CBSE Previous Year Question Papers Class 12 Chemistry 2018

CBSE Previous Year Question Papers Class 12 Chemistry 2018

Time allowed: 3 hours
Maximum Marks: 70

General Instructions

• All questions are compulsory.
• Section A: Questions number 1 to 5 are very short answer questions and carry 1 mark each.
• Section B: Questions number 6 to 12 are short answer questions and carry 2 marks each.
• Section C: Questions number 13 to 24 are also short answer questions and carry 3 marks each.
• Section D: Questions number 25 to 27 are long answer questions and carry 5 marks each.
• There is no overall choice. However, an internal choice has been provided in two questions of one mark, two questions of two marks, four questions of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions
• Use of log tables, if necessary. Use of calculators is not allowed.

Question 1.
The analysis shows that FeO has a non-stoichiometric composition with formula Fe_0.95O. Give reason. [1]

Question 2.
CO(g) and H₂(g) react to give different products in the presence of different catalysts. Which ability of the catalyst is shown by these reactions? [1]
Answer:
CO(g) and H₂(g) react in the presence of different catalysts to give different products, this shows that the action of a catalyst is highly selective in nature.

Question 3.
Write the coordination number and oxidation state of Platinum in the complex [Pt(en)₂Cl₂]. [1]
Answer:
Coordination number: 6;
Oxidation state: +2

Question 4.
Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why? [1]
Answer:
Benzyl chloride would be easily hydrolysed compared to chlorobenzene. In the given reaction condition, hydrolysis proceeds by nucleophilic substitution mechanism and the benzylic carbonium ion formed after losing the leaving group (-Cl) is better stabilized (through resonating structures) hence reacts easily.

Question 5.
Write the IUPAC name of the following: [1]

Answer:
The IUPAC name would be 3, 3- Dimethyl-pentane-2-ol.

Question 6.
Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol^-1) in 250 g of water.
(K_f of water = 1.86 K kg mol^-1) [2]
Answer:
Molality (m) of a given solution of Glucose:
m = [(60/180) g mol^-1/250 g] × 1000 = 1.33 mol kg^-1
Now, depression in freezing point is given by, ΔT_f = K_fm
Putting the given values,
ΔT_f = K_fm = 1.86 × 1.33 = 2.5
So, the freezing point of the solution would be = 273.15 K – 2.5 K = 270.65 K.

Question 7.
For the reaction [2]
2N₂O₅(g) → 4NO₂(g) + O₂(g)
the rate of formation of NO₂(g) is 2.8 × 10^-3 Ms^-1. Calculate the rate of disappearance of N₂O₅(g).
Answer:
Rate of reaction for the given reaction can be given as,
Rate = 1/2 {-Δ[N₂O₅]/Δt} or {-Δ[N₂O₅]/Δt} = 1/2 {[NO₂/ Δt]}
So, the rate of disappearance of N₂O₅ would be half of the rate of production of NO₂ (given 2.8 × 10^-3 Ms^-1).
So, the rate of disappearance of N₂O₅ is 1.4 × 10^-3 Ms^-1.

Question 8.
Among the hydrides of Group-15 elements, which have the [2]
(a) lowest boiling point?
(b) maximum basic character?
(c) highest bond angle?
(d) maximum reducing character?

Question 9.
How do you convert the following? [2]
(a) Ethanal to Propanone
(b) Toluene to Benzoic acid
OR
Account for the following:
(a) Aromatic carboxylic acids do not undergo Friedel- Crafts reaction.
(b) pK_a value of 4-nitrobenzoic acid is lower than that of benzoic acid.
Answer:
(a) Conversion of ethanol to Propanone:


OR
(a) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction because the carboxyl group is deactivating for electrophilic substitution reaction, secondarily, the catalyst aluminium chloride gets bonded to the carboxyl group.

(b) pK_a value of 4-Nitrobenzoic acid is lower than benzoic acid, which means 4-Nitrobenzoic acid is more acidic than the benzoic acid. Being an electron-withdrawing group, the -NO₂ group withdraws electrons towards itself resulting in ease of carboxylic proton release, hence increasing the acidity.

