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Chemistry (Theory)
Time Allowed: 3 hours Maximum Marks: 70
General Instructions:
(a) All questions are compulsory.
(b) Question numbers 1 to 5 are very short answer questions and carry 1 mark
each.
(c) Question numbers. 6 to 10 are short answer questions and carry 2 marks each.
(d) Question numbers 11 to 22 are also short answer questions and carry 3 marks
each.
(e) Question number 23 carries 4 marks.
(f) Question numbers. 24 to 26 are long answer questions and carry 5 marks each.
1. Agl crystallises in cubic close packed ZnS structure. What fraction of
tetrahedral sites are occupied by Ag+ ion?
Ans. Half of the voids of tetrahedral sites are occupied by Ag+ ion.
2. When dried fruits and vegetables are placed in water for some time, they
slowly swell up. Give reason.
Ans. When dried fruits and vegetables are placed in water for some time,
they slowly swell up as the process of osmosis takes place.
3. An element occurs in the BCC structure. How many atoms are present in its
unit cell.
Ans. 2
4. What happens when blood cells are placed in pure water? Write two ions
which are responsible for maintaining proper osmotic pressure inside and
outside the living cells.
Ans. If blood cells are placed in pure water ,water molecules move into the
blood cells and blood cells swell.Na+ and K+ ions are responsible for
maintaining proper osmotic pressure inside and outside the cells.
5. Based on the type of dispersed phase, what type of colloids are micelles?
Ans. Associated colloids
6. Calculate the molality of ethanol solution in which the mole fraction of
water is 0.88.
Ans. Mole fraction of water, H2o=0.88
Mole fraction of ethanol, C2H5OH= 1-0.88
=0.12
n2= number of moles of ethanol.
n1= number of moles of water. Molality of ethanol means the number of
moles of ethanol present in 1000 g of water.
n1= 1000/18 = 55.5 moles
Substituting the value of n1 in equation (1)
n2=7.57 moles
Molality of ethanol ( C2H5OH) = 7.57 m
7. Write the structures and IUPAC names of the cross aldol condensation
products only of ethanal and Propanal.
Ans.
i. 2-Methylbut-2-enal
ii. Pent-2-enal
CH3-CH2-CH=CH-CHO
8. Name the crystal defect which reduces the density of an ionic solid? What
type of ionic substances show this defect?
Ans. Schottky defect It is shown by ionic substances in which the cation
and anion are of almost similar sizes. / ionic substances having high
coordination number.
9. Draw the structure of the following compounds:
i. H2S2O7
ii. XeOF4
Ans.
10. Draw the molecular structures of the following:
i. Noble gas species which is isostructural with BrO3-
ii. Dibasic oxoacid of phosphorus
Ans. (i)
(ii)
11. State briefly the principles involved in the following operations in
metallurgy. Give an example.
i. Hydraulic washing.
ii. Zone refining.
Ans.
i. Hydraulic washing: Principle involved: differences in gravities of the
ore and the gangue particles e.g. oxide ores (haematite), native ores
Au, Ag.
ii. Zone refining: Principle involved: the impurities are more soluble in
the melt than in the solid state of the metal. e.g. germanium, silicon,
boron, gallium and indium.
12. Give reasons for the following:
i. When 2g of benzoic acid is dissolved in 25 g of benzene, the
experimentally determined molar mass is always greater than the true
value.
ii. Mixture of ethanol and acetone shows positive deviation from Raoult’s
Law.
iii. The preservation of fruits by adding concentrated sugar solution protects
against bacterial action.
Ans.
i. Molecules of benzoic acid dimerise in benzene, the number of
particles are reduced
ii. The intermolecular interactions between ethanol and acetone are
weaker/ the escaping tendency of ethanol and acetone molecules
increases on mixing / the vapour pressure increases.
iii. Due to osmosis, a bacterium on fruit loses water, shrivels and dies.
13. Explain what is observed when
i. Silver nitrate solution is added to potassium iodide solution.
ii. The size of the finest gold sol particles increases in the gold sol. Two
oppositely charged sols are mixed in almost equal proportions.
Ans.
i. When silver nitrate solution is added to potassium iodide
solution, a precipitate of silver iodide is formed which adsorbs
iodide ions from the dispersion medium and a negatively
charged colloidal solution is formed.
ii. As the size of the gold sol particles increases, the colour of the
solution changes from red to purple, then blue and finally
golden because the colour of colloidal solution depends on the
wavelength of the light scattered by the dispersed particles and
wavelength further depends on the size of the particles.
iii. When oppositely charged sols are mixed in almost equal
proportions, neutralization of their charges occur and
precipitation occurs.
14. Give reasons for the following
i. Use of aspartame as an artificial sweetener is limited to cold foods.
ii. Metal hydroxides are better alternatives than sodium
hydrogencarbonate for treatment of acidity.
iii. Aspirin is used in prevention of heart attacks.
