GURU HARKRISHAN PUBLIC SCHOOL
KAROL BAGH
CLASS 12
Pre mock examination (2014-15)
SUBJECT –CHEMISTRY
Time allowed: 3 hours
Maximum Marks: 70
General Instructions:
a) All the questions are compulsory.
b) There are 26 questions in total.
c) Questions 1 to 5 are very short answer type questions and carry one mark each.
d) Questions 6 to 10 carry two marks each.
e) Questions 11 to 22 carry three marks each.
f) Question 23 is value based question carrying four marks.
g) Questions 24 to 26 carry five marks each.
h) There is no overall choice. However, an internal choice has been provided in one question
of two marks, one question of three marks and all three questions in five marks each. You
have to attempt only one of the choices in such questions.
i) Use of calculators is not permitted. However, you may use log tables if necessary.
Q1
Schottky defect is formed when calcium chloride is added to silver chloride crystal.
Comment.
1
Q2
Q3
Q4
Explain the difference between a weak field ligand and a strong field ligand.
Give the application of colloids in electrical precipitation of smoke.
Explain how much portion of an atom located at a) the corner and b) body centre of a cubic
unit cell is part of its neighbouring unit cell?
Why does NH3 form hydrogen bond but PH3 does not?
Explain pseudo first order reaction with an appropriate example.
Give a chemical test to distinguish[ by giving equations]:
I .Ethylamine and Diethylamine
ii. Propanamine and Aniline
What is meant by positive and negative deviations from Raoult's law and how is the sign
of ∆solH related to positive and negative deviations from Raoult's law?
OR
Under what conditions Vant Hoffs factor ‘i’ is equal to unity and less than one and greater
than one?
1
1
1
Q9
Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10−8 cm and density is 10.5
g cm−3, calculate the atomic mass of silver.
2
Q10
Q11
Write the mechanism of hydration of Ethene to yield Ethanol.
a. Write a note on the following with an example each:
2
3
Q5
Q6
Q7
Q8
1
1
2
2
2
i. Williamson synthesis of ether
ii. Kolbe’s reaction
b. Write the IUPAC name of CHO-CH2-CH(CH3)2
Q12
Write structures of different isomers corresponding to the molecular formula:
i.C3H9N
ii. 1−Bromo−3, 3−dimethyl−1−phenylbutane
i.Using IUPAC norms write the formulas for the Tetrahydroxozincate(II).
ii. Discuss the geometry and magnetic property of [Fe(CN)6]4− on the basis of valence
bond theory.
3
Q14
Write a short note on Tyndall effect and its cause.
3
Q15
Outline the principles of refining of metals by the following methods:
(i) Zone refining
(ii) Electrolytic refining
(iii) Vapour phase refining
I) What is tincture of iodine? What is its use?
ii) Name two food preservatives?
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon
tetrachloride.
i. Where does the water present in the egg go after boiling the egg?
Ii.Why cannot vitamin C be stored in our body?
iii.What products would be formed when a nucleotide from DNA containing thymine
is hydrolysed .
OR
i.What do you understand by the term glycosidic linkage?
ii. Enumerate two reactions of D-glucose which cannot be explained by its open chain
structure.
The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing
wood had only 80% of the 14C found in a living tree. Estimate the age of the
sample.
i.
The HNH angle value is higher than HPH, HAsH and HSbH angles. Why?
ii.
Illustrate how copper metal can give different products on reaction with HNO3.
How are xenon fluorides XeF2, XeF4 and XeF6 obtained?
Give the reason for the following:
a. Ethyl iodide undergoes SN 2 reaction faster than ethyl bromide
b. (±) 2-Butanol is optically inactive
c. C – X bond length in halobenzene is smaller than C – X bond length in CH3 – X
3
Sara went to market to buy fruits and vegetables. The vendor put the fruits and vegetables
in the polythene bag but Sara ask the vendor to put the things in the jute bag which he
carried with him.
Now answer the following question :
a. Why did Sara refuse to use polythene bags?
b. As a student of chemistry why would you advocate the use of jute bags instead of
polythene bags? Which values are promoted through the use of jute bag?
4
Q13
Q16
Q17
Q18
Q19
Q20
Q21
Q22
Q23
2
3
3
3
3
3
3
3
3
c. Suggest two activities to promote these activities .
Q24
Q25
Q26
a)Arrange the following compounds in increasing order of their reactivity in nucleophilic
addition reactions.
(i)Ethanal, Propanal, Propanone, Butanone.
