FIRST PRE BOARD EXAM CLASS XII
BLUE PRINT
SL
NO
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
UNITS
VSA 1
SA II 3M
M( NO. Q)
-3(1)
3(1)
-3(1)
3(1)
LA 5M
M( NO. Q)
---5(1)
---
TOTAL
----1(1)
--
SA 1 2M
M( NO. Q)
4(2)
2(1)
2(1)
----
SOLID STATE
SOLUTIONS
ELECTROCHEM
CHEM KINETICS
SURFACE CHEM
ISOLATION OF
ELEMENTS
P BLOCK
D AND F BLOCK
CO-ORDINATION
COMPOUNDS
HALO ALKANES AND
HALO ARENES
ALCOHOLS,PHENOLS
AND ETHERS
ALDEHEYDES,KETONES
AND CARB ACIDS
ORGANIC COMPS CONT
NITROGEN
BIOMOLECULES
POLYMERS
CHEMISTRY IN
EVERYDAY LIFE
TOTAL
--1(1)
-2(1)
2(1)
3(1)
3(1)
--
5(1)
---
8(2)
5(2)
3(2)
--
4(2)
--
--
4(2)
1(1)
--
3(1)
--
4(2)
1(1)
--
--
5(1)
6(2)
1(1)
--
3(1)
--
4(2)
1(1)
1(1)
1(1)
-2(1)
2(1)
3(1)
---
----
4(2)
3(2)
3(2)
8(8)
20(10)
27(9)
15(3)
70(30)
4(2)
5(2)
5(2)
5(1)
4(2)
3(1)
KENDRIYA VIDYALAYA GANESHKHIND PUNE
Time: Three Hours
CLASS – XII
CHEMISTRY
I PRE BOARD EXAM
Max. Marks: 70
General Instructions
1. All questions are compulsory.
2. Question nos. 1 to 8 are very short answer questions and carry 1 mark each.
3. Question nos. 9 to 18 are short answer questions and carry 2 marks each.
4. Question nos. 19 to 27 are also short answer questions and carry 3 marks
each
5. Question nos. 28 to 30 are long answer questions and carry 5 marks each
6. Use log tables if necessary, use of calculators is not allowed.
Q1.Define peptisation.What is colloidon?
The process of converting the precipitate in to colloidal particles on adding a
suitable electrolyte.
1/2M
Colloidon is a sol of cellulose nitrate in ethyl alcohol 1/2M
Q2. How do the co-ordination compounds show colour?
Co-ordination compounds impart colour due to the presence of unpaired
electrons in the central metal atom or ion. 1/2M
The unpaired electrons show d-d transition by jumping between t2g and eg
orbitals.
1/2M
Q3. Write the structures of the each of the following alcohols.
1)3,3-dimethyl cyclopentanol
2)3-chloro-2-methyl-1-butanol
1)
1/2M
2) CH3-CH(Cl)-CH(CH3)-CH2OH
1/2M
Q4.What does PCC stand for?Give its one use.
PCC stands for pyridunuum chlorochromate.
1/2M
It is used to oxidise 10 alcohol to aldehydes and 20 alcohols to ketones.
1/2M
Q5. Why are alkyl cyanides more volatile than the corresponding acids?
Alkyl cyanides are more volatile because they can not associate via intermolecular hydrogen
bonding.
1M
Q6. What do you understand by reducing sugars?Give one example.
Carbohydrates which contain free aldehyde group and reduce Fehling
solution or Tollens reagent are called reducing sugars.
1/2M
Eg Glucose
1/2M
Q7. What are co-polymers? Give one example.
Co-Polymers are obtained by the polymerisation of two or more than two
different monomers.
1/2M
Eg Nylon 66.
1/2M
Q8. Why aspartame has a limited use?
Aspartame has a limited use because it is unstable at cooking temperture.
1M
Q9. Aluminium crystallises in a cubic close packed structure.Its metallic radius is
125pm.What is the length of the side of the unit cell?
CCP means fcc.
Z= 4 r=125pm
1/2M
For FCC
4r = 2 a
1M
a= 4x125/2
a= 353.6pm
1/2M
Q10.Explain the following terms
a)Schottky defect
b)Frenkel defect
Schottky defect:- It is a stoichiometric defect and is formed when one cation and one anion
is missing from the lattice.The lattice remains electrically neutral.
1M
Frenkel defect:- This also is a stoichiometric defect in which one ion ( preferably cation)
occupies an interstitial site instead of its correct crystal site.Electrical neutrality and
stoichiometry is maintained.