Question 10.
Complete and balance the following chemical equations: [2]
(a) Fe²⁺ + MnO^–₄ + H⁺ →
(b) MnO^–₄ + H₂O + I^– →
Answer:
(a) 5Fe²⁺ + MnO^–₄ + 8H⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
(b) 2MnO^–₄ + H₂O + I^– → 2MnO₂ + 2OH^– + IO^–₃

Question 11.
Give reasons for the following: [3]
(a) Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers.
(b) Aquatic animals are more comfortable in cold water than in warm water.
(c) Elevation of the boiling point of 1M KCl solution is nearly double than that of 1M sugar solution.
Answer:
(a) Molar masses of macromolecules like polymers and proteins are measured through osmotic pressure method. The osmotic pressure method uses ‘molarity’ of solution (instead of molality) which has a large magnitude even for dilute solutions, given that polymers have poor solubility, osmotic pressure measurement is used for determination of their molar masses. Macromolecules such as proteins are not stable at high temperatures and because measurement of osmotic pressure is done at around room temperature, it is useful for determination of molar masses of proteins.

(b) The solubility of gases in liquids decreases on increasing the temperature. Hence, the availability of dissolved oxygen in water is more at lower temperatures hence, the aquatic animals feel more comfortable at lower temperatures than at the higher temperatures.

(c) Elevation of boiling point is a colligative property and hence depends on the number of solute particles in the solution. Now, 1 M KCl would have twice the number of solute particles, as KCl dissociates into K⁺ and Cl^–, compared to sugar solution (as sugar does not undergo any dissociation). So, the elevation of boiling point is nearly double for 1M KCl solution compared to 1M sugar solution.

Question 12.
An element ‘X’ (At. mass = 40 g mol^-1) having f.c.c. the structure has a unit cell edge length of 400 pm. Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’. (NA = 6.022 × 10²³ mol^-1) [3]

Question 13.
A first-order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. (Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK^-1 mol^-1) [3]
Answer:
Rate constant for a first-order reaction is given by,

Question 14.
What happens when [3]
(a) a freshly prepared precipitate of Fe(OH)₃ is shaken with a small amount of FeCl₃ solution?
(b) persistent dialysis of a colloidal solution is carried out?
(c) an emulsion is centrifuged?
Answer:
(a) When FeCl₃ is added to a freshly prepared precipitate of Fe(OH)₃, a positively charged sol of hydrated ferric oxide is formed due to adsorption of Fe³⁺ ions.

(b) When persistent dialysis of the colloidal solution is carried out, traces of electrolytes present in the sol are removed almost completely leaving the colloids unstable and finally, coagulation takes place.

(c) Emulsions are centrifuged to separate them into constituent liquids.

Question 15.
Write the chemical reactions involved in the process of extraction of Gold. Explain the role of dilute NaCN and Zn in this process. [3]
Answer:
Extraction of gold involves leaching the metal with a dilute solution of NaCN or KCN in the presence of air (for O₂) from which the metal is obtained later by replacement method (using Zinc).
The reactions involved are:
4Au(s) + 8CN^–(aq) + 2H₂O(aq) + O₂(g) → 4[AU(CN)₂]^– (aq) + 4OH^– (aq)
2 [AU(CN)₂]^– (aq) + Zn(s) → 2Au (s) + [Zn(CN)₄]^2- (aq)

Question 16.
Give reasons:
(a) E⁰ value for Mn³⁺/Mn²⁺ couple is much more positive than that for Fe³⁺/Fe²⁺.
(b) Iron has a higher enthalpy of atomization than that of copper.
(c) Sc³⁺ is colourless in aqueous solution whereas Ti³⁺ is coloured. [3]
Answer:
(a) Mn²⁺ has a d⁵ configuration, and the extra stability of half-filled d-orbitals is compromised when another electron is taken out to give Mn³⁺, On the contrary, Fe³⁺ attains a half-filled orbital configuration when Fe²⁺ gets oxidized to Fe³⁺. Hence, the E⁰ value for Mn³⁺/ Mn²⁺ couple has more positive E⁰ value.

