Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2015 Comptt (Delhi) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2015 Comptt (Delhi) with Solutions
Time allowed : 3 hours
Maximum marks : 70
General Instructions:
- All questions are compulsory.
- Questions number 1 to 8 are very short answer questions and carry 1 mark each.
- Questions number 9 to 18 are short answer questions and carry 2 marks each.
- Questions number 19 to 27 are also short answer questions and carry 3 marks each.
- Questions number 28 to 30 are long answer questions and carry 5 marks each.
-
Use log tables if necessary. Use of calculators is not allowed.
Deleted from Syllabus.
Question 1.
What is the no. of atoms per unit cell (z) in a body-centred cubic structure? [1]
Answer:
Contribution by the atoms present at eight corners = 8 × \(\frac{1}{8}\) = 1
Contribution by the atoms present at centre = 1
Total number of atoms present in unit cell = 1 + 1 = 2
Question 2.
In reference to surface chemistry, define dialysis. [1]
Answer:
Dialysis : The process of removing the dissolved substances from a colloidal solution by means of diffusion through a suitable membrane is called dialysis.
Question 3.
What is the IUPAC name of the complex [Ni(NH
3
)
6
]Cl
2
? [1]
Answer:
[Ni(NH
3
)
6
]Cl
2
IUPAC name : Hexaamminenickel (II) chloride.
Question 4.
Draw the structure of 3-methylpentanal. [1]
Answer:
3-Methylpentanal
Question 5.
Complete the following reaction equation :
C
6
H
5
N
2
Cl + H
3
PO
2
+ H
2
0 →
Answer:
Question 6.
Define osmotic pressure of a solution. How is the osmotic pressure related to the concentration of a solute in a solution? [2]
Answer:
Osmotic pressure : It is the external pressure which is applied on the side solution which is sufficient to prevent the entry of the solvent through semi-permeable membrane. According to the Boyle-Van’t Hoff Law, the osmotic pressure (JT) of a dilute solution is directly proportional to its molar concentration provided temperature is constant.
π ∝ C (At constant temperature)
π ∝ CT (At constant concentration)
π = CRT (R = Solution constant)
or, π = \(\frac{n}{v}\) RT
Question 7.
Define the following terms : [2]
(i) Half-life of a reaction (t
1/2
) (ii) Rate constant (k)t
(i) Half-life of a reaction (t
1/2
) : Half-life period (t
1/2
) is the time in which half of the substance has reacted and its concentration is reduced to one-half of its initial concentration.
(ii) Rate constant (k): Rate constant may be defined as the rate of reaction when the molar concentration of each reactant is taken as unity.
Question 8.
Draw the structures of the following : [2]
(i) H
2
SO
4
(ii) XeF
2
Answer:
Question 9.
What is meant by ‘disproportionation’? Give an example of a disproportionation reaction in aqueous solution. 2
Answer:
Disproportionation : In a disproportionation reaction an element undergoes self-oxidation as well as self-reducdon forming two different compounds.
For example,
Or
Suggest reasons for the following features of transition metal chemistry :
(i) The transition metals and their compounds are usually paramagnetic.
(ii) The transition metals exhibit variable oxidation states.
Answer:
(i) The transition metals and their compounds are usually paramagnetic because of the presence of unpaired electrons in their d-orbitals.
(ii) The transition metals exhibit variable oxidation states because of very close energies of incompletely filled (n – 1) d-orbitals and ns orbitals due to which both can participate in bonding.
Question 10.
Explain the mechanism of dehydration steps of ethanol : [2]
Answer:
Acid Dehydration of ethanol :
Question 11.
Define the following :
(i) Schottky defect
(ii) Frenkel defect
(iii) F-centre
Answer:
(i) Schottky defect : If in an ionic crystal of type A
+
B
–
, equal number of cations and anions are missing from their lattice sites so that the electrical neutrality is maintained, it is called Schottky defect.
(ii) Frenkel defect : If an ion leaves its site from its lattice site and occupies the interstitial site and maintains electrical neutrality, then it is called Frenkel defect.
(iii) F-centre : The centres which are created by trapping of electrons in anionic vacancies and which are responsible for imparting colour to the crystals are called F-centres. (F = Fabre)
Question 12.