1M
Q11. When two solutions X and Y are mixed the solution becomes hot.When Y and Z are
mixed the solution becomes cold.Which of these solutions will exhibit negative deviation
from ideal behaviour?
Solution of X and Y exhibit negative deviation.Inthis X—Y interaction is much
stronger than the X—X and Y—Y interaction hence heat is evolved.
2M
Q12.What happens to conductivity of solution with dilution?Why?
The conductivity of solution decreases on dilution
1M
because the number of ions per unit volume that carry the current in a solution
decreases.
1M
Q13. Give oxidising property of KmnO4 in neutral medium with example.
KmnO4 is a moderate oxidising agent in neutral medium because it gives three oxygen
atom in water.
1M
2KmnO4 + H2O  2KOH + 2MnO2 + 3 [O]
In water KmnO4 oxidises manganese sulphate to manganese dioxide.
3MnSO4 3H2O + 3O  3MnO2 + 3H2SO4
1M
Q14.Write IUPAC names of the following co-ordination compounds.
a) [PtCl(NH2CH3)(NH3)2]Cl
b) [Co(en)3]3+
a)Diamminechloro (methylamine) platinum (II) chloride
1M
b)Tris (1,2-ethanediamine) cobalt (III) ion
1M
Q15. How would you convert 2-methyl propene to
a)Tertiary butyl bromide
b)iso-butyl bromide
a) CH3—C (CH3) = CH2 + HBr  CH3 – C (CH3) (Br) – CH3
1M
b) CH3 – C (CH3) = CH2 + HBr  (Benzoyl peroxide)  CH3—CH ( CH3) – CH2Br
1M
Q16.Convert
a)Chlorobenzene to bromobenzene
b)Chlorobenzene to toluene
C6H5—Cl + NH3  C6H5—NH2  NaNO2/HCl  C6H5N2Cl  CuBr/HBr 
C6H5—Br.
1M
C6H5—Cl  NaOH/High Temp/High Press  C6H5—OH  Zn  C6H6
C6H6 + CH3 Cl/FeCl3  C6H5—CH3
1M
Q17.Explain the structure of natural rubber.Name its monomeric unit.
The natural rubber is a linear 1,4-polymer of isopropene.
1M
Monomer is isoprene,
1M
Q18. What is an antibiotic? Name the first antibiotic discovered.Is this antibiotic narrow
spectrum or broad spectrum antibiotic?
Antibiotics are chemical substances which are produced by microorganisms
inorder to inhibit the growth or even to destroy other microorganisms.
1M
The first antibiotic discovered was penicillin.
1/2M
It is a narrow spectrum antibiotic.
1/2M
Q19.Calculate the mass of ascorbic acid (Vitamine C = C6H8O6) to be dissolved in 75 g
acetic acid to lower its melting point by1.50C .Kf=3.9K Kg mol—1.
Tf=1.5K
W1 = 75g
Molar mass of solute Mb = 72+8+96 =176 g mol—1
Kf = 3.9 K kg mol—1
Tf = kf x Wb x 1000/Wa x Mb
1+1/2M
Wb = Mb x Wa x Tf/1000 x kf
176 x 75 x 1.5/1000 x 3.9
= 5.077
1+ 1/2
Q20.How much electricity in coulombs is required for the oxidation of
a)
1 mol of H2O to O2
b)
1 mol of FeO to Fe2O3
H2O  2H+ + ½ O2 + 2e—
1M
Quantity of electricity required is 2F = 2 x 96500 C
1/2M
Fe 2+  Fe3+ e—1 ( FeO  ½ Fe2O3)
Quantity of electrons required is 1F = 96500 C
1M
1/2M
Q 21. Why
a) True solutions do not show Tyndall effect?
b) Medicines are more effective in colloidal state.
c) Alum is added to purify muddy water
a) Because the particle size is so small that no scattering of light is possible.
1M
b) A colloidal state has a larger surface area. Thus medicines in colloidal state
are effectively adsorbed and assimilated and give better results. 1M
c) Since the clay particles present are negatively charged alum provides Al3+
ions which neutralise the negatively charge of the particles which then
coagulate and get settled at the bottom.
1M
Q22.The value of fG0 for the formation of Cr2O3 is –540kj/mol and that of Al2O3 is –827
kj/mol.Is the reduction of Cr2O3 possible with Al?