(b) Fe has a 3d⁶4s² outer electronic configuration whereas Cu has 3d¹⁰4s¹ configuration. Now, more the number of impaired electrons in d-orbital, more favourable are interatomic attractions and thus higher atomization enthalpies. Hence, Fe having 4 unpaired d-electrons has more enthalpy of atomization than copper having no unpaired d-electron.

(c) Sc³⁺ has a 3d⁰ configuration whereas Ti³⁺ has a 3d¹ configuration. As there are no electrons in d orbital for Sc³⁺ ion, there is no transition of electrons by absorption of energy and hence no emission in visible range imparting colour to the Sc³⁺ ion.

Question 17.
(a) Identify the chiral molecule in the following pair: [3]

(b) Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether.
(c) Write the structure of the alkene formed by dehydrohalogenation of 1-Bromo-1 methylcyclohexane with alcoholic KOH.
Answer:
(a) The molecule (i) is a chiral molecule.

(b) Chlorobenzene reacts with methyl chloride in the presence of sodium metal and dry ether to give toluene. This reaction is known as Wurtz-Fitting reaction.

(c) In the 1-Bromo-1-methylcyclohexane, all β-hydrogen atoms are equivalent. Thus dehydrohalogenation takes place, in the reaction of this compound with KOH.

Question 18.
(A), (B) and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C₄H₈O. Isomers (A) and (C) give positive Tollen’s test whereas isomer (B) does not give Tollen’s test but gives positive Iodoform test. Isomers (A) and (B) on reduction with Zn (Hg)/conc. HCl, give the same product (D).
(a) Write the structures of (A), (B), (C) and (D).
(b) Out of (A), (B) and (C) isomers, which one is least reactive towards the addition of HCN? [3]
Answer:
(a) Compound A and C give positive Tollen’s test which indicates that they are aldehydes. Compound C gives Iodoform test which means it contains a carbonyl group with a methyl group attached to the carbonyl carbon so, with formula C₄H₈O the structure of compound would be CH₃COCH₂CH₃ (Butanone).

Now upon reduction with Zn(Hg)/conc. HCl, the corresponding alkanes are obtained, so a reduction of B gives Butane (D), so the isomer A has to be a linear chain aldehyde (Butanal), giving Butane (compound D) on reduction. So, the last isomer possible is compound C, 2-Methyl propionaldehyde. The reactions involved are shown below with the structures of compounds:

(b) Out of the three isomers A, B and C, compound B’ (Butanone) would be least reactive towards the addition of HCl as the carbonyl carbon is sterically hindered and most reactive would be compound A (Butanal) towards the addition of HCN.

Question 19.
Write the structures of the main products in the following reactions: [3]

Answer:
(i) Sodium borohydride doesn’t reduce esters, so the product would be,

Question 20.

1. Why is bithional added to soap? [3]
2. What is the tincture of iodine? Write its one use.
3. Among the following, which one acts as a food preservative?
   Aspartame, Aspirin, Sodium Benzoate, Paracetamol

Answer:

1. Bithional is added to soaps to impart antiseptic properties to soap.
2. Tincture of iodine is 2-3 per cent mixture of iodine in the alcohol-water mixture. It is used as an antiseptic.
3. Sodium benzoate is used as a food preservative.

Question 21.
Define the following with an example of each: [3]
(a) Polysaccharides
(b) Denatured protein
(c) Essential amino acids
OR
(a) Write the product when D-glucose reacts with conc. HNO₃.
(b) Amino acids show amphoteric behaviour. Why?
(c) Write one difference between α-helix and β-pleated structures of proteins.
Answer:
(a) Polysaccharides: Polysaccharides are food storage materials and most commonly found carbohydrates in nature. These are the compound which is formed of a large number of monosaccharide units joined together by glycosidic linkages. Example. Starch, main storage polysaccharide of plants.

(b) Denatured protein: Proteins have a unique three-dimensional structure in their native form. If the native form of protein is subjected to any physical change (such as temperature change) or any chemical change (such as a change in pH), the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled due to which protein loses its biological activity. This is called denaturation of the protein. During denaturation 2° and 3° structures of proteins are destroyed but 1° structure remains intact. Coagulation of egg white is an example of denaturation of the protein.