45 g of ethylene glycol (C
2
H
6
O
2
) is mixed with 600 g of water. Calculate
(i) the freezing point depression and
(ii) the freezing point of the solution (Given : k
f
water = 1.86 K kg mol
-1
) [3]
Answer:
(i) Given : w = 45 g, W = 600 g, K
f
= 1.86 K kg mol
-1
, ∆T
f
= ?
Using the formula for freezing point depression,
Question 13.
The rate constant of a reaction at 500 K and 700 K are 0.02 s
-1
and 0.07 s
-1
respectively. Calculate the value of activation energy, Efl. (R = 8.314 J K
-1
mol
-1
) [3]
Answer:
Given : k
2
= 0.07 s
-1
, k
1
= 0.02 s
-1
,
T
1
= 500 K, T
2
= 700 K, E
a
= ?
Question 14.
Define the following terms :
(i) Electrophoresis
(ii) Adsorption
(iii) Shape selective catalysis [3]
Answer:
(i) Electrophoresis : When electric current is passed through a colloidal solution, the positively charged particles move towards cathode while negatively charged particles move towards anode where they lose their charge and get coagulated. This phenomenon is known as electrophoresis.
(ii) Adsorption : The phenomenon of accumulation or higher concentration of molecular species (gases or liquids) at the surface rather than in the bulk of a solid or liquid is called adsorption.
(iii) Shape selective catalysis : The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape- selective catalysis.
Question 15.
Outline the principles of refining of metals by the following methods :
(i) Distillation
(ii) Zone refining
(iii) Electrolysis [3]
Answer:
(i) Principle of Distillation : The impure metal is evaporated to obtain the pure metal as distillate. The metals having low boiling points (Bi, Hg, Cd etc.) readily change into vapours leaving behind impurities which further get collected in receivers and upon cooling, pure metal is obtained.
(ii) Zone refining of metals : It is based on the principle that the impurities are more soluble in melt than in the solid state of the metal.
(iii) Electrolysis : Here the impure metal is made to act as anode and a strip of the same metal in pure form is used as cathode. When they both are put in a suitable electrolyte containing soluble salt of same metal, the more basic metal remains in the solution and the less basic ones go to the anode mud.
Example : In refining of Cu
At anode : (oxidation)
Cu → Cu
2+
+ 2e
–
At cathode : (reduction)
Cu
2+
+ 2e
–
→ Cu
Or
Write down the reactions taking place in different zones in the blast furnace during the extraction of iron. How is pig iron different from cast iron?
Answer:
Reactions occur in blast furnace in the extraction of iron from iron oxide ores. The following oxidations occur in different zones of furnace :
Pig iron differs from cast iron with respect to the carbon contents. Pig iron has nearly 4% carbon content and cast iron has nearly 3% carbon content. Cast iron is hard and brittle whereas pig iron is soft.
Question 16.
What is lanthanoid contraction? What are the consequences of lanthanoid contraction? [3]
Answer:
Lanthanoid contraction : The overall decrease in atomic and ionic radii with increasing atomic number from La to Lu due to imperfect shielding of 4f-orbital is known as lanthanoid contraction.
Cause : As we move along the lanthanoid series, the effective nuclear charge increases on addition of electrons and the electrons added in /-subshell causes imperfect shielding which is unable to counterbalance the effect of the increased nuclear charge. Hence the contraction in size occurs.
Consequences :
(i) Due to small change in atomic radii, the chemical properties of lanthanoids are very similar due to which separation of lanthanoids becomes very difficult.
(ii) There is similarity in size of elements belonging to same group of second and third transition series.
Example : Zr and Hf are known as chemical twins due to their almost identical radii.
Question 17.
Indicate the types of isomerism exhibited by the following complexes : [3]
(i) [CO(NH
3
)
5
(N0
2
)]
2+
(ii) [Co(en)
3
]Cl
3
(en = ethylene diamine)
(iii) [Pt(NH
3
)
2
Cl
2
]
Answer:
(i) Linkage isomerism is shown by (i) [CO(NH
3
)
5
(N0
2
)]
2+
(i) [CO(NH
3
)
5
(NO
2
)]
2+
Pentaamminenitro Cobalt (III)
[CO(NH
3
)
5
(O-NO
2
)]
2+
Pentaamminenitrito-N-Cobalt (III)
(ii) [Co(en)
3
]Cl
3
shows optical isomerism
(iii) [Pt(NH
3
)
2
Cl
2
] shows geometrical isomerism
Question 18.
Name the following according to IUPAC system :
Answer:
(i) Butan-2-ol
(ii) 2-Bromotoluene
(iii) 2,2-dimethylchloropropane
Question 19.
