Chemistry: Cbse Sample Paper For Class-12
Chemistry: Cbse Sample Paper For Class-12
Chemistry: Cbse Sample Paper For Class-12
CHEMISTRY
General Instructions
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.
5. Give a chemical test to distinguish between a primary and a secondary amine. [1]
7. Write the structures of monomers used and one used of each of the following polymers: [1]
a) Teflon
b) Buna-N
Cationic detergents
9. AgI crystallizes in cubic close packed ZnS structure. What fractions of tetrahedral sites are
occupied by Ag+ ions? [2]
11. The Henry law constant for oxygen dissolved in water is 4.34 ´ 104 atm. at 25°C. If the partial
pressure of oxygen in air is 0.2 atm. under ordinary atmospheric conditions. Calculate the
concentration (in moles per litre) of dissolved oxygen in water in equilibrium with air at 25°C.
[2]
13. Why in any transition series, melting points first increase and then decrease and also they show
a dip in the middle? [2]
2
15. What is a stereospecific reaction? Give one example of this reaction. [2]
17. Write the names and molecular structures of the monomers of the following polymers. Give one
use of each specifying the property responsible for this use [2]
i) Nylon-6
ii) Neoprene
18. a) How do antiseptics differ from disinfectants? Give example of each. [2]
OR
19. Urea forms an ideal solution in water. Determine the vapour pressure of an aqueous solution
containing 10% by mass of urea at 40°C. (Vapour pressure of wter at 40°C = 55.3 mm of Hg)
[3]
[2]
20. a) Why does the conductivity of a solution decrease with dilution? [3]
b) Calculate the emf of the cell in which the following reaction takes place:
21. a) Of NH3 and CO2 which will be absorbed more readily one the surface of charcoal and why?
[3]
c) Among the noble gases only xenon is well known to form chemical compounds.
3
23. Justifying the order of your choice, arrange the following in decreasing order of property
indicated. [3]
OR
24. a) Among the iron complexes, K3[Fe(CN)6] is weakly paramagnetism whereas K3[FeF6] is highly
paramagnetic explain.
25. How will you distinguish between 1-phenylethanol and 2-phenylethanol? [3]
26. Give a chemical test to distinguish between aniline and ethylamine. Give a chemical test to
distinguish between a primary and a secondary amine. [3]
27. a) Write two differences between vitamins and hormones. Give one example each. [3]
iv) Among the noble gases, only xenon is known to form true chemical compounds.
OR
i) Name chief ore of Lead. Write chemical reactions involving the extraction of Lead from this
ore.
4
29. For the chemical reaction A + 2B ¾® 2C + D, the experimentally determined information has
been tabulated below: [5]
OR
For the reaction A + B ¾® products, the following initial rates were obtained at various given
initial concentrations.
Write rate law and find the rate constant for the above reaction.
a) H C COOH b) H C COOH
¾¾¾
D2 /Pt
® Pr oduct ¾¾¾
D2 /Pt
® Pr oduct
H C COOH HOOC C H
OR
O Zn - Hg/HCl
O O ¾¾¾¾® (A)
5
SOLUTIONS
SOLUTIO
1.
They consists of aggregation of small They consist of large molecules, i.e., polymers.
atoms or molecules.
e.g., starch sol., albumin sol.
e.g., sulphur sol., As2S3 sol.
2. Ionisation isomers:
Test : Add BaCl2(aq), [Co(NH3)5Br]SO4 will give white precipitate insoluble in conc. HCl.
Add AgNO3 solution, [Co(NH3)5SO4]Br will give yellow ppt. due to AgBr which is partially soluble
in excess of NH4OH.
i) NaHCO3 Test: Add sodium bicarbonate (NaHCO3) to each compound. With benzoic acid,
CO2 gas is evolved.
HO O
COONa
+ NaHCO3 ¾¾
® + CO 2 + H 2 O
Phenol +NaHCO3 ¾¾
® do not product CO 2 i.e. no efferversence
ii) Mechanism
® H + + CN -
HCN ¾¾
R -
R O R OH
d+ d-
+ CN - ¾¾
+
O ® +H
¾¾¾ ®
H H CN H CN
Aldehyde cynohydrin
4. It is because there is +ve charge on o- and p-positions, therefore, electrophillic substitution takes
place at m-position as shown in following resonating structures.
