46. Convert Glycerol into Acrolein. When glycerol is heated with potassium bisulphate or conc.
Sulphuric acid or phosphorous pentoxide dehydration takes place. Two b-elimination reaction takes place to give acrolein or acrylic
aldehyde.
3
47. Why is glycol more viscous than ethanol ?
Because of the presence of two hydroxyl groups the intermolecular hydrogen bonding is made much stronger.
Hydrogen bond can be formed between both OH groups resulting in a polymeric structure. This leads to high viscosity.
2
1
3
48.
i)
ii)
How is benzophenone prepared by Friedel-Crafts method ?
By Friedel Crafts reaction : Benzoylation of benzene takes place in presence of anhydrous aluminium chloride as a catalyst.
Benzoyl cation (C6H5CO+) is the electrophile.
(OR)
By the reaction of excess of benzene with carbonyl chloride
3
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Q.No PART ─ II (Contd.) Marks
(phosgene) in presence of anhydrous aluminium chloride as a catalyst, Benzoylchloride may be formed initially. Thus involves two Friedel craft’s benzoylation reactions.
49. Mention the uses of formic acid.
(Any three uses or any three suitable uses) (a) In Textile Industry for preparing ‘‘mordants’’.
(b) In leather tanning for removing lime from the hides. (c) In coagulating rubber latex. (d) Nickel formate as hydrogenation catalyst.
(e) As a stimulant for the growth of yeast. (f) As an antiseptic and in preservation of fruits. (g) In the treatment of gout.
31 3
50. An yellow coloured liquid (A) called as oil of mirbane is
reduced with Sn /HCl to give B. Identify A and B and write the equations.
The compound (A) which is called oil of mirbane is Nitrobenzene Which on reduction with Sn/HCl gives Anline (B)
1
+
1
3
51. What are antioxidants ? Give example .
The substances that act against oxidants are called antioxidants and protect us against cardiovascular disease, cancer and cataract
also slow down the effect of ageing. Example : vitamin C, vitamin E and b-carotene.
Antioxidants act as radical inhibitors. and as food preservatives.
2
1
3
52.
PART ─ III
Section ─ A
Derive de-Broglie’s equation. The wavelength of the wave associated with any material particle was calculated by analogy with photon as follows :-
In case of a photon, if it is assumed to have wave character, its energy is given by
E = h (according to the Planck‟s quantum theory) ...(i)
where is the frequency of the wave and h is Planck‟s constant.
If the photon is supposed to have particle character, its energy is given by
E = mc2 (according to Einstein equation) ...(ii) where m is the mass of photon and c is the velocity of light.
From equations (i) and (ii), we get
1
1
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Q.No Section ─ A (Contd.) Marks
de Broglie pointed out that the above equation is applicable to any material particle. The mass of the photon is replaced by the mass of the material particle and the velocity “c” of the photon is
replaced by the velocity v of the material particle. Thus, for any material particle like electron, we may write
where m = p is the momentum of the particle.
The above equation is called de Broglie equation and „λ‟ is
called deBroglie wavelength.
1
5
53. Explain the extraction of Zinc from its chief ore.
The chief ore of Zinc is Zinc blende.
1. Concentration The ore is crushed and then concentrated by froth-floatation
process. 2. Roasting The concentrated ore is then roasted in the presence of excess
of air at about 1200 K.
3. Reduction Zinc oxide is mixed with powdered coke and heated to 1673 K
in a fire clay retort, in which ZnO is reduced to zinc metal.
Purification
Zinc is purified by electrolytic refining. In this process, Impure Zinc is anode and cathode is of pure thin sheet of Zinc. The electrolyte is ZnSO4 solution containing a little of dil.H2SO4. On
passing electric current, pure zinc get deposited at the cathode.
