Questions & explanations
1. Which of the following statements correctly compares Molecular Orbital Theory (MOT) and Valence Bond Theory (VBT) for diatomic molecules?
- VBT successfully explains the paramagnetic nature of O₂, while MOT fails.
- MOT predicts that all diatomic molecules have a net magnetic moment.
- VBT treats electrons in molecules as localized between two atoms, whereas MOT treats electrons as delocalized over the entire molecule.
- According to MOT, the bond order is always an integer, while VBT allows fractional bond orders.
Answer: VBT treats electrons in molecules as localized between two atoms, whereas MOT treats electrons as delocalized over the entire molecule.
In Valence Bond Theory (VBT), a chemical bond is formed by the overlap of atomic orbitals, resulting in localized electron pairs between two atoms. In Molecular Orbital Theory (MOT), atomic orbitals combine to form molecular orbitals that extend over the whole molecule, allowing electrons to be delocalized. Option A is incorrect because MOT, not VBT, explains the paramagnetic nature of O₂ (due to unpaired electrons in π* orbitals). Option B is false: MOT predicts magnetic moment only if unpaired electrons are present; many diatomic molecules are diamagnetic. Option D is wrong: MOT can produce fractional bond orders (e.g., He₂⁺ has bond order 0.5), while VBT typically gives integer bond orders for single, double, or triple bonds.
2. Which of the following compounds exhibits intramolecular hydrogen bonding?
- p-nitrophenol
- o-nitrophenol
- p-chlorophenol
- Phenol
Answer: o-nitrophenol
In o-nitrophenol, the –OH group and the –NO₂ group are ortho to each other. The hydrogen of –OH can form a hydrogen bond with an oxygen atom of –NO₂ within the same molecule, leading to intramolecular hydrogen bonding. In p-nitrophenol, the groups are para, so hydrogen bonding occurs intermolecularly with other molecules. p-chlorophenol and phenol lack a suitable acceptor (like the nitro group) to form an intramolecular hydrogen bond. Intramolecular hydrogen bonding often reduces boiling point because it prevents extensive intermolecular association.
3. Why is the molecular orbital energy order in B₂, C₂, and N₂ different from that in O₂ and F₂ (i.e., the σ2p orbital lies above the π2p orbitals)?
- In lighter diatomic molecules, the 2s and 2p orbitals have a large energy difference, preventing mixing.
- The size of the 2p orbitals is much larger than that of 2s orbitals in lighter elements.
- There is strong s-p mixing due to a small energy gap between 2s and 2p orbitals in B₂, C₂, and N₂.
- The p orbitals in O₂ and F₂ have higher energy, causing inversion.
Answer: There is strong s-p mixing due to a small energy gap between 2s and 2p orbitals in B₂, C₂, and N₂.
In B₂, C₂, and N₂ (lighter period-2 elements), the 2s and 2p orbitals are closer in energy, leading to significant s-p mixing that raises the σ2p orbital above the π2p orbitals. In O₂ and F₂, the larger energy gap reduces mixing, so σ2p lies below π2p. Option A states the opposite. B is unrelated to energy ordering. D is incorrect; inversion occurs in lighter, not heavier, molecules.
4. Which of the following is a postulate of VSEPR theory?
- The shape of a molecule is determined by the repulsion between bonding and non-bonding electron pairs around the central atom.
- The molecular geometry is determined by the number of atoms only.
- Electrons in the same orbital repel each other more than electrons in different orbitals.
- The bond angles are always 109.5° in all molecules.
Answer: The shape of a molecule is determined by the repulsion between bonding and non-bonding electron pairs around the central atom.
VSEPR theory states that electron pairs (bonding and lone pairs) repel each other and arrange to minimize repulsion, which determines the molecular shape. The other options are incorrect: molecular geometry depends on electron pair geometry, not just atom count; orbital repulsion is not a VSEPR postulate; bond angles vary based on electron pair repulsion.
5. Which of the following correctly distinguishes a bonding molecular orbital from an antibonding molecular orbital?
- Bonding MO has higher energy than the original atomic orbitals; antibonding MO has lower energy.
- Bonding MO has an electron density node between the nuclei; antibonding MO has no node.
- Bonding MO results from constructive interference and has lower energy; antibonding MO results from destructive interference and has higher energy.
- Bonding MO is formed by addition of wavefunctions; antibonding MO is formed by subtraction, and both have equal energy.
Answer: Bonding MO results from constructive interference and has lower energy; antibonding MO results from destructive interference and has higher energy.
In MOT, constructive interference (addition) gives a bonding MO with increased electron density between nuclei and lower energy, while destructive interference (subtraction) gives an antibonding MO with a node between nuclei and higher energy. Option A reverses the energy order. B reverses node presence. D is false because their energies are different.
