Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2017 (Outside Delhi) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2017 Outside Delhi with Solutions
Time allowed : 3 hours
Maximum marks : 70
General Instructions
(j) All questions are compulsory.
(ii) Questions number 1 to 5 are very short-answer questions and carry 1 mark each.
(iii) Questions number 6 to 10 are short-answer questions and carry 2 marks each.
(iv) Questions number 11 to 22 are also short-answer questions and carry 3 marks each.
(v) Question number 23 is a value based question and carries 4 marks.[*]
(vi) Questions number 24 to 26 are long-answer questions and carry 5 marks each.
(vii) Use log tables, if necessary, Use of calculators is not allowed.
Set
Question 1.
Write the formula of the compound of phosphorus which is obtained when cone. HNO
3
oxidises P
4
. [1]
Answer:
Question 2.
Write the IUPAC name of the following compound: [1]
Answer:
2-Bromo-3-methylbut-2-enol-l-ol
Question 3.
What is the effect of adding a catalyst on
(a) The activation energy (E
a
), and
(b) Gibbs energy (ΔG) of a reaction? [1]
Answer:
On adding a catalyst
(a) Activation energy of the reaction decreases.
(b) Gibbs energy doesn’t change.
Question 4.
Out of
and
, which is an example of allylic halide? [1]
Answer:
is an allylic halide.
Question 5.
What type of colloid is formed when a liquid is dispersed in a solid? Give an example. [1]
Answer:
When a liquid is dispersed in solid, ‘gel’ colloid is formed. Examples Jelly, butter, cheese, curd, etc.
Question 6.
(a) Arrange the following compounds in the increasing order of their acid strength: [2]
p-cresol, p-nitrophenol, phenol
(b) Write the mechanism (using curved arrow notation) of the following reaction:
Answer:
(a) Order of acidic strength:
p-cresol < phenol < p-nitrophenol
(b) Mechanism of acid catalysed hydration of alkene:
Step 1: Protonation of alkene to form carbocation by electrophilic attack of H
3
O
+
.
Step 2: Nucleophilic attack of water on carbocation
Step 3: Deprotonation to form an alcohol
OR
Write the structures of the products when Butan-2-ol reacts with the following: 1+1=2
(a) CrO
3
(b) SOCl
2
Answer:
Question 7.
Calculate the number of unit cells in 8.1g of aluminum if it crystallizes in a face-centered cubic (f.c.c.) structure. (Atomic mass of Al = 27 g mol
-1
) [2]
Answer:
Ans. 1 mole of Aluminium = 27 g = 6.022 × 10
23
Hence, No. of atoms present in 27 g of A1 = \(\frac{6.022 \times 10^{23}}{27}\)
As f.c.c. unit cell contains 4 atoms
∴ No. of f.c.c. unit cells present = \(\frac{6.022 \times 10^{23} \times 8.1}{27 \times 4}\)
= 0.45165 × 10 23 = 4.5165 × 10 22
Question 8.
Draw the structures of the following: 1+1=2
(a) H
2
SO
3
(b) HClO
3
Answer:
(a) Structures of H
2
SO
3
(Sulphurous acid)
(b) Structures of HClO
3
(Chloric acid)
Question 9.
Write the name of the cell which is generally used in hearing aids. Write the reactions taking place at the anode and the cathode of this cell. [2]
Answer:
Mercury cells are used in hearing aids.
Reaction at anode:
Zn (Hg) + 20H
–
→ ZnO (s) + H
2
0 + 2e
–
Reaction at cathode:
M
+
→ M + e
–
HgO + H
2
0 + 2e
–
→ Hg (l) + 20H
–
Question 10.
Using IUPAC norms write the formulae for the following:
(i) Sodium dicyanidoaurate (I)
(ii) Tetraamminechloridonitrito-N-platinum (IV) sulfate [2]
Answer:
(i) Na [Au (CN)
2
]
(ii) [Pt(NH
3
)
4
Cl(NO
2
)] (SO
4
)
Question 11.