6
HO O HO O HO O HO O HO O
5. Carbylamine reaction (Isocyanide test). Add CHCl3 and KOH. Primary amine will form offensive
smelling compound isocyanide whereas secondary amine does not react.
6. H-bonding in d-helix, formed between the C = O of one amino acid residue and the N-H of the
fourth amino acid residue in the chain.
9. In the face-centred unit cell, there are eight tetrahedral voids. Of these, half are occupied by
silver cations.
10. i) Ferromagnetism. When magnetic moments (electron spins) in a substance align in parallel
and antiparallel directions in unequal numbers so that there is net dipole moment, the
substance is ferromagnetic.
ii) n-type semiconductor: It is formed when impurity atom containing more valency
electrons than the parent insulator atom is introduced into it. The unbounded electrons are
the current carriers.
p O2 = K H ´ X O2
p O2 0.2atm
X O2 = = = 4.6 ´ 10–6
KH 4.34 ´ 104 (atm)
1000
n H2 O = = 55.5 moles
18
7
! X O2 is very small as compared to X H2 O
n O2 + n H 2 O = n H 2 O
n O2
x O2 =
n H2 O
X O2 + n H 2 O = n O2
b) Al3+ + 3e– ¾® Al
3
\ 40g of Al will require electricity = ´ 40 = 4.44F
25
13. Melting points first increase because the number of unpaired electrons increases and hence
strength of metallic bond increases. After reacting the maximum, the melting points decrease
because the pairing of electron starts in the d-subshell and number of unpaired electrons
decreases and so the strength of metallic bond decreases. The dip in the middle is due to exactly
half-filled configuration of d-subshell which has higher stability. Hence, electrons are held tightly
by the nucleus. As a result, metallic bond is weaker.
14. i) [Co(NH3)5Cl]2+
ii) Li(AlH4]
15. A reaction is stereospecific when a particular stereoisomereic form of the starting material reacts
in such a way that it gives stereoisomeric form of the product.
16. The three dimensional structure of the three compounds along with the direction of dipole
moment in each of their bonds are given below:
Cl H Cl Cl Cl Cl
C C C
Cl H Cl H Cl Cl
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CCl4 being symmetrical has zero dipole moments. In CHCl3.
It is used for making ropes, tyre cords because it has high tensile strength
H2C
Cl CH2
18. a) Antiseptics are chemical substances which prevent the growth of microorganisms but are
not harmful to human or animal tissues. For example, dettol and savlon.
Disinfectants are chemical substances which kill microorganisms but are not safe to be
applied to the living tissues. Some common examples of disinfectants are phenol (³ 1%
solution) and chlorine (0.2 to 0.4 ppm).
b) They can be used in hard water as well in acidic solution. The reason being that sulphonic
acids and their calcium and magnesium salts are soluble in water but the fatty acids and
their calcium and magnesium salts are insoluble.
OR
a) It is due to poor shielding effect of d-electrons due to which effective nuclear charge
increases.
b) It is because they are electron deficient, i.e., their octet is not complete.
PAo - PA
19. = xB
PAo
PA
1- = xB
PAo
10
P P 1 1 1
1 - A = 60 Þ 1 - A = ´ =
55.3 90 55.3 6 5 30
18
PA 1 29
Þ = 1- =
55.3 30 30
9
55.3 ´ 29 1603.7
Þ PA = = = 53.47 mm
30 30
20. a) Conductivity of a solution is the conductance of ions present in a unit volume of the solution.
On dilution, the number of ions per unit volume decreases. Hence, the conductivity
decreases.
0.0591 [Ni 2+ ]
E cell = E 0cell - log = 1.05
n [Ag + ]2
0.0591 0.160 0.0591
V- log 2
= 1.05 - log(4 ´104 )
2 (0.002) 2
0.0591
= 1.05 - (4.6021) = 1.05 - 0.14V = 0.91V
2
21. a) NH3 will be adsorbed more readily on the surface of charcoal because it has higher critical
temperature due to more Vander Waal’s forces of attraction.
b)
They consists of aggregation of small They consist of large molecules, i.e., polymers.
atoms or molecules.
e.g., starch sol., albumin sol.
e.g., sulphur sol., As2S3 sol.
c) Electrodialysis. It is based on the principle that ions can pass through parchment paper
faster in electric field than colloidal particles and thus get separated.
22. a) It is due to formation of lead oxide and finally lead carbonate and lead hydroxide on their
surface.
c) It is due to bigger size or ‘Xe’ due to which it has lower ionization energy and high polarizing
power.
