TS Intermediate Chemistry 1st Year Previous Paper 2023

SECTION-A

Note ; (i) Answer ALL Questions

(ii) Each Question carries TWO marks

(iii) All are very short answer type questions.

Here are the answers to your questions:

1. What is Biochemical Oxygen Demand (BOD)?

Answer:
Biochemical Oxygen Demand (BOD) is the amount of dissolved oxygen required by aerobic microorganisms to break down organic matter in water. It is an indicator of the organic pollution level in water. A high BOD suggests a high level of pollution, as it means microorganisms require more oxygen to decompose the organic matter.

2. What happens when magnesium metal is burnt in air?

Answer:
When magnesium metal is burnt in air, it reacts with oxygen to form magnesium oxide (MgO), a white powder. The reaction is highly exothermic:

$\text{2Mg} + \text{O}_2 \rightarrow \text{2MgO}$

3. Describe the important uses of sodium carbonate.

Answer:
Sodium carbonate (Na₂CO₃) has several important uses, including:

As a cleaning agent and detergent (washing soda).
In the manufacture of glass.
In the production of soaps and detergents.
As a pH regulator in swimming pools and in some chemical processes.
In water treatment for softening hard water.

4. Why is a pressure cooker used for cooking food on hills?

Answer:
A pressure cooker is used for cooking food on hills because, at higher altitudes, the atmospheric pressure is lower, which reduces the boiling point of water. This means food takes longer to cook. A pressure cooker increases the pressure inside, which raises the boiling point of water and helps cook food faster.

5. Give the hybridization of carbon in:

a. CO₂ – sp hybridization (Linear geometry). b. diamond – sp³ hybridization (Tetrahedral geometry). c. graphite – sp² hybridization (Trigonal planar geometry). d. fullerene – sp² hybridization (Mostly trigonal planar geometry for the carbon atoms forming the hexagonal rings).

6. What is PAN? What effect is caused by it?

Answer:
PAN (Peroxyacetyl Nitrate) is a harmful air pollutant, primarily formed as a secondary pollutant in photochemical smog. It is a component of the pollutant mixture in urban air. PAN can irritate the eyes, skin, and respiratory system and can cause damage to vegetation.

7. The empirical formula of a compound is CH₂O. Its molecular weight is 90. Calculate the molecular formula of the compound.

Answer:
The empirical formula is CH₂O, and its molecular weight is 90.

The molar mass of the empirical formula (CH₂O) = 12 + (2 × 1) + 16 = 30 g/mol.
To find the molecular formula, divide the molecular weight by the empirical formula mass: $\frac{90}{30} = 3$
Therefore, the molecular formula is C₃H₆O₃.

8. All Lewis acids are not Bronsted acids. Why?

Answer:
A Lewis acid is a substance that can accept a pair of electrons, whereas a Bronsted acid is a substance that can donate a proton (H⁺). Not all Lewis acids are Bronsted acids because a Lewis acid doesn’t necessarily have to donate a proton. For example, AlCl₃ is a Lewis acid because it can accept electron pairs, but it does not donate protons, so it is not a Bronsted acid.

9. Diamond has a high melting point – explain.

Answer:
Diamond has a high melting point because it has a very strong covalent bond between carbon atoms. Each carbon atom in diamond forms four strong covalent bonds in a tetrahedral structure, creating a rigid three-dimensional network. This structure requires a significant amount of energy to break the bonds and melt the substance, resulting in diamond having an extremely high melting point.

10. Write the IUPAC name of:

a. CH₃-CH₂-CH₂-CH=CH₂
Answer:
The IUPAC name of this compound is Pent-2-ene.

Explanation:

The compound has 5 carbon atoms, so it is a pentane derivative.
The double bond (C=C) is located between the 2nd and 3rd carbon atoms, so it is named pent-2-ene.
The numbering starts from the end nearest to the double bond to give the smallest possible number to the double bond.

SECTION-B

Note:

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

11. State Fajan’s Rules, and give suitable examples.

Answer: Fajan’s rules describe the polarizing power of cations and the polarizability of anions in ionic bonding. These rules help explain the degree of covalent character in an ionic bond. The rules are:

Smaller cations have a higher charge density and are more polarizing (they tend to distort the electron cloud of the anion).
Larger anions are more easily polarized by cations.
Higher charge on the cation increases its polarizing power.
The anions with high charge and large size are more easily polarized.

