CBSE Previous Year Question Papers Class 12 Chemistry 2019 Delhi

CBSE Previous Year Question Papers Class 12 Chemistry 2019 Delhi

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.

CBSE Previous Year Question Papers Class 12 Chemistry 2019 Delhi Set I

Section – A

Question 1.
Out of NaCl and AgCl, which one shows Frenkel defect and why? [1]

Question 2.
Arrange the following in increasing order of boiling points: [1]
(CH ₃ ) ₃ N, C ₂ H ₅ OH, C ₂ H ₅ NH ₂
Answer:
Increasing order of boiling point would be:
(CH ₃ ) ₃ N < C ₂ H ₅ NH ₂ < C ₂ H ₅ OH

Question 3.
Why are medicines more effective in the colloidal state? [1]
OR
What is the difference between an emulsion and a gel?
Answer:
Medicines are effective in the colloidal state because in the colloidal state they have a large surface area and are easily assimilated in the body.
OR
Emulsions are the colloids made up of liquids dispersed in liquid dispersion medium whereas gels are liquids dispersed in solid dispersion medium.
For example, milk is an emulsion whereas butter is a gel.

Question 4.
Define ambidient nucleophile with an example. [1]
Answer:
Ambidient nucleophiles are the ones which can attack through two different atoms present in the same nucleophilic species, such as NO _2- . This can attack the electrophilic center either through nitrogen or the oxygen.

Question 5.
What is the basic structural difference between glucose and fructose? [1]
OR
Write the products obtained after hydrolysis of lactose.
Answer:
Both glucose and fructose have the molecular formula C ₆ H ₁₂ O ₆ but Glucose has an aldehydic functional group at C-1 (in its open-chain structure) and Fructose has a ketonic functional group at C-2. Glucose is an aldohexose whereas fructose is a ketohexose.

Section – B

Question 6.
Write balanced chemical equations for the following processes:

1. XeF ₂ undergoes hydrolysis.
2. MnO ₂ is heated with conc. HCl. [2]

OR
Arrange the following in order of property indicated for each set:

1. H ₂ O, H ₂ S, H ₂ Se, H ₂ Te – increasing acidic character
2. HF, HCl, HBr, HI – decreasing bond enthalpy

Answer:

1. XeF ₂ undergoes hydrolysis to give Xe, HF and O ₂
   2XeF ₂ (s) + 2H ₂ O (l) → 2Xe (g) + 4HF (aq) + O ₂ (g)
2. MnO ₂ is heated with conc. HCl to give MnCl ₂ , Cl ₂ and H ₂ O
   MnO ₂ + 4HCl → MnCl ₂ + Cl ₂ + 2H ₂ O

OR

1. Increasing order of acidity- H ₂ O < H ₂ S < H ₂ Se < H ₂ Te
2. Decreasing bond enthalpy- HF > HCl > HBr > HI

Question 7.
State Raoult’s law for a solution containing volatile components. Write two characteristics of the solution which obeys Raoult’s law at all concentrations. [2]
Answer:
Raoult’s law states that for any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.
The solutions which obey Raoult’s law at all concentrations are known as ideal solutions. The two important properties of ideal solutions are as follows:

1. Enthalpy of mixing of the pure components to form the solution is zero
   ΔH _mix = 0
2. The volume of mixing of the pure components to form the solution is zero
   ΔV _mix = 0

Question 8.
For a reaction 2H ₂ O ₂

the proposed mechanism is as given below:

1. H ₂ O ₂ + I ^– → H ₂ O + IO ^– (slow)
2. H ₂ O ₂ + IO ^– → H ₂ O + I ^– + O ₂ (fast)

(i) Write the rate law for the reaction.
(ii) Write the overall order of a reaction.
(iii) Out of steps (1) and (2), which one is rate-determining step? [2]
Answer:
(i) Rate law of the reaction is given by:
Rate = -d [H ₂ O]/dt = k [H ₂ O ₂ ][I ^– ]
(ii) As the rate law shows that the reaction is first order with H ₂ O ₂ and I ^– both, hence the overall order of the reaction becomes 1 + 1 = 2.
(iii) Step no. 1 is the slowest among two steps, hence this is the rate-determining step.