2Cr + 3/2 O2  Cr2O3 fG0 = -540kj/mol
Cr2O3  2Cr + 3/2 O2 fG0 = +540kj/mol………1
2Al + 3/2O2  Al2O3 fG0 = -827kj/mol……………..2
1+ 2 represents reduction of Cr2O3 with Al.
1M
1M
Since fG0 of this reaction is negative the reduction is possible.
1M
Q23.Starting from elemental sulphur how would you prepare the following
a)H2SO4
b)SCl2
c)SF6
a) 1/8 S8 + O2 SO2
2SO2 + O2 V2O5 2SO3
SO3 + H2SO4  H2S2O7H2O  H2SO4
1M
b)1/8S8 + Cl2  SCl2
1M
c)1/8S8 + 3F2  SF6
1M
Q24.Describe the preparation of potassium dichromate from ferrochrome.
I Step
Fusion of chomite ore with sodium carbonate and O2
4FeCr2O4 + 8Na2CO3 + 7O2  8Na2CrO4 + 2Fe2O3 + 8CO2
II Step
Acidification of chromate ore
2Na2CrO4 + 8H2SO4  Na2Cr2O7 + Na2SO4 + H2O
III Step
Conversion of sodium dichromate to potassium dichromate
Na2Cr2O7 + 2KCl  K2Cr2O7 + 2NaCl
A
Q25.How are the following conversions carried out?
1M
1M
1M
a)Toluene to benzyl alcohol
b)Ethyl magnesium chloride to propan-1-ol
c)Methyl magnesium bromide to 2-methyl propane-2-ol
a)C6H5—CH3  Cl2/UV LIGHT  C6H5—CH2Cl  Aq KOH  C6H5—CH2 OH
1M
b)CH3—CH2 MgCl + H—CHO  H—CH(CH2—CH3)—OMgCl  H2O 
CH3—CH2-CH2—OH
1M
c)CH3 Mg Br + CH3—CO—CH3  H2O  CH3—CH(CH3)(OH) – CH3
1M
Q26.Explain why does nitrobenzene on nitration with nitric acid and sulphuric acid forms
only meta dinitro benzene?
NO2 group is electron withdrawing .The electron displacement is away from the
benzene.As a result O and P carbon acquires positive charge.So meta position is more
electron rich and is favourable for an electrophile (NO2+).
1M
Show resonance hybrids.
2M
Q27.Give any three points of differences between DNA and RNA.
3M
DNA
RNA
1.It is mainly found in chromosomes
It is mainly found in cytoplasm but also
found in chromosomes
2.Has de oxy ribose sugar
2.Has ribose sugar
3.Four nitrogenous bases in DNA are
3.The four nitrogenous bases are A,G,U and
A,G,T,C
C.
4.It has double helix structure.
4.It has single helix structure.
Q28.From the following data for the reaction between A + B
A+BM+N
Expt no
[A] mol/l
[B] mol/l
Initial Rate mol/l/sec
At 300K
1
2.5 x 10--4
3.0 x 10--5
5.0 x 10—4
2
5.0 x 10--4
6.0 x 10--5
4.0 x 10—3
3
1.0 x 10--3
6.0 x 10--5
1.6 x 10—2
1.Calculate order of reaction with respect to A and B
2.Rate constant at 300K
3.Energy of activation if the rate constant at 320 K is 1.07 x 109
Assume the rate law to be r= k[A]x [B]y……………………1
5.4 x 10—4 = )x (3.0 x 10—5 )y…………………2
4.0 x 10 –3 = (5.0 x 10—4)x (6.0x 10—5)y……………………3
1.6 x 10—2 = (1.0 x 10—3)x (6.0 x 10—5)y……………………4
4/3 …………………1.6x 10—2/4.0 x 10—3 = (1.0 x 10—3)x/(5.0 x 10—4)y
2x = 4
x= 2
1M
3/2 ,
4.0 x 10
–3/5.4
x 10—4 = 2x 2y
4 x 2y =8
y=1
1M
order with respect to A = 2 and with respect to B = 1.
Rate law is r = k [A]2 [B]
K300 = r/[A]2[B]
K300 = 5.4 x 10—4/5.4 x 10—4 x3.0 x 10—5
= 2.7 x 108
1M
1M
k320 = 1.07 x 109
Ea = 2.303 x 8.314 x 300x 320 /20 log (10.7/2.7)
Ea = 54962j/mol
Ea = 54.962
1M
OR
The activation energy for the reaction
2HI (g)  H2(g) + I2 (g)
is 209.5 kj/mol. At 581 K.Calculate the fraction of molecules of reactants having energy
equal to or greater than activation energy?
