TS Inter 2nd Year – Chemistry Previous Paper 2023

CHEMISTRY

Paper -2

English Version

Time: 3 hours

Max.marks: 60

SECTION – A

Note:

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

Question 1:

What are isotonic solutions?

Answer:

Isotonic solutions are solutions that have the same osmotic pressure. This means that the concentration of solute particles in these solutions is equal.

Key points:

Osmotic pressure: It is the pressure required to prevent the flow of solvent molecules across a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration.
Significance of isotonic solutions:
- Biological systems: Red blood cells are isotonic with the surrounding plasma. If a cell is placed in a hypotonic solution (lower solute concentration), water will enter the cell, causing it to swell and burst (hemolysis). Conversely, if a cell is placed in a hypertonic solution (higher solute concentration), water will leave the cell, causing it to shrink (crenation).
- Medical applications: Isotonic solutions are used in intravenous fluids and eye drops to maintain the osmotic balance of body fluids.

Question 2:

Explain the terms gangue and slag.

Answer:

Gangue: Gangue refers to the unwanted impurities present in the ore along with the desired mineral. These impurities can include substances like sand, clay, and other minerals.
Slag: Slag is the byproduct formed during the smelting (extraction of metal from its ore) process. It is a molten mixture of impurities like silica (SiO2), alumina (Al2O3), and other metal oxides.

Question 3:

Calculate the ‘spin only’ magnetic moment of Fe2+ (aq) ion.

Answer:

Electronic configuration of Fe2+: [Ar] 3d6
Number of unpaired electrons (n): 4
Spin-only magnetic moment (μs) = √n(n+2) BM where μs is the spin-only magnetic moment in Bohr Magnetons (BM) and n is the number of unpaired electrons.
μs = √4(4+2) BM = √24 BM = 4.90 BM

Therefore, the spin-only magnetic moment of Fe2+ (aq) ion is 4.90 BM.

Question 4:

What are food preservatives? Give an example.

Answer:

Food preservatives are substances that are added to food to prevent spoilage and extend their shelf life. They work by inhibiting the growth of microorganisms like bacteria, yeast, and mold.

Example: Sodium benzoate.

Question 5:

NH3 forms hydrogen bonds but PH3 does not – Why?

Answer:

NH3 (ammonia) can form hydrogen bonds because it has a highly electronegative nitrogen atom bonded to hydrogen atoms. This creates a significant difference in electronegativity, leading to a strong dipole moment and the formation of hydrogen bonds between the lone pair on the nitrogen of one molecule and the hydrogen of another molecule.

PH3 (phosphine), on the other hand, has a much smaller electronegativity difference between phosphorus and hydrogen. This results in a weaker dipole moment, and hydrogen bonding is not significant in PH3.

Question 6:

What is addition polymer? Give an example.

Answer:

An addition polymer is formed by the repeated addition of monomer units to a growing polymer chain without the elimination of any atoms or molecules.

Example: Polyethylene is an addition polymer formed from the polymerization of ethylene monomers.

Question 7:

What are antacids? Give an example.

Answer:

Antacids are substances that neutralize excess stomach acid (hydrochloric acid, HCl) and provide relief from heartburn, indigestion, and acid reflux.

Example: Milk of Magnesia (magnesium hydroxide)

Question 8:

What is metallic corrosion? Give one example.

Answer:

Metallic corrosion is the gradual deterioration of metals due to chemical reactions with their environment. It often involves the oxidation of the metal.

Example: Rusting of iron.

Question 9:

What is Ziegler-Natta catalyst?

Answer:

Ziegler-Natta catalysts are a class of organometallic compounds used to polymerize alkenes (like ethylene and propylene) into high-density polyethene (HDPE) and polypropylene. They are typically composed of a transition metal compound (like titanium tetrachloride) and an organoaluminum compound (like triethylaluminum).

Question 10:

Write the IUPAC names of the following compounds and classify them into primary, secondary, and tertiary amines.

i) CH3(CH2)2NH2

IUPAC name: Propan-1-amine
Classification: Primary amine

ii) (CH3CH2)2NH

IUPAC name: Diethylamine
Classification: Secondary amine

iii) (CH3CH2)2NCH3

IUPAC name: N-Ethyl-N-methylethanamine
Classification: Tertiary amine

SECTION-B

Note:

(i) Answer ANY SIX questions. (ii) Each question carries FOUR marks. (iii) All are of short answer type questions.

Question 11:

Derive Bragg’s equation.

Answer:

Bragg’s equation is a fundamental equation in X-ray crystallography that relates the angle of incidence of X-rays on a crystal lattice to the spacing between the planes of atoms in the crystal.

Derivation:

Consider a beam of X-rays incident on a crystal lattice. The X-rays are scattered by the atoms in the crystal lattice. If the path difference between the scattered waves from adjacent planes of atoms is an integral multiple of the wavelength of the X-rays, constructive interference occurs. This results in a diffracted beam of X-rays.