How are the following conversions carried out?
(i) Propene to propane-2-ol
(ii) Benzyl chloride to Benzyl alcohol
(iii) Anisole to p-Bromoanisole
Ans.
(i) Propene to propane-2-ol
(ii) Benzyl chloride to Benzyl alcohol
(iii) Anisole to p-Bromoanisole
Question 20.
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br
2
and KOH forms a compound ‘C’ of molecular formula C
6
H
7
N. Write the structures and IUPAC names of compounds A, B and C. [3]
Answer:
The data shows that C
6
H
7
N may be C
6
H
5
NH
2
i.e. Aniline. Since it is obtained by heating with Br
2
and KOH (Hoffmann bromamide reaction), then the compound ‘B’ is Benzamide C
6
H
5
CONH
2
which is in turn obtained ‘by reaction with aqueous ammonia then the compound ‘A’ can be Benzoic acid i.e. C
6
H
5
COOH
Thus, A → Benzoic acid, B → Benzamide, C → Aniline
Question 21.
How are vitamins classified? Name the vitamin responsible for the coagulation of blood. [3]
Answer:
Vitamins are classified into two types :
(i) Water insoluble vitamins : These are fat soluble substances E.g. Vitamin A, D, E and K.
(ii) Water soluble vitamins : These include Vitamin B-Complex and Vitamin C (except B
12
). Vitamin K or phylloquinone is responsible for the coagulation of blood.
Question 22.
Write the names and structures of the monomers of the following polymers : 3
(i) Buna-S (ii) Neoprene (iii) Teflon
Answer:
Question 23.
Ramesh went to a departmental store to purchase groceries. On one of the shelves he noticed sugar-free tablets. He decided to buy them for his grandfather who was a diabetic. There were three types of sugar-free tablets. Ramesh decided to buy sucrolose which was good for his grandfather’s health.
(i) Name another sugar free tablet which Ramesh did not buy.
(ii) Was it right to purchase such medicines without doctor’s prescription?
(iii) What quality of Ramesh is reflected above? [4]
Answer:
(i) Other tablets are Aspartame and Saccharin.
(ii) No, it was not right because medicines should always be taken under medical supervision as a doctor knows best about your body reactivity towards medicines.
(iii) As per latest CBSE curriculum, Value Based Questions will not be asked in the examination.
Question 24.
(a) Define the following terms :
(i) Molar conductivity (Λ
m
)
(ii) Secondary batteries
(iii) Fuel cell
Answer:
(b) State the following laws :
(i) Faraday first law of electrolysis
(ii) Kohlrausch’s law of independent migration of ions [5]
Answer:
(a)
(i) Molar conductivity (Λ
m
) : Molar conductivity can be defined as the conductance of the volume V of electrolytic solution kept between tjkvo electrodes of a conducting cell at distance of unit length but having area of cross section large enough to accomodate sufficient volume of solution that contains one mole of the electrolyte.
Λ
m
= KV
(ii) Secondary batteries : Those cells which can be recharged on passing electric current through them in opposite, direction and can be used again are called secondary batteries, e.g. Lead-acid storage cell.
(iii) Fuel cell : Galvanic cells that are designed to convert the chemical energy of combustion of fuels like hydrogen, methane etc. into electrical energy are called fuel cells, e.g. H 2 – O 2 fuel cell
(b) (i) Faraday first law of electrolysis : According to this law the mass of the substance deposited or liberated at any electrode during electrolysis is directly proportional to the quantity of charge passed through the electrolyte.
ω ∝ Q
ω = ZIt
(ii) Kohlrausch’s law of independent migration of ions : According to this law limiting molar conductivity of an electrolyte can be represented as the sum of the limiting ionic conductivities of the cation and the anion each multiplied with the number of ions present in one formula unit of electrolyte.
\(\Lambda_m^o\) for AxBy = x\(\lambda_{+}^0\) + y\(\lambda_{-}^0\)
Or
(a) Define the term degree of dissociation. Write an expression that relates the molar conductivity of a weak electrolyte to its degree of dissociation.
(b) For the cell reaction
Ni
(s)
| Ni
2+
(aq) || Ag
+
(aq) | Ag
(s)
Calculate the equilibrium constant at 25 °C. How much maximum work would be obtained by operation of this cell?
