Hybridization & Polarity

You’ll be able to

Assign sp, sp², or sp³ hybridization from the number of electron domains
Distinguish sigma (σ) from pi (π) bonds in single, double, and triple bonds
Determine molecular polarity by combining bond dipoles with geometry

Hybridization mixes orbitals to match the geometry

To point bonds in the directions VSEPR demands, an atom blends its atomic s and p orbitals into equivalent hybrid orbitals. The count of hybrids must equal the number of electron domains. So hybridization reads straight off the domain count — the same number that gave you the geometry.

Hybridization from electron domains

2 domains → sp (linear) · 3 → sp² (trigonal planar) · 4 → sp³ (tetrahedral)

Superscript = number of p orbitals mixed with the one s orbital; total hybrids = domains.

Sigma and pi bonds

A sigma (σ) bond is an end-to-end overlap along the axis joining two nuclei; every single bond is one σ bond. A pi (π) bond is a side-by-side overlap of unhybridized p orbitals, above and below that axis. The pattern: a single bond = 1 σ; a double bond = 1 σ + 1 π; a triple bond = 1 σ + 2 π. Only the first bond between two atoms is ever a sigma bond.

From bond dipoles to molecular polarity

A polar bond forms when two atoms differ in electronegativity — the shared electrons sit closer to the more electronegative atom, giving it a partial negative charge (δ⁻) and its partner a partial positive charge (δ⁺). Each polar bond is a vector (a bond dipole). A molecule is polar only if these vectors do not cancel. Symmetric shapes cancel their dipoles and are nonpolar even with polar bonds; an asymmetric shape — often caused by lone pairs — leaves a net dipole.

Worked example

Compare the polarity of CO₂ and H₂O, and give the hybridization of the central atom in each.

1.CO₂: carbon has 2 domains (two double bonds, no lone pairs) → sp hybridized, linear. Each C=O bond is polar toward O.
2.Because CO₂ is linear and symmetric, the two equal C→O dipoles point exactly opposite and cancel → net dipole zero → nonpolar.
3.H₂O: oxygen has 4 domains (2 bonds + 2 lone pairs) → sp³ hybridized, bent. Each O–H bond is polar toward O.
4.Because H₂O is bent, the two O–H dipoles do not oppose; they add to a net dipole pointing toward the oxygen → polar.

Answer: CO₂ is nonpolar (sp, linear — dipoles cancel); H₂O is polar (sp³, bent — dipoles add). Geometry, not bond polarity alone, decides.

Watch out

Polar bonds do not guarantee a polar molecule. CO₂, CCl₄, and BF₃ all contain polar bonds yet are nonpolar because their symmetric shapes cancel every dipole. Always check the geometry before deciding.

Checkpoint

What is the hybridization of the carbon atom in ethyne (acetylene), H–C≡C–H?

Checkpoint

How many sigma (σ) and pi (π) bonds are in a nitrogen molecule, N≡N?

Checkpoint

Which of these molecules is nonpolar despite containing polar bonds?

Checkpoint

In an H–Cl bond (EN: H = 2.2, Cl = 3.0), where do the partial charges lie?

On the exam

The exam's favorite two-step: first use domain count for both geometry AND hybridization (they come from the same number), then judge polarity by asking whether the shape lets the bond dipoles cancel. Symmetry with identical outer atoms → nonpolar; asymmetry or lone pairs → polar.

Answer the 4 checkpoints as you read.