Question 9.
When MnO ₂ is fused with KOH in the presence of KNO ₃ as an oxidizing agent, it gives a dark green compound (A). Compound (A) disproportionates in an acidic solution to give a purple compound (B). An alkaline solution of compound (B) oxidises KI to compound (C) whereas an acidified solution of compound (B) oxidises KI to (D). Identify (A), (B), (C) and (D). [2]
Answer:
The reaction sequence can be written as-

Question 10.
Write IUPAC name of the complex [Pt(en) ₂ Cl ₂ ]. Draw structures of geometrical isomers for this complex. [2]
OR
Using IUPAC norms write the formulae for the following:
(i) Hexaamminecobalt (III) sulphate
(ii) Potassium tri oxalato chromate (III)
Answer:
Dichlorido bis(ethane-1, 2-diamine platinum(II)

OR
(i) Hexaamminecobalt(III) sulphate
[C0(NH ₃ ) ₆ ] ₂ (SO ₄ ) ₃
(ii) Potassium trioxalatochromate(III)
K ₃ [Cr(C ₂ O ₄ ) ₃ ]

Question 11.
Out of [CoF ₆ ] ^3- and [Co(en) ₃ ] ³⁺ , which one complex is:

1. paramagnetic
2. more stable
3. inner orbital complex and
4. high spin complex
   (Atomic no. of Co = 27) [2]

Answer:

1. [CoF ₆ ] ^3- is paramagnetic as it has four unpaired electrons.
2. [Co(en) ₃ ] ³⁺ is more stable.
3. [Co(en) ₃ ] ³⁺ is inner orbital complex.
4. [CoF ₆ ] ^3- is high spin complex.

Question 12.
Write structures of compound A and B in each of the following reactions: [2]

Answer:

Section – C

Question 13.
The decomposition of NH ₃ on a platinum surface is the surface is a zero-order reaction. If the rate constant (k) is 4 × 10 ^-3 Ms ^-1 , how long will it take to reduce the initial concentration of NH ₃ from 0.1 M to 0.064 M. [3]
Answer:
Rate law for a zero-order reaction can be given as k = [R] ⁰ – [R] / t
Substituting the given values-
4 × 10 ^-3 Ms ^-1 = (0.1 – 0.064) M / t
t = 0.036 M/4 × 10 ^-3 Ms ^-1 = 9 s
So, the time taken would be 9 seconds.

Question 14.
(i) What is the role of activated charcoal in a gas mask?
(ii) A colloidal sol is prepared by the given method in the figure. What is the charge on hydrated ferric oxide colloidal particles formed in the test tube? How is the sol represented? [3]

(iii) How does chemisorption vary with temperature?
Answer:
(i) Activated charcoal present in gas mask adsorbs the harmful suffocating gases which surround the user and protects them from the particulate matter and ashes present in coal mines.

(ii) In the given method, FeCl ₃ is being added to the NaOH solution to give a sol of hydrated ferric oxide, Fe ₂ O ₃ .xH ₂ O. The sol formed preferentially adsorbs the negatively charged OH ^– present in the solution and hence acquire a negative charge on sol particles. The negatively charged sol formed can be represented as Fe ₂ D ₃ . xH ₂ O/OH ^– (negatively charged)

(iii) Chemisorption increases with temperature up to a certain extent as then after it starts decreasing, chemisorption occurs due to a chemical reaction between adsorbate and adsorbent, and chemical reactions increase with the increase in temperature.

Question 15.
An element crystallizes in fee lattice and with a cell edge of 300 pm. The density of the element is 10.8 g cm-3. Calculate the number of atoms in 108 g of the element. [3]

Question 16.
A 4% solution (w/w) of sucrose (M = 342 g mol ^-1 ) in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 g mol ^-1 ) in water. (Given: Freezing point of pure water = 273.15 K) [3]
Answer:
Depression in freezing point can be shown by the given formula-

So, the freezing point of 5% solution of glucose is 268.35 K.

Question 17.
(a) Name the method of refining which is
(i) used to obtain semiconductor of high purity.
(ii) used to obtain low boiling metal.
(b) Write chemical reactions taking place in the extraction of copper from Cu ₂ S. [3]
Answer:
(a) (i) Zone refining is being used for the production of semiconductors like germanium, silicon, boron, gallium etc. of high purity.
(ii) Distillation method is used to obtain low boiling metals such as zinc and mercury.

(b) Chemical reactions involved in the extraction of Cu from Cu ₂ S are:
Cu ₂ S ore is heated in a reverberatory furnace after mixing with silica. In the furnace, iron oxide slags of as iron silicate and copper are produced in the form of copper matte. This contains Cu ₂ S and FeS.
The reactions undergo as given below.

