4 Most SAQ’s of Hydrogen and its Compounds Chapter in Inter 1st Year Chemistry (TS/AP)

4 Marks

SAQ-1 : Write a few lines on the utility of hydrogen as a fuel.

Introduction

Hydrogen has emerged as a significant alternative fuel source due to its unique properties and environmental benefits. Its utility in various fields demonstrates its potential in shaping a sustainable energy future.

Utility of Hydrogen as a Fuel

1. Rocket Fuel: Hydrogen, when used in combination with oxygen, serves as a major propellant in rocket engines due to its high calorific value and clean combustion, producing water as the only byproduct.
2. Fuel Cells: Hydrogen can be used in fuel cells to generate electricity. In this electrochemical process, hydrogen and oxygen combine to produce water, and in the process, electricity is generated. This makes fuel cells an attractive option for powering vehicles and other applications without the harmful emissions of conventional combustion engines.
3. High Energy Content: The energy produced by burning hydrogen is three times greater than that produced by burning an equivalent amount of petrol. This high energy content, combined with its clean combustion, makes hydrogen a compelling choice for transportation applications.
4. Additive in CNG: A small percentage (around 4%) of hydrogen is mixed with Compressed Natural Gas (CNG) to improve its combustion efficiency in vehicles.
5. Gaseous Fuels: Hydrogen is an integral component of certain gaseous fuels, including coal gas and water gas, enhancing their calorific value and combustion properties.
6. Industrial Fuel: Dihydrogen is utilized directly as an industrial fuel in various processes and applications, given its clean-burning characteristics and high energy content.

Summary

Hydrogen’s potential as a clean and efficient fuel source, especially in the context of reducing greenhouse gas emissions and countering climate change, has led to increased interest and investment in hydrogen-based technologies and infrastructure.

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SAQ-2 : Explain the terms hard water and soft water. Write a note on the Calgon method for the removal of hardness of water.

Introduction

Understanding the terms hard water and soft water is crucial in the context of water quality and its suitability for various uses, especially in domestic and industrial settings.

Hard Water

Hard water is water that contains high levels of dissolved minerals, particularly calcium and magnesium ions. It can cause problems like scaling in boilers, heaters, and household appliances, and reduces the lathering ability of soaps.

Soft Water

In contrast, soft water has a low concentration of calcium and magnesium ions. It is more effective for cleaning and lathering purposes and is generally more desirable for household use.

Calgon Method for Removing Hardness

The Calgon method is a popular method for softening hard water, particularly for removing temporary hardness.

1. Mechanism: The process involves adding sodium hexametaphosphate, commonly known as Calgon, to hard water. Calgon chemically binds with calcium and magnesium ions, forming soluble complexes.
2. Reaction: The chemical reaction can be represented as: $$(NaPO_3)_6 + Ca^{2+} \rightarrow [Ca(PO_3)_6]^{2-} + 2Na^+$$ Similar reactions occur with magnesium ions.
3. Effectiveness: This method is effective because it prevents the minerals from precipitating and forming scale. The resulting complexes are soluble in water and don’t interfere with cleaning processes.

Summary

The distinction between hard water and soft water is based on the concentration of calcium and magnesium ions. The Calgon method of water softening is an efficient technique to remove hardness, especially useful in industrial applications and in areas where water has high mineral content.

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SAQ-3 : Write any two oxidizing & reducing properties of hydrogen peroxide.

Introduction

Hydrogen peroxide (H₂O₂) exhibits both oxidizing and reducing properties. This dual nature allows it to either accept electrons (in reduction) or donate electrons (in oxidation), making it a versatile chemical in various reactions.

Oxidizing Properties of Hydrogen Peroxide

1. Conversion of Hydrogen Chloride to Chlorine: $$2HCl + H_2O_2 \rightarrow Cl_2 + 2H_2O$$ Here, H₂O₂ oxidizes HCl to chlorine.
2. Oxidation of Lead Sulphide to Lead Sulphate: $$PbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2O$$ Lead sulphide turns white when converted to lead sulphate by H₂O₂.
3. Oxidation of Iron (II) to Iron (III): $$2Fe^{2+} + H_2O_2 \rightarrow 2Fe^{3+} + 2OH^-$$ In a basic medium, H₂O₂ oxidizes Fe²⁺ to Fe³⁺.
4. Oxidation of Manganese (II) to Manganese (IV): $$Mn^{2+} + H_2O_2 \rightarrow Mn^{4+} + 2OH^-$$ Here, H₂O₂ acts as an oxidizing agent, turning Mn²⁺ into Mn⁴⁺.

Reducing Properties of Hydrogen Peroxide

1. Conversion of Silver Oxide to Silver: $$Ag_2O + H_2O_2 \rightarrow 2Ag + H_2O + O_2$$ H₂O₂ reduces silver oxide to silver in this reaction.
2. Reduction of Acidified Potassium Permanganate: $$2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$$ Potassium permanganate is reduced to colorless manganese sulphate by H₂O₂.
3. Conversion of Hypochlorous Acid to Chloride Ion: $$HOCl + H_2O_2 \rightarrow Cl^- + H_3O^+ + O_2​$$ In this reaction, H₂O₂ acts as a reducing agent for hypochlorous acid.
4. Reduction of Permanganate Ion in Basic Medium: $$2MnO_4^- + 3H_2O_2 \rightarrow 2MnO_2 + 3O_2 + 2OH^-$$ H₂O₂ reduces MnO₄⁻ to MnO₂ in a basic environment.

Summary

Hydrogen peroxide is a versatile chemical agent, displaying both oxidizing and reducing characteristics. Its ability to either donate or accept electrons allows it to participate in various chemical reactions, making H₂O₂ an essential compound in chemistry.

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SAQ-4 : Explain, with suitable examples, the following:
i. Electron deficient ii. Electron-precise iii. Electron-rich hydrides.

Introduction

Hydrides, compounds formed by the combination of hydrogen with other elements, can be classified into three categories based on their electron count: electron-deficient, electron-precise, and electron-rich hydrides.

1. Electron-Deficient Hydrides : These are hydrides in which the total number of electrons is less than the sum of the valence electrons of the constituent atoms. They often form from elements of Group 13.
   • Example: Diborane (B₂H₆): In diborane, each boron atom contributes three valence electrons and each hydrogen atom contributes one, totaling 12 electrons. However, to satisfy the standard bonding requirements for all atoms, 14 electrons would be needed. This deficiency leads to the formation of three-center two-electron bonds.
2. Electron-Precise Hydrides: These hydrides have a sufficient number of electrons to form conventional two-center two-electron bonds. They are typically formed by elements of Group 14.
   • Example: Methane (CH₄): Methane has four C-H bonds. Carbon brings four electrons, and each hydrogen contributes one, totaling eight electrons which are just enough to form four two-center two-electron bonds.
3. Electron-Rich Hydrides: These hydrides possess more electrons than required for forming two-center two-electron bonds. Such excess electrons are present as lone pairs. Elements of Group 15, 16, and 17 often form electron-rich hydrides.
   • Example: Ammonia (NH₃): In ammonia, nitrogen contributes five electrons, and each hydrogen atom contributes one, making a total of eight electrons. Three bonds are formed using six electrons, and the remaining two electrons form a lone pair on the nitrogen atom.

Summary

Understanding the electron count in hydrides is crucial in chemistry. Electron-deficient hydrides, like diborane, lack sufficient electrons for standard bonding; electron-precise hydrides, like methane, have just the right number of electrons; and electron-rich hydrides, such as ammonia, have extra electrons that form lone pairs.