Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2015 Comptt (Outside Delhi) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2015 Comptt (Outside Delhi)
General Instructions:
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 type of stoichiometric defect is shown by AgCl? [1]
Answer:
AgCl shows Frenkel defect.
Question 2.
What are emulsions? Give an example. [1]
Answer:
Emulsions : An emulsion is a colloidal dispersion in which both the dispersed phase and dispersion medium are liquids. For example, milk, cream.
Question 3.
What is meant by chelate effect? [1]
Answer:
Chelate effect : When a bidentate or a polydentate ligand contains donor atoms positioned in such a way that when they coordinate with the central metal ion, a five or a six membered ring is formed. This effect is called Chelate effect. As a result, the stability of the complex increases.
Example : the complex of Ni
2+
with ‘+ion’ is more stable than NH
3
.
Question 4.
Write the IUPAC name of the following: [1]
CH
3
– CH
2
– CHO
Answer:
IUPAC name : Propan-1-al
Question 5.
Arrange the following in increasing order of basic strength
Aniline, p-Nitroaniline and p-Toluidine
Answer:
Question 6.
Define the following terms : [2]
(i) Mole fraction (x)
(ii) Molality of a solution (m)
Answer:
(i) Mole fraction : Mole fraction of a constituent is the fraction obtained by dividing number of moles of that constituent by the total number of moles of all the constituents present in the solution. It is denoted by ‘x’.
Example : x
1
= \(\frac{\text { No. of moles of } x_1}{\text { Total no. of moles }}\)
= \(\frac{n x_1}{n x_1+n x_2}\)
(ii) Molality of a solution : Molality of a solution is defined as the number of moles of the solute dissolved in 1000 grams (1 kg) of the solvent. It is denoted by’m’.
m = \(\frac{w \times 1000}{\mathrm{M} \times ω}\)
[where ω = Weight of solute in grams
M = Molecular mass of solute
W = Weight of solvent in grams
Question 7.
Write units of rate constants for zero order and for the second order reactions if the concentration is expressed in mol L
-1
and time in second.
Answer:
Using formula of rate constant,
K = [mol L
-1
]
1
S
-1
Unit for zero order reaction,
K = [mol L
-1
]
1-0
S
-1
K = [mol L
-1
] S
-1
= mol L
-1
s
-1
Unit for second order reaction,
K = [mol L
-1
]
1-2
S
-1
= [mol L
-1
]
-1
s
-1
= mol
-1
L s
-1
Question 8.
Explain the following :
(i) Nitrogen is much less reactive than phosphorus,
(ii) NF3 is an exothermic compound but NCl
3
is an endothermic compound. [2]
Answer:
(i) Due to presence of weak single bond in P – P than N = N phosphorous is more reactive than nitrogen and also because of high bond dissociation enthalpy of N = N.
(ii) Due to smaller size of F as compared to Cl, the N-F bond is much stronger than N-Cl bond while bond dissociation energy of F
2
is much lower than that of Cl
2
. Therefore, energy released during the formation of NF
3
molecule is more than the energy needed to break N
2
and F
2
molecules into individual atoms. In other words, formation of NF
3
is an exothermic reaction.
The energy released during the formation of NCl 3 molecule is less than the energy needed to break N 2 and Cl 2 molecules into individual atoms. Thus formation of NCl 3 is an endothermic reaction.
Question 9.
Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with oxalic acid? Write the ionic equations for the reactions. [2]
Answer:
Potassium Permangante (KMnO
4
) is prepared from pyrolusite ore (MnO
2
). The ore (MnO
2
) is fused with an alkali metal hydroxide like KOH in the presence of air or an oxidising agent like KNO
3
to give dark green potassium manganate (K
2
MnO
4
). K
2
MnO
4
disproportionates in a neutral or acidic solution to give potassium permanganate.
Or
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with (i) iodine (ii) H
2
S.
Answer:
Potassium dichromate (K
2
Cr
2
O
7
) acts as a strong oxidising agent in acidic medium using H
2
SO
4
.