• 2FeS + 3O ₂ → 2FeO + 2SO ₂
• FeO + SiO ₂ → FeSiO ₃
• 2Cu ₂ S + 3O ₂ → 2Cu ₂ O + 2SO ₂
• 2Cu ₂ O + Cu ₂ S → 6Cu + SO ₂

Question 18.
Give reasons for the following:

1. Transition elements and their compounds act as catalysts.
2. E ⁰ value for (Mn ²⁺ / Mn) is negative whereas for (Cu ²⁺ / Cu) is positive.
3. Actinoids show irregularities in their electronic configuration. [3]

Answer:

1. Transition elements and their compounds act as a catalyst due to their ability to adopt multiple oxidation states and to form complexes.
2. Mn ²⁺ has d ⁵ configuration (stable half-filled configuration) and it prefers to stay in +2 oxidation state, so, E ⁰ value for Mn ²⁺ / Mn is negative. Whereas the high energy to transform Cu(s) to Cu ²⁺ (aq) is not balanced by its hydration enthalpy, hence Cu ²⁺ is not favoured over Cu ⁰ state and it has positive E ⁰ value.
3. Actinoids show irregularities in their oxidation states due to extra stability of empty, half-filled and fully filled/subshells.

Question 19.
Write the structures of monomers used for getting the following polymers:
(i) Nylon-6,6
(ii) Glyptal
(iii) Buna-S [3]
OR

(ii) Write the monomers of the following polymer:

(iii) What is the role of Sulphur in the vulcanization of rubber?
Answer:
(i) Monomers of Nylon-6,6 are adipic acid and hexamethylenediamine.

(ii) Monomers of glyptal are ethylene glycol and phthalic acid.

(iii) Monomers of Buna-S are Butadiene and Styrene.

OR
(i) -[CH ₂ -CH ₂ (CH ₃ )]- is a homopolymer and the monomer from which it is obtained is CH ₂ = CH(CH ₃ ).
(ii) The given polymer is melamine polymer and the monomers are Melamine and Formaldehyde

(iii) Natural rubber becomes soft at high temperatures. In the vulcanisation process, sulphur forms cross-links at the reactive sites of double bonds in rubber monomers and thus the rubber gets stiffened.

Question 20.

1. What type of drug is used in sleeping pills?
2. What type of detergents are used in toothpaste?
3. Why the use of alitame as an artificial sweetener is not recommended? [3]

OR
Define the following terms with a suitable example in each:

1. Broad-spectrum antibiotics
2. Disinfectants
3. Cationic detergents

Answer:

1. Tranquillizers are used in sleeping pills.
2. Anionic detergents are used in toothpaste.
3. It is difficult to control the sweetness of food by using alitame hence its use is not recommended.

OR

1. Broad-spectrum antibiotics: Antibiotics which kill or inhibit a wide range of Gram-positive and Gram-negative bacteria are said to be broad-spectrum antibiotics. Ampicillin is an example.
2. Disinfectants: These are chemicals which are applied to inanimate objects such as floors, drainage system, instruments etc. An example is a 1% solution of phenol.
3. Cationic detergents: These are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions. An example is Cetyl trimethyl ammonium bromide.

Question 21.
(i) Out of (CH ₃ ) ₃ C-Br and (CH ₃ ) ₃ C-I, which one is more reactive towards SNI and why?
(ii) Write the product formed when p-nitro-chlorobenzene is heated with aqueous NaOH at 443 K followed by acidification.
(iii) Why dextro and laevorotatory isomers of Butan-2-ol are difficult to separate by fractional distillation? [3]
Answer:
(i) (CH ₃ ) ₃ C-I is more reactive than (CH ₃ ) ₃ C—Br towards SNI reaction because, C-I bond being weaker than C-Br bond (due to larger size of I- compared to Br-) forms the tertiary carbocation easily.

(ii) 4-Nitrophenol is formed as a product when p-nitro chlorobenzene is heated with NaOH at 443 K and acidified later.

(iii) Dextro- and laevorotatory isomers of Butan-2-oI are stereoisomers of each other and have the same physical properties. As they have the same boiling points, thus it is difficult to isolate them through fractional distillation.

Question 22.
An aromatic compound ‘A’ on heating with Br ₂ and KOH forms a compound ‘B’ of molecular formula C ₆ H ₇ N which on reacting with CHCl ₃ and alcoholic KOH produces a foul-smelling compound ‘C’. Write the structures and IUPAC names of compounds
A, B and C. [3]
Answer:
As the chemical combination Br ₂ and KOH are used for Hofmann Bromamide reaction where an amide is reduced to amine, the compound B, C ₆ H ₇ N seems to be Aniline. The reaction can be suggested as follows:

IUPAC names are as follows:
Compound A – Benzamide
Compound B – Aniline or Benzenamine
Compound C – Isocyanobenzene or phenyl- isocyanide.