(a) Based on the nature of intermolecular forces, classify the following solids: Silicon carbide, Argon
(b) ZnO turns yellow on heating. Why?
(c) What is meant by groups 12-16 compounds? Give an example. [3]
Answer:
(a) Silicon carbide is a covalent or network solid while Argon is a non-polar molecular solid.
(b) ZnO shows metal excess defect due to presence of extra cations, i.e., Zn
2+
ions in interstitial sites which on heating changes into yellow due to loss of oxygen.
(c) Group 12-16 compounds are imperfect covalent compounds in which the ionic character depends on the electronegativities of the two elements, e.g., ZnS, CdS, etc.
Question 12.
(a) The cell in which the following reaction occurs:
2Fe
3+
(aq) + 2I
–
(aq) → 2Fe
2+
(aq) + I
2
(s)
has E°
cell
= 0.236 V at 298 K. Calculate the standard Gibbs energy of the cell reaction.
(Given: 1F = 96,500 C mol
-1
)
(b) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours?
(Given: 1 F = 96,500 C mol
-1
) [3]
Answer:
(a) 2Fe
3+
(aq) + 2I
–
(aq) → 2Fe
2+
(aq) + I
2
(s)
For the given reaction, n = 2, E° = 0.236 V
Using formula
ΔG° = -nFE
cell
= -2 × 96500 C mol
-1
× 0.236
∴ ΔG° = -45.548 kJ/mol
(b) Given:
I = 0.5 A
t = 2 hrs. = 2 × 60 × 60s = 7,200 s
Q = I × t = 0.5 × 7200 = 3600 C
96,500 C eletricity flows to produce = 6.022 × 10
23
electrons
∴ 1 C electricity flows to produce = \(\frac{6.022 \times 10^{23}}{96,500}\)
∴ 3600 C electricity flows to produce = \(\frac{6.022 \times 10^{23}}{96,500}\) × 3,600 = 22.46 × 10 21 electrons
Question 13.
(a) What type of isomerism is shown by the complex [Co(NH
3
)
5
(SCN)]
2+
?
(b) Why is [NiCl
4
]
2-
paramagnetic while [Ni(CN)
4
]
2-
is diamagnetic?
(Atomic number of Ni = 28)
(c) Why are low spin tetrahedral complexes rarely observed? [3]
Answer:
(a) [Co(NH
3
)
5
(SCN)]
2+
shows linkage isomerism.
(b) Since in [NiCl 4 ] 2- , Cl is a weak field ligand, it forms outer orbital complex and has unpaired electrons which imparts paramagnetic character to complex while in [Ni(CN) 4 ] 2- , CN – is a strong field ligand, forms inner orbital complex and has paired electrons which imparts diamagnetic character to it.
(c) Low spin tetrahedral complexes are rarely observed because orbital splitting energies for tetrahedral complexes are not sufficiently large for forcing pairing.
Question 14.
Write one difference in each of the following:
(a) Multimolecular colloid and Associated colloid
(b) Coagulation and Peptization
(c) Homogeneous catalysis and Heterogeneous catalysis [3]
Answer:
(a) Multimolecular colloid and Associated colloid. Multimolecular colloids are formed by the aggregation of a large number of atoms or molecules which generally have diameters less than 1 nm, e.g., sols of gold, etc. while Associated colloids are formed by the aggregation of a large number of ions in concentrated solutions, e.g., micelles in soap.
(b) Coagulation and Peptization. Coagulation is a process of aggregating together the colloidal particles into large sized particles to form their precipitate while peptization is a process of converting fresh precipitate into colloidal particles by shaking it with the dispersion medium in the presence of a small amount of a suitable electrolyte.
(c) Homogeneous catalysis and Heterogeneous catalysis. Homogeneous catalysis is the phenomenon of changing the rate of reaction when catalyst has same phase as the reactants while in heterogeneous catalysis, the catalyst has different phase than that of the reactants.