K
2
Cr
2
O
7
+ 4H
2
SO
4
→ K
2
SO
4
+ Cr
2
(SO
2
)
3
+ 4H
2
O + 3[O]
Ionic reactions :
Question 10.
Write the mechanism of acid dehydration of ethanol to yield ethene. [2]
Answer:
The mechanism of dehydration of ethanol involves the following steps :
Step 1 : Formation of protonated alcohol
Step 2 : Formation of Carbocation : It is the slowest and rate determining step.
Step 3 : Formation of ethene by elimination of a proton
Question 11.
Silver crystallises in fee lattice. If edge length of the unit cell is 4.077 × 10
-8
cm, then calculate the radius of silver atom. [3]
Solution:
Given : a = 4.077 × 10
-8
cm r = ? for fee lattice
Using formula,
Radius (r) = \(\frac{a}{2 \sqrt{2}}\)
r = \(\frac{4.077 \times 10^{-8}}{2 \times 1.414}\) cm
= \(\frac{4.077 \times 10^{-8}}{2.828}\)
r = 1.441 × 10
-8
cm
Question 12.
A 5 percent solution (by mass) of cane-sugar (M.W. 342) is isotonic with 0.877% solution of substance X. Find the molecular weight of X. [3]
Solution:
Given : W (mass) of cane-sugar = 5% means 5 g
Molar mass of cane-sugar (M) = 342 g mol
-1
Mass of isotonic substance X = 0.877% means 0.877 g
Molar mass of X = ?
Using formula
Question 13.
The rate constant for a first order reaction is 60 s
-1
. How much time will it take to reduce the initial concentration of the reactant to its 1/10th value? [3]
Solution:
Given : k = 60 s
-1
, t = ?
If initial concentration is \(\mathrm{A}_0\right\)
P Then \(\frac{1}{10}\)th of initial concentration is \(\frac{\left[\mathrm{A}_0\right]}{10}\).
Question 14.
Describe the following processes : [3]
(i) Dialysis
(ii) Electrophoresis
(iii) Tyndall effect
Answer:
(i) Dialysis : The process of removing the dissolved substance from a colloidal solution by diffusion of the mixture through a semi-permeable membrane is known as dialysis.
(ii) 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.
(iii) Tyndall effect : Colloidal particles scatter light in all directions in space. When a beam of light is passed through a colloidal solution and viewed perpendicular to the path of the incident light, the path of light becomes visible as a bright streak. The illuminated path is called Tyndall cone and the phenomenon is called Tyndall effect.
Question 15.
Answer the following : [3]
(i) What is the role of cryolite in the metallurgy of aluminium?
(ii) Difference between roasting and calcination.
(iii) What is meant by the term ‘chromatography’?
Answer:
(i) The role of cryolite (Na
3
AlF
6
) is to lower the melting point of the mixture and brings conductivity
(ii)
(iii) The term chromatography was originally derived from the Greek word ‘chroma’ meaning colour and ‘graphy’ meaning writing because the method was first used for the separation of coloured substances into individual components. Chromatography is a widely used process for separation, purification, identification and characterization of the different components of a mixture which are differently adsorbed on a suitable adsorb.
Or
Write the reactions taking place in different zones of the blast furnace to obtain Iron.
Answer:
Extraction of iron from iron ore in different zones of blast furnace :
(i) Lower zone (zone of heat absorption) : Temperature 1423 – 1673 K.
The COz formed near upper layers reacts with the cdke which reduces it into CO.
CO
2
(g) + C(s) → 2CO (g)
(ii) Middle zone (zone of slag formation) : Temperature 1123 K.
Here limestone (CaCO
3
) decomposes into CaO and CO
2
. The CaO acts as a flux which combines with silica to form fusible calcium silicate slag.
(ifi) Upper zone (zone of reduction) : Temperature 500 – 900 K.
Here ores are reduced to FeO by CuO.
Question 16.