(c) Essential amino acids: The amino acids which are not synthesized in our body and have to be obtained through diet are known as essential amino acids. Example: Tryptophan
OR
(a) D-Glucose gets oxidized to give saccharic acid, a dicarboxylic acid on reacting with nitric acid.

(b) Amino acids show amphoteric behaviour due to the presence of both acidic (carboxylic group) and basic (amino group) in the same molecule. So, in the basic medium, the carboxyl group can lose a proton and in acidic medium, the amino group can accept a proton.

(c) In α-helix structure the polypeptide chain forms all possible hydrogen bonds by twisting into a right-handed screw (helix) with the -NH group of each amino acid residue gets hydrogen-bonded to the -C = O of an adjacent turn of the helix (Intra. molecular bonding), whereas in β-structure all peptide chains are stretched out to nearly maximum extension and then laid side by side which are held together by intermolecular hydrogen bonds (intermolecular bonding).

Question 22.
(a) Write the formula of the following coordination compound: Iron (III) hexacyanoferrate (II)
(b) What type of isomerism is exhibited by the complex [Co(NH₃)₅ Cl]SO₄?
(c) Write the hybridisation and number of unpaired electrons in the complex [CoF₆]^3-.
(Atomic number of Co = 27) [3]
Answer:
(a) The molecular formula of Iron(III) α-cyanoferrate(II) is Fe₄[Fe(CN)₆]₃
(b) [CO(NH₃)₅Cl]SO₄ will show Ionisation isomerism and the possible isomers are [CO(NH₃)₅Cl]SO₄ and [Co (NH₃)₅SO₄]Cl
(c) Electronic configuration of Co³⁺ ion is,

Electronic configuration of sp³d² hybridized (as F is a weak field ligand) orbitals of Co³⁺, with six pairs of electrons from six F ions.

There are 4 impaired electrons in [CoF₆]₃.

Question 23.
Shyam went to a grocery shop to purchase some food items. The shopkeeper packed all the items in polythene bags and gave them to Shyam. But Shyam refused to accept the polythene bags and asked the shopkeeper to pack the items in paper bags. He informed the shopkeeper about the heavy penalty imposed by the government for using polythene bags. The shopkeeper promised that he would use paper bags in future in place of polythene bags. [4]
Answer the following:
(a) Write the values (at least two) shown by Shyam.
(b) Write one structural difference between low-density polythene and high-density polythene.
(c) Why did Shyam refuse to accept the items in polythene bags?
(d) What is a biodegradable polymer? Give an example.
Answer:
(b) Low-density polythene has a branched-chain structure, whereas the high-density polythene has a linear chain structure.
(c) Shyam refused to take the items in polythene bags as polythene is non-biodegradable neither recyclable,
(d) Biodegradable polymers contain functional groups similar to functional groups present in biopolymers, so they get degraded in the environment by certain microorganisms and thus are environment-friendly.
For example Poly β -hydroxybutyrate-co-β-hydroxy valerate (PHBV).

Question 24.
(a) Give reasons: [5]
(i) H₃PO₃ undergoes disproportionation reaction but H₃PO₄ does not.
(ii) When Cl₂ reacts with an excess of F₂, ClF₃ is formed and not FCl₃.
(iii) Dioxygen is a gas while Sulphur is a solid at room temperature.
(b) Draw the structures of the following:
(i) XeF₄
(ii) HClO₃
OR
(a) When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas intensified when copper turnings were added to this test tube. On cooling, the gas (A) changed into a colourless solid (B).
(i) Identify (A) and (B).
(ii) Write the structures of (A) and (B).
(iii) Why does gas (A) change to solid on cooling?
(b) Arrange the following in the decreasing order of their reducing character: HF, HCl, HBr, HI
(c) Complete the following reaction:
XeF₄ + SbF₅ →
Answer:
(a) (i) In H₃PO₃ (orthophosphoric acid) oxidation state of phosphorus is +3 and it contains one P-H bond in addition to P = O and P-OH bonds. These type of oxoacids tend to undergo disproportionation to give orthophosphoric acid (P has +5 state) and phosphine (P has +3 state). Whereas in H₃PO₄ (orthophosphoric acid), Phosphorus is in +5 state hence no disproportionation takes place in H₃PO₄.