OR
(a) Write the dispersed phase and dispersion medium of milk.
(b) Write one similarity between physisorption and chemisorption.
(c) Write the chemical method by which Fe(OH)
3
sol is prepared from FeCl
3
. 1×3=3
Answer:
(a) In milk dispersed phase is liquid fat and dispersion medium is water.
(b) Both physisorption and chemisorption increase with increase in pressure.
(c) Fe(OH)sub>3 is prepared from FeCl by hydrolysis.
Question 15.
A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed.
Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021) [3]
Answer:
Question 16.
The following compounds are given to you:
2-Bromopentane, 2-Bromo-2-methyl butane, 1-Bromopentane
(i) Write the compound which is most reactive towards S
N
2 reaction.
(ii) Write the compound which is optically active.
(iii) Write the compound which is most reactive towards ß-elimination reaction. [3]
Answer:
(i) 1-Bromo pentane CH
3
CH
2
CH
2
CH
2
Br is most reactive towards S
N
2 reaction.
(ii) 2-Bromo pentane
is optically active.
(iii) 2-Bromo-2-methylbutane
is most reactive towards ß-elimination reaction.
Question 17.
Write the principle of the following:
(a) Zone refining
(b) Froth floatation process
(c) Chromatography [3]
Answer:
(i) 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.
(b) Froth floatation process. This method is used for removing gangue from sulphide ores. In this powdered ore is mixed with collectors (e.gpine oils, fatty acids, etc.) and froth stabilisers (e.g., cresols, aniline) which enhance non-wettability of the mineral particles and froth stabilisation respectively. As a result of which ore comes with froth and gangue remains in the solution.
(c) Chromatography:
This is the method used for the separation and purification of elements. It can also be used for testing the purity of a compound. The principle behind the chromatography is that different components of a mixture are differently adsorbed on an adsorbent.
Question 18.
Write the structures of compounds A, B and C in the following reactions:
Answer:
Question 19.
Write the structures of the monomers used for getting the following polymers:
(a) Nylon-6,6
(b) Melamine-formaldehyde polymer
(c) Buna-S [3]
Answer:
(b) Monomers of melamine-formaldehyde polymer are melamine and formaldehyde.
Img 21
(c) Buna-S
Question 20.
Define the following: [3]
(a) Anionic detergents
(b) Limited spectrum antibiotics
(c) Antiseptics
Answer:
(a) Anionic detergents. Those detergents in which large part of their molecules are anions and used in cleansing action, are called anionic detergents.
Example : Sodium alkyl suiphates
These are obtained from long straight chain alcohols containing 12-48 carbon atoms by treatment with conc. H2S04 followed by neutralization with NaOH.
Example: Sodium lauryl sulphate.
(b) Limited spectrum antibiotics are effective against a single organism or disease, eg., Streptomycin.
(c) Antiseptics are the chemicals which either kill or prevent growth of microbes on living tissues, eg., Penicillin.
Question 21.
Give reasons for the following:
(a) Red phosphorus is less reactive than white phosphorus.
(b) Electron gain enthalpies of halogens are largely negative.
(c) N
2
O
5
is more acidic than N
2
O
3
. [3]
Answer:
(a) Red phosphorus is less reactive than white phosphorus because white phosphorus possess angle strain where long angles are only 60° making it more reactive. Also, red phosphorus being polymeric is less reactive than white phosphorus which has discrete tetrahedral structure.
(b) Electron gain enthalpies of halogens are largely negative due to high effective nuclear charge and smaller size among period. They readily accept an electron to attain noble gas configuration.
(c) N2O 5 is more acidic than N 2 O 3 because higher the oxidation state, higher will be acidic character. N 2 O 5 has +5 oxidation state and N 2 O 3 has +3 oxidation state.
Question 22.
Give reasons for the following:
(a) Acetylation of aniline reduces its activation effect.
