TS Intermediate Chemistry 1st Year Model Paper 2024

Section-A

Note: (i) Answer ALL questions. (ii) Each question carries TWO marks. (iii) All are very short answer type questions.

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1. What is Chemical Oxygen Demand (COD)?
Chemical Oxygen Demand (COD) is a measure of the amount of oxygen required to oxidize organic matter in water, such as that found in wastewater, to carbon dioxide and water. It is an important parameter in water quality testing and indicates the level of pollution in water. A high COD value suggests high levels of pollutants and a lower quality of water.

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2. Lithium salts are mostly hydrated. Why?
Lithium salts are mostly hydrated because lithium ions (
$\text{Li}^+$

) are small in size and have a high charge density. This allows lithium to form strong ionic bonds with water molecules. As a result, lithium salts readily attract and hold onto water molecules, forming hydrates. This property is more pronounced in lithium salts compared to other alkali metals.

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3. Name any two man-made silicates.
Two examples of man-made silicates are:

• Portland cement: A mixture of silicates and aluminosilicates that hardens over time.
• Glass: A non-crystalline solid made primarily from silica (silicon dioxide) and other compounds like sodium carbonate and calcium oxide.

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4. Which oxides cause acid rain? What is its pH value?
Oxides that cause acid rain include:

• Sulfur dioxide (SO₂), which reacts with water to form sulfuric acid (H₂SO₄).
• Nitrogen oxides (NO and NO₂), which react with water to form nitric acid (HNO₃).

The pH value of acid rain is typically below 5.6, often ranging from 4.0 to 5.0, depending on the concentration of acidic oxides.

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5. Describe the importance of Plaster of Paris.
Plaster of Paris (CaSO₄·1/2H₂O) is a versatile material with several important uses:

• Medical field: It is used for making casts to immobilize broken bones.
• Construction: It is used for creating decorative moldings, coatings, and as a building material.
• Art: It is used for sculptures and other artistic creations.
• Fireproofing: It is used to make fire-resistant surfaces.

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6. Give the hybridization of carbon in:

a. CO₂:
In carbon dioxide (CO₂), carbon is sp hybridized. The two bonds formed with oxygen are sigma bonds formed by overlap of sp hybrid orbitals from carbon and p orbitals from oxygen.

b. Diamond:
In diamond, each carbon atom is sp³ hybridized, forming four sigma bonds with other carbon atoms in a tetrahedral arrangement.

c. Graphite:
In graphite, carbon atoms are sp² hybridized, forming three sigma bonds with neighboring carbon atoms in a planar hexagonal arrangement. The fourth electron remains in a p orbital, forming a pi bond that is delocalized across the layers.

d. Fullerene:
In fullerene (C₆₀), the carbon atoms are sp² hybridized, forming hexagonal and pentagonal rings. Similar to graphite, the carbon atoms form three sigma bonds, and the remaining electron participates in delocalized pi bonding.

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7. State Graham’s law of diffusion.
Graham’s law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it is expressed as:

$$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$$

Where:

• 
  $r_1$

  and
  $r_2$

  are the rates of diffusion of gases 1 and 2,

• 
  $M_1$

  and
  $M_2$

  are the molar masses of gases 1 and 2.

This means that lighter gases diffuse faster than heavier gases.

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8. What is a redox concept? Give an example.
A redox reaction (reduction-oxidation reaction) is a chemical reaction in which there is a transfer of electrons between two species. One species undergoes oxidation (loses electrons) while the other undergoes reduction (gains electrons).

Example:
In the reaction between hydrogen and oxygen to form water:

$$2H_2 + O_2 \rightarrow 2H_2O$$

• Hydrogen (
  $H_2$

  ) is oxidized (loses electrons) to form
  $H^+$

  .

• Oxygen (
  $O_2$

  ) is reduced (gains electrons) to form
  $O^{2-}$

  .

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9. What is meant by the ionic product of water?
The ionic product of water, denoted as
$K_w$

, is the product of the concentrations of hydrogen ions (
$[H^+]$

) and hydroxide ions (
$[OH^-]$

) in pure water. At 25°C, the ionic product of water is:

$$K_w = [H^+][OH^-] = 1 \times 10^{-14}$$

This constant is important because it governs the balance between
$H^+$

and
$OH^-$

in aqueous solutions. In pure water,
$[H^+] = [OH^-] = 1 \times 10^{-7} \, \text{mol/L}$

.

