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Hybridization & MO Theory Questions

Home Chemistry Hybridization & MO Theory
Q. The hybridization of carbon in methane (CH₄) is:
sp³
sp
sp³d
sp²

Explanation:

Carbon in CH₄ has four single bonds and no lone pairs, so it uses sp³ hybridization with tetrahedral geometry.

Q. The hybridization of carbon in ethene (C₂H₄) is:
sp³
sp²
sp
sp³d

Explanation:

Each carbon in C₂H₄ forms three sigma bonds and one pi bond, using sp² hybridization with trigonal planar geometry.

Q. The hybridization of carbon in ethyne (C₂H₂) is:
sp³
sp
sp³d²
sp²

Explanation:

Each carbon in C₂H₂ forms two sigma bonds and two pi bonds, using sp hybridization with linear geometry.

Q. The hybridization of nitrogen in ammonia (NH₃) is:
sp³
sp³d
sp²
sp

Explanation:

Nitrogen in NH₃ has 3 sigma bonds and 1 lone pair, giving a steric number of 4 and sp³ hybridization.

Q. The hybridization of oxygen in water (H₂O) is:
sp²
sp³
sp
sp³d

Explanation:

Oxygen in H₂O has 2 sigma bonds and 2 lone pairs, giving a steric number of 4 and sp³ hybridization.

Q. The hybridization of phosphorus in PCl₅ is:
sp³d
sp²
sp³d²
sp³

Explanation:

Phosphorus in PCl₅ has 5 sigma bonds and 0 lone pairs, giving a steric number of 5 and sp³d hybridization.

Q. The hybridization of sulfur in SF₆ is:
sp³
sp³d²
sp²
sp³d

Explanation:

Sulfur in SF₆ has 6 sigma bonds and 0 lone pairs, giving a steric number of 6 and sp³d² hybridization.

Q. The hybridization of carbon in benzene (C₆H₆) is:
sp³d
sp²
sp
sp³

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.

Q. The number of sigma bonds in benzene is:
12
18
3
6

Explanation:

Benzene has 6 C-C sigma bonds and 6 C-H sigma bonds = 12 sigma bonds, plus 6 delocalized pi electrons.

Q. The number of pi electrons in benzene is:
3
12
4
6

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.

Q. Molecular orbital theory describes the formation of a covalent bond in H₂ as:
Combination of two 1s orbitals to form a bonding σ and antibonding σ* MO
Overlap of p orbitals only
Formation of an ionic bond
Transfer of one electron to the other atom

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.

Q. According to MO theory, the bond order of O₂ is:
3
2
2.5
1

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.

Q. According to MO theory, the bond order of N₂ is:
1
3
2.5
2

Explanation:

N₂ has 10 bonding and 4 antibonding electrons, giving bond order = (10−4)/2 = 3, consistent with a triple bond.

Q. The paramagnetism of O₂ is best explained by:
Two unpaired electrons in sigma orbitals
Loss of an electron to form O₂⁺
Presence of a triple bond
Two unpaired electrons in antibonding π* orbitals

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.

Q. The correct energy ordering of molecular orbitals for O₂ and F₂ is:
π_2p < σ_2p < π*_2p < σ*_2p
π*_2p < σ*_2p < π_2p < σ_2p
σ_2p < σ*_2p < π_2p < π*_2p
σ_2p < π_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.

Q. The bond order of the superoxide ion (O₂⁻) is:
2
2.5
1
1.5

Explanation:

O₂⁻ has 8 bonding and 5 antibonding electrons, giving bond order = (8−5)/2 = 1.5.

Q. The HOMO of the N₂ molecule is:
π_2p
π*_2p
σ*_2p
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).

Q. For a stable diatomic molecule, which statement is TRUE?
There are more electrons in bonding MOs than in antibonding MOs
MO theory does not apply to stable molecules
There are more electrons in antibonding MOs than in bonding MOs
Bonding and antibonding MOs are equally filled

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.

Q. The number of unpaired electrons in the O₂ molecule is:
3
2
1

Explanation:

According to MO theory, O₂ has two unpaired electrons in the degenerate π*_2p orbitals (one in each). This makes O₂ paramagnetic.

Q. A σ bond is formed by:
Delocalization over three or more atoms
Head-on (axial) overlap of atomic orbitals
Complete transfer of electrons
Side-on (lateral) overlap of p orbitals

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.

Q. A π bond is formed by:
End-to-end overlap of hybrid orbitals
Complete electron transfer
Head-on overlap of s orbitals
Side-on (lateral) overlap of parallel p orbitals

Explanation:

Pi bonds result from side-on overlap of parallel p orbitals, creating electron density above and below the internuclear axis.

Q. Which is a stronger bond, sigma or pi?
Both are equal in strength
Pi bonds are stronger only in triple bonds
Sigma bonds are generally stronger
Pi bonds are generally stronger

Explanation:

Sigma bonds are stronger than pi bonds due to greater overlap along the internuclear axis, which concentrates electron density directly between the nuclei.

Q. The hybridization of sulfur in SO₂ is:
sp³
sp
sp³d
sp²

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.

Q. The hybridization of the central atom in the carbonate ion (CO₃²⁻) is:
sp
sp³
sp³d
sp²

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.

Q. In molecular orbital theory, the antibonding MO has:
No nodes and lower energy than the bonding MO
The same energy as the bonding MO
Only s-character
A node between the nuclei and higher energy than the bonding MO

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 (*).

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