What is meant by ‘disproportionation’? Give one example of disproportionation reaction in aqueous solutions. [3]
Answer:
Disproportionation : When in a reaction, the oxidation of an element in a compound increases in one of the products and decreases in the other product, it is said to undergo disproportionation of oxidation state. In other words, the reaction in which an element undergoes self-oxidation and self-reduction simultaneously.
Example : In acidic solution Mn (VI) in MnO changes to Mn (VII) in the product MnOj and to Mn (IV) in the product Mn0
2
.
Question 17.
Write the IUPAC name of the following :
(j) [Co(NH
3
)
6
]Cl
3
(ii) [NiCl
4
]
2-
(iii) K
3
[Fe(CN)
6
]
Answer:
(i) [Co(NH
3
)
6
]Cl
3
IUPAC name : Hexaan,n,jne cobalt (III) chloride.
(ii) [NiCl
4
]
2-
IUPAC name : TL’frnchlorido iiickelate (II) ion.
(iii) K
3
[Fe(CN)
6
]
IUI’AC name : Potassiuni lzcxaci,ano f’rrate (III).
Question 18.
Give the IUPAC names of the following compounds : [3]
Answer:
Question 19.
How are the following conversions carried out?
(i) Benzyl chloride to Benzyl alcohol
(ii) Ethyl magnesium chloride to Propan-1-ol
(iii) Propene to Propan-2-ol
Answer:
(i) Benzyl chloride to benzyl alcohol :
(ii) Ethyl magnesium chloride to propan-1-ol : When ethyl magnesium chloride is treated with methanol, an adduct is produced which gives propan-1-ol on hydrolysis.
(iii) Propene to propan-2-ol
Question 20.
Write the major product in the following equations :
Answer:
Question 21.
Define the following as related to proteins : [3]
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation
Answer:
(i) Peptide linkage : It is an amide linkage formed between – COOH group of one a-amino acid and NH
2
, group of another a-amino acid by loss of a molecule of water. The – CO – NH – bond formed is called Peptide linkage.
(ii) Primary structure : Proteins may have one or more polypeptide chains. Each polypeptide in a protein molecule has amino acids linked with each other in a specific sequence which is known as primary structure of protein.
(iii) Denaturation : Due to coagulation of globular protein under the influence of change in temperature, change in pH etc., the native shape of the protein is destroyed and biological activity is lost and the formed protein is called denaturated protein and the phenomenon is denaturation.
Question 22.
Explain the term ‘copolymerization’ and give tzvo examples of copolymerization. [3]
Answer:
When two or more different monomers are allowed to polymerize together, the product formed is called a copolymer and the process is called copolymerization. Co-polymerization is a polymerization reaction in which a mixture of more than one monomeric species is allowed to polymerize and form a copolymer. e.g. Buna- S, Buna-N.
Equations :
Question 23.
Neeraj went to the 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. He decided to buy sucrolose which was good for his grandfather’s health.
(i) Name another sugar free tablet which Neeraj did not purchase.
(ii) Was it right to purchase such medicines without doctor’s prescription?
(iii) What quality of Neeraj is reflected above? [4]
Answer:
(i) Neeraj did not purchase Aspartame and Saccharin.
(ii) It was not right to purchase and take any medicine without doctor’s prescription.
(iii) As per latest CBSE Curriculum, Value Based Questions will not be asked in the examination.
Question 24.
Calculate ∆
r
.G° and e.m.f. (E) that can be-obtained from the following cell under the standard conditions at 25°C :
Zn (s) | Zn
2+
(aq) || Sn
2+
(aq) | Sn (s)
Given: \(\mathrm{E}_{\mathrm{Zn}^{2+/ Z n}}^{\mathrm{o}}\) = -0.76 V; \(\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\mathrm{o}}\) = -0.14 V and F = 96500 C mol
-1
Solution:
Given: \(\mathrm{E}_{\mathrm{Zn}^{2+/ Z n}}^{\mathrm{o}}\) = -0.76 V; \(\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\mathrm{o}}\) = -0.14 V
F = 96500 C mol
-1
, ∆
r
.G° = ?, E = ?
