MODEL TEST PAPER –1
CHEMISTRY- CLASS: XII SC.
TIME : 3 Hrs.
M.M-70
General Instructions:
i. All Questions are compulsory
ii. Marks for each question are indicated against it.
iii. Questions no. 1-8 are very short answer questions each of 1 mark.
Answer these in one word or about one sentence each.
iv. Questions no. 9-18 are short answer type each carrying 2 marks.
v. Questions no. 19-27 are short answer questions each carrying 3
marks.
vi. Questions no. 28-30 are long answer type each carrying 5 marks.
vii. Use log tables if necessary. Use of calculator is not permitted.
Q.1 Why does ZnO appears yellow on heating ?
Q.2 Define peptization.
Q.3 Give the IUPAC name of the compound
H
CH3 CH NH
Cl
CH3
Q.4 Convert: Toluene to Benzaldehyde.
O
Q.5 Identify the no. of chiral C-atoms in
Cl
Q.6 What is vant Hoff’s factor for AlCl3 ?
Q.7 Give an example of zero order reaction.
Q.8 The rate constant of a reaction is 5 x 10-2 moles-3/2 lit. 3/2 min-1 . what
is the order ?
Q.9 Draw the structure of Cr2 O7-2
Q.10 Write notes on
i. Hoffman Bromo amide reaction
ii. Sandmeyer’s reaction.
i.
ii.
OR
Coupling reaction.
Diazotization.
Q.11 Identify A and B in the following reaction
O
CH3
C
HCN
(A)
H2O/H+
(B)
H
Q.12 Define Zwitter ion with example.
Q.13 What are reducing and non reducing sugar? Give example.
Q.14 Aluminium has f.c.c unit cell with atomic radius 105pm. What is
the cell edge? How many unit cells are present in 10cc of Al.
Q.15 Assuming K2SO4 completely dissociated in aq solution, Calculate
the boiling point , if 1.74 gm of it is present in 500cc of solution.
(kb for H2O = 0.52 Km-1 )
Q.16 What do you mean by activated complex?
Q.17 For the reaction, at constant volume the following data are
obtained
SO2Cl2
SO2 (g) + Cl2 (g)
Expt.
1
2
Time/s-1
0
100
Total pressure
0.5
0.6
Calculate the rate of the reaction if total pressure is 0.65 atm
Q.18 (a) What is Hardy Schuldze rule?
(b) what happens when equimolar concentrations of Arsenic
Sulphide sol is mixed with Fe(OH) 3 sol.
Q.19 Outline the principle of refining of metals by
i. Electrolytic refining
ii. Zone refining
Q.20 (a).Write the balanced equation
P4 + NaOH + H2O
(b). Arrange in decreasing order:
i. HClO3, HClO, HClO4, HClO2 . (acidic character)
ii. NH3, PH3, AsH3, SbH3, BiH3
(boiling point)
Q.21 Explain why
(a) Eo Mn+3 /Mn is much higher than EoCr +3 /Cr+2 .
(b) Hexa aqua scandium(III) is colourless whereas hexa aqua
titanium (III) is coloured.
(c) Ce shows stable +4 oxidation state.
OR
(a) What is Lanthanoid contraction?
(b) Give its consequences.
Q.22 Write the IUPAC name of
(a) (NH4 )3[Cr(CN)6 ]
(b) How is the magnitude of
0 is affected by
nature of the ligand and oxidation state of
metal.
Q.23 Explain why:
(a) 3o haloalkane undergo SN1 reaction readily and why 1o
haloalkane gives SN2 readily.
(b) haloarenes are less reactive than haloalkane.
(c) p-dichloro benzene has more melting point than ortho or
meta isomers.
Q.24
(a) Distinguish between 2-propanol and 3o –butanol .
(b) write note on Williamson synthesis.
(c) Phenol becomes red when it is exposed to air. Explain.
Q.25 (a) Write the monomers of
(i) Glyptal
(ii) Nylon-26
(b) explain with example cationic polymerization.
Q.26 Give example of each
(i) Food preservative
(ii) Antiseptic
(iii) Tranquilizer.
Q.27 (a) What is selectivity of catalyst.
(b) What are shape selective catalyst ?
Q.28 (a) for the cell
(Pt) Br2 (g) | Br - (ad) || Au +3 (aq) | Au(s)
write cell reaction
EoBr +3 /2Br- = 1.06 V
EoAu +3 /Au = 1.4 V
Calculate EO cell and electrical work.
2
(b) Write reactions involved in Ni-Cd cell
(c) The resistance 0.1(M) KCl is 200 ohm. The electrodes are
separated by 0.1 cm. and area of cross section is 1 cm2 . Calculate the
molar conductance of the solution.
OR
(a) Calculate the charge in faraday and coulomb required for the
(i)
2 moles H2O into O2
(ii)
4 moles of Cr2 O7 –2 to Cr+3 .
(b) Calculate the molar conductivity at infinite dilution of
CH3COOH (given Limiting conductance of
HCl = 427.5 ohm-1cm2
CHCOOK = 91 ohm-1cm2
KCl = 130 ohm-1cm2 )
If the molar concentration of the CH3COOH solution be 0.1 (M)
and conductivity 0.003 ohm-1cm2 what will be dissociation
constant of acid.