1
1
1
5
54. Differenciate Lanthanides and Actinides . (Any five differences)
Lanthanides Actinides
i) Binding energies of 4f
electrons are higher.
i) Binding energies of 5f
electrons arelower.
ii) Maximum oxidation state exhibited by lanthanides is
+4 e.g. Ce4+
ii) Due to lower binding energies they show higher
oxidation states such as +4, +5 and +6. Uranium
exhibits +6 oxidation state
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Q.No Section ─ A (Contd.) Marks
54…
Lanthanides Actinides
ii) in UF6 and UO2Cl2
iii) 4f electrons have greater shielding effect.
iii) 5f electrons have poor shielding effect.
iv) Most of their ions are colourless.
iv) Most of their ions are coloured U3+ (red), U4+
(green) and UO22+ (yellow)
v) They are paramagnetic
but magnetic properties can be easily explained.
v) They are also
paramagnetic but their magnetic properties are very difficult to interpret.
vi) They do not form
complexes easily.
vi) They have much greater
tendency to form complexes.
vii) Except promethium, they are non-radioactive.
All of them are radioactive.
viii) Their compounds are less basic.
Their compounds are more basic.
ix) They do not form
oxocations
They form oxocations such
as UO22+, UO+, NpO2
+, PuO2
+.
51
5
55. Write a note on Haemoglobin.
Haemoglobin in the red blood cells carries oxygen from the lungs to the tissues. It delivers the oxygen molecule to myoglobin in the tissues. When the oxygen has been released for cell respiration,
haemoglobin loses its bright red colour and becomes purple. It then combines with the waste carbon dioxide produced by the cells
and deposits in the lungs so that the gas can be exhaled. Nature of haemoglobin and myoglobin Both are having the same structure excepting the fact that
myoglobin is a monomer and haemoglobin is a tetramer. Both are iron-porphyrin complex. These are biocoordination complexes
formed between porphyrin and iron in its +2 oxidation state (Fe2+). The iron-porphyrin complex is called the heme group, which is a part of haemoglobin. Each haemoglobin molecule consists of four
subunits, each unit is being a folded chain. The working part of haemoglobin is a hemegroup containing an Fe2+ cation coordinated to four nitrogen atoms of porphyrin group and one nitrogen atom
of histidine group. The sixth octahedral site is available to bind oxygen molecule.
1
1
1
1
1
5
56.
Section ─ B
State the various statements of second law of thermodynamics.
i) “It is impossible to construct an engine which operated in a complete cycle will absorb heat from a single body and convert it completely to work without leaving some changes in the working
system”.This is called as the Kelvin – Planck statement of II law of thermodynamics.
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Q.No Section ─ B (Contd) Marks
ii) “It is impossible to transfer heat from a cold body to a hot body by a machine without doing some work”. This is called as the clausius statement of II law of
thermodynamics. iii) „A process accompanied by increase in entropy tends to be
spontaneous”. This statement is called as the entropy statement of II law of thermodynamics.
Entropy is a measure of randomness or disorder of the molecules of a system and it is a thermodynamic state function. A system
always spontaneously changes from ordered to a disordered state. Therefore entropy of a spontaneous process is constantly increasing.
iv) “Efficiency of a machine can never be cent percent”. v) The heat Efficiency of any machine is given by the value of ratio of output to input energies. Output can be in the form of any
measurable energy or temperature change while input can be in the form of heat energy or fuel amount which can be converted to
heat energy.
The machine can be a heat engine also. Consider a heat engine
which has an initial temperature T1 and final temperature as T2, then if T1>T2 then when some amount of heat is being converted into work, T2 is the lowered temperature. The efficiency „η‟ is given by,
According to II law of thermodynamics it is impossible to have a
machine or heat engine which converts the input energy completely into output energy or output work without any amount of heat or energy being absorbed by the machine.
Hence, % efficiency can never be achieved as cent percent.
51
5
57. State Le Chatelier’s principle. Discuss the effect of pressure,
concentration and temperature on the following reaction. N2(g) + O2(g) 2NO(g)
Le Chatelier’s principle: According to this principle, if a system at equilibrium is subjected to a disturbance or stress, then the equilibrium shifts in the
direction that tends to nullify the effect of the disturbance or stress.
Effect of change of concentration: At the equilibrium conditions the reaction mixture contains both the reactant and product molecules, that is, N2, O2 and NO
molecules. The concentrations of reactant and product molecules are constant and remain the same as long as the equilibrium
2
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Q.No Section ─ B (Contd) Marks
conditions are maintained the same. If a change is imposed on the system by purposely adding NO into the reaction mixture then the product concentration is raised. Since the system possesses
equilibrium concentrations of reactants and products, the excess amount of NO react in the reverse direction to produce back the
reactants and this results in the increase in concentrations of N2
and O2. Similarly if the concentration of reactants such as N2 and O2 are purposely raised when the system is already in the state of
equilibrium, the excess concentrations of N2 and O2 favour forward reaction. Concentration of NO is raised in the reaction mixture.