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10. Write the IUPAC name of the given compound:

Analysis of the Structure

• The longest carbon chain is 4 carbons long, making it a butane derivative.
• There is a double bond between the first and second carbon atoms, indicating it is a 1-butene.
• There is an aldehyde functional group (CHO) on the first carbon atom.

IUPAC Name

Combining these features, the IUPAC name of the compound is 3-butenal.

Section-B

Note:

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

Continuing with the next set of questions starting from Question 11:

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11. Explain the hybridization involved in SF₆ molecule.

In SF₆ (sulfur hexafluoride), sulfur is the central atom, and it forms six bonds with six fluorine atoms. The sulfur atom has 6 valence electrons, and each fluorine atom contributes 1 electron. The bonding in SF₆ involves sp³d² hybridization.

• sp³d² hybridization means that one
  $s$

  -orbital, three
  $p$

  -orbitals, and two
  $d$

  -orbitals from sulfur mix to form six hybrid orbitals.

• These six hybrid orbitals overlap with the
  $p$

  -orbitals of the fluorine atoms to form six sigma bonds.

• The molecular geometry of SF₆ is octahedral, with bond angles of 90° between adjacent bonds and 180° between opposite bonds.

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12. Deduce (a) Boyle’s law and (b) Charles’ law from the Kinetic gas equation.

The Kinetic Gas Equation is given by:

$$PV = \frac{1}{3}m \frac{v^2}{V}$$

Where:

• 
  $P$

  = pressure

• 
  $V$

  = volume

• 
  $m$

  = mass of gas

• 
  $v$

  = average velocity of gas molecules

(a) Boyle’s Law:
Boyle’s law states that the pressure of a gas is inversely proportional to its volume at constant temperature.

$$P \propto \frac{1}{V} \quad \text{(at constant temperature)}$$

From the Kinetic Gas Equation, if temperature (and hence velocity,
$v^2$

) is constant, the pressure
$P$

is inversely proportional to the volume
$V$

. Therefore, Boyle’s Law is derived.

(b) Charles’ Law:
Charles’ law states that the volume of a gas is directly proportional to its temperature at constant pressure.

$$V \propto T \quad \text{(at constant pressure)}$$

From the Kinetic Gas Equation, when the pressure is constant, the volume
$V$

is proportional to the temperature
$T$

, as the kinetic energy (and thus the velocity
$v^2$

) increases with temperature.

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13. Write the general properties of Ionic Compounds.

The general properties of ionic compounds are:

1. High Melting and Boiling Points: Due to strong electrostatic forces (ionic bonds) between the oppositely charged ions, ionic compounds have high melting and boiling points.

2. Electrical Conductivity: Ionic compounds conduct electricity when molten or dissolved in water because their ions are free to move.

3. Solubility: Ionic compounds are generally soluble in water, a polar solvent, because water molecules can surround and stabilize the ions.

4. Brittleness: Ionic compounds are brittle because when stress is applied, like charges are forced together, causing the crystal to shatter.

5. Formation of Crystals: Ionic compounds usually form solid crystalline structures, with a regular and repeating arrangement of ions.

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

$$\text{H}_2\text{O}_2(aq) + \text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + \text{H}_2\text{O}(l) \quad \text{(in acidic solution)}$$

Step-by-step balancing:

1. Write half-reactions for oxidation and reduction:

   • Oxidation (Hydrogen Peroxide to Water):

     
     $$\text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} + \text{O}_2$$

   • Reduction (Fe²⁺ to Fe³⁺):

     
     $$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$$

2. Balance the atoms:

   • Oxygen atoms are already balanced in the oxidation half-reaction.
   • Hydrogen atoms are balanced in the oxidation half-reaction by adding
     $H^+$

     ions on the left side.