(ii)Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.
b. How will you bring about the following conversions in not more than two steps?
(i) Propanone to Propene
(ii) Benzoic acid to Benzaldehyde
(iii) Ethanol to 3-Hydroxybutanal
OR
An organic compound with the molecular formula C9H10O forms2,4-DNP derivative,
reduces Tollens reagent and undergoes cannizzaro reaction. On Vigorous oxidation, it
gives 1,2 benzene dicarboxylic acid. Identify the compound and give equations.
a)A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5
amperes for 20 minutes. What mass of Ni is deposited at the cathode?
(Given atomic mass of Ni=58.7 u)
b) What are the observations made in a galvanic cell after the circuit is completed?
OR
a)Define conductivity and molar conductivity for the solution of an electrolyte. Discuss
their variation with concentration.
b) The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω.
What is the cell constant if conductivity of 0.001M KCl solution at 298 K is
0.146 × 10−3 S cm−1.
Describe the preparation of potassium dichromate from iron chromite ore.
What is the effect of increasing pH on a solution of potassium dichromate?
OR
a) Differentiate actinoids and lanthanoids [3 points each].
b) How does the acidified permanganate solution react with (i) iron(II) ions (ii) SO2
5
5
5
3
GURU HARKRISHAN PUBLIC SCHOOL (KAROL BAGH)
CLASS 12
Pre mock examination (2014-15)
SUBJECT –CHEMISTRY
Marking Scheme
Time allowed: 3 hours
Maximum Marks: 70
General Instructions:
a) All the questions are compulsory.
b) There are 26 questions in total.
c) Questions 1 to 5 are very short answer type questions and carry one mark each.
d) Questions 6 to 10 carry two marks each.
e) Questions 11 to 22 carry three marks each.
f) Question 23 is value based question carrying four marks.
g) Questions 24 to 26 carry five marks each.
h) There is no overall choice. However, an internal choice has been provided in one question
of two marks, one question of three marks and all three questions in five marks each. You
have to attempt only one of the choices in such questions.
i) Use of calculators is not permitted. However, you may use log tables if necessary.
A1
When calcium chloride is added to silver chloride crystal, an impurity defect is formed. The
addition of one calcium ion will replace two silver ions to maintain electrical conductivity.
One of the positions of silver ion will be occupied by one calcium ion and other will be left
as a hole similar to Schottky defect.
Some ligands are able to produce strong fields in which case, the splitting will be large
whereas others produce weak fields and consequently result in small splitting of d orbitals
Smoke being a colloidal solution ,before it comes out from the chimney, comes in contact
with electrical plates lose their charge and get precipitated in electrical precipitation of
smoke.
1
A4
A5
a)1/8
1
1
A6
The order of a reaction is sometimes altered by conditions. In a chemical reaction between
two substances when one reactant is present in large excess and the reaction behaves as
first order reaction. Such reactions are called pseudo first order reactions..
EX. During the hydrolysis of 0.01 mol of ethyl acetate with 10 mol of water,
A2
A3
b) no part
Hydrogen bonding occurs when hydrogen is bonded to a small, highly electronegative
element. N being most electronegative in NH3 form hydrogen bond but PH3 does does not?
1
1
2
𝐻+
A7
CH3COOC2H5 + H2O
→ CH3COOH + C2H5OH
The concentration of water does not get altered much during the course of the reaction.
I .Ethylamine and Diethylamine
Ethylamine and Dimethylamine can be distinguished by the carbylamine test.
4
2
Carbylamine test: Aliphatic and aromatic primary amines on heating with chloroform and
ethanolic potassium hydroxide form foul-smelling isocyanides or carbylamines.
Methylamine (being an aliphatic primary amine) gives a positive carbylamine test, but
dimethylamine does not.
Note: write C2H5 NH2 INSTEAD OF CH3NH2
ii.Propanamine and Aniline
Primary aliphatic amines react with nitrous acid to form aliphatic diazonium
salts which being unstable, liberate nitrogen gas quantitatively and alcohols.
𝑁𝑎𝑁𝑂2,𝐻𝐶𝑙
𝐻2 𝑂
RNH2+HNO2
→ RN2+Cl−
→ ROH+N2+HCl
(b) Aromatic amines react with nitrous acid at low temperatures (273-278 K) to
form stable diazonium sals
𝑁𝑎𝑁𝑂2,𝐻𝐶𝑙
ArNH2+HNO2
→ RN2+Cl− +NaCl +H2O
273−278 𝐾
A8
The vapour pressure of a non-ideal solution is either higher or lower than that predicted by 2
Raoult’s law . If it is higher, the solution exhibits positive deviation and if it is lower, it
exhibits negative deviation from Raoult’s law.
the sign of ∆solH for positive deviations from Raoult's law= +ve
and the sign of ∆solH for negative deviations from Raoult's law= -ve
OR
If non-volatile solute is neither associated nor dissociated ‘i’ is equal to unity.
