CBSE Class 12 Chemistry Question Paper 2018 (Delhi & Outside Delhi) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2018 Delhi to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Chemistry Question Paper 2018 (Delhi & Outside Delhi) with Solutions

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.
  \(\text { † }\) Deleted from syllabus

\(\text { † }\)Question 1.
The analysis shows that FeO has a non-stoichiometric composition with formula Fe_0.95O. Give reason. [1]
Answer:
Because in the crystals of FeO, some of the Fe²⁺ ions are replaced by Fe³⁺ ions.

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:
It shows the selectivity of the catalyst in which a substance which acts as a catalyst in one reaction may fail to catalyse another reaction.

Question 3.
Write the coordination number and oxidation state of Platinum in the complex [Pt(en)₂Cl₂]. [1]
Answer:
Coordination no. = 6
Oxidation state of platinum = Pt⁺²

Question 4.
Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why? [1]
Answer:
Benzyl chloride is easily hydrolysed with aqueous NaOH because benzyl chloride is stabilised by resonance and can be easily ionised to form chloride ion.

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

Answer:
3, 3 Dimethylpentan-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:
Given:
w = 60 g; M = 180 g mol^-1; = 1.86 K Kg mol^-1 ; W = 250 g
Using formula ∆T_f = K_fm
∆T_f = K_f.\(\frac{w \times 1000}{\mathrm{M} \times \mathrm{W}}\)
∆T_f = \(\frac{1.86 \times 60 \times 1000}{180 \times 250}\)
∆T_f = 2.48 K
Temperature at which it will freeze = 273 – 2.48 K = 270.52 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:

\(\text { † }\)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?
Answer:
Hydrides of Group-15 elements
NH₃, PH₃, ASH₃, SbH₃, BiH₃
(a) Phosphine i.e., PH₃ has lowest boiling point.
(b) Ammonia i.e., NH₃ has maximum basic character.
(c) NH₃ has the highest bond angle.
(d) BiH₃ has the 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) Ethanal to Propanone

(b) Toluene to Benzoic acid

Or

(a) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction because the carboxyl group is deactivating and meta-directing. Moreover, the catalyst aluminium chloride (Lewis acid) gets bonded to the carboxyl group.

(b) Nitro group is an electron withdrawing group (-I effect) s anion and strengthens the acid.

Question 10.
Complete and balance the following chemical equations: [2]
(a) Fe²⁺ + MnO^–₄ + H⁺ →
(b) MnO^–₄ + H₂O + I^– →
Answer:
(a) 5Fe²⁺ + Mn\(\mathrm{O}_4^{-}\) + 8H⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
(b)

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) Because proteins and polymers are generally not stable at higher temperatures and polymers have poor solubility.
(b) Because solubility of dissolved oxygen in warm water is less than that of cold water. Therefore, aquatic species respire more comfortably in cold water than in warm water.
(c) Because KCl is an electrolyte and will dissociate completely into 2 ions while sugar solution is non electrolyte and does not dissociate.

\(\text { † }\)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]
Answer:
Given:
a = 400 pm = 400 × 10^-10 cm
z = 4(For fcc)
M = 40 g mol^-1

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:

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 a freshly prepared precipitate of Fe(OH)₃ is shaken with a small amount of FeCl₃ solution then a reddish brown coloured colloidal solution is obtained by peptization.
(b) On prolonged dialysis, the electrolyte present in colloidal solution is completely removed and the colloidal solution becomes unstable and gets coagulated.
(c) When an emulsion is centrifuged it breaks down into its 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:
NaCN is used to leach the metal present in the ore by oxidising gold to Au⁺ which combines with CN^– ions to form their respective soluble complexes.

The metal is then recovered from this complex by using more electropositive reducing agent i.e. Zinc
2[Au(CN)₂]^– + Zn → 2Au + [Zn(CN)₄]^2-
The zinc being highly reactive than gold replaces it from its complex by reduction.

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⁺ exists in half-filled d5 state which is very stable while Mn³⁺ is d⁴ which is not so stable. Mn³⁺ can be easily reduced to Mn²⁺. Conversion from d⁴ to d⁵ will be quick and have negative value. Hence, because of the stability factor the E° value is high for this process. While Cr³⁺ is d³ is half-filled (t_2g3) is stable in nature and Cr²⁺ is d⁴, has one extra electron which it would like to donate to
attain the stable half-filled (t_2g3) configuration. Hence for the process Cr³⁺ to Cr²⁺, the value of E° is less.
(b) Iron has higher enthalpy of atomization than that of copper due to strong metallic bonding in iron because of presence of higher number of unpaired electrons.
(c) Sc³⁺(3d°4s°) is colourless in aqueous solution whereas Ti³⁺ (3d¹4s°) is coloured because of d-d transistion by one unpaired electron in Ti³⁺ while in Sc³⁺ no unpaired electron is present.