Example:

NaCl (Sodium chloride) – Ionic bond (less covalent character).
Covalent character increases in compounds like CuCl (Copper(I) chloride) due to the high polarizing power of Cu⁺ (small, high charge).

12. Deduce (a) Boyle’s law and (b) Charles’s law from Kinetic Gas Equation.

Answer:

(a) Boyle’s Law (at constant temperature):

Boyle’s law states that the pressure of a gas is inversely proportional to its volume, at constant temperature. From the kinetic gas equation:

$PV = \frac{1}{3} n m v_{\text{avg}}^2$

$P V = \frac{1}{3} nm v_{avg 2}$

At constant temperature, the average kinetic energy of the gas molecules remains the same, so

$v_{\text{avg}}^2$

$v_{avg 2}$ is constant. Therefore, for a fixed amount of gas,

$P \propto \frac{1}{V}$

This leads to Boyle’s Law:

$P \cdot V = \text{constant}$

$P \cdot V = constant$

(b) Charles’s Law (at constant pressure):

Charles’s law states that the volume of a gas is directly proportional to its temperature, at constant pressure. From the kinetic gas equation, at constant pressure:

$P = \frac{n m v_{\text{avg}}^2}{3V}$

$P = \frac{nm v ^{avg 2}}{3 V}$

Since

$P$

$P$ is constant, we get:

$V \propto T$

This leads to Charles’s Law:

$\frac{V}{T} = \text{constant}$

$\frac{V}{T} = constant$

13. State and explain the Hess’s law of constant Heat summation.

Answer:
Hess’s Law states that the total enthalpy change of a chemical reaction is the sum of the enthalpy changes of the steps into which the reaction can be divided. This is true regardless of the path taken.

Explanation:
If a reaction can be expressed as the sum of two or more reactions, the total enthalpy change for the reaction is the sum of the enthalpy changes for each individual step.

Example:
For the reaction:

$C(s) + O_2(g) \rightarrow CO_2(g)$

$C (s) + O_{2} (g) \to C O_{2} (g)$

If it can be written in two steps, the total enthalpy change for the reaction is the sum of the enthalpy changes for each step.

14. Derive the relation between Kp and Kc for the equilibrium reaction:

$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 NH_{3} (g)$

Answer:
The relationship between

$K_p$

$K_{p}$ and

$K_c$

$K_{c}$ can be derived from the ideal gas law:

The equilibrium expression for

$K_c$

$K_{c}$ is:

$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$

$K_{c} = \frac{[ NH ^{3} ] ^{2}}{[ N ^{2} ] [ H ^{2} ] ^{3}}$

Now, using the ideal gas law,

$P = \frac{nRT}{V}$

$P = \frac{n RT}{V}$ , we can express partial pressures in terms of concentrations.

For gases at equilibrium:

$K_p = K_c \left( RT \right)^{\Delta n}$

$K_{p} = K_{c} (RT)^{Δ n}$

Where:

$R$
$T$
$\Delta n$

For the reaction:

$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 NH_{3} (g)$

$\Delta n = (2) – (1 + 3) = -2$

$Δ n = (2) - (1 + 3) = - 2$

Thus, the relation becomes:

$K_p = K_c (RT)^{-2}$

$K_{p} = K_{c} (RT)^{-2}$

15. Balance the following redox reaction by ion-electron method:

$\text{MnO}_4^-(aq) + \text{SO}_2(g) \rightarrow \text{Mn}^{2+}(aq) + \text{HSO}_4^-(aq)$

$MnO_{4 -} (a q) + SO_{2} (g) \to Mn^{2+} (a q) + HSO_{4 -} (a q)$
(in acidic solution)

Answer: The steps to balance the reaction by the ion-electron method:

Write the oxidation half-reaction for SO₂:

$\text{SO}_2 \rightarrow \text{HSO}_4^-$

$SO_{2} \to HSO_{4 -}$

Write the reduction half-reaction for MnO₄⁻:

$\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$

$MnO_{4 -} \to Mn^{2+}$

Balance atoms and charge using electrons and H⁺ ions (acidic medium).
Multiply the half-reactions to make the electrons equal.
Add the half-reactions.

