Hybridization & MO Theory
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All Questions for: Hybridization & MO Theory
Question 1: The hybridization of carbon in methane (CH₄) is:
- A sp³
- B sp²
- C sp
- D sp³d
Explanation: Carbon in CH₄ has four single bonds and no lone pairs, so it uses sp³ hybridization with tetrahedral geometry.
Question 2: The hybridization of carbon in ethene (C₂H₄) is:
- A sp²
- B sp³
- C sp
- D sp³d
Explanation: Each carbon in C₂H₄ forms three sigma bonds and one pi bond, using sp² hybridization with trigonal planar geometry.
Question 3: The hybridization of carbon in ethyne (C₂H₂) is:
- A sp
- B sp²
- C sp³
- D sp³d²
Explanation: Each carbon in C₂H₂ forms two sigma bonds and two pi bonds, using sp hybridization with linear geometry.
Question 4: The hybridization of nitrogen in ammonia (NH₃) is:
- A sp³
- B sp²
- C sp
- D sp³d
Explanation: Nitrogen in NH₃ has 3 sigma bonds and 1 lone pair, giving a steric number of 4 and sp³ hybridization.
Question 5: The hybridization of oxygen in water (H₂O) is:
- A sp³
- B sp²
- C sp
- D sp³d
Explanation: Oxygen in H₂O has 2 sigma bonds and 2 lone pairs, giving a steric number of 4 and sp³ hybridization.
Question 6: The hybridization of phosphorus in PCl₅ is:
- A sp³d
- B sp³
- C sp³d²
- D sp²
Explanation: Phosphorus in PCl₅ has 5 sigma bonds and 0 lone pairs, giving a steric number of 5 and sp³d hybridization.
Question 7: The hybridization of sulfur in SF₆ is:
- A sp³d²
- B sp³d
- C sp³
- D sp²
Explanation: Sulfur in SF₆ has 6 sigma bonds and 0 lone pairs, giving a steric number of 6 and sp³d² hybridization.
Question 8: The hybridization of carbon in benzene (C₆H₆) is:
- A sp²
- B sp³
- C sp
- D sp³d
Explanation: Each carbon in benzene has 3 sigma bonds and 1 pi bond, making it sp² hybridized. The remaining p orbitals combine to form delocalized pi molecular orbitals.
Question 9: The number of sigma bonds in benzene is:
- A 12
- B 6
- C 18
- D 3
Explanation: Benzene has 6 C-C sigma bonds and 6 C-H sigma bonds = 12 sigma bonds, plus 6 delocalized pi electrons.
Question 10: The number of pi electrons in benzene is:
- A 6
- B 12
- C 3
- D 4
Explanation: Benzene has 6 pi electrons (one from each carbon's unhybridized p orbital), delocalized over the ring, satisfying Hückel's 4n+2 rule with n=1.
Question 11: Molecular orbital theory describes the formation of a covalent bond in H₂ as:
- A Combination of two 1s orbitals to form a bonding σ and antibonding σ* MO
- B Transfer of one electron to the other atom
- C Formation of an ionic bond
- D Overlap of p orbitals only
Explanation: In H₂, the two 1s atomic orbitals combine to form a lower-energy bonding sigma MO (σ_1s) and a higher-energy antibonding sigma* MO (σ*_1s). The two electrons fill the bonding MO.
Question 12: According to MO theory, the bond order of O₂ is:
- A 2
- B 1
- C 3
- D 2.5
Explanation: O₂ has 8 bonding and 4 antibonding electrons in MOs, giving bond order = (8−4)/2 = 2. O₂ also has two unpaired electrons in the π* antibonding orbitals, making it paramagnetic.
Question 13: According to MO theory, the bond order of N₂ is:
- A 3
- B 2
- C 2.5
- D 1
Explanation: N₂ has 10 bonding and 4 antibonding electrons, giving bond order = (10−4)/2 = 3, consistent with a triple bond.