Q.29 (a) What happens when NaCl is heated with sulphuric acid in
presence of MnO2
(b) Noble gases have very low boiling point. Why?
(c) H3PO3 is dibasic and H3PO2 is monobasic; justify?
(d) Fill it:
Ca 3 P2
+
H2 O
(A)
+ Ca(OH) 2
(B)
Q.30. I. Bring about the following conversions:
(a) 2-Butanone to Butane.
(b) Benzyl chloride to Benzoic acid.
(c) Benzoyl chloride to Benzyl amine.
II. A compound ‘A’ with molecular formula C7 H6 O2 , gives
effervescence with NaHCO3 solution .’A’ on nitration followed by
treatment with NH3 gives ‘B’. ‘B’ on treatment with NaNO2 and HCl
and then Br2 + KOH gives ‘C’ which responds to as dye test.
Identify A,B,C and give reactions.
------------------------
Hints and solution
Model Question Paper - 1
1. Due to formation of “F-Centre”.
2. Conversion of a freshly prepared precipitate in colloidal solution. As FeCl3 is added to
Fe(OH)3 pat.
3. 1-Chloro N-methyl ethanamine.
4.
CH3
CHO
CrO2Cl2
CrO3
5. 2-Chiral C-atom.
6. Van Hoff’s factor
i=1+(n-1)α
=1+(4-1)α
=1+3α
=1+3=4
AlCl3  Al+3 + 3Cl‾
7. N2 + 3H2  2NH3
8. Order is 5/2.
9.
O
ˉO
O
O
O
O
Oˉ
Cr2O7ˉ2 + 3Sˉ2 + 14H+  2Cr+3 + 3S + 7H2O
10. (i) CH3-CONH2 + Br2 + KOH  CH3NH2 + KBr + H2O + K2CO3
(ii)
CuCl
N=N-Cl
Cl + N2
HCl
(i)
OH
N=N-Cl +
N=N
OH
+ HCl
(ii)
NH2
N=N-Cl
NaNO2+HCl
0 - 5˚C
OH
11. A = CH3-CH
CN
B = CH3-CH(OH)-COOH (Lactic acid)
12. Zwitter ion : In α – amino acid or Sulphanilic acid there exists an equilibrium between neutral
and dipolar ion of the acid.
H2N-CH2-COOH  H3N+-CH2COOˉ
OR
..
NH2
NH3
SO3H
SO3ˉ
+
13. Reducing sugars : Monosaccharides.
Eg.
Glucose, Fructose.(free aldehyde or ketone present)
Non reducing sugar : Disaccharides
Eg.
14. For F.C.C
r =
Sucrose(free aldehyde or ketone absent)
a
2√2
a = 105 x 2√2 = 210√2 pm.
a³ = (210√2)3 x (10ˉ10)3 cm3
= (210)3 x 2√2 x 10ˉ30 cc
= 26190 x 10ˉ27
No. of unit cell in 10 cc = 10 x 1027
26190
= 3.8 x 1023 unit cell.
15. For K2SO4 ; vantHoff’s factor i=3.
∆Tb = it . Kb . m
= 3 x .52 x 1.74 x 1000
174 x 500
= 3 x .52 x 2 x 10‾2
= .0312
∆Tb = Tb (Soln) - Tb (Solvent)
Tb (solution) = .3012 + 100
= 100.0312ºC
16. The most unstable arrangement of atoms of reactant at the activated stable is called activated
complex.
H2 + I2  2HI
H
I
H
H
I

+
I
 HI + HI
H
I
Activated complex
17.
SO2Cl2(g)

When t=0: 0.5 arm
t=100 (0.5 – x)
PT = 0.5 – x + x + x = 0.6
x=0.1
K= 2.303 log 0.5
100
0.4
= 2.303 (0.6990 – 0.6020)
100
= 2.303 x 0.0970
100
SO2(g) +
0
x
Cl2(g)
0
x
= 0.0022 Sˉ1
ATQ
PT = 0.65
x=0.15
Rate = k[0.5 - 0.15]1
= 0.0022 x 0.35 atm Sˉ1
= 7.7 x 10ˉ4
18. a) Hardy Schulz rule:
(i)
The ions opposite to that of the colloidal sol are responsible for the coagulation.
(ii)
Greater the valency of the ions greater will be the coagulation power.
b) Flocculation / Coagulation takes place.
19. (i) Metal to be purified is made as anode and pure strip of metal is made as cathode and
electrolysis is carried out by using metal-salt soln.
(ii) Zone refining : Based on the principle that the impurity are mere soluble in the melt than
in the solid state of the metal.
Eg. Ge , Si , In etc. are purified by the process.
20. (a) P4 + 3NaOH + 3H2O  PH3 + 3NaH2PO2
(b) HClO4 > HClO3 > HClO2 > HClO
(c) BiH3 > NH3 > SbH3 > AsH3 > PH3
(b.pt)
21. (a) Mn+2 has stable d5 e-system and Cr+3 has stable d3 system.(t2g level half filled).
