Engineering Chemistry Assignment (Unit 1: Atomic and Molecular Structure)

Assignment – Answers

Subject: Engineering Chemistry (BAS102)
Unit – 01 (Atomic and Molecular Structure)

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1. Define bond order as per Molecular Orbital Theory.

Bond order is defined as half the difference between the number of electrons in bonding molecular orbitals (Nb) and antibonding molecular orbitals (Na).

\text{Bond Order (B.O.)} = \frac{N_b - N_a}{2}

• If bond order = 0 → molecule does not exist.

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2. Explain on the basis of MO theory, why O₂ is paramagnetic and N₂ is diamagnetic.

• O₂ (16 electrons): Electronic configuration (M.O.):
  
  Here, two unpaired electrons are present → paramagnetic.

• N₂ (14 electrons): Electronic configuration (M.O.):
  
  No unpaired electrons → diamagnetic.

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3. Calculate bond order of O₂, O₂⁺, O₂⁻, O₂²⁻ and arrange in increasing order of bond length.

• O₂ → BO = (10–6)/2 = 2
• O₂⁺ → BO = (10–5)/2 = 2.5
• O₂⁻ → BO = (10–7)/2 = 1.5
• O₂²⁻ → BO = (10–8)/2 = 1

Bond length ∝ 1/Bond order
So, order of bond length:
O₂⁺ < O₂ < O₂⁻ < O₂²⁻

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4. Examine the reason why H₂⁺ is more stable than H₂⁻?

• H₂⁺: One bonding electron, no antibonding → BO = 0.5 → stable.
• H₂⁻: Two bonding, one antibonding → BO = 0.5 too, but presence of antibonding electron reduces stability.
  Thus, H₂⁺ is more stable than H₂⁻.

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5. Draw molecular energy level diagram of CO and HF and find out bond order, magnetic property, also write down electronic configuration.

• CO (14 electrons):
  M.O. configuration → (σ1s)² (σ1s*)² (σ2s)² (σ2s*)² (π2px)² (π2py)² (σ2pz)²
  Bond order = (10–4)/2 = 3 → strong triple bond, diamagnetic.

• HF (8 + 1 = 9 electrons):
  Mainly overlap between H(1s) and F(2pz).
  Bond order = 1 → single bond, diamagnetic.

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6. State the hybridization of carbon atoms in graphite and fullerene.

• Graphite: Each C is sp² hybridized, forming planar hexagonal sheets.
• Fullerene (C₆₀): Each C is also sp² hybridized, arranged in pentagons and hexagons forming a spherical structure.

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7. Explain why graphite conducts electricity?

• In graphite, each C atom uses only 3 electrons for σ-bonds (sp²).
• The fourth electron remains delocalized in π-orbitals.
• These delocalized electrons move freely → graphite conducts electricity.

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8. Discuss the reason for the lubricating property of graphite.

• Graphite layers are held together by weak van der Waals forces.
• Layers can slide over each other easily.
• Hence, graphite is used as a lubricant.

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9. Discuss the structure of graphite.

• Graphite is composed of layers of hexagonal rings of carbon.
• Each carbon is sp² hybridized and bonded to three others.
• Layers are held by weak forces → soft and slippery.
• Good conductor due to delocalized π-electrons.

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10. Write the conventional and green method for the synthesis of paracetamol.

• Conventional method: Acetylation of p-aminophenol with acetic anhydride gives paracetamol.
• Green method: Using safer solvents like water/ethanol and greener catalysts (enzymes, solid acids) to reduce hazardous waste.

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11. Write the conventional and green method for the synthesis of adipic acid.

• Conventional method: Oxidation of cyclohexanone or cyclohexanol using nitric acid → produces N₂O (harmful greenhouse gas).
• Green method: Using bio-catalysts or hydrogen peroxide as oxidant → avoids toxic byproducts.

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12. Explain atom economy with suitable examples.

• Atom economy measures efficiency of a chemical reaction in terms of how many atoms from reactants are incorporated into the desired product.

\text{Atom Economy} = \frac{\text{Molar mass of desired product}}{\text{Molar mass of reactants}} \times 100

• Addition reaction (high atom economy): CH₂=CH₂ + Cl₂ → CH₂Cl–CH₂Cl (100%).
• Substitution reaction (low atom economy).

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13. Define liquid crystals. Explain classification of liquid crystals on the basis of temperature.

• Liquid crystals: Substances that exhibit properties of both liquids and solids (intermediate phase).
• Classification (thermotropic):
  • Nematic: Molecules aligned in one direction, free to move.
  • Smectic: Molecules arranged in layers.
  • Cholesteric: Helical arrangement of molecules.

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14. Difference between top-down and bottom-up approaches in nanotechnology.

• Top-down: Large bulk material → reduced to nanoscale by lithography, etching, grinding. (Less precise, more waste).
• Bottom-up: Atoms/molecules assemble themselves into nanostructures (chemical vapor deposition, sol-gel). (Precise, less waste).

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15. Summarize the role of carbon nanotubes in nanotechnology.

• Carbon nanotubes (CNTs) are cylindrical nanostructures with sp² carbon atoms.
• Properties: High tensile strength, excellent electrical & thermal conductivity, lightweight.
• Applications:
  • Nanoelectronics (conductors, transistors).
  • Reinforcement in composite materials.
  • Drug delivery in medicine.
  • Sensors and energy storage devices.