TS Inter 2nd Year – Chemistry Previous Paper 2022

CHEMISTRY

Paper -2

English Version

Time: 3 hours

Max.marks: 60

SECTION-A

Note:

(i) Answer ANY TEN Questions (ii) Each Question carries TWO marks (iii) All are very short answer type questions.

Question 1:

What is Schottky defect?

Answer:

A Schottky defect is a type of point defect in crystalline solids, where an equal number of cations and anions are missing from their lattice sites, creating vacancies. This maintains the overall electrical neutrality of the crystal.

Question 2:

What are f-centers?

Answer:F-centers, also known as color centers, are anion vacancies in ionic crystals that are occupied by an electron. These trapped electrons absorb light in the visible region of the electromagnetic spectrum, giving the crystal a characteristic color.

Question 4:

Define osmotic pressure.

Answer:

Osmotic pressure is the minimum pressure that must be applied to a solution to prevent the inward flow of solvent molecules across a semipermeable membrane when the solution is placed in contact with pure solvent.

Question 5:

What is an ideal solution?

Answer:

An ideal solution is a solution that obeys Raoult’s Law at all concentrations and temperatures. In an ideal solution, the interactions between solute and solvent molecules are identical to the interactions between solute-solute and solvent-solvent molecules.

Key characteristics of an ideal solution:

ΔHmix = 0 (enthalpy of mixing is zero)
ΔVmix = 0 (volume change on mixing is zero)

Question 6:

State Henry’s Law.

Answer:

Henry’s Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid at a constant temperature.

Mathematically, it can be expressed as:

S = kH * P

where:

S is the solubility of the gas in the liquid
kH is Henry’s law constant
P is the partial pressure of the gas

Question 7:

What is coagulation?

Answer:

Coagulation is the process of destabilizing a colloidal solution and causing the dispersed particles to aggregate and settle down. This can be achieved by adding electrolytes, heating, or by adding oppositely charged sols.

Question 8:

Name any two applications of colloidal solutions.

Answer:

Medicine: Colloidal solutions like milk of magnesia are used as antacids.
Industry: Colloidal solutions like paints, inks, and dyes are widely used in various industries.

Question 9:

What is inert pair effect?

Answer:

The inert pair effect is the tendency of the valence electrons of heavier elements of group 13, 14, 15, and 16 to remain paired in the s-orbital and not participate in bonding. This is due to the poor shielding of the s-electrons by the intervening d and f electrons, resulting in a higher effective nuclear charge.

Question 10:

What is tailing of mercury? How is it removed?

Answer:

Tailing of mercury: Tailing of mercury refers to the loss of mercury during its extraction and processing due to its high vapor pressure. Mercury vaporizes easily at room temperature, leading to its escape into the atmosphere.
Removal of mercury tailing:
- Condensation: The mercury vapor can be condensed by cooling the air in contact with it.
- Absorption: Mercury vapor can be absorbed by activated carbon or other adsorbents.
- Amalgamation: Mercury vapors can be collected by amalgamation with gold or silver.

Question 11:

CuSO4.5H2O is blue in colour whereas anhydrous CuSO4 is colourless. Why?

Answer:

CuSO4.5H2O (copper sulfate pentahydrate) is blue due to the presence of water molecules coordinated to the copper ion. These water molecules interact with the d-orbitals of the copper ion, causing a splitting of the d-orbitals and leading to the absorption of light in the red region of the spectrum, resulting in the blue color.

In anhydrous CuSO4, the water molecules are absent. Therefore, the d-orbitals of the copper ion are not split, and it does not absorb light in the visible region, appearing colorless.

Question 12:

Define denaturation as related to proteins.

Answer:

Denaturation is the process of unfolding or disruption of the native three-dimensional structure of a protein molecule. This can be caused by factors such as heat, extreme pH, organic solvents, and heavy metal ions.

When a protein is denatured, its secondary, tertiary, and quaternary structures are disrupted, leading to a loss of its biological activity.

Question 13:

What are essential and non-essential amino acids? Give one example for each.