Question 14: The paramagnetism of O₂ is best explained by:
- A Two unpaired electrons in antibonding π* orbitals
- B Presence of a triple bond
- C Two unpaired electrons in sigma orbitals
- D Loss of an electron to form O₂⁺
Explanation: MO theory predicts two unpaired electrons in the π* antibonding orbitals of O₂. This explains why liquid O₂ is attracted to a magnetic field (paramagnetism), which VSEPR and valence bond theory cannot explain.
Question 15: The correct energy ordering of molecular orbitals for O₂ and F₂ is:
- A σ_2p < π_2p < π*_2p < σ*_2p
- B π_2p < σ_2p < π*_2p < σ*_2p
- C σ_2p < σ*_2p < π_2p < π*_2p
- D π*_2p < σ*_2p < π_2p < σ_2p
Explanation: For O₂ and F₂, the σ_2p orbital is lower than the π_2p orbitals. For N₂ and lighter diatomics, the ordering is reversed.
Question 16: The bond order of the superoxide ion (O₂⁻) is:
- A 1.5
- B 2
- C 2.5
- D 1
Explanation: O₂⁻ has 8 bonding and 5 antibonding electrons, giving bond order = (8−5)/2 = 1.5.
Question 17: The HOMO of the N₂ molecule is:
- A σ_2p
- B π_2p
- C π*_2p
- D σ*_2p
Explanation: In N₂, the σ_2p orbital is higher in energy than the π_2p orbitals. With 10 valence electrons, the HOMO is σ_2p (the highest occupied molecular orbital).
Question 18: For a stable diatomic molecule, which statement is TRUE?
- A There are more electrons in bonding MOs than in antibonding MOs
- B There are more electrons in antibonding MOs than in bonding MOs
- C Bonding and antibonding MOs are equally filled
- D MO theory does not apply to stable molecules
Explanation: A stable molecule requires more electrons in bonding than in antibonding MOs, giving a positive bond order. If BO ≤ 0, the molecule does not exist as a stable species.
Question 19: The number of unpaired electrons in the O₂ molecule is:
- A 2
- B 1
- C 3
Explanation: According to MO theory, O₂ has two unpaired electrons in the degenerate π*_2p orbitals (one in each). This makes O₂ paramagnetic.
Question 20: A σ bond is formed by:
- A Head-on (axial) overlap of atomic orbitals
- B Side-on (lateral) overlap of p orbitals
- C Complete transfer of electrons
- D Delocalization over three or more atoms
Explanation: Sigma bonds form by head-on (axial) overlap of orbitals, concentrating electron density along the internuclear axis. Pi bonds form by side-on overlap.
Question 21: A π bond is formed by:
- A Side-on (lateral) overlap of parallel p orbitals
- B Head-on overlap of s orbitals
- C End-to-end overlap of hybrid orbitals
- D Complete electron transfer
Explanation: Pi bonds result from side-on overlap of parallel p orbitals, creating electron density above and below the internuclear axis.
Question 22: Which is a stronger bond, sigma or pi?
- A Sigma bonds are generally stronger
- B Pi bonds are generally stronger
- C Both are equal in strength
- D Pi bonds are stronger only in triple bonds
Explanation: Sigma bonds are stronger than pi bonds due to greater overlap along the internuclear axis, which concentrates electron density directly between the nuclei.
Question 23: The hybridization of sulfur in SO₂ is:
- A sp²
- B sp³
- C sp
- D sp³d
Explanation: Sulfur in SO₂ has 2 sigma bonds to oxygen and 1 lone pair (resonance structures contribute additional bonding character). Steric number = 3, so sp² hybridization with bent geometry.
Question 24: The hybridization of the central atom in the carbonate ion (CO₃²⁻) is:
- A sp²
- B sp³
- C sp
- D sp³d
Explanation: The carbonate ion has a central carbon bonded to three oxygens with one delocalized pi bond. This gives three sigma bonds and no lone pairs on carbon: sp² hybridization with trigonal planar geometry.
Question 25: In molecular orbital theory, the antibonding MO has:
- A A node between the nuclei and higher energy than the bonding MO
- B No nodes and lower energy than the bonding MO
- C The same energy as the bonding MO
- D Only s-character
Explanation: Antibonding orbitals have a node between the nuclei, reducing electron density in the bonding region and weakening the bond. They are marked with an asterisk (*).