Ans.
i. It is unstable at cooking temperature.
ii. Excessive hydrogencarbonate can make the stomach alkaline
and trigger the production of even more acid. Metal hydroxides
being insoluble do not increase the pH above neutrality.
iii. Aspirin has anti blood clotting action.
15.
i. Name the branched chain component of starch.
ii. Ribose in RNA and deoxyribose in DNA differ in the structure around
which carbon atom?
iii. How many peptide linkages are present in a tripeptide?
Ans.
i. Amylopectin.
ii. C-2
iii. Two peptide linkages.
16.
i. Which of the following biomolecule is insoluble in water? Justify.
Insulin, Haemoglobin, Keratin.
ii. Draw the Haworth structure for α-D-Glucopyranose.
iii. Write chemical reaction to show that glucose contains aldehyde as
carbonyl group.
Ans.
i. Keratin is insoluble in water. It is a fibrous protein in which the
polypeptide chains are held together by strong intermolecular
forces, hence insoluble in water.
ii. α-D-Glucopyranose
iii. The sequence in the complimentary strand is
ATGCTTGA
17. Explain the following observations giving appropriate reasons:
a. Ozone is thermodynamically unstable with respect to oxygen.
b. The HEH bond angle of the hydrides of group 15 elements decrease as
we move down the group.
c. Bleaching effect of chlorine is permanent.
Ans.
i. Decomposition of ozone into oxygen is an exothermic process (∆H = -ve)
and results in an increase in entropy (∆S = +ve), resulting in large
negative Gibbs energy change (∆G = -ve). Decomposition of ozone to
oxygen is a spontaneous process.
ii. As we move down the group, the size of the central atom increases and
the electronegativity decreases. Therefore the bond pair of electrons lies
away from the central atom, the force of repulsion between the adjacent
bond pairs decreases. So the bond angle decreases.
iii. The bleaching action of Cl2 is due to the oxidation of colored substance
to colourless by nascent oxygen.
Cl2 + H2O → 2 HCl + [O] Nascent oxygen.
18.
a. Predict the number of unpaired electrons in the tetrahedral [MnBr4]2-
ion.
b. Draw structures of geometrical isomers of [Co (NH3)4Cl2]+.
c. Write the formula for the following coordinate compound:
Amminebromidochloridonitrito-N-platinate(II)
Ans.
i. Mn has the configuration 3d54s2. Hence the configuration of Mn2+ in
[MnBr4]2- is 3d5.
Mn2=
Since it is tetrahedral in shape, the hybridization is sp3.
There are five unpaired electrons.
ii.
iii. [Pt(NH3)BrCl(NO2)]-
19. A metal complex having composition Cr (NH3)4Cl2Br has been isolated in
two forms A and B. The form A reacts with AgNO3to give a white
precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale
yellow precipitate soluble in concentrated ammonia.
i. Write the formulae of isomers A and B.
ii. State the hybridisation of chromium in each of them.
iii. Calculate the magnetic moment (spin only value) of the isomer A
Ans.
i. Isomer A: [Cr(NH3)4BrCl]Cl
Isomer B: [Cr (NH3)4 Cl2] Br
ii. Hybridisation of Cr in isomer A and B is d2sp3.
iii. Number of unpaired electrons in Cr3+(3d3) is 3
Magnetic moment =√𝑛(𝑛 + 2)
√3(3 + 2) =3.87BM
20. Complete the following reactions :
Ans.
a. b. HOOC -CH2-CH2-CH2-CH2–COOH
c.
21.
a. A colloidal sol is prepared by the given method in figure. What is the
charge of AgI colloidal particles in the test tube? How is the sol formed,
represented?
b. Explain how the phenomenon of adsorption finds application in
Heterogeneous catalysis.
c. Which of the following electrolytes is the most effective for the
coagulation of Fe (OH) 3 sol which is a positively charged sol?
NaCl, Na2SO4, Na3PO4
Ans.
a. Negative charge is developed on the sol.
Sol is represented as AgI/ I-
b. Adsorption of reactants on the solid surface of the catalysts increases the rate
of reaction.
c. Na3PO4 Hardy-Schulze rule
22. Niobium crystallises in body-centred cubic structure. If the atomic radius is
143.1 pm, calculate the density of Niobium. (Atomic mass = 93u)
Ans.
.
23. John had gone with his mother to the doctor as he was down with fever. He
then went to the chemist shop with his mother to purchase medicines
prescribed by the doctor. There he observed a young man pleading with the
chemist to give him medicines as he had nasal congestion. The chemist gave
him cimetidine. John advised and also explained to the young man that he
should only take the medicines prescribed by the doctor.