In general, in a chemical equilibrium increasing the concentrations of the reactants results in shifting the equilibrium in favour of the products while increasing the concentrations of the products
results in shifting the equilibrium in favour ofthe reactants. Effect of change of pressure:
In the above equilibrium Dng = 0 So the equilibrium does not
dependent on pressure and volume. Therefore pressure has no effect
on the equilibrium. Hence change of pressure does not affect the equilibrium.
2
1
5
58. Write notes on (i) consecutive reactions, (ii) parallel reactions
(i) consecutive reactions: The reactions in which the reactant forms an intermediate and the
intermediate forms the product in one or many subsequent reactions are called as consecutive or sequential reactions.
In such reactions the product is not formed directly from the reactant. Various steps in the consecutive reaction are shown as below :
A = reactant ; B = intermediate ; C = product. Initially only the reactant A will be present. As the reaction starts, A produces an intermediate B through k1 rate constant. As and when B is formed,
it produces the product C through k2 rate constant. After the completion of reaction only „C‟ is present and concentrations of A
and B will be zero. Example of consecutive reactions
Saponification of a diester in presence of an alkali :
(ii) Parallel reactions In these group of reactions, one or more reactants react
simultaneously in two or more pathways to give two or more products. The parallel reactions are also called as side reactions.
1
1
1
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Q.No Section ─ B (Contd) Marks
The reactant A reacts to give products B,C,D separately by following three different reaction pathways involving different k1,
k2, k3 rate constants respectively. Among the many side reactions, the reaction in which maximum yield of the product obtained is called as the main or major reaction while the other reactions are
called as side or parallel reactions Examples of parallel reaction (Any one Example)
(i) Bromination of bromobenzene :
(ii) Dehydration of 2-methyl-2-butanol :
1
5
59. Calculate the potential of the following cell at 298 K Zn│Zn2+ (a = 0.1) ║Cu2+ (a = 0.01) │Cu
EoZn
2+│Zn = – 0.762 V & EoCu
2+│Cu = + 0.337 V
The overall cell reaction is Zn + Cu2+ (a = 0.01) Zn2+ (a = 0.1) + Cu
The cell potential given by nernst equation
1
or
1
1
1+
+
5
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Q.No Section ─ C Marks
60.
(i)
(ii)
(iii)
How are the following reactions effected ? (i) diethylether with Grignard reagent (ii) diethylether with mineral acid
(iii) ethylmethylether with excess hot Conc. HI acid.
An ether solution of Grignard reagent contains the following complex of ether. Thus the Grignard reagent is stabilised in dry ether. Hence ether is used as a solvent for Grignard‟s reagent.
Strong mineral acids protonate the ethereal oxygen forming
oxonium salts. In this reaction diethyl ether acts as Lewis base.
With excess hot concentrated hydroidic acid, alkyl iodides are formed. This reaction is used in the Zeisel’s method of detection
and estimation of alkoxy (especially methoxy) group in natural products like alkaloids.
1
2
1
5
61.
(i)
(ii)
Give the following reactions: (i) Benzoin condensation (ii) Knoevenagal reaction
This is another reaction, which is characteristic of aromatic aldehydes. When benzaldehyde is refluxed with aqueous alcoholic
potassium cyanide -a-hydroxy ketone called benzoin is formed.
Cyanide ion (CN–) is the specific catalyst in this reaction. Benzoin can be considered as dimer of benzaldehyde. In presence of cyanide ion, one molecule of benzaldehyde behaves
as a carbanion. That brings forth nucleophilic attack on the carbonyl group of another molecule of benzaldehyde. This results
to Benzoin.
Benzaldehyde condenses with malonic acid in presence of pyridine
forming cinnamic acid, pyridine is the basic catalyst here. This mechanism is quite similar to the previous reaction, involving
2
5
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Q.No Section ─ C (Contd) Marks
carbanion, from the malonic acid.
2
62. Write the mechanism involved in the bromination of salicylic
acid.