In case of association, value of i is less than unity
while for dissociation it is greater than unity.
A9
ANS.It is given that the edge length, a = 4.077 × 10−8 cm Density, d = 10.5 g cm−3
As the lattice is fcc type, the number of atoms per unit cell, z = 4 We also know that, N A =
6.022 × 1023 mol−1
Using the relation: 𝑀 =
2
d NA a3
Z
= 107.13 g mol−1
Therefore, atomic mass of silver = 107.13
u
A10
2
The mechanism of hydration of ethene to form ethanol involves three steps.
Step 1:
Protonation of ethene to form carbocation by electrophilic attack of H3O+:
5
Step 2:
Nucleophilic attack of water on carbocation:
Step 3:
Deprotonation to form ethanol:
A11
a.
i. This reaction is used to prepare both symmetrical and unsymmetrical ethers by treating
alkyl halide with either sodium alkoxide or sodium phenoxide.
CH3Br
+ C2H5ONa
→ CH3OC2H5 + NaBr
Methylbromide
Sod.ethoxide
3
Methoxyethane
ii.
(i) Kolbe’s reaction:
When phenol is treated with sodium hydroxide, sodium phenoxide is produced. This sodium
phenoxide when treated with carbon dioxide, followed by acidification, undergoes electrophilic
substitution to give ortho-hydroxybenzoic acid as the main product. This reaction is known as
Kolbe’s reaction.
b.
A12
A13
3-Methylbutanal
i.C3H9N( Pls .do yourself)
CH3CH2CH2NH2,
ii. 1−Bromo−3, 3−dimethyl−1−phenylbutane…………………….
i.
Tetrahydroxozincate(II).
3
3
6
[Zn(OH)4]2ii. the geometry is octahedral( d2sp3)and magnetic property of [Fe(CN)6]4− is dimagnetic on
the basis of valence bond theory.
A14
A15
Write a short note on Tyndall effect and its cause.
If a homogeneous solution placed in dark is observed in the direction of light, it appears
clear and, if it is observed from a direction at right angles to the direction of light beam, it
appears perfectly dark. Colloidal solutions viewed in the same way may also appear
reasonably clear or translucent by the transmitted light but they show a mild to strong
opalescence, when viewed at right angles to the passage of light, i.e., the path of the beam
is illuminated by a bluish light. This effect was first observed by Faraday and later studied in
detail by Tyndall and is termed as Tyndalleffect. The bright cone of the light is called
Tyndallcone. The Tyndall effect is due to the fact that colloidal particles scatter light in all
directions in space. This scattering of light illuminates the path of beam in the colloidal
dispersion.
Outline the principles of refining of metals by the following methods:
(i)
Zone refining
This method is based on the principle that impurities are more soluble in the molten
state of metal (the melt) than in the solid state.
(ii)
Electrolytic refining
3
3
the electrolysis of aqueous solutions or salt melts, yields metals of high purity at cathode.
A16
A17
(iii)
Vapour phase refining
In this method, the metal is converted into its volatile compound. It is then decomposed
to give pure metal.
i. tincture of iodine.
2-3 per cent solution of Iodine in alcohol- water mixture is known as tincture of iodine.
use :Iodine is a powerful antiseptic .It is applied on wounds
ii) Name two food preservatives:
table salt, sugar, vegetable oils and sodium benzoate, C6H5COONa.
Let the total mass of the solution be 100 g and the mass of benzene be 30 g. ∴
Mass of carbon tetrachloride = (100 − 30)g = 70 )
Molar mass of benzene (C6H6) = (6 × 12 + 6 × 1) g mol−1 = 78 g mol−1
Number of moles of benzene=30/78 = 0.3846 mol
Molar mass of carbon tetrachloride (CCl4) = 1 × 12 + 4 × 355 = 154 g mol−1 ∴
Number of moles of CCl4 = 70/154 =0.4545 mol
Thus, the mole fraction of C6H6 is given as:
Number of moles of benzene
mole fraction of C6H6 = Number of moles of benzene+ Number of moles of CCl4
=
=
30/78
= 0.458
30 70
78 + 154
0.458
7
3
3
A18
i.When an egg is boiled, the proteins present inside the egg get denatured and
coagulate. After boiling the egg, the water present in it is absorbed by the coagulated
protein through hydrogen bonding .