Question 23.
Complete the following reactions: [3]

Write chemical equations for the following reactions:
(i) Propanone is treated with dilute Ba(OH) ₂ .
(ii) Acetophenone is treated with Zn(Hg) / conc. HCl.
(iii) Benzoyl chloride is hydrogenated in presence of Pd/BaSO ₄ .
Answer:

Question 24.
Differentiate between the following:
(i) Amylose and Amylopectin
(ii) Peptide linkage and Glycosidic linkage
(iii) Fibrous proteins and Globular proteins. [3]
OR
Write chemical reactions to show that an open structure of D-glucose contains the following:
(i) Straight chain
(ii) Five alcohol groups
(iii) Aldehyde as a carbonyl group.
Answer:
(i) Difference between Amylose and Amylopectin

S.No.	Amylose	Amylopectin
1.	Water-soluble.	Water-insoluble.
2.	Constitutes about 15-20% of starch.	Constitutes about 80-85% of starch.
3.	It has the un-branched chain	It has branched-chain.

(ii) Difference between Peptide linkage and Glycosidic linkage

S.No.	Peptide linkage	Glycosidic linkage
1.	It is an amide formed between -COOH group of one amine and -NH 2 group of the second amino acid molecule.	It is an oxide linkage, that is two monosaccharides are joined through an oxygen atom.
2.	It is found in protein molecules.	It is found in carbohydrate molecules.

(iii) Difference between Fibrous proteins and Globular Proteins

S.No.	Fibrous protein	Globular protein
1.	In these polypeptide chains run parallel and are held together by hydrogen and disulphide bonds.	In these chains of polypeptide coil around to give a spherical shape.
2.	Usually insoluble in water. Examples are Keratin and myosin.	Usually soluble in water. Examples are albumin and insulin.

OR
(i) Open chain structure of Glucose is straight-chain can be shown as follows
On prolonged heating with HI, it forms n-hexane, this shows that glucose has a straight open-chain structure.

(ii) On acetylation with acetic anhydride, glucose gives glucose pentaacetate, this shows that glucose has five alcoholic groups.

(iii) Glucose gets oxidized to six carbon carboxylic acid (Gluconic acid) on reaction with a mild oxidising agent like bromine water. This indicates that the carbonyl group is present as an aldehydic group.

Section – D

Question 25.
E0cell for the given redox reaction is 2.71 V
Mg(s) + Cu ²⁺ (0.01 M) → Mg ²⁺ (0.001 M) + Cu(s), Calculate Eceii for the reaction. Write the direction of flow of current when an external opposite potential applied is
(i) less than 2.71 V and
(ii) greater than 2.71 V. [5]
OR
(a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO ₄ and ZnSO ₄ until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass of Zn deposited at the cathode of cell Y.
(Molar mass : Fe = 56 g mol ^-1 , Zn = 65.3 g mol ^-1 , 1F = 96500 C mol ^-1 )
(b) In the plot of molar conductivity (∧ _m ) vs square root of concentration (c ^1/2 ) following curves are obtained for two electrolytes A and B:

Answer the following:
(i) Predict the nature of electrolytes A and B.
(ii) What happens on extrapolation of ∧ _m to concentration approaching zero for electrolytes A and B?
Answer:
The cell can be represented as:


(i) When the external opposite potential is less than 2.71 V then electron flows from Mg rod to Cu rod hence current flows from Cu to Mg.
(ii) When the external opposite potential is greater than 2.71 V then electron flows from Cu rod to Mg rod and current flows from Mg to Cu.
OR
(a) Charge = Current × Time
Now, in the given experiment, 2.8 g of iron was deposited or \(\frac { 2.8 }{ 5.6 }\) = 0.05 moles of iron were deposited.
Now, as it is a 2 electron transfer process:
1 mol of iron is deposited by 2 × 96500 C of charge
Hence, 0.05 mol of iron will need 0.05 × 2 × 96500 C of charge = 9650 C
So, 9650 C = 2 A × t
Time = 4825 seconds = 80.41 minutes
Similarly,
So, the same amount of charge will flow to deposit Zn as well, keeping that in mind
2 × 96500 C of charge can deposit 1 mol of Zn
Hence, 9650 C of charge would deposit
= (\(\frac { 1 }{ 2 }\) × 96500 C) × 9650 = 0.05 mol
Weight of Zinc deposited = 0.05 mol × 65.3 g mol ^-1 = 3.26 g

(b) (i) The electrolyte A is a strong electrolyte, and the electrolyte B is a weak electrolyte.
(ii) On extrapolation, for electrolyte, A limiting value of conductance is obtained that is conductance at zero concentration or infinite dilution.

