Questions: Orbital Hybridization and Bonding Models

Question 1

Ammonia (NH₃) has three N–H bonds. A student predicts it is sp² hybridized and has a flat, trigonal planar shape. What is the error in this reasoning?

• A. Nitrogen cannot form sp² hybrid orbitals — only carbon can
• B. The lone pair on nitrogen counts as an electron group and occupies a fourth sp³ orbital, giving a pyramidal shape
• C. NH₃ is actually linear because nitrogen has only one lone pair
• D. sp² hybridization would give bond angles of 90°, not 120°

Answer: (B)

This is the classic hybridization misconception: counting only bonds to predict hybridization and ignoring lone pairs. The rule is to count all electron groups — bonds AND lone pairs. Nitrogen in NH₃ has three bonds plus one lone pair, giving four electron groups. Four electron groups → sp³ hybridization → tetrahedral electron geometry. The three N–H bonds form the three corners of a trigonal pyramid; the lone pair occupies the fourth sp³ orbital and pushes the bonds downward, producing the observed pyramidal molecular shape.

Question 2

Which statement best describes the relationship between hybridization and observed molecular geometry?

• A. Hybridization is a physical process atoms undergo before bonding, rearranging their electrons into new orbitals
• B. Hybridization is a mathematical model that correctly predicts bond angles and equivalent bond lengths, connecting electron configuration to molecular geometry
• C. Hybridization applies only to carbon; other atoms use their unmodified atomic orbitals for bonding
• D. Hybridization and VSEPR are competing theories; only one can be correct for a given molecule

Answer: (B)

Hybridization is a mathematical description, not a physical event. Atoms do not literally remix their orbitals before bonding occurs. The model's value is predictive: given electron configuration and electron group count, it correctly predicts bond angles, bond equivalence, and three-dimensional geometry. For example, it explains why methane's four C–H bonds are identical (all sp³) and why they point toward tetrahedral corners — something the unhybridized orbital picture cannot explain. Hybridization and VSEPR are complementary, not competing.

Question 3

A carbon atom in acetylene (C₂H₂) uses sp hybridization, leaving two unhybridized p orbitals per carbon available for pi bonding.

• T. True
• F. False

Answer: True

In sp hybridization, one s orbital mixes with one p orbital to produce two sp hybrid orbitals pointing in opposite directions (180°). This leaves two p orbitals per carbon untouched — one for each pi bond. In acetylene, the C≡C triple bond consists of one sigma bond (sp–sp overlap) and two pi bonds (from the two pairs of parallel p orbitals). This is why acetylene is linear: the two sp orbitals point in exactly opposite directions.

Question 4

Hybridization is a physical process in which an atom's electrons reorganize into new orbitals before forming bonds with another atom.

• T. True
• F. False

Answer: False

Hybridization is a mathematical model, not a real physical process. Atoms do not literally go through a remixing step before bonding. Rather, hybridization is a convenient theoretical framework that uses linear combinations of atomic orbital wavefunctions to produce hybrid orbitals that better match observed molecular geometries. Its value is entirely predictive — it works because it correctly describes the electron distribution in the bonded molecule, not because the atom physically rearranged itself.

Question 5

Why does ammonia (NH₃) have a pyramidal shape rather than a flat trigonal arrangement, even though it has only three N–H bonds?

Model answer: Ammonia has sp³ hybridization because nitrogen has four electron groups: three N–H bonds plus one lone pair. The lone pair occupies the fourth sp³ hybrid orbital, making the electron geometry tetrahedral. Since molecular shape describes only bond positions (not lone pairs), the three N–H bonds form a trigonal pyramid rather than a flat triangle. The lone pair still occupies space and exerts repulsion, compressing the H–N–H bond angles slightly below the ideal tetrahedral 109.5°.

The key insight is that lone pairs count as electron groups when predicting hybridization and geometry. A student who counts only bonds (three) might incorrectly predict sp² hybridization and a flat shape. Including the lone pair gives four groups → sp³ → tetrahedral electron geometry → pyramidal molecular geometry. This is why hybridization prediction always begins with counting ALL electron groups around the central atom.