Hybridisation in Organic Molecules - UNSOLVED PRACTICE SET

Q7. Explain the concept of hybridisation using the example of methane (CH₄). How does sp³ hybridisation account for its tetrahedral shape?

Q8. Compare the hybridisation, bond angles, and geometry of ethane, ethene, and ethyne.

Q9. What is the difference between a sigma (σ) bond and a pi (π) bond? In ethene, how many sigma and pi bonds are present?

Q10. Explain why the C–C bond length in ethyne (120 pm) is shorter than in ethene (134 pm), which is shorter than in ethane (154 pm).

Q11. Your teacher asks you to predict the shape of a molecule with the formula CH₂O (formaldehyde). Identify the hybridisation of the carbon atom and draw its structure showing bond angles.

Q12. In allene (C₃H₄), the central carbon atom is sp hybridised. Explain how the two terminal CH₂ groups are oriented in space relative to each other.

Q13. Discuss the concept of hybridisation in organic molecules with reference to:

(a) sp³ hybridisation — formation, geometry, examples (methane, ethane)

(b) sp² hybridisation — formation, geometry, examples (ethene, benzene)

(c) sp hybridisation — formation, geometry, examples (ethyne, allene)

(d) How hybridisation explains bond angles and bond lengths in these molecules

Q14. Explain the structure of ethene (C₂H₄) and ethyne (C₂H₂) using the concept of hybridisation:

(a) Draw orbital overlap diagrams for both molecules

(b) Show the formation of sigma and pi bonds

(c) Explain why ethene is planar and ethyne is linear

(d) Compare the reactivity of the C=C double bond and C≡C triple bond based on their electronic structure

Q15. Hybridisation is fundamental to understanding modern materials and drug design. Discuss:

(a) Why graphene (single layer of graphite) has extraordinary electrical conductivity due to sp² hybridisation

(b) How the sp³ hybridisation in diamond makes it the hardest natural substance

(c) Why understanding carbon hybridisation is crucial for designing new drugs in Indian pharmaceutical research

(d) The role of hybridisation in determining the shape and biological activity of molecules like penicillin

Q16. Consider the following molecules: CH₄, C₂H₄, C₂H₂, C₂H₆, and C₆H₆.

(a) Identify the hybridisation of each carbon atom in these molecules.

(b) Calculate the percentage of s-character in each type of hybridisation (sp³, sp², sp).

(c) Based on the s-character, predict the order of C–H bond lengths in these molecules and justify your answer.

Q17. A molecule contains 8 carbon atoms arranged in a ring, with alternating single and double bonds.

(a) Identify the hybridisation of each carbon atom.

(b) Calculate the total number of sigma bonds and pi bonds in the molecule.

(c) If one hydrogen atom is replaced by a chlorine atom, how does the hybridisation at that carbon change? Explain.

Q18. In a research lab in Bangalore, scientists are studying a new organic semiconductor material with the formula C₁₀H₈ (naphthalene).

(a) Determine the hybridisation of each carbon atom in naphthalene.

(b) Calculate the total number of sigma and pi bonds in the molecule.

(c) Explain why naphthalene is planar and how this planarity affects its electrical conductivity compared to a saturated hydrocarbon like decane (C₁₀H₂₂).