The curve obtained for a strong electrolyte shows that there is a small decrease in molar conductivity with increase in concentration. In other words, the molar conductivity is increased only slightly on dilution (for observing dilution effects, go towards zero on X-axis). A strong electrolyte is completely dissociated in solution and thus, furnishes all ions for conductance. However, at higher concentrations, the dissociated ions are close to each other and thus, the inter-ionic attractions are greater. These forces retard the motion of the ions and thus, conductivity is low. With a decrease in concentration (dilution), the ions move away from each other thereby feeling less attractive forces from the counterions. This results in an increase in molar conductivity with dilution. The molar conductivity approaches a maximum limiting value at infinite dilution designated as ∧ _m ⁰ .

For electrolyte B:
The curve obtained for B shows that there is a large increase in the value of molar conductivity with dilution, especially near-infinite dilution. This is because as the solution of a weak electrolyte is diluted, its ionization is increased. This results in more number of ions in solution and thus, there is an increase in molar conductivity. However, the conductance of a weak electrolyte never approaches a limiting value. Or in other words, it is not possible to find conductance at infinite dilution (zero concentration).

Question 26.
(a) How do you convert the following:
(i) Phenol to Anisole
(ii) Ethanol to Propan-2-ol
(b) Write the mechanism of the following reaction:

(c) Why phenol undergoes electrophilic substitution more easily than benzene? [5]
OR
(a) Account for the following :
(i) o-nitrophenol is more steam volatile than p-nitrophenol.
(ii) f-butyl chloride on heating with sodium methoxide gives 2-methylpropene instead of f-butyl methyl ether.
(b) Write the reaction involved in the following:
(i) Riemer Tiemann reaction
(ii) Friedel Crafts Alkylation of Phenol
(c) Give a simple chemical test to distinguish between Ethanol and Phenol.
Answer:
(a) (i) Conversion of phenol to anisole can be done as follows.

(ii) Ethanol can be converted to propane-2-ol through the following sequence:

(b) The mechanism of dehydration of ethanol to give ethene can be written as follows:

(c) The -OH group in phenol increases the electron density on the benzene ring hence electrophilic substitution reaction is more prominent in phenol compared to benzene. The lone pair of oxygen on phenolic-OH group takes part in the resonance and makes the ring electron-rich, hence activating the ring for incoming electrophiles.

OR
(a) (i) This difference arises due to the difference in the extent of association of molecules with each other. In o-nitrophenol, the nitro and hydroxyl group present in the same molecule forms a hydrogen bond (intra-molecular) and thus have least association with nearby molecules, whereas in p-nitrophenol the nitro and hydroxyl groups of adjacent molecules form hydrogen bonds (intermolecular) which result in the long-range association of molecules. Hence, a large amount of energy is required to break the intermolecular hydrogen bonds hence, high boiling point than the 2-nitrophenol molecules.


(ii) Sodium methoxide acts as a strong base and extracts a proton from one of the methyl groups of t-butyl chloride giving rise to a primary carbanion which quickly loses Cl – to give a double bond, hence the product formed is 2-methyl propene.

Question 27.
Give reasons for the following:
(i) Sulphur in vapour state shows paramagnetic behaviour.
(ii) N-N bond is weaker than the P—P bond.
(iii) Ozone is thermodynamically less stable than oxygen.
(b) Write the name of gas released when Cu is added to
(i) dilute HNO ₃ and
(ii) conc. HNO ₃ [5]
OR
(a) (i) Write the disproportionation reaction of H ₃ PO ₃ .
(ii) Draw the structure of XeF ₄ .
(b) Account for the following :
(i) Although Fluorine has less negative electron gain enthalpy yet F ₂ is a strong oxidizing agent.
(ii) Acidic character decreases from N ₂ O ₃ to Bi ₂ O ₃ in group 15.
(c) Write a chemical reaction to test sulphur dioxide gas. Write the chemical equation involved.
Answer:
(a) (i) In the vapour phase, sulphur partly exists as S ₂ molecule and S ₂ molecule like O ₂ molecule has two unpaired electrons in antibonding n orbital, hence it shows paramagnetic behaviour.
(iii) Ozone (O ₃ ) is thermodynamically less stable than dioxygen (O ₂ ) because decomposition of ozone into dioxygen results in the liberation of heat (ΔH is negative) and increase in entropy (ΔS is positive). These two effects reinforce each other.