Let’s consider two parallel planes of atoms in the crystal lattice. The distance between these planes is ‘d’. Let the angle of incidence of the X-ray beam be θ.

The path difference between the X-rays scattered from the two planes is given by:

Path difference = AB + BC = 2d sinθ

For constructive interference to occur, this path difference must be equal to an integral multiple of the wavelength of the X-rays (λ):

2d sinθ = nλ

where n is an integer (1, 2, 3, …).

This equation is known as Bragg’s equation.

In summary, Bragg’s equation states that:

2d sinθ = nλ

where:

d is the distance between the planes of atoms in the crystal lattice

θ is the angle of incidence of the X-ray beam
λ is the wavelength of the X-rays
n is an integer

Bragg’s equation is used to determine the crystal structure of materials by analyzing the diffraction pattern of X-rays scattered by the crystal.

Question 12:

Explain the terms

(i) Ligand

(ii) Coordination number

(iii) Coordination entity

(iv) Central metal atom/ion

Answer:

(i) Ligand: A ligand is a molecule or ion that can donate a pair of electrons to a central metal atom or ion to form a coordinate covalent bond. Ligands act as Lewis bases.

(ii) Coordination number: The coordination number of a central metal atom/ion is the number of ligands directly bonded to it.

(iii) Coordination entity: A coordination entity is a complex formed by the association of a central metal atom/ion with a fixed number of ligands.

(iv) Central metal atom/ion: The central atom/ion in a coordination compound is the atom or ion to which the ligands are coordinated. It is usually a transition metal ion.

Question 13:

Give the sources of the following vitamins and name the diseases caused by their deficiency (a) A (b) D (c) E and (d) K

Answer:

(a) Vitamin A

Sources: Carrots, spinach, sweet potatoes, milk, eggs
Deficiency: Night blindness, xerophthalmia (dryness of the eyes)

(b) Vitamin D

Sources: Sunlight, fortified foods (milk, cereals), fatty fish
Deficiency: Rickets (in children), osteomalacia (in adults)

(c) Vitamin E

Sources: Vegetable oils, nuts, seeds, whole grains
Deficiency: Anemia, muscle weakness, nerve damage

(d) Vitamin K

Sources: Green leafy vegetables, broccoli, kale
Deficiency: Bleeding disorders, poor blood clotting

Question 14:

A solution of glucose in water is labeled as 10% w/w. What would be the molarity of the solution?

Answer:

10% w/w solution means 10 g of glucose is present in 100 g of solution.

Mass of glucose = 10 g
Mass of water = 100 g – 10 g = 90 g
Density of water = 1 g/mL
Volume of water = 90 mL = 0.09 L
Molar mass of glucose (C6H12O6) = 180 g/mol
Number of moles of glucose = (10 g) / (180 g/mol) = 0.0556 moles
Molarity = Number of moles of solute / Volume of solution (in liters)
Molarity = 0.0556 moles / 0.09 L = 0.618 M

Therefore, the molarity of the solution is 0.618 M.

Question 15:

Explain the formation of micelles with a neat sketch.

Answer:

Micelles are spherical structures formed by the aggregation of surfactant molecules in water. Surfactants are molecules that have both hydrophilic (water-loving) and hydrophobic (water-hating) parts.

Formation: When surfactant molecules are added to water, they tend to arrange themselves in a way that minimizes contact between the hydrophobic tails and water. This leads to the formation of micelles, where the hydrophobic tails are clustered together in the interior of the sphere, while the hydrophilic heads are exposed to the water.
Sketch: (A simple sketch showing the spherical structure of a micelle with hydrophilic heads on the surface and hydrophobic tails inside)

Question 16:

Explain the extraction of zinc from zinc blende.

Answer:

Zinc blende (ZnS) is the most common ore of zinc. The extraction of zinc involves the following steps:

Concentration:
- Froth floatation method is used to concentrate the zinc blende ore.
- The ore is finely ground and mixed with water and pine oil.
- Air is blown through the mixture, and the zinc sulfide particles get attached to the air bubbles and form a froth, which is skimmed off.
Roasting:
- The concentrated zinc blende is roasted in the presence of air to convert it into zinc oxide.
- ZnS + 3/2 O2 → ZnO + SO2
Reduction:
- Zinc oxide is reduced to metallic zinc by heating it with carbon at high temperature.
- ZnO + C → Zn + CO

Question 17:

Explain the structures of a) XeF4 and b) XeOF4

Answer:

(a) XeF4

Shape: Square planar
Hybridization: sp3d2
Lone pairs on Xe: 2

(b) XeOF4

Shape: Square pyramidal
Hybridization: sp3d2
Lone pairs on Xe: 1

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.

Answer:

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

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

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