Answer the following questions:
i. Did the chemist give an appropriate medicine? Justify your answer.
ii. John’s action was appreciated by his mother. List any two reasons.
a. Write the mechanism of hydration of ethene to form ethanol.
Ans.
i. No the chemist did not give the appropriate medicine. Cimetidine is
an antihistamine, but it is an antacid and not an antiallergic drug.
Antacid and antiallergic drugs work on different receptors. Therefore
cimetidine cannot be used to treat nasal congestion.
ii. Critical thinking
iii. Social responsibility
24.
a. Account for the following:
i. Direct nitration of aniline yields significant amount of meta
derivative.
ii. Primary aromatic amines cannot be prepared by Gabriel phthalimide
synthesis.
b. Carry out the following conversions:
i. Ethanoic acid into methanamine.
ii. Aniline to p-Bromoaniline.
c. Arrange the following in increasing order of basic strength: Aniline, p-
nitroaniline and p-toludine.
Ans.
a. i. In strongly acidic medium, aniline is protonated to form the anilium ion
which is meta directing.
ii. Aryl halides do not undergo nucleophilc substitution with the anion
formed by phthalimide.
b. i.
ii.
iii. p-Nitroaniline < Aniline < p-Toludine.
OR
a. Identify A-D
b. Distinguish between the following pair of compounds:
i. Aniline and Benzylamine.
ii. Methylamine and Dimethylamine.
c. Complete the following:
Ans. a.
b.
i.
Experiment Aniline Benzylamine
Azo dye test: Dissolve
the amine in HCl, cool
it and then add cold
aqueous solution of
NaNO2and then
solution of -naphthol.
A brilliant orange red
dye is observed.
No dye is formed
ii.
Experiment Methylamine Dimethylamine
Carbylamine test: To
the organic compound
add chloroform and
ethanolic potassium
hydroxide and heat.
A foul smelling
substance
(isocyanide).
No reaction.
c. A = CH3CH2CH2NH2
B = CH3CH2CH2OH
25.
i. Graphically explain the effect of temperature on the rate constant of
reaction? How can this temperature effect on rate constant be
represented quantitatively?
ii. The decomposition of a hydrocarbon follows the equation
Calculate Ea
Ans.
(i) For a chemical reaction with rise in temperature by 100, the rate
constant is nearly doubled.
Increasing the temperature of the substance increases the fraction of
molecules which collide with energies greater than Ea (activation
energy)
As the temperature increases (T2), the fraction of molecules having
energy equal to or greater than activation energy gets doubled leading
to doubling the rate of reaction.
k = A e-Ea/RT(Arrhenius equation)
ii. k = A e-Ea/RT
26.
a. Account for the following observations:
i. SF4 is easily hydrolysed whereas SF6is not easily hydrolysed.
ii. Chlorine water is a powerful bleaching agent.
iii. Bi(V) is a stronger oxidising agent than Sb(V)
b. What happens when
i. White phosphorus is heated with concentrated NaOH solution in an inert
atmosphere of CO2.
ii. (XeF6 undergoes partial hydrolysis.
(Give the chemical equations involved).
Ans. a.
i. S atom in SF4is not sterically protected as it is surrounded by
only four F atoms, so attack of water molecules can take place
easily. In contrast, S atom in SF6 is protected by six F aoms.
Thus attack by water molecules cannot take place easily.
ii. Chlorine water produces nascent oxygen (causes oxidation)
which is responsible for bleaching action.
CI2+ H2O 2HCL+[O]
iii. Due to inert pair effect Bi (V) can accept a pair of electrons to
form more stable Bi (III). (+3 oxidation state of Bi is more
stable than its +5 oxidation state).
b.
i. Phosphorus undergoes disproportionation reaction to form phosphine gas.
ii. On partial hydrolysis, XeF6 gives oxyfluoride XeOF4 and HF.
OR
What inspired N.Bartlett for carrying out reaction between Xe and PtF6?
Arrange the following in the order of property indicated against each set:
i. F2, I2, Br2, Cl2 (increasing bond dissociation enthalpy)
ii. NH3, AsH3, SbH3, Bi H3, P H3 (decreasing base strength
c. Complete the following equations
i. CI2+NaOH (cold and dilute)
ii. Fe3+ + SO2 + H2O
Ans.
a. N.Bartlett first prepared a red compound O2+PtF6
-. He then realised
that the first ionisation enthalpy of molecular oxygen was almost
identical with Xenon. So he carried out reaction between Xe and PtF6.
b.
i. I2< F2< Br2< Cl2
ii. NH3> PH3> AsH3> SbH3> BiH3
c.
a. 2NaOH+Cl2 NaCl+NaOCI+H2O
b. 2Fe3++SO2+2H2O 2Fe2+ + SO2-4 + 4H+