2
1
1
5
63. Mention the significance of rocket propellants. Rockets have been in use since early fifty‟s. Russians used
powerful rockets to put their space vehicles in space. United States of America used Saturn rockets for their Apollo space missions.
India has recently launched its satellite launch vehicle SLV-3 from Sri harikota. Rocket motors are used both in space vehicles and in offensive
weapons such as missiles. The propulsion system in most space vehicles consists of rocket engines powered by chemical propellants. These also called rocket propellants.
Propellants are combustible compounds which on ignition undergo rapid combustion to release large quantities of hot
gases. A propellant is a combination of an oxidiser and a fuel.
Working of a propellant. When a propellant is ignited, it burns to produce a large quantity of hot gases. These gases then come out
through the nozzle of the rocket motor. The passage of gases through the nozzle of the rocket motor, provides the necessary thrust for the rocket to move forward according to the Newton’s
Third law of Motion (to every action, there is an equal and opposite reaction).
Some of the examples for propellents are Hydrazine, Liquid hydrogen, Polyurethane, etc.
1
1
1
1
1
5
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Q.No PART ─ IV Marks
64a.
Explain the variation of IE along the group and period. In a period, the value of ionisation potential increases from left to
right with breaks where the atoms have somewhat stable configurations. This is due to the reason that the nuclear charge
increases whereas atomic radius decreases. In a group, the ionisation potential decreases from top to bottom. This is due to the effect of the increased atomic radius.
2
2
5
64b. Write any five uses of Fluorine.
1. Fluorine is used in the manufacture of a series of compounds known as freons. These non-toxic, non-combustible and volatile liquids are used as refrigerants in refrigerators, deep freezers and
air conditioners. The most common, freon is known as dichlorodifluoro methane CF2 Cl2.
2. CaF2 is used as flux in metallurgy. 3. NaF is used as a preservative to prevent fermentation and also for preventing dental cavities.
4. SF6 is used as an insulating material in high voltage equipment. 5. Teflon is used as container to store hydrofluoric acid.
6. UF6 is used in the separation of U235 from U238.
51
5
65a. Give the postulates of Werner’s theory of Co-ordination
Compounds.
1) Every metal atom has two types of valencies i) Primary valency or ionisable valency ii) Secondary valency or non ionisable valency
2) The primary valency corresponds to the oxidation state of the metal ion. The primary valency of the metal ion is always satisfied
by negative ions. 3) Secondary valency corresponds to the coordination number of the metal ion or atom. The secondary valencies may be satisfied by
either negative ions or neutral molecules. 4) The molecules or ion that satisfy secondary valencies are called ligands.
5) The ligands which satisfy secondary valencies must project in definite directions in space. So the secondary valencies are
directional in nature whereas the primary valencies are non-directional in nature. 6) The ligands have unshared pair of electrons. These unshared
pair of electrons are donated to central metal ion or atom in a compound. Such compounds are called coordination compounds.
Werner’s representation Werner represented the first member of the series [Co(NH3)6]Cl3 as follows. In this representation, the primary valency (dotted lines)
are satisfied by the three chloride ions. The six secondary valencies (solid lines) are satisfied by the six ammonia molecules.
51 5
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Q.No PART ─ IV (Contd) Marks
65b. Chemical reactions Nuclear reactions
1 These reaction involve some loss, gain or overlap of outer orbital electrons of the reactant atoms.
1 Nuclear reactions involve emission of alpha, beta and gamma particles from the nucleus.
2 A chemical reaction is balanced in terms of mass only
2 Nuclear reaction is balanced in terms of both mass and energy.
3 The energy changes in any chemical reaction is very much less when compared with nuclear reaction.
3 The energy changes are far exceed than the energy changes in chemical reactions.
4 In chemical reactions, the energy is expressed in terms of
kilojoules per mole.
4 In nuclear reactions, the energy involved is expressed
in MeV (Million electron volts) per individual nucleus.
5 No new element is produced since nucleus is unaffected.
5 New element / isotope may be produced during the nuclear reaction.
51 5
66a. Write about the most common point defects.
The most common point defects are the Schottky defect and Frenkel defect. Schottky defects
This defect is caused if some of the lattice points are unoccupied.
The points which are unoccupied are called lattice vacancies. The number of missing positive and negative ions is the same in this case and thus, the crystal remains neutral. The existence of two
vacancies, one due to a missing Na+ ion and the other due to a missing Cl- ion in a crystal of NaCl is shown in Figure.