(b) Ti+3 has one unpairedelectron. So d-d transition takes place. Sc+3 has no unpaired electron
in d-orbital( 4s˚3d˚).
(c) Ce+4 is stable due to 6s˚4f˚ configuration.
Or
(a) Lanthanoid contraction : There is a gradual decrease in size of lanthanoid elements and
their ion with increasing atomic no. Electrons are filled successively in 4f orbital and as 4f
orbital is more diffused, its shielding effect is very poor. Therefore, this decrease is very small.
(La+3 to Lu+3 ; 106 – 84 pm).
(b) Consequences :
(i)
Similarity among lanthanoids.
(ii)
Basic strength of hydroxides decreases from La(OH)3 to Lu(OH)3
(iii) Similar properties of corresponding 4d and 5d block elements.
22. (a) Ammonium hexa Cyano Chromate(III).
(b) Magnitude of CFSE (∆ 0 ) depends on
(i)
Change density of the central metal ions; More the charge density more is the ∆ 0
value.
(ii)
More basic the ligand more will be crystal field splitting.
23. (a) In 3˚ haloalkane, carbonium ion is readily formed due to +I effect of alkyl group.
In 1˚ haloalkane steric hindrance is less.
(b) In haloarenes, resonance takes place due to electron withdrawing nature of benzene ring
and partial double bond developes.
Whereas in haloalkanes carbonium ion can be readily formed and are more reactive
towards SN1 and also SN2 .
(c) p-dichloro benzene has symmetric structure and the molecules can fit into the crystal lattice
strongly.
24. (a) Iodoform Test
NaOH
CH3-CH(OH)-CH3
I2
CHI3 + CH3COONa
Yellow
(b) Williamson’s synthesis
R-OˉNa+ + RX  R-OR + NaX
(c) Phenol is converted to quinone and then phenoquinone is formed when exposed to air.
25. (a) Glyptal : Monomer : Phthalic acid and Glycol
Nylon 2-6 : Glycine and caprolactoic acid
(b) When electron releasing group is attached to vinyl carbon atom cationic polymerization
takes place using H+ or electron deficient species.
26. (i) Food preservative : Na-benzoate.
(ii) Antiseptic: Chloxyleneol (Dettol).
(iii) Seconal
(a) Selectivity : Ability of a catalyst to direct a particular reaction.
Ni
CO + 3H2
CH4 + H2O
CuO/ZnO
CO + 2H2
CH3OH
CrO3
Cu
CO + H2
HCHO
(b) Shape selective catalysts are alumino silicates in which Si-O-Al- linkage is present. The
pore size vary from 260-740 pm. Eg ZSM-5 used to convert straight chain compound into
gasoline.
27. (a) 2Brˉ - 2e  Br2 (anode)
Au+3 + 3e  Au (cathode)
Net reaction
2Al+3 + 6Brˉ  3Br2 + 2Au
E0 cell = 1.4 – 1.06 = 0.34 V
∆G0 = -6 x 96500 x 0.34 = -196860 J
(b) Cd + Ni(OH)3  CdO + 2Ni(OH)2 + H2O
(c) κ = 1 x l/a = (1/200) x (0.1/1) = 5 x 10ˉ4 S cmˉ1
R
λm = κ x 1000
c
= (5 x 10ˉ4 )/0.1 = 5 S cm2 / mole
or
(a) (i) H2O – 2e  ½ O2 +
2F
For 2 moles 4F
2H+
(ii) Cr2O7ˉ2 + 14H+ + 6e  2Cr+3 + 7H2O
6F
For 4 moles 24F
(b) λ˚m CH3COOH = 427.5 + 91 – 230
= 518.50 – 230
= 288.5 S cm2 / mole
λm = 0.0031 x 1000
0.1
= 3 x 10 = 30 S cm2 / mole
α = 30
= 0.104
288.5
kd = 0.1 x (0.104)2
= 0.0011
28. (a) NaCl + H2SO4 + MnO2  Na2SO4 + MnSO4 + Cl2 + H2O
(b) Noble gases are monoatomic and weak Vander wall’s force exists.
(c) H3PO3 has 2 O-H bonds and H3PO2 has one O-H bond. (Replaceable H-1)
(d) Ca3P2 + 6H2O  3Ca(OH)2 + 2PH3
HBr
PH4+Brˉ
Zn/Hg
29. (a) CH3- C-CH2-CH3
O
CH3-CH2-CH2-CH3
HCl
Clemensen
(b)
CH2-Cl
Aq. KOH
CH2OH
COOH
[O]
KmnO4
(c)
O
C
CONH2
Cl
CH2NH2
NH3
H2
LAH
II:
C7H6O2
(A) HNO3
∆ H2SO4
CO2
NH3
(B)
NaNO2
HCl
(C)
responds azo dye test
COOH
NaHCO3
COONa
CO2 +
+ H2 O
(A)
HNO3
H2SO4
COOH
CONH2
NH3
(B)
NO2
NO2
Br2 / KOH
N=N-Cl
NH2
(C)
NaNO2
HCl O˚C
NO2
NO2