Final balanced reaction:

$2 \text{MnO}_4^-(aq) + 5 \text{SO}_2(g) + 4 \text{H}^+ \rightarrow 2 \text{Mn}^{2+}(aq) + 5 \text{HSO}_4^-(aq) + 2 \text{H}_2O$

$2 MnO_{4 -} (a q) + 5 SO_{2} (g) + 4 H^{+} \to 2 Mn^{2+} (a q) + 5 HSO_{4 -} (a q) + 2 H_{2} O$

16. Explain the hybridization involved in SF₆ molecule.

Answer:
In SF₆ (sulfur hexafluoride), sulfur (S) is the central atom and forms six bonds with six fluorine atoms. The sulfur atom undergoes sp³d² hybridization, where one 3s, three 3p, and two 3d orbitals combine to form six sp³d² hybrid orbitals. These orbitals arrange themselves in an octahedral geometry.

17. Write a few lines on the utility of hydrogen as a fuel.

Answer:
Hydrogen is a clean and efficient fuel source. When hydrogen is used in fuel cells, it reacts with oxygen to produce electricity, with water as the only byproduct. This makes hydrogen a sustainable alternative to fossil fuels, which emit harmful pollutants and greenhouse gases. Hydrogen fuel can be used in transportation (hydrogen-powered vehicles), power generation, and in industrial processes. It is a promising energy carrier for a sustainable future.

18. Explain the borax bead test with a suitable example.

Answer:
The borax bead test is used to detect the presence of metal ions in salts. A small amount of borax is heated in a flame until it melts to form a bead. The bead is then dipped into the sample and reheated. The color of the bead changes based on the metal ions present.

Example:

Copper: When copper salts are heated with the borax bead, the bead turns blue due to the formation of copper borate.
Iron: Iron salts cause the bead to turn green.

The test is useful in qualitative inorganic analysis to identify various metal ions.

SECTION-C

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

19. Postulates of Bohr’s Model and Importance for Line Spectra

Postulates:

Electrons revolve in fixed, quantized orbits without radiating energy.
Energy is emitted or absorbed when electrons jump between orbits.
Angular momentum is quantized: $L = n \hbar$

Importance for Line Spectra:

Explains discrete spectral lines (Lyman, Balmer, etc.) through electron transitions between quantized orbits.
Spectral lines result from the energy difference between these orbits.

20. Ionization Energies (IE₁ and IE₂)

Definitions:

IE₁: Energy required to remove the first electron from a neutral atom.
IE₂: Energy required to remove the second electron from the cation.

Why IE₂ > IE₁:

After removing the first electron, the nucleus exerts a stronger pull on the remaining electrons, making it harder to remove the second electron.

Factors Affecting IE₁:

Atomic size (smaller size, higher IE).
Nuclear charge (higher charge, higher IE).
Electron shielding (more shielding, lower IE).

21. Reactions of Acetylene (C₂H₂)

a. With Water (Hydration):

$\text{C}_2\text{H}_2 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{CHO}$

$C_{2} H_{2} + H_{2} O H^{+} CH_{3} CHO$

Product: Acetaldehyde

b. With Hydrogen (Hydrogenation):

To ethylene:

$\text{C}_2\text{H}_2 + \text{H}_2 \rightarrow \text{C}_2\text{H}_4$

$C_{2} H_{2} + H_{2} \to C_{2} H_{4}$

To ethane:

$\text{C}_2\text{H}_4 + \text{H}_2 \rightarrow \text{C}_2\text{H}_6$

$C_{2} H_{4} + H_{2} \to C_{2} H_{6}$

Products: Ethylene, Ethane

c. With Halogens (Halogenation):

$\text{C}_2\text{H}_2 + \text{Br}_2 \rightarrow \text{C}_2\text{H}_2\text{Br}_2$

$C_{2} H_{2} + Br_{2} \to C_{2} H_{2} Br_{2}$

Product: 1,1-Dibromoethene

d. With Hydrogen Halide (Hydrohalogenation):

$\text{C}_2\text{H}_2 + \text{HCl} \rightarrow \text{CH}_2\text{=CHCl}$

$C_{2} H_{2} + HCl \to CH_{2} =CHCl$

Product: Vinyl chloride (C₂H₃Cl)