Y = e –Ea/RT
1M
Logy = --Ea / 2.303RT
--209500/2.303x8.314 x581
= --18.8323
2M
y = antilog(--18.8323)
y = 1.47 x 10—19
2M
Q29.Discuss oxides of nitrogen.Give structure and calculate the oxidation number of
nitrogen in each of them.Arrange the oxides in order of increasing acid strength.
Nitrogen forms a wide range of oxided .The oxidation number of nitrogen in these oxides
varies from +1 to +5.The oxides are
1)N2O dinitrogen oxide
It is formed by gently heating ammonium nitrate.
NH4NO3  N2O + 2H2O
N2O is the oside of nitrogen with the lower oxidation number +1.
The structure of N2 O is
1M
2)Nitrogen oxide NO.
NO is also known as nitric oxide.The oxidation number of nitrogen is +2.NO can be
prepared by in the laboratory by reducing anitrite with mild reducing agent such as I—1
2HNO2 + 2HI  2NO2 +I2 + H2O
The structure of NO is
1M
3)Dinitrogen trioxide N2O3.
This can be obtained by condensing NO and NO2 together Its structure is 1M
4)Nitrogen dioxide
NO2.It is obtained by thermal decomposition of lead nitrate.
2Pb(NO3)2  4000C 4NO2 + PbO + O2
The oxidation number of nitrogen in NO2 is +4.
Its structure is
5)Dinitrogen pentoxide N2O5
Oxidation number of nitrogen in +5.
It is obtained by the reaction of P2O5 on HNO3.
2HNO3 + P2O5  N2O5 + 2HPO3
Its structure is
1M
The order of increasing acidic strength is
N2O NO,N2O3,NO2,N2O5
1M
OR
Illustrate how nitrogen compounds provide good examples of multiple bonding and
resonance.
Show the resonance hybrids
5M
Q30.a)Explain the mechanism of a nucleophilic aattack on carbonyl group of an aldehyde
or a ketone.
b)An organic compound A molecular formula C8H16O2 was hydrolysed with dilute
sulphuric acid to give a carboxylic acid B and an alcohol C.Oxidation of C with chromic
acid also produced B.On dehydration C gives but-1-ene.Write equations of reactions
involved.
a) =C=O + Nu—  =C(Nu) – O
--
 E+  Nu—C—OE
2M
b)There are two possible structures of A
CH3-CH2-CH2-CO-O-CH2-CH2-CH2-CH3
CH3-CH(CH3)-CO-O-CH2-CH(CH3)-CH3
CH3-CH2-CH2-CO-O-CH2-CH2-CH2-CH3 + H2O/H+ CH3-CH2-CH2-COOH +
(B)
CH3-CH2-CH2-CH2-OH (
(C)
CH3-CH2-CH2-CH2-OH  Chromic acid  CH3-CH2-CH2-COOH
(C)
CH3CH(CH3)COOCH2CH(CH3)CH3 H2O/H+  CH3CH(CH3)CH2OH +
(C )
CH3CH(CH3)COOH
(B)
CH3—CH(CH3) CH2OH  Chromic acid CH3 CH(CH3) COOH
OR
a)Give chemical test to distinguish between the following pairs of compounds.
1) ethanal and propanal
2) Phenol and benzoic acid.
b)How will you bring about the following conversions
1.Benzoic acid to benzaldehyde
2.Ethanal to but-2-ene
3.Propanone to propene.
Give complete reaction in each case.
3M
a) Ethanal answers iodoform test where as propanal does not give iodoform test.
1M
CH3CH2OH  I2/NaOH  CHI3 + Na2CO3
b)C6H5-OH + NaHCO3  No effervescence
1M
C6H5-COOH + NaHCO3  Brisk effervescence due to liberation of CO2.
1. Benzoic acid to benzaldehyde
C6H5-COOH + SOCl2  C6H5-COCl  H2/Pd/BaSO4  C6H5—CHO
1M
2.CH3 CHO + CH3—CH2MgBr/H2O  CH3—CH(OH)—CH2—CH3 Alc KOH
CH3—CH=CH—CH3
1M
3. CH3—CO—CH3  [H]  CH3—CH(OH)—CH3Alc KOH  CH3-CH=CH2 1M
…………………………………………………………………………………………………….