3. Balance the charges by adding electrons:

   • The oxidation half-reaction involves the release of 2 electrons, so we need to add 2 electrons to balance charges:

     
     $$\text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} + \text{O}_2 + 2e^-$$

   • The reduction half-reaction already has 1 electron. Multiply the reduction half-reaction by 2 to balance electrons:

     
     $$2\text{Fe}^{2+} \rightarrow 2\text{Fe}^{3+} + 2e^-$$

4. Add the two half-reactions:

   • Cancel the electrons, and the balanced equation becomes:
     $$\text{H}_2\text{O}_2 + 2\text{Fe}^{2+} \rightarrow 2\text{Fe}^{3+} + 2\text{H}_2\text{O}$$

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15. Explain extensive and intensive properties.

• Extensive Properties: These properties depend on the amount or quantity of substance present. Examples include mass, volume, energy, and enthalpy. These properties change as the size or amount of the system changes.

• Intensive Properties: These properties are independent of the amount of substance in the system. Examples include temperature, pressure, density, and refractive index. These properties remain the same regardless of the size or quantity of the system.

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16. What is a conjugate acid-base pair? Illustrate with one example.

A conjugate acid-base pair consists of two species that differ by one proton (H⁺). When an acid donates a proton, it forms its conjugate base; when a base accepts a proton, it forms its conjugate acid.

Example:
In the reaction between ammonia and water:

$$\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$$

• NH₃ is the base, and NH₄⁺ is its conjugate acid.
• H₂O is the acid, and OH⁻ is its conjugate base.

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17. Name the isotopes of hydrogen. What is the ratio of the masses of these isotopes?

The isotopes of hydrogen are:

1. Protium (¹H): This isotope has 1 proton and 0 neutrons.
2. Deuterium (²H or D): This isotope has 1 proton and 1 neutron.
3. Tritium (³H or T): This isotope has 1 proton and 2 neutrons.

The mass ratio of Protium : Deuterium : Tritium is approximately 1 : 2 : 3.

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18. Explain borax bead test with a suitable example.

The borax bead test is a qualitative test used to detect the presence of metal ions by heating a mixture of the metal salt and borax (sodium tetraborate) in a flame.

Procedure:

1. A small amount of borax is heated on a platinum wire until it forms a clear, colorless bead.
2. The metal salt is added to the borax bead, and the bead is reheated in a flame.
3. The metal ions form colored complexes with borax, and the color of the bead helps identify the metal.

Example:

• Copper (Cu²⁺): When copper salt is heated in a borax bead, the bead turns blue due to the formation of a copper-borate complex.
• Chromium (Cr³⁺): When chromium salts are tested, the bead turns green.

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Section-C

Note:

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

19: What are the postulates of Bohr’s model of hydrogen atom? Discuss the importance of this model to explain various series of line spectra in hydrogen atom.

Postulates of Bohr’s Model of the Hydrogen Atom:

1. Quantized Orbits: Bohr postulated that the electron in a hydrogen atom moves in certain discrete, stable orbits around the nucleus. These orbits are called stationary orbits. The electron in these orbits does not radiate energy, and it remains in these orbits without any loss of energy.

2. Energy Quantization: The energy of an electron in a particular orbit is quantized. This means that only specific orbits are allowed, and the energy of each orbit depends on its principal quantum number
   $n$

   n. The energy associated with an orbit is given by the formula:

   
   $$E_n = – \frac{13.6 \, \text{eV}}{n^2}$$

   En​=−n213.6eV​

   where
   $n$

   n is the principal quantum number (n = 1, 2, 3,…).

3. Radiation of Energy: When an electron transitions from a higher orbit to a lower orbit, it emits a photon of energy equal to the difference in energy between the two orbits. The energy of the emitted photon is given by:

   
   $$\Delta E = E_{\text{initial}} – E_{\text{final}}$$

   ΔE=Einitial​−Efinal​

   This results in the emission of light, which has specific wavelengths corresponding to the energy difference between the two orbits.

4. Angular Momentum Quantization: The angular momentum of the electron in the nth orbit is quantized. It is given by:

   
   $$L = n \hbar$$

   L=nℏ

   where
   $\hbar$

   ℏ is the reduced Planck’s constant.