Answer:

Essential amino acids: These are amino acids that the human body cannot synthesize on its own and must be obtained from the diet.
- Example: Lysine
Non-essential amino acids: These are amino acids that the human body can synthesize from other molecules.
- Example: Alanine

Question 14:

What is Tollens’ reagent? Explain its reaction with Aldehydes.

Answer:

Tollens’ reagent:

It is a chemical reagent used to distinguish between aldehydes and ketones.
It is prepared by dissolving silver nitrate (AgNO3) in dilute ammonia solution.
The active species in Tollens’ reagent is the diamminesilver(I) ion, [Ag(NH3)2]+.

Reaction with Aldehydes:

Aldehydes are oxidized to carboxylic acids by Tollens’ reagent. In this reaction, the silver ions (Ag+) are reduced to metallic silver, which forms a silver mirror on the inner surface of the test tube.

General reaction: RCHO + 2[Ag(NH3)2]+ + 3OH- → RCOO- + 2Ag↓ + 4NH3 + 2H2O

Question 15:

Write the reaction showing α-halogenation of carboxylic acid and give its name (HVZ reaction).

Answer:

The α-halogenation of carboxylic acids is known as the Hell-Volhard-Zelinsky (HVZ) reaction. In this reaction, α-hydrogen atoms of a carboxylic acid are substituted with halogen atoms (usually chlorine or bromine) in the presence of phosphorus and halogen.

Reaction:

Formation of acyl halide: RCOOH + PCl3 → RCOCl + H3PO3
Halogenation: RCOCl + Cl2 → RCHClCOCl + HCl
Hydrolysis: RCHClCOCl + H2O → RCHClCOOH + HCl

Overall reaction:

RCOOH + Cl2 + P → RCHClCOOH + POCl3 + HCl

SECTION-C

Note:

(i) Answer ANY TWO questions. (ii) Each question carries EIGHT marks. (iii) All are long answer type questions.

Question 19:

(a) State and explain Kohlrausch’s law of independent migration of ions.

(b) What is “molecularity” of a reaction? How is it different from the ‘order’ of a reaction? Name one bimolecular and one trimolecular gaseous reactions.

(8 Marks)

Answer:

(a) State and explain Kohlrausch’s law of independent migration of ions.

Kohlrausch’s Law of Independent Migration of Ions states that at infinite dilution, the molar conductivity of an electrolyte is the sum of the individual contributions of its constituent ions. In other words, the molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its cation and anion at infinite dilution.

Explanation:

At infinite dilution, the ions are so far apart that they do not interact with each other.
Therefore, the conductivity of the solution is solely due to the independent movement of the ions.
The molar conductivity of an electrolyte at infinite dilution (Λ°) can be expressed as:

Λ° = λ°+ + λ°-

where λ°+ and λ°- are the molar conductivities of the cation and anion at infinite dilution, respectively.
This law is useful for determining the molar conductivities of weak electrolytes, which cannot be measured directly at infinite dilution.

(b) What is “molecularity” of a reaction? How is it different from the ‘order’ of a reaction? Name one bimolecular and one trimolecular gaseous reactions.

Answer:

Molecularity: The molecularity of a reaction is the number of reacting molecules that collide simultaneously to form the activated complex or transition state. It is a theoretical concept and can only be determined from the balanced chemical equation.
Order of a reaction: The order of a reaction is an experimentally determined quantity that indicates how the rate of reaction depends on the concentration of the reactants. It is determined from the rate law of the reaction.
Differences:
- Molecularity is a theoretical concept based on the balanced chemical equation, while the order of a reaction is an experimentally determined quantity.
- Molecularity can only be a whole number (1, 2, 3, etc.), while the order of a reaction can be a whole number, a fraction, or even zero.
Examples:
- Bimolecular reaction:
  - H2(g) + I2(g) → 2HI(g)
- Trimolecular reaction:
  - 2NO(g) + O2(g) → 2NO2(g)

Question 20:

How is ozone prepared from oxygen? Explain its reaction with

(a) C2H4

(b) KI

(c) Hg

(d) PbS

(8 Marks)

Answer:

Preparation of Ozone:

Ozone (O3) can be prepared from oxygen (O2) by passing a silent electric discharge through dry oxygen gas. This process is known as the Siemens-Bosch ozonizer process.