3
Ii.Because it is a water soluble vitamin which readily excreted in urine and cannot be stored
in our body.
iii.What products would be formed when a nucleotide from DNA containing thymine
is hydrolysed .
Complete hydrolysis of nucleotide from DNA yields a β-D-2-deoxyribose pentose sugar,
phosphoric acid and thymine base.
OR
i.What do you understand by the term glycosidic linkage?
The two monosaccharides are joined together by an oxide linkage formed by the loss of a
water molecule. Such a linkage between two monosaccharide units through oxygen atom is
called glycosidic linkage.
ii
1. Despite having the aldehyde group, glucose does not give Schiff’s test and it does not
form the hydrogensulphite addition product with NaHSO3.
2. The pentaacetate of glucose does not react with hydroxylamine indicating the absence of
free —CHO group
A19
The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing
wood had only 80% of the 14C found in a living tree. Estimate the age of the
sample.
Here,
It is known that,
𝒕𝟏 𝟎.𝟔𝟗𝟑
𝟐
K
=
𝑡=
=
3
𝑲
𝑂.693
5730
2.303
𝑘
log
[𝐴0 ]
[𝐴]
, 𝑡=
2.303
𝑘
log
100
80
,𝑡 =
2.303
𝑂.693
5730
log
100
80
= 1845 years
A20
I.
The above trend in the H−M−H bond angle can be explained on the basis of the
electronegativity of the central atom. Since nitrogen is highly electronegative, there is high
electron density around nitrogen. This causes greater repulsion between the electron pairs
around nitrogen, resulting in maximum bond angle. We know that electronegativity
decreases on moving down a group. Consequently, the repulsive interactions between the
electron pairs decrease, thereby decreasing the H−M−H bond angle.
II. 3Cu + 8 HNO3(dilute) → 3Cu(NO3)2 + 2NO + 4H2O
Cu + 4HNO3(conc.) → Cu(NO3)2 + 2NO2 + 2H2O
A21
3
3
Xe (g) +
F2 (g)
673𝐾,1 𝑏𝑎𝑟
→
XeF2(s)
(xenon in excess)
8
Xe (g) + 2F2 (g)
(1:5 ratio)
Xe (g) + 3F2 (g)
(1:20 ratio)
873 K,7 bar
→
573 K,60 70bar
XeF4(s)
→ XeF6(s)
A22
a. Iodide is a better leaving group because of its larger size, than bromide, therefore, ethyl
iodide undergoes SN2 reaction faster than ethyl bromide
b. ( ) ± 2-butanol is a racemic mixture. It is a mixture which contains two enantiomers in
equal proportion and thus, has zero optical rotation. Therefore it is optically inactive.
c. Due to the delocalization of lone pairs of electrons of the X atom over the benzene ring C
– X bond in halobenzene acquires some duble bond character while in CH3 – X , C – X bond
is a pure single bond. Therefore C – X bond in halobenzene is shorter than in CH3 – X
3
A23
a. Polythene is non-biodegradable hence causes environmental pollution
b. Jute bag are biodegradable revenue, hence do not cause any environmental pollution.
Promoted Values Reducing environmental pollution, concern for environmental protection
c. Use paper bags instead of polythene bags. Organizing mass campaigns for spreading
awareness.
4
A24
a) (i)
Butanone ˂ Propanone ˂ Propanal ˂ Ethanal
(ii) Acetophenone ˂ p-Tolualdehyde ˂ Benzaldehyde ˂ p-Nitrobenzaldehyde
b. How will you bring about the following conversions in not more than two steps?
(i) Propanone to Propene
𝑁𝑎𝐵𝐻4
𝐶𝑜𝑛.𝐻2𝑆𝑂4
CH3CH2CHO
─ ─→ CH3CH2CH2OH
→ CH3CH=CH2
Propene
(ii) Benzoic acid to Benzaldehyde
𝑆𝑂𝐶𝑙2
𝐻2
C6H5COOH
→ C6H5COCl
→ C6H5CHO
𝑃𝑑−𝐵𝑎𝑆𝑂4
Benzaldehyde
5
iii.Ethanol to 3-Hydroxybutanal
CH3CH20H
𝑃𝐶𝐶
→ CH3CHO
CH3CHO,dil.NaOH
→3-Hydroxybutanal
OR
It is given that the compound (with molecular formula C9H10O) forms 2, 4-DNP derivative and
reduces Tollen’s reagent. Therefore, the given compound must be an aldehyde.
