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Hybridisation sp sp2 sp3 sp3d sp3d2 - UNSOLVED PRACTICE SET

Class 11

Chapter: Chemical Bonding and Molecular Structure | Topic: Hybridisation sp sp2 sp3 sp3d sp3d2

Study Material.
Class 11

HYBRIDISATION SP SP2 SP3 SP3D SP3D2 - UNSOLVED PRACTICE SET

Topic: Hybridisation sp sp2 sp3 sp3d sp3d2

Time: 40 mins | Marks: 30 | Difficulty: Medium

Multiple Choice Questions

Q1. Hybridisation involves:

  1. Mixing of atomic orbitals of different energies to form new orbitals of equivalent energy
  2. Mixing of atomic orbitals of the same energy
  3. Separation of atomic orbitals
  4. Destruction of atomic orbitals

Q2. The shape of a molecule with sp hybridisation is:

  1. Tetrahedral
  2. Trigonal planar
  3. Linear
  4. Octahedral

Q3. In sp² hybridisation, the orbitals are oriented at:

  1. 109.5° to each other
  2. 120° to each other
  3. 90° to each other
  4. 180° to each other

Q4. The hybridisation of carbon in methane (CH₄) is:

  1. sp
  2. sp²
  3. sp³
  4. sp³d

Q5. The hybridisation of phosphorus in PCl₅ is:

  1. sp³
  2. sp³d
  3. sp³d²
  4. dsp²

Q6. The number of hybrid orbitals formed is always:

  1. Less than the number of atomic orbitals mixed
  2. Equal to the number of atomic orbitals mixed
  3. More than the number of atomic orbitals mixed
  4. Independent of the number of atomic orbitals

Short Answer Questions

Q7. Define hybridisation. Why do atoms undergo hybridisation?...

Q8. Predict the hybridisation of the central atom in:

(a) BeCl₂

(b) BF₃

(c) CH₄

Q9. Differentiate between sp, sp², and sp³ hybridisation on the basis of:

(a) Number of orbitals mixed

(b) Geometry

(c) Bond angle

Q10. Why does carbon in ethene (C₂H₄) use sp² hybridisation instead of sp³?

Q11. Your art teacher mixes red, blue, and yellow paints to create new colours for a painting. She explains that hybridisation is similar — atomic orbitals mix to create new orbitals with different shapes and properties. If she mixes one s and one p orbital, what "colour" (geometry) does she get? What if she mixes one s and three p orbitals? 

Q12. What is the hybridisation of sulphur in SF₆? Explain how sulphur can expand its octet to accommodate this hybridisation.

Long Answer Questions

Q13. Explain the concept of hybridisation with suitable examples. Describe the formation of:

(a) sp hybrid orbitals (BeCl₂)

(b) sp² hybrid orbitals (BF₃)

(c) sp³ hybrid orbitals (CH₄)

(d) sp³d hybrid orbitals (PCl₅)

(e) sp³d² hybrid orbitals (SF₆)

For each, show the orbital diagram and predict the geometry.

Q14. Discuss the relationship between hybridisation and molecular geometry. Explain how to determine the hybridisation of a central atom using:

(a) The steric number method

(b) The number of sigma bonds and lone pairs

Predict the hybridisation, shape, and bond angles of:

(a) NH₃

(b) H₂O

(c) CO₂

(d) SO₂

Q15. During a chemistry workshop, students are debating the structure of ethyne (C₂H₂).

(a) One student draws the structure H-C≡C-H and claims both carbons are sp hybridised. Verify this using the steric number method.

(b) Explain why the C-C bond in ethyne is shorter and stronger than in ethene or ethane, using hybridisation concepts.

(c) In benzene (C₆H₆), all carbon atoms are sp² hybridised. Explain how this leads to the planar hexagonal structure and the delocalized π electron system.

(d) A student asks: "If diamond has sp³ hybridised carbon and graphite has sp² hybridised carbon, can we have a form of carbon with sp hybridised atoms?" How would you answer, considering carbon allotropes?

Numerical / Application-Based Problems

Q16. Determine the hybridization of the central atom in the following molecules/ions using the steric number method. Show your calculation.

(a) CO₂

(b) NH₄⁺

(c) SO₂

(d) IF₇

(e) XeO₃F₂

Q17. The C-C bond lengths in different hydrocarbons are:

Ethane (C-C single): 154 pm

Ethene (C=C double): 134 pm

Ethyne (C≡C triple): 120 pm

(a) Explain the trend in bond lengths using hybridisation (sp³, sp², sp).

(b) Calculate the percentage decrease in bond length from ethane to ethene, and from ethene to ethyne.

(c) The C-H bond length in ethane is 109 pm, in ethene is 108 pm, and in ethyne is 106 pm. Explain this trend.

(d) Why does greater s-character in the hybrid orbital lead to shorter and stronger bonds?

Q18. Consider the following molecules and answer the questions:

(a) In CO₂, the carbon atom is sp hybridised. Draw the orbital overlap diagram showing how the two sigma bonds and two pi bonds form.

(b) In ethene (C₂H₄), each carbon is sp² hybridised. Calculate the percentage of s-character and p-character in each hybrid orbital.

(c) In NH₃, nitrogen is sp³ hybridised but the bond angle is 107° instead of 109.5°. Explain this deviation using the concept of lone pair-bond pair repulsion.

(d) In PCl₅, phosphorus uses sp³d hybridisation. Explain why the axial P-Cl bonds (240 pm) are longer than the equatorial P-Cl bonds (202 pm).


Total: 30 Marks | Time: 40 mins

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