Importance of Bohr’s Model in Explaining the Spectral Lines of Hydrogen Atom:

Bohr’s model explained the discrete line spectra observed in the hydrogen atom, a phenomenon that could not be explained by classical mechanics. According to Bohr’s model:

• The electron in the hydrogen atom is confined to discrete orbits around the nucleus.
• When the electron transitions from a higher energy orbit (level) to a lower energy orbit, energy is emitted in the form of light. This light has specific wavelengths, leading to the formation of spectral lines.

These transitions lead to distinct series of lines:

• The Lyman series corresponds to transitions where the electron falls to the first energy level (n=1), emitting ultraviolet radiation.
• The Balmer series corresponds to transitions to the second energy level (n=2), emitting visible light.
• The Paschen, Brackett, and Pfund series correspond to transitions to the third, fourth, and fifth energy levels, respectively, emitting infrared radiation.

Thus, Bohr’s model provided a clear explanation for the observed spectral lines in the hydrogen atom, showing that the energy levels are quantized and related to the discrete wavelengths of light emitted during electron transitions.

20: Write an essay on s, p, d, and f-block elements.

Essay on s, p, d, and f-block elements:

The periodic table is divided into different blocks based on the electron configuration of the elements. These blocks are the s-block, p-block, d-block, and f-block. Each block contains elements that have characteristic properties, largely due to their electron configurations.

1. S-Block Elements: The s-block elements are found in Groups 1 and 2 of the periodic table. They have their outermost electrons in s-orbitals.

   • Group 1 (Alkali Metals): The alkali metals include lithium, sodium, potassium, rubidium, cesium, and francium. These elements are highly reactive, especially with water, forming hydroxides and hydrogen gas. They are soft metals with low melting points and low densities.
   • Group 2 (Alkaline Earth Metals): The alkaline earth metals include beryllium, magnesium, calcium, strontium, barium, and radium. They are less reactive than alkali metals but still react with water (less vigorously) and form basic oxides and hydroxides. These metals are harder and have higher melting points than alkali metals.

2. P-Block Elements: The p-block elements are located in Groups 13 to 18 of the periodic table. These elements have their outermost electrons in p-orbitals.

   • Group 13 to 15: Elements in these groups, such as boron, carbon, nitrogen, oxygen, and fluorine, exhibit a wide range of properties. Carbon (Group 14) is a non-metal, silicon is a metalloid, and metals like aluminum (Group 13) are also present.
   • Group 16 (Chalcogens): The chalcogens, such as oxygen, sulfur, selenium, and tellurium, form important compounds like water (H₂O) and sulfur dioxide (SO₂). These elements can be non-metals or metalloids.
   • Group 17 (Halogens): The halogens, including fluorine, chlorine, bromine, iodine, and astatine, are highly reactive nonmetals. They readily form salts when combined with metals.
   • Group 18 (Noble Gases): The noble gases, such as helium, neon, argon, krypton, xenon, and radon, are inert due to their filled valence electron shells. They are used in various applications, including lighting and as refrigerants.

3. D-Block Elements (Transition Metals): The d-block elements, or transition metals, are found in Groups 3 to 12 of the periodic table. These elements have their outermost electrons in d-orbitals.

   • Transition metals such as iron, copper, and gold are known for their ability to form multiple oxidation states. They are good conductors of heat and electricity and are typically hard and dense.
   • Many transition metals are used as catalysts in industrial processes. They also form various alloys (e.g., steel, which is an alloy of iron and carbon), and they are used in the production of jewelry, electronics, and machinery.

4. F-Block Elements (Lanthanides and Actinides): The f-block elements are located at the bottom of the periodic table and consist of two series: the lanthanides and the actinides.

   • Lanthanides: These elements, also called rare earth elements, include metals like cerium, neodymium, and europium. They are used in the production of strong permanent magnets, phosphors for color television tubes, and catalysts.
   • Actinides: The actinides, such as uranium, thorium, and plutonium, are mostly radioactive. They are important in nuclear power generation and the production of nuclear weapons.

In conclusion, the s, p, d, and f-block elements have distinct chemical and physical properties due to the electron configurations in different orbitals. Their study helps explain the behavior of elements in chemical reactions and their various applications in industries such as electronics, metallurgy, and energy production.

21.How does acetylene react with the following reagents? Give the corresponding equations and name the products formed in the reactions. 

a. Water b. Hydrogen c. Halogens d. Hydrogen halide