(b) CH
3
NH
2
is more basic than C
6
H
5
NH
2
.
(c) Although-NH
2
is olp directing group, yet aniline on nitration gives a significant amount of m-nitroaniline. [3]
Answer:
(i) Acetylation of aniline reduces its activation effect because acetyl group being electron withdrawing group attracts the lone pair of electrons of the N-atom towards carboxyl group and the lone pair of electrons on N is less available for donation to benzene ring by resonance.
(ii) CH 3 NH 2 is more basic than aniline due to availabilitÿ of lone pair of electrons for donation while in aniline lone pair of electrons on the nitrogen atom is delocalised over benzene ring and thus unavailable for donation.
(iii) Because of nitration in an acidic medium, aniline gets protonated to give anilinium ion which is rn-directing. . .
Question 23.
After watching a program on TV about the presence of carcinogens (cancer causing agents) Potassium bromate and potassium iodate in bread and other bakery products, Rupali a Class XII student decided to make others aware about the adverse effects of these carcinogens in foods. She consulted the school principal and requested him to instruct the canteen contractor to stop selling sandwiches, pizzas, burgers and other bakery products to the students. The principal took immediate action and instructed the canteen contractor to replace the bakery products with some protein and vitamin rich food like fruits, salads, sprouts, etc. The decision was welcomed by the parents and the students.
After reading the above passage, answer the following questions:
(a) What are the values (at least two) displayed by Rupali?
(b) Which polysaccharide component of carbohydrates is commonly present in bread?
(c) Write the two types of secondary structures of proteins.
(d) Give two examples of water soluble vitamins. [4]
Answer:
*(a) As per latest CBSE curriculum, Value Based Questions will not be asked in the examination.
(b) Starch is present in bread.
(c) ct-helix and (3-pleated sheet structures
(d) Vitamin-B and Vitamin-C are water soluble.
Question 24.
(a) Account for the following:
(i) Transition metals show variable oxidation states.
(ii) Zn, Cd, and Hg are soft metals.
(iii) E
0
value for the Mn
3+
/Mn
2+
couple is highly positive (+1.57 V) as compared to Cr
3+
/Cr
2+
.
(b) Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements. [5]
Answer:
(i) The variability of oxidation state of transition elements is due to incompletely filled d-orbitals. In transitional elements ns, and (n – 1) d electrons have a approximate equal energies hence in addition to ns electrons, (n – 1) d electrons are also taking part in chemical bonding.
(ii) Zn, Cd and Hg are soft metals because they do not exhibit covalency due to completely filled d-orbitals. Absence of unpaired d electrons causes weak metallic bonding.
(iii) Mn 2+ exists in half-filled d 5 state which is very stable while Mn 3+ is d 4 which is not so stable. Mn 3+ can be easily reduced to Mn 2+ Conversion from d 4 to d 5 will be quick and have negative ∆G value. Hence, because of the stability factor the E° value is high for this process. While Cr 3+ is d 3 is half-filled (t 2g 3) is stable in nature and Cr 2+ is d 4 , has one extra electron which it would like to donate to attain the stable half-filled (t 2g 3) configuration. Hence for the process Cr 3+ to Cr 2+ , the value of E° is less.
OR
(a) Following are the transition metal ions of 3d series:
Ti
4+
, V
2+
, Mn
3+
, Cr
3+
(Atomic numbers: Ti = 22, V = 23, Mn = 25, Cr = 24)
Answer the following:
(i) Which ion is most stable in an aqueous solution and why?
(ii) Which ion is a strong oxidising agent and why?
(iii) Which ion is colourless and why?
(b) Complete the following equation:
(i) 2MnO
4
+ 16H
+
+ 5S
2-
→
(ii) KMnO
4
→
Answer:
(i) Cr
3+
is most stable because of its small size and t
3
2g
configuration.