(b) (i) When Cu is added to dil.HNO ₃ Nitrogen monoxide (NO) is released-
3Cu + 8HNO ₃ (dil.) → 3CU(NO ₃ ) ₂ + 2NO + 4H ₂ O
(ii) When Cu is added to conc. HNO ₃ Nitrogen dioxide (NO ₂ ) is released:
Cu + 4HNO ₃ (conc.) → Cu(NO ₃ ) ₂ + 2NO ₂ + 2H ₂ O
OR
(a) (ii) Structure of XeF ₄

(b) (i) Although Fluorine has less electron gain enthalpy yet it is a strong oxidising agent due to its low enthalpy of dissociation of F—F bond and high enthalpy of hydration of F- ion.

(c) Sulphur dioxide behaves as a reducing agent when moist, this property is being used for its test in the laboratory. It decolorizes the purple coloured potassium permanganate(VII) solution:
5SO ₂ + 2KMnO ₄ + 2H ₂ O → 2H ₂ SO ₄ + 2MnSO ₄ + K ₂ SO4

CBSE Previous Year Question Papers Class 12 Chemistry 2019 Delhi Set II

Note: Except for the following questions, all the remaining questions have been asked in the previous set.

Section – A

Question 2.
Arrange the following in increasing order of base strength in the gas phase: [1]
(CH ₂ H ₅ ) ₃ N, C ₂ H ₅ NH ₂ , (C ₂ H ₃ ) ₂ NH
Answer:
In the gas phase, the basicity order will be
(C ₂ H ₅ ) ₃ N > (C ₂ H ₅ ) ₂ NH > C ₂ H ₅ NH ₂

Question 3.
Why the conductivity of silicon increases on doping with phosphorus? [1]

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

Answer:

Section – B

Question 8.
Write two differences between an ideal solution and a non-ideal solution. [2]
Answer:
Differences between Ideal and non-ideal solutions

S.No.	Ideal solution	Non-ideal solution
1.	Follows Raoult’s law.	Does not follow Raoult’s law.
2.	ΔH mix = 0, ΔV mix = 0	ΔH mix ≠ 0, ΔV mix ≠ 0

Question 10.
Write IUPAC of the complex [Cr(NH ₃ ) ₄ Cl ₂ ] ⁺ . Draw structures of geometrical isomers for this complex. [2]
OR
Using IUPAC norms write the formulae for the following:
(i) Pentamminenitrito-O-cobalt(III) chloride
(ii) Potassium tetracyanidonickelate (II)
Answer:
The IUPAC name for [Cr(NH ₃ ) ₄ Cl ₂ ] ⁺ is tetraammine dichlorochromium(III):

OR
(i) Pentamminenitritocobalt(III)chloride: [Co(NH ₃ ) ₅ ONO]Cl ₂
(ii) Potassium tetracyanidonickelate (II): K ₂ [Ni(CN) ₄ ]

Question 11.
Out of [CoFe ₆ ] ^3- and [Co(C ₂ O ₄ )] ^3- , which one complex is:

1. diamagnetic
2. more stable
3. outer orbital complex and
4. low spin complex?
   (Atomic no. of (Co = 27) [2]

Answer:
[Co(C ₂ O ₄ ) ₃ ] ^3- has d ² sp ³ hybridisation (low spin complex) and [CoF ₆ ] ^3- has sp ³ d ² hybridisation (high spin complex).

1. [Co(C ₂ O ₄ ) ₃ ] ^3- is diamagnetic
2. [Co(C ₂ O ₄ ) ₃ ] ^3- is more stable.
3. [CoF ₆ ] ^3- is outer orbital complex
4. [Co(C ₂ O ₄ ) ₃ ] ^3- is low spin complex.

Section – B

Question 17.
(i) Write the role of ‘CO’ in the purification of nickel.
(ii) What is the role of silica in the extraction of copper?
(iii) What type of metals are generally extracted by electrolytic method? [3]
Answer:
(i) Role of CO in the purification of Nickel: Mond process is the technique used to purify nickel. The impure nickel reacts with carbon monoxide at 50-60° C to form the gas nickel carbonyl, leaving the impurities as solids. Nickel carbonyl is subjected to a higher temperature so that it is decomposed giving the pure metal.

(ii) Role of silica in the extraction of Copper: The role of silica in the metallurgy of copper is to remove the iron oxide obtained during the process of roasting. If the sulphide ore of copper contains iron, then silica (SiO ₂ ) is added as flux before roasting.
FeO + SiO ₂ → FeSiO ₃ (slag)

(iii) Metals whose ions easily get reduced in solution or molten state are generally extracted by the electrolytic method.