Again, the compound undergoes cannizzaro reaction and on oxidation gives 1, 2benzenedicarboxylic acid. Therefore, the −CHO group is directly attached to a benzene ring and
this benzaldehyde is ortho-substituted. Hence, the compound is 2-ethylbenzaldehyde.
9
The given reactions can be explained by the following equations.
A25
a)A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5
amperes for 20 minutes. What mass of Ni is deposited at the cathode?
(Given atomic mass of Ni=58.7 u)
Given,
Current = 5A
Time = 20 × 60 = 1200 s
Charge = current × time
= 5 × 1200
= 6000 C
According to the reaction,
Ni2+(aq) + 2e-
→ Ni(s)
58.7 u
Nickel deposited by 2 × 96487 C = 58.71 g
58.71∗6000
Therefore , nickel deposited by 6000 C = 2∗96487
= 1.825 g
b) What are the observations made in a galvanic cell after the circuit is completed?
 Electron transfer from one species to another takes place indirectly through
electrodes.
 Energy is liberated in the form electrical energy.
10
5

Redox reaction takes place separately at the Anode and cathode surface.
OR
a)Define conductivity and molar conductivity for the solution of an electrolyte. Discuss
their variation with concentration.
The conductivity of a solution at any given concentration is the conductance (G) of one
unit volume of solution kept between two platinum electrodes with the unit area of
cross-section and at a distance of unit length.
Conductivity always decreases with a decrease in concentration, both for weak and
strong electrolytes. This is because the number of ions per unit volume that carry the
current in a solution decreases with a decrease in concentration.
Molar conductivity:
Molar conductivity of a solution at a given concentration is the conductance of volume V
of a solution containing 1 mole of the electrolyte kept between two electrodes with the
area of cross-section A and distance of unit length.
Molar conductivity increases with a decrease in concentration. This is because the total
volume V of the solution containing one mole of the electrolyte increases on dilution.
b)
Given, Conductivity, κ = 0.146 × 10−3 S cm−1
Resistance, R = 1500 Ω
Cell constant = κ × R = 0.146 × 10−3 × 1500 = 0.219 cm−1
A26
Describe the preparation of potassium dichromate from iron chromite ore.
What is the effect of increasing pH on a solution of potassium dichromate?
Potassium dichromate is prepared from chromite ore
5
in the following steps.
Step (1): Preparation of sodium chromate
Step (2): Conversion of sodium chromate into sodium dichromate
Step(3): Conversion of sodium dichromate to potassium dichromate
The chromates and dichromates are interconvertible in aqueous solution depending upon
pH of the solution. The oxidation state of chromium in chromate and dichromate is the
same.
2 CrO42– + 2H+ → Cr2O72– + H2O
Cr2O72– + 2 OH- → 2 CrO42– + H2O
the dichromate ion exists in equilibrium with chromate ion at pH 4.
However, by changing the pH, they can be interconverted.
OR
a) Differentiate actinoids and lanthanoids [3 points each].
) Electronic configuration
11
The general electronic configuration for lanthanoids is [Xe]54 4f0-14 5d0-1 6s2 and that for actinoids
is [Rn]86 5f1-14 6d0-1 7s2. Unlike 4f orbitals, 5f orbitals are not deeply buried and participate in
bonding to a greater extent.
(ii) Oxidation states
The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter
oxidation states of + 2 and + 4. This is because of extra stability of fully-filled and half-filled
orbitals.
Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d, and 7s levels are
of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such
as lanthanoids have more compounds in +3 state than in +4 state.
(iii) Atomic and lonic sizes
Similar to lanthanoids, actinoids also exhibit actinoid contraction (overall decrease in atomic and
ionic radii). The contraction is greater due to the poor shielding effect of 5f orbitals.
iv.
Chemical reactivity
In the lanthanide series, the earlier members of the series are more reactive. They have
reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start
behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when
they are finely divided. When they are added to boiling water, they give a mixture of oxide and
hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have
no action on these actinoids. In case of acids, they are slightly affected by nitric acid (because of
the formation of a protective oxide layer).
(i) Acidified KMnO4 solution oxidizes Fe (II) ions to Fe (III) ions i.e., ferrous ions to ferric ions.
(ii) Acidified potassium permanganate oxidizes SO2 to sulphuric acid.
12