6. Which of the following is a postulate of Molecular Orbital Theory?
- In a molecule, atomic orbitals retain their individual identities.
- The number of molecular orbitals formed is always less than the number of atomic orbitals combined.
- Atomic orbitals of the same energy combine linearly to form an equal number of molecular orbitals that belong to the entire molecule.
- Electrons in a molecule are always associated with a particular bond between two atoms.
Answer: Atomic orbitals of the same energy combine linearly to form an equal number of molecular orbitals that belong to the entire molecule.
MOT postulates that atomic orbitals combine linearly (LCAO) to form the same number of molecular orbitals, which are delocalized over the whole molecule. Option A is incorrect because atomic orbitals lose their identity upon combination. B is false because the number of MOs equals the number of AOs. D describes the localized bond concept, not MOT.
7. The hybridization of the central atom in sulfuric acid (H2SO4) is:
- sp3
- sp2
- sp3d
- sp3d2
Answer: sp3
In H2SO4, sulfur is the central atom and forms four sigma bonds (two with OH groups and two with oxygen atoms via double bonds, but only one sigma bond each). It has no lone pair, so the steric number is 4, corresponding to sp3 hybridization. The other options would require a different number of sigma bonds or lone pairs.
8. Which of the following molecules has zero dipole moment due to its symmetric shape?
- H2O
- NH3
- BF3
- CHCl3
Answer: BF3
BF3 has a trigonal planar shape with three identical B–F bonds arranged symmetrically. The bond dipoles cancel each other out, resulting in a net zero dipole moment. H2O (bent), NH3 (trigonal pyramidal), and CHCl3 (tetrahedral with different substituents) are all asymmetric and have non-zero dipole moments.
9. Which of the following statements correctly relates bond order with bond length and bond energy?
- Higher bond order corresponds to longer bond length and higher bond energy.
- Higher bond order corresponds to shorter bond length and lower bond energy.
- Higher bond order corresponds to shorter bond length and higher bond energy.
- Bond order has no correlation with bond length or bond energy.
Answer: Higher bond order corresponds to shorter bond length and higher bond energy.
A higher bond order indicates a greater number of bonding electrons relative to antibonding electrons, resulting in a stronger bond (higher bond energy) and a shorter bond length. Options A and B incorrectly describe the relationship. D is false because bond order is directly related to these properties.
10. Which of the following is NOT a characteristic feature of hybridisation?
- Hybrid orbitals have equal energy.
- The number of hybrid orbitals formed equals the number of atomic orbitals mixed.
- Hybrid orbitals are formed by the mixing of orbitals of different atoms.
- Hybridisation is a theoretical concept used to explain molecular geometry.
Answer: Hybrid orbitals are formed by the mixing of orbitals of different atoms.
Hybridisation involves mixing of atomic orbitals of the same atom, not of different atoms. The other options are correct: hybrid orbitals have equal energy, their number equals the number of atomic orbitals mixed, and hybridisation is a theoretical concept to explain geometry.
11. In the molecular orbital energy level diagram of N₂ (nitrogen molecule), which orbital is the highest occupied molecular orbital (HOMO)?
- π2pₓ and π2pᵧ (degenerate pair)
- σ2p_z
- σ*2p_z
- π*2pₓ and π*2pᵧ (degenerate pair)
Answer: σ2p_z
For N₂, due to s-p mixing, the MO energy order is σ2s < σ*2s < π2pₓ = π2pᵧ < σ2p_z < π*2pₓ = π*2pᵧ < σ*2p_z. N₂ has 10 valence electrons, filling up to σ2p_z, which is therefore the HOMO. Option A (π2p) are lower filled orbitals. C and D are unoccupied (higher energy).
12. Which of the following Lewis structures is correct for the carbonate ion (CO₃²⁻)?
- Carbon with a double bond to one oxygen (O has two lone pairs) and single bonds to two oxygens (each O has three lone pairs and a negative charge)
- Carbon with single bonds to all three oxygens (each O has three lone pairs) and no formal charges
- Carbon with a triple bond to one oxygen and single bonds to two oxygens
- Carbon with double bonds to all three oxygens (each O has two lone pairs)
Answer: Carbon with a double bond to one oxygen (O has two lone pairs) and single bonds to two oxygens (each O has three lone pairs and a negative charge)
The most stable Lewis structure for CO₃²⁻ uses one C=O bond and two C–O⁻ bonds, satisfying the octet for all atoms and matching the total valence electrons (24). The other options either violate octet, have incorrect formal charges, or exceed the electron count.