\(\mathrm{E}_{\mathrm{Ni}^2 / \mathrm{Ni}}^{\mathrm{o}}\) = 0.25 V and \(\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\mathrm{o}}\) = 0.80 V
(a) Degree of dissociation : It is the measure of the extent to which an electrolyte gets dissociated into its constitutent ions.
Thus higher the degree of dissociation, higher will be its molar conductance.
Mathematically it can be expressed as :
(b) Ni
(s)
| Ni
2+
(aq) || Ag
+
(aq) | Ag
(s)
Question 25.
(a) Complete the following chemical equations :
(i) Cu + HN
3
(dilute) →
(ii) P
4
+ NaOH + H
2
O →
(b) (i) Why does R
3
P = O exist but R
3
N = O does not? (R = alkyl group)
(ii) Why is dioxygen a gas but sulphur a solid?
(iii) Why are halogens coloured? [5]
Answer:
(a)
(b) (i) Nitrogen in R 3 N = O cannot form pπ – dπ multiple bonds because it cannot expand its covalency beyond 4 due to absence of d-orbital, so it will not exist. But R 3 P = O can do so due to presence of d-orbitals and formation of pπ – dπ multiple bonds which can expand its covalency up to 5.
(ii) Because of bigger size and the strong forces of attraction holding 8 atoms, their bonds cannot be broken easily and hence sulphur exists as solid while oxygen due to high electro-negativity and tendency to form pπ – dπ multiple bonds through Vander-waals forces of attraction can be broken easily and hence exists as gas.
(iii) All halogens are coloured due to absorption of light in the visible region as a result of which their electrons get excited to higher energy levels and while returning to lower level transmit energv of corresponding colour.
Or
(a) Write balanced equations for the following reactions :
(i) Chlorine reacts with dry slaked lime.
(ii) Carbon reacts with concentrated H
2
SO
4
(b) Describe the contact process for the manufacture of sulphuric acid with special reference to the reaction conditions, catalysts used and the yield in the process.
Answer:
(a)
(b) Contact process of sulphuric acid :
It involves the following steps :
(i) Formation of sulphur dioxide by burning either sulphur or iron pyrites in excess of air.
s + O
2
→ SO
2
4FeS
2
+ 11O
2
→ 2Fe
2
O
3
+ 8SO
2
(ii) Catalytic oxidation of SO
2
into SO
3
by using V
2
O
5
as catalyst
According to Le-Chatelier’s principle the reaction conditions are :
(a) High concentration of reactants
(b) Low temperature (623-723 K)
(c) High pressure (2 bar)
By obeying above conditions the yield of H2S04 will be 96-98%
(iii) Absorption of SOs in 98% H2S04 to give Olgum (H
2
S
2
O
7
)
(iv) Dilution of Oleum to give sulphuric acid
H
2
S
2
O
7
+ H
2
O → 2H
2
SO
4
Question 26.
(a) Describe the following giving chemical equations :
(i) De-carboxylation reaction (ii) Friedel-Crafts reaction
(b) How will you bring about the following conversions?
(i) Benzoic acid to Benzaldehyde
(ii) Benzene to m-Nitroacetophenone
(iii) Ethanol to 3-Hydroxybutanal [5]
Answer:
(a) (i) De-carboxylation reaction : When sodium salt of carboxylic acid is heated with soda lime (NaOH + CaO), then corresponding alkane and CO
2
will be evolved
(ii) Friedel-Crafts reaction : The introduction of alkyl or acetyl group in the presence of anhydrous almuninium chloride (AlCl
3
) as catalyst to ortho and para positions of an aromatic compound is called Friedel-Crafts reaction.
(b)
Or
(a) Describe the following actions :
(i) Acetylation (ii) Aldol condensation
(b) Write the main product in the following equations
Answer:
(a) (i) Acetylation : The addition of acyl (-COCH
3
) group on the ortho and para positions of acyl halide in the presence of a Lewis acid i.e. anhydrous aluminium chloride (AlCl
3
) which acts as a catalyst.
(ii) Aldol condensation : Those aldehyde and ketones with atleast one α-hydrogen atom condense together in presence of dil. alkali e.g. NaOH etc. to form a ß- hydroxy aldehyde or ß-hyroxy ketone respectively which gets dehydrated in the presence of acid upon heating to form α, ß-unsaturated compound.
(b)
Please Note : The questions asked in Set II and Set III are identical to Set I. Simply the serial numbers of questions were changed.