Question 18.
Give reasons for the following:
(i) Transition metals form alloys.
(ii) Mn ₂ O ₃ is basic whereas Mn ₂ O ₇ is acidic.
(iii) EU ²⁺ is a strong reducing agent. [3]
Answer:
(i) Transition metal forms alloys because the atomic sizes of transition metals are very similar to each other. As the atomic sizes are very similar, one metal can replace the other metal from its lattice and form a solid solution. This solid solution is known as an alloy.

(ii) In Mn ₂ O ₃ manganese has +3 oxidation state, it has lone pairs of electrons which can be donated (Lewis base), hence it is basic in nature. Whereas in Mn ₂ O ₇ manganese has +7 oxidation state. Higher oxidation states are short of electrons, meaning they can accept electrons and thus function as Lewis acids. So, Mn ₂ O ₇ is acidic.

(iii) A reducing agent is that which can reduce other species and itself gets oxidized. Eu ²⁺ readily changes to the common +3 oxidation state shown by the lanthanides by losing one more electron. So, Eu ²⁺ is regarded as a strong reducing agent.

Question 20.
(i) Why bithional is added in soap?
(ii) Why magnesium hydroxide is a better antacid than sodium bicarbonate?
(ii) Why soaps are biodegradable whereas detergents are non-biodegradable? [3]
OR
Define the following terms with a suitable example in each:
(i) Antibiotics
(ii) Artificial sweeteners
(iii) Analgesics
Answer:
(i) Bithional is added in soap to impart antiseptic properties to soap.

(ii) Sodium bicarbonate if taken in excess can make the stomach alkaline, in turn, stimulating more acid release, hence magnesium hydroxide is better antacid than sodium bicarbonate because being insoluble, it does not increase the pH above neutrality.

(iii) Soaps are sodium or potassium salts of long-chain fatty acids whereas the hydrocarbon portion of synthetic detergents contain highly branched hydrocarbon chain which is not easy for the bacteria to degrade. Hence, soaps are biodegradable but synthetic detergents are not.
OR
(i) Antibiotics: They are the compounds (produced by microorganisms or synthetically) which either inhibit the growth of bacteria or kill bacteria.
Example: Penicillin.

(ii) Artificial sweeteners: They are the compounds which make the food sweet in taste without adding calories to the food.
Example: Aspartame.

(iii) Analgesics: These are the compounds which reduce or abolish pain without causing impairment of consciousness, mental confusion or any other disturbances to the central nervous system. They are of two types, Narcotic (Morphine) and non-narcotic (Example: Paracetamol).

Question 21.
Write the structures of main products when benzene diazonium chloride reacts with the following reagents.
(i) CuCN
(ii) CH ₃ CH ₂ OH
(iii) KI [3]
Answer:
The products will be as follows:

CBSE Previous Year Question Papers Class 12 Chemistry 2019 Delhi Set III

Note: Except for the following questions, all the remaining questions have been asked in previous sets.

Section – A

Question 1.
Arrange the following in decreasing order of solubility in water:
(CH ₃ ) ₃ N, (CH ₃ ) ₂ NH, CH ₃ NH ₂ [1]
Answer:
Decreasing order of solubility in water is-
CH ₃ NH ₂ > (CH ₃ ) ₂ NH > (CH ₃ ) ₃ N

Question 2.
What type of stoichiometric defect is shown by ZnS and why? [1]

Question 3.
Write one stereochemical difference between SN ₁ and SN ₂ reactions. [1]
Answer:
Products of SN ₁ reactions are usually racemic in nature, whereas Products of SN ₂ reactions have inverted configuration compared to the starting reactant.

Section – B

Question 7.
State Henry’s law and write its two applications. [2]
Answer:
Henry’s law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution.

The most commonly used form of above law can be put as The partial pressure of the gas in the vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution
p = K _H x
Applications:

1. To increase the solubility of CO ₂ in soft drinks and soda water, the sealing is done under high pressure.
2. Tanks used by scuba divers are filled with air diluted with helium gas in order to avoid an accumulation of nitrogen in bubbles in their blood. As increased pressure underwater increases the solubility of nitrogen in the blood.

Question 11.
Write the hybridization and magnetic character of the following complexes:
(a) [Fe(H ₂ O) ₆ ] ²⁺
(b) [Fe(CO) ₅ ]
(Atomic no. of Fe = 26) [2]
Answer:
(a) The hybridisation in [Fe(H ₂ O) ₆ ] ³⁺ is sp ³ d ² . As there are five unpaired electrons, it is strongly paramagnetic in nature.

