10 Hybridization and Molecular Orbital Theory Online Quiz Questions

Valence-bond theory describes orbitals spread over the whole molecule, while molecular orbital theory confines bonds to pairs of neighboring atoms.

Valence-bond theory emphasizes overlaps between orbitals associated mainly with particular atoms, while molecular orbital theory uses orbitals that extend over the molecule.

Both models require electrons to follow fixed paths around individual nuclei.

Both models describe covalent bonds only as end-to-end overlaps.

End-to-end overlap along the internuclear axis forms a sigma bond; side-by-side overlap of parallel p orbitals forms a pi bond.

Side-by-side overlap along the internuclear axis forms a sigma bond; end-to-end overlap of parallel p orbitals forms a pi bond.

Both sigma and pi bonds form only through end-to-end overlap along the internuclear axis.

Both sigma and pi bonds form only through side-by-side overlap of parallel p orbitals.

Two sp hybrids and one unhybridized s orbital

One sp hybrid and two unhybridized p orbitals

Three sp² hybrids

Four sp³ hybrids

Three regions; sp² hybridization; trigonal planar arrangement

Two regions; sp hybridization; linear arrangement

Four regions; sp³ hybridization; tetrahedral arrangement

Two regions; sp² hybridization; linear arrangement

Each carbon is sp³ hybridized, and all four hybrids on each carbon form sigma bonds.

Each carbon is sp hybridized, and two unhybridized p orbitals form two pi bonds.

Each carbon is sp² hybridized, and an unhybridized p orbital on each carbon forms the pi part of the C=C bond.

Each carbon is sp² hybridized, and its three hybrid orbitals form the pi part of the C=C bond.

Two hybrids hold lone pairs and two form bonds; unequal repulsions make the angle about 104.5° rather than 109.5°.

All four hybrids form bonds, so the angle remains exactly 109.5°.

Two hybrids hold lone pairs and two form bonds; the angle is about 120° because oxygen is sp² hybridized.

One hybrid holds a lone pair and three form bonds; the angle is about 104.5°.

They must have identical energies, even if their shapes and symmetry are incompatible.

They must be on the same atom, regardless of energy or symmetry.

They must have opposite symmetry and widely different energies.

They must have suitable energies and compatible shapes and symmetry.

A bonding MO is higher in energy with a node between nuclei; an antibonding MO is lower in energy with increased density between nuclei.

A bonding MO is lower in energy with increased density between nuclei; an antibonding MO is higher in energy with a node between nuclei.

Both types are lower in energy than the atomic orbitals and increase electron density between nuclei.

Both types have a node between nuclei; the asterisk marks the one with lower energy.

Place one electron in each orbital with parallel spins.

Pair both electrons in one orbital before using the other.

Place one electron in each orbital with opposite spins.

Place both electrons in the higher-energy orbital.

The sigma(2p) orbital lies below the pi(2p) orbitals for B₂ through N₂.

The pi(2p) orbitals lie below the sigma(2p) orbital only for O₂ and F₂.

The pi(2p) orbitals lie below the sigma(2p) orbital for B₂ through N₂.

The sigma(2p) and pi(2p) orbitals have the same energy for B₂ through N₂.

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$\mathrm{O_2}$ is diamagnetic because its electrons pair in the $\pi^*$ orbitals.

$\mathrm{O_2}$ is paramagnetic because it has one unpaired electron in a bonding $\sigma$ orbital.

$\mathrm{O_2}$ is diamagnetic because a double bond requires all electrons to be paired.

$\mathrm{O_2}$ is paramagnetic because its last two electrons occupy separate equal-energy $\pi^*$ orbitals, leaving two unpaired electrons; a basic Lewis structure does not account for these unpaired electrons.

sp

sp²

sp³

unhybridized p

Three sigma bonds

Two sigma bonds and one pi bond

One sigma bond and two pi bonds

One sigma bond and one pi bond

Its angle is 109.5° because sp³ hybridization requires that exact angle.

Its angle is about 120° because oxygen has two bonds.

Its angle is 180° because the two O–H bonds are equivalent.

Its angle is about 104.5° because electron regions do not repel equally.

A bonding MO is lower in energy with increased density between the nuclei; an antibonding MO is higher in energy with a node between them.

A bonding MO is higher in energy with a node between the nuclei; an antibonding MO is lower in energy with density between them.

Both are lower in energy, but only the antibonding MO extends over both atoms.

Both have a node between the nuclei, but only the bonding MO is marked with an asterisk.

They must have identical energies; their shapes and symmetry do not matter.

They must have opposite spins and belong to different atoms.

They must have suitable energies and compatible shapes and symmetry.

They must be hybrid orbitals directed along the internuclear axis.

Both electrons pair in one orbital because it is lowest in energy.

The electrons occupy separate equal-energy orbitals before either orbital is paired.

One electron occupies an antibonding orbital only if the bonding orbitals are full.

The electrons must occupy orbitals of different energies.

In O₂, π(2p) lies below σ(2p), as it does in B₂ through N₂.

In O₂, σ(2p) and π(2p) have the same energy.

In O₂, σ(2p) lies above π(2p), as it does in B₂ through N₂.

In O₂, σ(2p) lies below π(2p), unlike the ordering in B₂ through N₂.

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