Reactions of Ozone:

(a) With C2H4 (Ethylene):

Ozone reacts with ethylene to form ethylene ozonide, which upon hydrolysis gives formaldehyde and formic acid.

C2H4 + O3 → C2H4O3 (Ethylene ozonide)

C2H4O3 + H2O → 2HCHO + H2O2

(b) With KI:

Ozone oxidizes potassium iodide (KI) to iodine (I2) and potassium hydroxide (KOH).

2KI + O3 + H2O → I2 + 2KOH + O2

(c) With Hg:

Ozone reacts with mercury (Hg) to form mercuric oxide (HgO).

2Hg + 2O3 → 2HgO + 2O2

(d) With PbS:

Ozone oxidizes lead sulfide (PbS) to lead sulfate (PbSO4).

PbS + 4O3 → PbSO4 + 4O2

Question 21:

Describe the following:

(i) Acetylation

(ii) Cannizzaro reaction

(iii) Cross aldol condensation

(iv) Decarboxylation

(8 Marks)

Answer:

(i) Acetylation:

Acetylation is the process of introducing an acetyl group (CH3CO-) into a molecule. It is commonly used to modify organic compounds, especially alcohols and amines.

Reagent: Acetic anhydride or acetyl chloride is commonly used as the acetylating agent.
Example: Acetylation of an alcohol (e.g., ethanol) to form ethyl acetate:

CH3CH2OH + CH3COCl → CH3COOCH2CH3 + HCl

(ii) Cannizzaro reaction:

The Cannizzaro reaction is a disproportionation reaction of aldehydes that do not have an alpha-hydrogen atom. In this reaction, one molecule of the aldehyde is reduced to the corresponding alcohol, while another molecule is oxidized to the corresponding carboxylic acid.

Example:

Reaction of benzaldehyde:

2C6H5CHO + NaOH → C6H5CH2OH + C6H5COONa

(iii) Cross aldol condensation:

Cross aldol condensation is a reaction between two different aldehydes or ketones to form a β-hydroxy carbonyl compound.

Example:

Reaction between acetaldehyde and benzaldehyde:

CH3CHO + C6H5CHO → CH3CH(OH)CH2C6H5

(iv) Decarboxylation:

Decarboxylation is the process of removing a carboxyl group (-COOH) from a carboxylic acid, often resulting in the formation of an alkane or alkene.

Example:

Decarboxylation of sodium acetate:

CH3COONa (heated) → CH4 + Na2CO3

Question 20:

How are XeF2 and XeF4 prepared? Give their structures.

Answer:

XeF2 Preparation:

XeF2 is prepared by fluorinating xenon with excess of fluorine gas at high temperatures (300°C) and low pressures.

Xe + F2 (excess) → XeF2

XeF4 Preparation:

XeF4 is prepared by reacting xenon with an excess of fluorine gas at higher temperatures (400°C) and higher pressures.

Xe + 2F2 → XeF4

Structures:

XeF2:
- Shape: Linear
- Hybridization: sp3d
- Lone pairs on Xe: 3
XeF4:
- Shape: Square planar
- Hybridization: sp3d2
- Lone pairs on Xe: 2

Question 21:

How is chlorine obtained in the laboratory? How does it react with

(a) cold dil. NaOH

(b) excess NH3

Answer:

Laboratory Preparation of Chlorine:

In the laboratory, chlorine gas is prepared by the oxidation of concentrated hydrochloric acid with a strong oxidizing agent like manganese dioxide (MnO2).

Reaction:

MnO2 + 4HCl → MnCl2 + 2H2O + Cl2

Reactions of Chlorine:

(a) with cold dilute NaOH:

Chlorine reacts with cold dilute NaOH to form sodium chloride (NaCl), sodium hypochlorite (NaOCl), and water. This reaction is known as the disproportionation reaction of chlorine.

Cl2 + 2NaOH → NaCl + NaOCl + H2O

(b) with excess NH3:

Chlorine reacts with excess ammonia to form nitrogen gas, ammonium chloride, and nitrogen trichloride.