Schottky defects appears generally in ionic crystals in which the positive and negative ions do not differ much in size. Frenkel defects
This defect arise when an ion occupies an interstitial position between the lattice points. This defect occurs generally in ionic
crystals in which the anion is much larger in size than the cation. AgBr is an example for this type of defect. One of the Ag+ ion occupies a position in the interstitial space rather than its own
appropriate site in the lattice is shown in Figure
1
1
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QNo PART ─ IV (Contd) Marks
The crystal remains neutral since the number of positive ions is the same as the number of negative ions.
66b. How colloids are prepared by Mechanical dispersion method and Electro-dispersion method ?
Mechanical dispersion using colloidal mill The solid along with the liquid is fed into a colloidal mill. The
colloidal mill consists of two steel plates nearly touching each other androtating in opposite directions with high speed. The solid particles are ground down to colloidal size and then dispersed in
the liquid. Colloidal graphite and printing inks are made by this method.
Electro-dispersion method: (Bredig’s Arc Method)
This method is suitable for the preparation of colloidal solution of metals like gold, silver, platinum etc. An arc is struck between the
metal electrodes under the surface of water containing some stabilising agent such as trace of alkali. The water is cooled by immersing the container in a cold bath. The intense heat of the arc
vapourises some of the metal which condenses under cold water.
1
1
1
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67a. Describe the action of an acid buffer solution with an example.
Buffer action : Let us illustrate buffer action by taking example of a common buffer system consisting of a solution of acetic acid and
sodium acetate (CH3COOH/CH3COONa). since the salt is completely ionised, it provides the common ions CH3COO– in excess. The common ion effect suppresses the
ionisation of acetic acid. This reduces the concentration of H+ ions which means that pH of the solution is raised. It is stated that a buffer solution containing equimolar amounts (0.10 M) of acetic
acid and sodium acetate has pH 4.74. Now we proceed to discuss how the addition of a small amount of HCl or NaOH to the buffer
solution affects its pH. The pH of the buffer is governed by the equilibrium
CH3COOH H+ + CH3COO─ ….(1)
1
1
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QNo PART ─ IV (Contd) Marks
The buffer solution has a large excess of CH3COO– ions produced by complete ionisation of sodium acetate, CH3COONa Na+ + CH3COO─ ….(2)
1. Addition of HCl. Upon the addition of HCl, the decrease of H+
ions is counteracted by association with the excess of acetate ions to form unionised CH3COOH. Thus the added H+ ions are neutralised and the pH of the buffer solution remains unchanged.
However owing to the increased concentration of CH3COOH, the equilibrium (1) shifts slightly to the right to increase H+ ions. This
explains the marginal increase of pH of the buffer solution on addition of HCl.
Addition of OH─ H2O Buffer solution CH3COOH H+ + CH3COO─
CH3COONa CH3COO─ + Na+
Addition of H+ CH3COOH
1
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67b. Write any five common terms in cell Terminology.
Current is the flow of electrons through a wire or any conductor. Electrode is the material : a metallic rod/bar/strip which
conducts electrons into and out of a solution. Anode is the electrode at which oxidation occurs. It sends
electrons into the outer circuit. It has negative charge and is shown as (–) in cell diagrams. Cathode is the electrode at which electrons are received from the
outer circuit. It has a positive charge and is known as (+) in the cell diagrams.
Electrolyte is the salt solution in a cell. Anode compartment is the compartment of the cell in which oxidation half-reaction occurs. It contains the anode.
Cathode compartment is the compartment of the cell in which reduction half-reaction occurs. It contains the cathode. Half-cell. Each half of an electrochemical cell, where oxidation
occurs and the half where reduction occurs, is called the half cell.
51 5
68a. Explain Geometrical Isomerism found in organic compounds With suitable examples.
Isomerism that arises out of difference in the spatial arrangement of atoms or groups about the doubly bonded carbon
atoms is called Geometrical isomerism. These isomers are not mirror images of each other. Rotation about C=C is not possible at normal conditions and hence the isomers are isolable.
If different atoms or groups are bonded to the „C=C‟ bond in a molecule, more than one spatial arrangement is possible. For
example, 2-butene exists in two isomeric forms.