23. i) NH3 > PH3 > AsH3 > SbH3 because atomic size goes on increasing therefore, lone pair of
electron becomes less available.
ii) HClO > HClO2 > HClO3 > HClO4 because HClO is least stable whereas HClO4 is most stable
and cannot give oxygen.
iii) M – F > M – Cl > M – Br > M – I because F is more electronegative whereas iodine is least
electronegative.
OR
10
a) It is due to formation of oxide layer on its surface which makes it passive.
c) It is because nitric acid decomposes to NO2 which is brown in colour and makes HNO3 yellow.
24. a) Since F– ion is a weak ligand, d-electrons remain unpaired whereas CN– ion is a strong ligand
and electrons get paired up.
b) The energy difference (D) between the lower and higher orbitals obtained as a result of
splitting d-orbitals in a crystal field is known as splitting energy.
c) Trigonal bipyramidal.
25. By iodoform test. 1-phenyethanol contains the grouping –CHOH – CH3 and hence gives iodoform
test while 2-phenylethanol does not contain the grouping –CHOH – CH3 and hence does not give
iodoform test.
C6 H 5 - CHOH - CH 3 ¾¾¾¾
I2 / NaOH
® CHI3
1-Phenylethanol Iodoform
26. Add NaNO2 and conc. HCl. Cool it to 0 to 5°C. Then add alkaline solution of phenol. Orange dye
is formed in case of aniline whereas ethyl amine does not give orange dye. Carbylamine reaction
(Isocyanide test). Add CHCl3 and KOH. Primary amine will form offensive smelling compound
isocyanide whereas secondary amine does not react.
27. a) Hormones
i) Molecules that transfer information from one group of cells to distant tissue or organ.
Vitamins
Vitamins : Vitamin D
b) Functions:
11
3) For regulation of metabolism.
28. i) There are two axial and three equatorial bonds in PCl5 molecules as a result of sp3d
hybridisation. There is repulsive force between the electrons in the axial plane and equatorial
plane which results in the axial elongation of bonds.
ii) Sulphur in vapour state forms some S2 molecules which like O2 molecules contains unpaired
electrons and hence paramagnetic.
iii) Because of the lower bond energy of F – F bond and higher hydration energy of F– ions.
iv) Xenon has the lowest ionization energy among the noble gases except radon which is
however radioactive.
v) The lower oxidation state gets stabilised with increase in atomic number in the same group
of p-block element (inert pair effect), hence PbO2 is a stronger oxidant than SnO2.
OR
ii) Chlorine dioxide ClO2 as prepared by reduction of ClO3 with SO2 in strongly acidic medium.
2NaClO3(aq) + SO2(g) ¾¾
acid
¾® 2ClO2(g) + Na2SO4(aq)
Xenon tetrafluride XeF4 is prepared by reaction of Xenon and F2 in the ratio of 1 : 5 at 873
K and 7 bar.
Let the rate law equation for the reaction be: rate = k [A]x[B]y.
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0.384 = k [0.60]x [0.30]y (ii)
(2)2 = (2)x or x = 2
(2)1 = (2)y or y = 1
Order w.r.t. A = 2; B = 1.
-d[A] 1 [C]
d) Rate of reaction in terms of A and C : Rate = = .
dt 2 dt
OR
13
Dividing equation (iii) by equation (i)
0.05
K[0.1]p [0.1]q = 0.05; k[0.1]1 [0.1]0 = 0.05; k = = 0.5
0.1
30. The addition is syn therefore cis alkene will give meso product (use CSM rule) and trans alkene
will give (dl) mixture.
a) COOH
H C COOH H C D
¾¾¾
D2 /Pt
®
H C COOH H C D
cis COOH
meso
b) COOH COOH
H C COOH H C D D C H
¾¾¾
D2 /Pt
®
HOOC C H D C H H C D
COOH COOH
dl mixture
OR
O
O O O +
O
Zn - Hg/HCl
O ¾¾¾¾® ¾¾¾¾
H /H 2 O
® Zn - Hg/HCl
¾¾¾¾®
Zn-Hg reduces only >C = O group to > CH2 group and does not affect any other functional group.
But in acidic medium keto group is deprotected (i.e., cyclic ketal hydrolyses) and the keto group
formed is further reduced to methylene (>CH2) group.
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