3Cl2 + 8NH3 → N2 + 6NH4Cl + NCl3

Question 22:

How does SO2 react with the following:

(a) Na2SO3(aq)

(b) Cl2

(c) Fe+3 ions

(d) KMnO4

Answer:

(a) SO2 with Na2SO3(aq):

SO2 reacts with sodium sulfite (Na2SO3) in aqueous solution to form sodium thiosulfate (Na2S2O3).

SO2 + Na2SO3 + H2O → 2NaHSO3

Note: The reaction may proceed further under specific conditions to form sodium dithionite (Na2S2O4).

(b) SO2 with Cl2:

SO2 reacts with chlorine gas (Cl2) in the presence of water to form sulfuric acid (H2SO4).

SO2 + Cl2 + 2H2O → H2SO4 + 2HCl

(c) SO2 with Fe+3 ions:

SO2 reduces ferric ions (Fe+3) to ferrous ions (Fe+2) in acidic medium.

2Fe+3 + SO2 + 2H2O → 2Fe+2 + SO42- + 4H+

(d) SO2 with KMnO4:

SO2 decolorizes potassium permanganate (KMnO4) solution due to its reducing properties.

5SO2 + 2KMnO4 + 2H2O → K2SO4 + 2MnSO4 + 2H2SO4

Question 23:

Write the characteristic properties of transition elements.

Answer:

Variable Oxidation States: Transition elements exhibit a wide range of oxidation states due to the availability of (n-1)d and ns electrons for bonding.
Formation of Colored Compounds: Many transition metal compounds are colored due to the presence of unpaired electrons in their d-orbitals, which can absorb visible light and undergo d-d transitions.
Catalytic Properties: Many transition metals and their compounds exhibit catalytic activity due to their ability to form intermediate complexes.
Formation of Complex Compounds: Transition metals have a strong tendency to form complex compounds with ligands due to their small size, high charge density, and availability of vacant d-orbitals.
Magnetic Properties: Transition metals and their compounds exhibit paramagnetic or ferromagnetic properties due to the presence of unpaired electrons.
Interstitial Compounds: Many transition metals form interstitial compounds by trapping small atoms like hydrogen, carbon, and nitrogen within their crystal lattices.

Question 24:

Explain Werner’s theory of coordination compounds with suitable examples.

Answer:

Werner’s theory of coordination compounds proposes the following:

Primary Valence: This refers to the oxidation state of the central metal ion.
Secondary Valence: This refers to the coordination number of the central metal ion, which is the number of ligands directly bonded to it.
Coordination Sphere: The central metal ion and the ligands directly bonded to it are enclosed within a coordination sphere, represented by square brackets [ ].

Example:

In the complex [Co(NH3)6]Cl3, the primary valence of cobalt is +3, and the secondary valence (coordination number) is 6. The coordination sphere is [Co(NH3)6]3+.

Question 25:

Using IUPAC norms write the formulas for the following:

(i) Tetrahydroxozincate(II)

(ii) Hexamminecobalt (III) sulphate

(iii) Potassium tetrachloropalladate(II)

(iv) Potassium tri(oxalato)chromate(III)

Answer:

(i) Tetrahydroxozincate(II): [Zn(OH)4]2-

(ii) Hexamminecobalt (III) sulphate: [Co(NH3)6]2(SO4)3

(iii) Potassium tetrachloropalladate(II): K2[PdCl4]

(iv) Potassium tri(oxalato)chromate(III): K3[Cr(C2O4)3]

Question 26:

What are nucleic acids? Mention their two important functions.

Answer:

Nucleic acids are biopolymers that store and transmit genetic information. They are composed of nucleotide monomers, which are made up of a nitrogenous base, a pentose sugar, and a phosphate group.

Two important functions of nucleic acids:

Storage of genetic information: DNA (deoxyribonucleic acid) stores the genetic information of an organism in the form of a double-stranded helix.
Protein synthesis: RNA (ribonucleic acid) plays a crucial role in protein synthesis. mRNA (messenger RNA) carries the genetic information from DNA to ribosomes, where tRNA (transfer RNA) brings the appropriate amino acids to form the protein chain.

Question 27:

Explain the mechanism of Nucleophilic bimolecular substitution (SN2) reaction with one example.