1
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QNo PART ─ IV (Contd) Marks
The isomer in which similar groups lie on the same side is called ‘cis isomer’ (I). The other in which similar groups lie in opposite direction is called ‘Trans isomer’ (II). This isomerism is called
„Cis-Trans‟ isomerism. The two groups attached to the carbon atoms need not be same, it
may be different also. e.g., 2-pentene
This isomerism arises out of the hindrance to rotation about the C=C bond in such molecules.
The cis-trans isomers do not differ much in chemical properties. They differ in physical properties like boiling point, melting point,
crystal structure, solubility and refractive index. Highly substituted olefin is more stable than less substituted olefin. Among substituted olefins, trans olefin is more stable than cis
olefin. In the cis isomer because similar groups are very near each other, Vander Waals repulsion and steric hindrance make the molecule much unstable. In the trans isomer, similar groups are
diagonally opposite to each other. Hence there is no such steric interaction. Generally trans isomer is more stable than cis isomer.
Hence reactivity of cis isomer may be little higher than the trans isomer. The energy of the cis isomer is greater than that of trans isomer.
1 1
2
5
68b. What is the action of heat on Oxalic acid and Succinic acid ?
Oxalic acid on heating at 373 K – 378 K loses water of hydration. On further heating it decomposes to formic acid and carbon dioxide.
Succinic acid on heating to 300oC loses a molecule of water to form anhydride.
2
2
5
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QNo PART ─ IV (Contd) Marks
69a. Write a note on the reduction of nitrobenzene under alkaline medium.
In alkaline medium, Nitro benzene on reduction forms the intermediate products nitrosobenzene (C6H5NO) and phenyl
hydroxylamine (C6H5NHOH). These undergo bimolecular condensation reaction. According to the appropriate reducing agent in alkaline medium different products are obtained.
2
1
1
5
69b. Outline the classification of carbohydrates With suitable examples.
3+2 5
70a. An organic compound (A) of molecular formula C6H6O gives violet colour with neutral ferric chloride. Compound (A) reacts
with metallic sodium and gives compound (B). When compound (B) is heated with CO2 at 400K under pressure gives
compound (C). With dilute HCl compound (C) reacts and gives compound (D). Identify (A), (B), (C) and (D) and explain the reactions.
Since compound A gives violet colour with neutral ferric chloride it is Phenol
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QNo PART ─ IV (Contd) Marks
Compound Name
A Phenol
B Sodium phenolate
C Sodium salicylate
D Salicylic acid
1
1
1
4
5
70b. An element (A) belongs to group number – 11 and period number – 4 reacts with conc H2SO4 to give its salt (B) with the liberation of SO2 gas. Compound (B) reacts with hydrogen
sulphide gas gives compound (C) which is black in colour. Identify (A) , (B) and (C) explain the reactions.
The element of group number 11 and period number 4 is Copper
Element /Compound Formula Name
A Cu Copper
B CuSO4 Copper sulphate
C CuS Copper sulphide
1
1
31
5
70c. An organic compound with molecular formula C3H6O (A) does not reduce Tollen’s reagent but undergoes Iodoform reaction.
Compound (A) reacts wth Zn/Hg – HCl gives the compound (B) , which is a hydrocarbon . in the presence of conc. H2SO4
compound (A) condenses to give compound (C) of molecular formula C9H12. Indentify (A), (B), (C) and explain the reactions.
Since compound (A) does not reduce Tollen‟s reagent but undergoes iodoform test it should be a ketone ie. Acetone
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QNo PART ─ IV (Contd) Marks
Compound Formula Name
A CH3COCH3 Acetone
B CH3CH2CH3 Propane
C C6H3(CH3)3 Mesitylene (or) 1,3,5 – trimethylbenzene
1
1
31
5
70d. An electric current is passed through three cells in series containing respectively solutions of copper sulphate, silver
nitrate and potassium iodide. What weights of silver and iodine will be liberated while 1.25 g of copper is being
deposited ?
1
(or)
+
1
(or)
+
Prepared by:
J.STEPHEN JAMES M.Sc., M.Ed., M.Phil., D.C.P.I.C
PG. ASSISTANT IN CHEMISTRY
TOWN Hr. Sec. SCHOOL
KUMBAKONAM
brighton1112@gmail.com
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