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:

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) The probable compounds with molecular formula C₄H₈O are :

Since isomers (A) and (C) give positive Tollens’ test so they are aldehydes while B which does not give this test is Butanone.
Reduction of (A) and (B) with Zn(Hg)/conc. HCl (Clemmensen reduction)

(b) B is least reactive.

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

Answer:

Question 20.
(a) Why is bithional added to soap? [3]
(b) What is the tincture of iodine? Write its one use.
(c) Among the following, which one acts as a food preservative?
Aspartame, Aspirin, Sodium Benzoate, Paracetamol
Answer:
(a) Bithional is added to soaps to impart antiseptic properties and to reduce the odour produced by bacterial decomposition of organic matter on skin.
(b) Tincture of iodine is a 2-3% solution of iodine in alcohol-water mixture. It is applied to wounds as an antiseptic.
(c) Sodium Benzoate is a 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: Carbohydrates that give a large number of molecules of monosaccharides on hydrolysis. Example, cellulose.
(b) Denatured protein: 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 denatured proteins and the phenomenon is denaturation.
(c) Essential amino acids: Amino acids which the body cannot synthesize are called essential amino acids. Example, Valine, leucine etc. Therefore, they must be supplied in the diet.

Or

(a)

(b) Due to the presence of both —NH₂ and —COOH group in the same molecule, amino acids show amphoteric behaviour.
(c) α-Helix structure : The polypeptide chains are held together (stabilized) by intramolecular H-bonding.
β-Pleated sheet structure : The two neighbouring polypeptide chains are held together by intermolecular H—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) Fe₄[Fe(CN)₆]₃
(b) Ionisation isomerism is exhibited by the complex [Co(NH₃)₅Cl]SO₄.
(c) Hybridisation = sp³d²; No. of unpaired electrons = 4

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:
*(a) As per latest CBSE curriculum, Value Based Questions will not be asked in the examination.
(h) Low-density polythene (LDPE) has a molecular mass of about 20,000 and has highly branched structure while high-density polythene (HOPE) consists of linear chains of polymer molecules.
(c) Shyam refuses to accept the items in polythene bags because he knows that polythene bags are non-biodegradabIe and caus hrm to the environment and animals who eat them.
(d) Biodegradable polymers are those biopolymers which can be degraded by micro organisms during a certain period of time. They degrade by enzymatic hydrolysis and oxidation. Example, 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, P atom exists in +3 oxidation state and it can expand its octet and thus can undergo both oxidation and reduction simultaneously and form orthophosphoric acid and phosphine. While in H₃PO₄, P atom exists in +5 oxidation state which is the highest oxidation state and thus it cannot undergo disproportionation reaction.
(ii) As fluorine is a small sized atom, it cannot pack three large sized Cl atoms around it. Also fluorine cannot expand its covalency.
(iii) Due to small size and high electronegativity, oxygen forms pπ-pπ multiple bonds and thus forms diatomic. O₂ molecule are held together by weak van der Waals forces of attraction which can be easily overcome by collisions at room temperature. Therefore O₂ is gas at room temperature. On the other hand due to higher tendency for catenation and lower tendency for pπ-pπ multiple bonds. Intra-atomic S₈ has strong forces of attraction which cannot be overcome by collisions. Therefore S is solid at room temperature.
(b) (i) XeF₄

(ii) HClO₃

Or

(a) The unknown salt is a metal nitrate which produces NO₂ which is a brown gas and it intensified on adding Cu turnings.

(b) 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) Cell reaction :

(b) (i) Cl₂ gas is liberated instead of °2 during electrolysis of aqueous NaCl because Cl₂ has low discharge potential than that of O₂.
(ii) Conductivity of CH₃COOH decreases on dilution because number of H⁺ ions decreases per unit volume of a solution.

Or

(a)

(b) Fuel cells : These cells are the devices which convert the energy produced during combustion of fuels like hydrogen, methane etc. directly into electrical energy.
Advantages of fuel cells:
(i) They have good efficiency i.e., 60% – 70%.
(ii) They do not cause any pollution.

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) Hoffmann’s bromamide reaction : In this reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom. Therefore the amine so formed has one carbon atom less than that of amide.

(ii) Diazotisation

(iii) Gabriel phthalimide synthesis: It is used to prepare 1° amine but aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
Example :

(b) In (CH₃)N there is maximum steric hindrance and least solvation but in (CH₃)₂NH the solvation is more and the steric hindrance is less than in (CH₃)₃NH; although + I effect is less, since there are two methyl groups; di-methyl amine is still a stronger base than tri-methyl.
(ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts because aromatic salts can be stablised by dispersal of positive charge over the benzene ring through resonance while alkadiazonium salt readily decomposes to form carbocation and nitrogen gas.

Or

(a)

(b) Distinction between Aniline and N,N dimethylaniline by Carbylamine test.
Aniline being 1° aromatic amine gives this test while N,N-dimethylaniline cannot.

(c) Increasing order of PK_b values
C₂H₅NH₂ < Ç₆H₅NHCH₃ < C₆H₅NH₂