Answer:

The SN2 (Substitution Nucleophilic Bimolecular) reaction is a type of substitution reaction in which a nucleophile attacks the substrate from the backside, leading to the simultaneous bond breaking and bond formation.

Mechanism:

Backside attack: The nucleophile attacks the substrate from the opposite side of the leaving group, leading to a transition state where the¹ nucleophile, the substrate carbon, and the leaving group are in a linear arrangement.
Inversion of configuration: The attack of the nucleophile from the backside leads to an inversion of configuration at the reaction center. This is known as Walden inversion.

Example:

The reaction of bromoethane with hydroxide ion to form ethanol:

CH3CH2Br + OH- → CH3CH2OH + Br-

Question 28:

With a suitable example write equations for the following:

(i) Reimer-Tiemann reaction.

(ii) Williamsons ether synthesis.

Answer:

(i) Reimer-Tiemann reaction:

This reaction involves the ortho-formylation of phenols using chloroform in the presence of a strong base like NaOH.

Example:

Reaction of phenol with chloroform and NaOH to form salicylaldehyde:

C6H5OH + CHCl3 + 3NaOH → C6H4(OH)CHO + 3NaCl + 2H2O

(ii) Williamsons ether synthesis:

Williamson ether synthesis is a method for the preparation of ethers by the reaction of an alkoxide ion with an alkyl halide.

Example:

Reaction of sodium ethoxide with bromoethane:

CH3CH2ONa + CH3CH2Br → CH3CH2OCH2CH3 + NaBr

Question 29:

Accomplish the following conversions:

(i) Benzoic acid to benzamide

(ii) Aniline to p-bromoaniline

Answer:

(i) Benzoic acid to benzamide

Reagents: Ammonia (NH3) and heat

C6H5COOH + NH3 → C6H5CONH2 + H2O

(ii) Aniline to p-bromoaniline

Reagents: Bromine (Br2) in the presence of iron (Fe)

C6H5NH2 + Br2 (Fe catalyst) → p-BrC6H4NH2 + HBr

SECTION – C

Note:

(i) Answer any TWO questions.

(ii) Each question carries EIGHT marks.

(iii) All are long answer type questions.

Question 30:

(a) State Raoult’s Law.

Answer:

Raoult’s Law states that the vapor pressure of a solvent in a solution is directly proportional to its mole fraction.

Mathematically:

Psolution = Xsolvent * P°solvent

where:

Psolution is the vapor pressure of the solution

Xsolvent is the mole fraction of the solvent
P°solvent is the vapor pressure of the pure solvent

(b) Define mole fraction. Calculate the mole fraction of H2SO4 in a solution containing 98% (w/w) H2SO4 by mass.

Answer:

Mole fraction: The mole fraction of a component in a solution is the ratio of the number of moles of that component to the total number of moles of all components in the solution.
Calculation:
- Assume 100 g of solution.
- Mass of H2SO4 = 98 g
- Mass of water = 100 g – 98 g = 2 g
- Molar mass of H2SO4 = 98 g/mol
- Molar mass of water = 18 g/mol
- Moles of H2SO4 = (98 g) / (98 g/mol) = 1 mol
- Moles of water = (2 g) / (18 g/mol) = 0.111 mol
- Mole fraction of H2SO4 (XH2SO4) = moles of H2SO4 / (moles of H2SO4 + moles of water)
- XH2SO4 = 1 mol / (1 mol + 0.111 mol) = 0.9
Therefore, the mole fraction of H2SO4 in the solution is 0.9.

Question 31:

(a) State and explain Kohlrausch’s law of independent migration of ions.

(b) Define Order of a reaction. Illustrate your answer with an example. Define molecularity of a reaction. Illustrate with an example.

(a) State and explain Kohlrausch’s law of independent migration of ions.

Explanation:

At infinite dilution, the ions are so far apart that they do not interact with each other.
Therefore, the conductivity of the solution is solely due to the independent movement of the ions.
The molar conductivity of an electrolyte at infinite dilution (Λ°) can be expressed as:

Λ° = λ°+ + λ°-

where λ°+ and λ°- are the molar conductivities of the cation and anion at infinite dilution, respectively.
This law is useful for determining the molar conductivities of weak electrolytes, which cannot be measured directly at infinite dilution.