(ii) When Cl₂ reacts with an excess of F₂, ClF₃ is formed-and not FCl₃ because Fluorine can’t expand its valency and can show only -1 oxidation state, whereas Cl can expand its valency due to the availability of d-orbitals.

(iii) Dioxygen is a gas while sulphur is a solid at room temperature this is because sulphur have S₈ molecules and these are packed to give different crystal structure, whereas dioxygen is a diatomic molecule (O₂) and it does not have enough intermolecular attraction and thus exists in gaseous form.

(a) (i) The brown gas A is NO₂ or nitrogen dioxide. On cooling, it dimerises to N₂O₄ and solidifies as a colourless solid.

(iii) Compound A, that is, NO₂ contains the odd number of valence electrons. It behaves as a typical odd molecule. On dimerization, it is converted to stable N₂O₄ molecule with even number of electrons (thus colourless) and have better intermolecular forces to get solidified. Thus, it changes to solid on cooling.
(b) Decreasing order of reducing character: HI > HBr > HCl > HF
(c) XeF₄ + SbF₅ → [XeF₃]⁺ + [SbF₆]^–

Question 25.
(a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K: [5]
Sn(s) | Sn²⁺ (0.004 M) || H⁺ (0.020 M) | H₂(g) (1 bar) | Pt(s)
(Given: E° Sn²⁺/Sn = – 0.14V)
(b) Give reasons:
(i) On the basis of E° values, O₂ gas should be liberated at anode but it is Cl₂ gas which is liberated in the electrolysis of aqueous NaCl.
(ii) The conductivity of CH₃COOH decreases on dilution.
OR
(a) For the reaction
2AgCl(s) + H₂(g) (1 atm) → 2Ag(s) + 2H⁺ (0.1M) + 2Cl^– (0.1M), ΔG° = -43600 J at 25°C.
Calculate the e.m.f. of the cell. [log 10^-n = -n]
(b) Define fuel cell and write its two advantages.
Answer:
(a) The half cell reactions can be written as;

The reaction at the anode with a lower value of Ecell is preferred and therefore, water should get oxidized to give O₂ but on account of overpotential of oxygen, Cl^– gets oxidized preferably, liberating Cl₂ gas.
(ii) The conductivity of CH₃COOH decreases on dilution because the number of ions per unit volume that carry the current in a solution decreases on dilution.
OR


(b) Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol etc. directly into electrical energy are called fuel cells.
Advantages of fuel cells are:

• Fuel cells produce electricity with an efficiency of about 70% compared to thermal plants whose efficiency is about 40%.
• Fuel cells are pollution-free.

Question 26.
(a) Write the reactions involved in the following: [5]
(i) Hofmann bromamide degradation reaction
(ii) Diazotisation
(iii) Gabriel phthalimide synthesis
(b) Give reasons:
(i) (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution.
(ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts.
OR
(a) Write the structures of the main products of the following reactions:

(b) Give a simple chemical test to distinguish between Aniline and N, N-dimethylaniline.
(c) Arrange the following in the increasing order of their pK_b values:
C₆H₅NH₂, C₂H₅NH₂, C₆H₅NHCH₃
Answer:
(a) (i) Hofmann bromamide degradation reaction: Acetamide can be considered for example. In this reaction, Acetamide (CH₃CONH₂) undergoes Hofmann degradation in the presence of Bromine and NaOH to give Methanamine.
CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O

(ii) Diazotisation: The conversion of primary aromatic amines into diazonium salts is known as diazotisation.

(iii) Gabriel phthalimide synthesis: This reaction is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine.

(b) (i) (CH₃)₂ NH is more basic than (CH₃)₃N in aqueous solutions because in (CH₃)₃N the lone pair of electrons on the nitrogen atom is responsible for its basicity are quite hindered by the three methyl groups, hence are less available. Due to which it is less basic as compared to (CH₃)₂NH.

(ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts because the positive charge on the nitrogen atom is stabilized by the resonance with an attached phenyl group.

(b) Aniline can be distinguished from N, N-dimethyl aniline by diazo coupling reaction. Aniline would react with benzene diazonium chloride to give a yellow dye, whereas N, N-dimethyl aniline won’t undergo this reaction.

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