(ii) Mn
3+
is a strong oxidising agent because after gaining one electron it is converted into Mn
2+
which has stable d
5
configuration.
(iii) Ti
4+
is colourless due to d° configuration, i.e., no unpaired electrons.
(b) (i) 2MnO 4 – + 16H + + 5S 2- → 2Mn 2+ + 8H 2 O + 5S
(ii) 2KMnO 4 → K 2 MnO 4 + MnO 2 + O 2
Question 25.
(a) A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15 K.
Given:
The molar mass of sucrose = 342g mol
-1
The molar mass of glucose = 180 g mol
-1
(b) Define the following terms:
(i) Molality (m)
(ii) Abnormal molar mass [5]
Answer:
Molar mass of sucrose = C
12
H
22
O
11
= 12 × 12 + 22 + 11 × 16 = 342
10% solution (by mass) of sucrose in water means 10 g of sucrose is present in (100 – 10) = 90 g of water
10% solution of sucrose means, w = 10 g
Mass of water W = 90 g
(b) (i) Molality (m). Number of moles of solute dissolved per kg of the solvent.
(ii) Abnormal molar mass. If the molar mass calculated by using any of the colligative properties comes to be different than theoretically expected molar mass.
OR
(a) 30 g of urea (M = 60g mol
-1
) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg.
(b) Write two differences between ideal solutions and non-ideal solutions.
Answer:
(a) Given:
(b) (i) Molality: It is the number of moles of the solute dissolved per 1000 g of the solvent. It is denoted by m.
(ii) Abnormal molar mass: Those solute that dissociates or associate in solution, show an abnormal molar mass in solution, for example, Molar mass of ethanoic acid is greater than normal molar mass.
The molar mass of KCl in solution is reduced than normal molar mass.
KCl → K
+
+ Cl
–
OR
(a) W
B
= 30 g
M
B
= 60 g mol
-1
W
A
= 846 g
M
A
= 18 g mol
-1
P
0
= 23.8mm Hg
P
s
= x
Relative lowering of vapour pressure
So, the vapour pressure of water for this solution = 23.597 mm Hg
(b)
| Ideal Solutions | Non-ideal Solutions |
| (i) They obey Raoult’s law over the entire range of concentration. | (i) They do not obey Raoult’s law over the entire range of concentration. |
| (ii) Neither the heat is evolved or absorbed during dissolution. | (ii) Heat is evolved or absorbed during dissolution. |
| (iii) Δ mix H = 0, Δ mix V = 0 | (iii) Δ mix H is not equal to 0., Δ mix V is not equal to 0 |
Question 26.
(a) Write the product(s) in the following reactions:
(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Butanal and Butan-2-one
(ii) Benzoic acid and Phenol
Answer:
(a)
(b) (i) Tollett’s reagent test. Add ammoniacal solution of sliver nitrate (Tollen’s Reagent) in both the solutions. Butanal gives silver mirror whereas Butan-2-one does not. Therefore Butanal gives Tollen’s test.
(ii) Ferric chloride test. Add neutral FeCl3 in both the solutions, phenol reacts with neutral FeCl3 to form an iron-phenol complex giving violet colour but benzoic acid does not.
OR
(a) Write the reactions involved in the following:
(i) Etard reaction
(ii) Stephen reduction
(b) How will you convert the following in not more than two steps:
(i) Benzoic acid to Benzaldehyde
(ii) Acetophenone to Benzoic acid
(iii) Ethanoic acid to 2-Hydroxyethanoic acid
Answer:
(i) Etard reaction.
(b) (i) Butanal and Butan-2-one
Set II
Note: Except for the following questions, all the remaining questions have been asked in previous sets.
Question 2.
What type of colloid is formed when a solid is dispersed in a liquid? Give an example. [1]
Answer:
Sol is formed when a solid is dispersed in a liquid, e.g., paints.
Question 3.
Write the IUPAC name of the following compound. [1]
Answer:
3-phenylprop-2-en-1-ol