(b) Define Order of a reaction. Illustrate your answer with an example. Define molecularity of a reaction. Illustrate with an example.

Order of a reaction: The order of a reaction is an experimentally determined quantity that indicates how the rate of reaction depends on the concentration of the reactants. It is determined from the rate law of the reaction.
- Example: For the reaction 2NO(g) + O2(g) → 2NO2(g), if the rate law is found to be Rate = k[NO]²[O2], the order of the reaction with respect to NO is 2, the order with respect to O2 is 1, and the overall order of the reaction is 3.
Molecularity of a reaction: The molecularity of a reaction is the number of reacting molecules that collide simultaneously to form the activated complex or transition state. It is a theoretical concept and can only be determined from the balanced chemical equation.
- Example: In the reaction H2(g) + I2(g) → 2HI(g), the molecularity is 2 (bimolecular reaction) because two molecules (one H2 and one I2) collide to form the transition state.

Question 32:

How is nitric acid manufactured by Ostwald’s process? How does it react with the following:

(a) Copper

(b) Zn

(c) S

(d) P

Answer:

Ostwald’s Process:

Nitric acid is manufactured by the Ostwald’s process, which involves the following steps:

Oxidation of ammonia: Ammonia is oxidized to nitric oxide (NO) in the presence of a platinum-rhodium catalyst at high temperature (800-900°C) and pressure.

4NH3 + 5O2 → 4NO + 6H2O
Oxidation of nitric oxide: Nitric oxide reacts with oxygen from the air to form nitrogen dioxide (NO2).

2NO + O2 → 2NO2
Absorption and reaction with water: Nitrogen dioxide is absorbed in water to form nitric acid (HNO3) and nitrous acid (HNO2).

4NO2 + 2H2O → 2HNO3 + NO + NO2

The nitrous acid (HNO2) further reacts with nitric oxide (NO) to produce more nitric acid.

HNO2 + NO + H2O → 2HNO3

Reactions of Nitric Acid:

(a) With Copper:

Concentrated nitric acid reacts with copper to form copper nitrate, nitrogen dioxide, and water.

Cu + 4HNO3 (conc.) → Cu(NO3)2 + 2NO2 + 2H2O

(b) With Zn:

Dilute nitric acid reacts with zinc to form zinc nitrate, ammonium nitrate, and water.

4Zn + 10HNO3 (dil.) → 4Zn(NO3)2 + NH4NO3 + 3H2O

(c) With S:

Concentrated nitric acid reacts with sulfur to form sulfuric acid and nitric oxide.

S + 6HNO3 (conc.) → H2SO4 + 6NO2 + 2H2O

(d) With P:

Concentrated nitric acid reacts with phosphorus to form phosphoric acid and nitric oxide.

P4 + 20HNO3 (conc.) → 4H3PO4 + 20NO2 + 4H2O

Question 33:

Describe the following:

(i) Acetylation

(ii) Cannizzaro reaction

(iii) Cross aldol condensation

(iv) Decarboxylation

Answer:

(i) Acetylation:

Acetylation is the process of introducing an acetyl group (CH3CO-) into a molecule. It is commonly used to modify organic compounds, especially alcohols and amines.

Reagent: Acetic anhydride or acetyl chloride is commonly used as the acetylating agent.
Example: Acetylation of an alcohol (e.g., ethanol) to form ethyl acetate:

CH3CH2OH + CH3COCl → CH3COOCH2CH3 + HCl

(ii) Cannizzaro reaction:

Example:

Reaction of benzaldehyde:

2C6H5CHO + NaOH → C6H5CH2OH + C6H5COONa

(iii) Cross aldol condensation:

Cross aldol condensation is a reaction between two different aldehydes or ketones to form a β-hydroxy carbonyl compound.

Example:

Reaction between acetaldehyde and benzaldehyde:

CH3CHO + C6H5CHO → CH3CH(OH)CH2C6H5

(iv) Decarboxylation:

Decarboxylation is the process of removing a carboxyl group (-COOH) from a carboxylic acid, often resulting in the formation of an alkane or alkene.

Example:

Decarboxylation of sodium acetate:

CH3COONa (heated) → CH4 + Na2CO3