(b) [Fe(CO) ₅ ] has dsp ³ hybridisation and no impaired electron, hence it is diamagnetic in nature.

Question 12.
Write structures of main compounds A and B in each of the following reactions: [2]

Answer:

Section – C

Question 17.
How will you convert the following:
(i) Impure nickel to pure nickel
(ii) Zinc blende to zinc metal
(ii) [Ag(CN) ₂ ] ^– to Ag [3]
Answer:
(i) Impure nickel can be converted to pure nickel by Mond’s process. In this process, Nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetracarbonyl

The carbonyl is subjected to a higher temperature so that it is decomposed giving the pure metal

(ii) Zinc can be obtained from zinc blende as follows:

(iii) [Ag(CN) ₂ ] ^2- can be converted into Ag by treating with zinc (a more electropositive element than silver). Hence, on treatment with zinc metal, Zn oxidizes to furnish Zn ²⁺ ions, which go into solution replacing the Ag ⁺ ions, which are in turn reduced to metallic silver and deposited in pure form.
2[Ag(CN) ₂ ] ^2- (aq) + Zn (s) → [Zn(CN) ₄ ] ^2- (aq) + 2Ag(s)

Question 18.
Give reasons for the following:
(i) The transition metals generally form coloured compounds.
(ii) E° value for (Mn ³⁺ /Mn ²⁺ ) is highly positive than that for (Cr ³⁺ /Cr ²⁺ ) couple.
(iii) The chemistry of actinoids elements is not so smooth as that of the lanthanoids. [3]
Answer:
(i) Transition metals have partly filled d-orbitals. So, the single electrons available in d-orbitals absorb energy and go to higher unoccupied electronic energy levels. When they come back, they emit energy in the visible range and hence impart colour.

(ii) E° value for Mn ³⁺ /Mn ²⁺ is highly positive because Mn ²⁺ has a stable d ⁵ configuration and it is reluctant to lose one electron to achieve the 3+ state. Cr ²⁺ has 3d ⁴ configuration and losing another electron to achieve 3d ³ configuration is not that difficult, hence the E ⁰ value is not more positive compared to Mn ³⁺ /Mn ²⁺ .

(iii) Chemistry of actinoid elements is not so smooth in view of their ability to exist in different oxidation states. Also, many of the actinoid elements are radioactive which makes the study of these elements difficult.

Question 22.
Write equations of the following reactions:
(i) Acetylation of aniline
(ii) Coupling reaction
(iii) Carbonyl amine reaction [3]
Answer:
(i) Acetylation of aniline:

(ii) Coupling reaction: Benzene diazonium chloride reacts with other suitable aromatic compounds to give azo compounds. This reaction is known as coupling reaction.

(iii) Carbylamine reaction: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines.

Question 24.
Define the following with a suitable example in each:
(i) Oligosaccharides
(ii) Denaturation of protein
(iii) Vitamins [3]
OR
Write the reactions involved when D-glucose is treated with the following reagents:
(i) Br ₂ water
(ii) H ₂ N-OH
(iii) (CH ₃ CO) ₂ O
Answer:
(i) Oligosaccharide: Carbohydrates that yield two to ten monosaccharide units on hydrolysis are known as oligosaccharides. They are further classified as disaccharides, trisaccharides, tetrasaccharides etc., depending upon the number of monosaccharides they provide upon hydrolysis. Disaccharides are the most common, e.g., maltose.

(ii) Denaturation of protein: Protein found in a biological system with a unique three-dimensional structure and biological activity is called a native protein. When a native protein is subjected to physical change like temperature, chemical or pH change, the hydrogen bonds within the protein structure are disturbed causing globules to unfold and α-helix to uncoil. The secondary and tertiary structures of a protein molecule are destroyed, in turn, losing their biological activity, this is known as denaturation of the protein. An example is coagulation of egg white on boiling.

(iii) Vitamins: These are the organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth and health of organisms.
Example – Vitamins A, B, C, D etc.
OR
(i) The reaction of glucose with Br ₂ water (a mild oxidising agent) gives gluconic acid, this reaction indicates that a carbonyl group of glucose is an aldehydic group:

(ii) The reaction of glucose with H ₂ N—OH (hydroxylamine) gives oxime, this reaction confirms that there is a carbonyl group present in glucose

(iii) The reaction of glucose with (CH ₃ CO) ₂ O (acetic anhydride) gives glucose pentaacetate, this reaction confirms that there are five hydroxyl groups present in glucose:

CBSE Previous Year Question Papers