Or
(a) Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
(b) Calculate the standard cell potential of the galvanic cell in which the following reaction takes place :
Fe
2+
(aq) + Ag
+
(aq) → Fe
3+
+ Ag (s)
Calculate the ∆
r
.G° and equilibrium constant of the reaction also.
\(\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\mathrm{o}}\) = 0.80 V; \(\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\mathbf{o}}\) = 0.77 V
Answer:
(i) Conductivity : The conductance of the solution of an electrolyte enclosed in a cell between two electrodes of unit area of cross section separated by 1 cm. It is represented as K with unit ohm
-1
cm
-1
.
Molar conductivity : It is the conductance of the volume V of solution containing one mole of electrolyte kept between two electrodes with area of cross section A and distance of unit length.
Molar conductivity increases with decrease in concentration of solute for both weak and strong electrolytes.
(b) Given : \(\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\mathrm{o}}\) = 0.80 V; \(\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\mathbf{o}}\) = 0.77 V
∆
r
.G° = ?, K
C
= ?
Cell Reaction
Question 25.
(a) Elements of Gr. 16 generally show lower value of first ionization enthalpy compared to the corresponding periods of Gr. 15. Why?
(b) What happens when
(i) Concentrated H
2
SO
4
is added to CaF
2
?
(ii) Sulphur dioxide reacts with chlorine in the presence of charcoal?
(iii) Ammonium chloride is treated with Ca(OH)
2
? [5]
Answer:
(a) Elements of group 16, i.e., oxygen family have general electronic configuration of ns
2
np
4
while elements of group 15, i.e., nitrogen family have general electronic configuration of ns
2
np
3
which is a relatively stable half-filled configuration with high exchange energy and therefore require more ionization energy to release electrons from this stable configuration.
(b)
Or
(a) Draw the structure of the following
(i) BrF3 (ii) XeO3
(b) Answer the following :
(i) Why is NH3 more basic than PH3?
(ii) Why are halogens strong oxidising agents?
(iii) Draw the structure of XeOF,.
Answer:
(a)
(b) (i) Both P and N contains lone pairs of electrons but due to small size and high electronegativity of nitrogen in NH
3
, the electron density is much higher than PH
3
, therefore it can easily donate electrons and acts as a strong Lewis base than PH
3
.
(ii) Halogens have strong tendency to accept an electron due to high negative electron gain enthalpies. Hence halogens act as strong oxidising agents.
(iii) XeOF4 : See Q. 16 (b), 2017 Comptt. (I Delhi). [Page 196
Question 26.
(a) Draw the structure of the following :
(i) p-Methylbenzaldehyde
(ii) 4-Methylpent-3-en-2-one
(b) Give chemical tests to distinguish between the following pairs of compounds :
(i) Benzoic acid and Ethyl benzoate.
(ii) Benzaldehyde and Acetophenone.
(iii) Phenol and Benzoic acid. [5]
Answer:
(a)
(b) (i) Distinction between Benzoic acid and Ethyl benzoate :
By Sodium bicarbonate test : Acids react with NaHCO
3
to produce brisk effervescence due to the evolution of CO
2
gas. Benzoic acid being an acid responds to this test, but ethyl benzoate does not.
(ii) Benzaldehyde and acetophenone :
By Iodoform test : Acetophenone being a methyl ketone on treatment with I2 and NaOH (NaOI) undergoes iodoform test to give yellow ppt. of iodoform on heating benzaldehyde does not.
(iii) Phenol and Benzoic acid : On addition of NaHC03 to both solutions carbon dioxide gas is evolved with benzoic acid while phenol does not form CO
2
.
Or
(a) Draw the structures of the following derivatives :
(i) Propanone oxime
(ii) Semicarbazone of CH
3
CHO
(b) How will you convert ethanal into the following compounds? Give the chemical equations involved.
(i) CH
3
– CH
3
(ii)
(iii) CH
3
CH
2
OH [5]
Answer:
(a)
(b)
