Questions: Orbital Hybridization: sp, sp², and sp³

Question 1

Nitrogen in ammonia (NH₃) forms three bonds to hydrogen atoms. What is nitrogen's hybridization in ammonia?

• A. sp — nitrogen uses one s and one p orbital for its three bonds
• B. sp² — three bonds require three hybrid orbitals
• C. sp³ — three bonds plus one lone pair give four electron groups requiring four hybrid orbitals
• D. sp³d — nitrogen requires an expanded octet to form three bonds

Answer: (C)

Hybridization is determined by the total number of electron groups around the central atom — including lone pairs, not just bonds. Nitrogen in NH₃ has three N-H bonds and one lone pair, giving four electron groups. Four electron groups require four hybrid orbitals, so nitrogen is sp³ hybridized. The common mistake is counting only bonds (three) and concluding sp² — but lone pairs occupy hybrid orbitals just as bonding pairs do.

Question 2

Which statement correctly distinguishes sp² from sp³ hybridization in carbon?

• A. sp² carbon has one unhybridized p orbital available for π bonding; sp³ carbon does not
• B. sp² carbon forms more total bonds than sp³ carbon
• C. sp³ hybridization requires more energy to form than sp² hybridization
• D. sp² hybridization occurs as a prior step before carbon bonds to three atoms

Answer: (A)

sp² carbon mixes one s + two p orbitals, leaving one p orbital unhybridized and perpendicular to the plane of the three hybrid orbitals. This leftover p orbital is what forms π bonds in double bonds. sp³ carbon mixes all three p orbitals, leaving none for π bonding. Option B is wrong — sp² carbon in ethylene forms the same number of total bonds as sp³ carbon in methane (four). Option D reverses causation: hybridization describes the result of bonding, not a preparatory process.

Question 3

An atom's hybridization can be determined by counting the number of atoms bonded to it, since each bond requires one hybrid orbital.

• T. True
• F. False

Answer: False

This is a critical misconception. Hybridization is determined by the total number of electron groups — both bonds AND lone pairs. Water (H₂O) has two O-H bonds but also two lone pairs on oxygen, giving four electron groups and sp³ hybridization, not sp. Counting only bonded atoms would incorrectly predict sp hybridization for water and give the wrong geometry. Always count lone pairs when determining hybridization.

Question 4

A carbon atom in a C=C double bond is sp² hybridized, with the π bond formed by lateral overlap of two unhybridized p orbitals that are not part of the hybrid orbital set.

• T. True
• F. False

Answer: True

sp² hybridization mixes one s + two p orbitals to produce three sp² hybrid orbitals arranged at 120° in a plane, which form the three σ bonds. The remaining unhybridized p orbital on each carbon sticks out perpendicular to this plane and overlaps sideways with its counterpart on the adjacent carbon to form the π bond. This division — sp² hybrids for σ bonds, unhybridized p for π — correctly explains both the 120° bond angles and the nature of double bonds.

Question 5

Why does the number of hybrid orbitals equal the number of atomic orbitals mixed, and what does this imply about where lone pairs are located?

Model answer: Orbital hybridization conserves the number of orbitals: mixing N atomic orbitals always produces exactly N hybrid orbitals. This means lone pairs must occupy hybrid orbitals — there are no separate 'unhybridized' slots for them. In ammonia, four orbitals mix (one s + three p) to give four sp³ hybrid orbitals; three hold bonding pairs and one holds the lone pair. Lone pairs count toward hybridization just as bonds do.

Orbital arithmetic is the key: start with N atomic orbitals, end with N hybrid orbitals, each holding either a bonding pair or a lone pair. The energy cost of mixing is paid by forming stronger, more directional bonds. Placing lone pairs in hybrid orbitals minimizes electron-electron repulsion by pointing them toward the corners of a tetrahedron (in sp³) rather than leaving them in compact spherical s orbitals. This is why water's bond angle is 104.5° rather than 90° — the lone pairs in sp³-like orbitals push the bonding pairs closer together than unhybridized p orbitals would.