Sigma and Pi Bonds & Hybrid Orbitals – Vocabulary

Required Documents

• Summary handout: “Geometry and Polarity for Molecules and Polyatomic Ions 2” (posted on the module’s Learning Activities page)

Covalent-Bond Formation Basics

• A covalent bond forms when each bonding atom supplies one unpaired valence electron.
  • These singly occupied orbitals must overlap to share the two electrons.
• For every covalent-bond type you should be able to state:
  • Which specific orbitals overlap.
  • The spatial (geometric) nature of the overlap.
  • The relative bond strength.
  • Whether the bond is found in single, double or triple bonds.

Theories of Covalent Bonding

• Valence Bond (VB) Theory
  • Basis for VSEPR shapes.
  • Explains actual molecular geometries reasonably well.
• Molecular Orbital (MO) Theory
  • More mathematically involved; explains magnetic & other electronic properties.
  • Not examined in this course.

Sigma ($\sigma$) vs Pi ($\pi$) Bonds

• Two fundamental covalent-bond classes:
  • $\sigma$ bond
  • Formed by direct (head-on) overlap of orbitals.
  • Overlap region lies along the internuclear axis.
  • $\pi$ bond
  • Formed by lateral (side-to-side) overlap of two unhybridized $p$ orbitals.
  • Overlap region is above & below (or in front of & behind) the internuclear axis.

Relative Strength

• Bond strength ≈ attraction of both nuclei for the shared e⁻ pair.
• $\sigma$ bonds are stronger than $\pi$ bonds because head-on overlap allows greater electron density between nuclei.

Relationship to Bond Order

• Single bond = 1 $\sigma$
• Double bond = 1 $\sigma$ + 1 $\pi$
• Triple bond = 1 $\sigma$ + 2 $\pi$
  • Thus a double bond is stronger than a single bond but < 2 × as strong, because the extra $\pi$ contribution is weaker.
  • Triple bond likewise < 3 × single-bond strength.

Which Orbitals Overlap?

• $\pi$ bond: strictly $p$–$p$ sideways overlap.
• $\sigma$ bond:
  • $1s$ vs hybrid orbital when H is involved.
  • Hybrid orbital vs hybrid orbital (or vs unhybridized $p$) for other atoms.
• Non-bonding (lone-pair) electrons also occupy hybrid orbitals.

Hybridization – Core Ideas

• Shapes/orientations of isolated-atom orbitals do not predict observed molecular angles → orbitals must reorganize when bonding.
• Hybrid orbitals form only in atoms that are covalently bonded; they do not exist in isolated atoms.
• Generating $n$ hybrid orbitals consumes $n$ atomic orbitals; $\Sigma n$ is conserved.
• All hybrids in a set are equivalent in shape & energy.
• The hybridization adopted is dictated by electron-pair geometry from VSEPR.

Mapping Domains → Hybrids → Angles

• 4 electron domains → ideal $109.5^{\circ}$ → $sp^{3}$
• 3 electron domains → $120^{\circ}$ → $sp^{2}$
• 2 electron domains → $180^{\circ}$ → $sp$

Example 1 – CFH₃ (Tetrahedral, $sp^{3}$)

• Lewis structure: central C bonded to F + 3 H.
• C: four $\sigma$ bonds → needs 4 hybrid orbitals.
  • One $2s$ electron is promoted to the vacant $2p$ to yield four unpaired e⁻.
  • $1s + 3p \rightarrow 4\,sp^{3}$ hybrids oriented tetrahedrally (109.5°).
• F: one $\sigma$ bond + three lone pairs = 4 domains → also $sp^{3}$.
  • Even though the original three $2p$ orbitals are at 90°, hybridization re-orients them to 109.5°.
• Resulting overlaps
  • C–F $\sigma$: $sp^{3}<em>{\text{C}}$ ∩ $sp^{3}</em>{\text{F}}$
  • C–H $\sigma$ (×3): $sp^{3}<em>{\text{C}}$ ∩ $1s</em>{\text{H}}$
  • F lone pairs: occupy remaining $sp^{3}_{\text{F}}$ hybrids.

Example 2 – COH₂ (Formaldehyde, Trigonal Planar $sp^{2}$)

• Lewis: O=CH₂ (double bond C=O).
• C: 3 $\sigma$ + 1 $\pi$ → needs 3 hybrids + leave 1 $p$ unhybridized.
  • $1s + 2p \rightarrow 3\,sp^{2}$ hybrids (planar, 120°).
• O: 1 $\sigma$ + 2 lone pairs + 1 $\pi$ → also 3 hybrids + 1 $p$.
• Overlaps
  • C–O $\sigma$: $sp^{2}<em>{\text{C}}$ ∩ $sp^{2}</em>{\text{O}}$
  • C–H $\sigma$ (×2): $sp^{2}<em>{\text{C}}$ ∩ $1s</em>{\text{H}}$
  • C=O $\pi$: lateral overlap $p<em>{\text{C}}$ ∩ $p</em>{\text{O}}$ (above & below plane).
  • O lone pairs: fill remaining two $sp^{2}_{\text{O}}$ hybrids.

Example 3 – C₂H₂ (Acetylene, Linear $sp$)

• Lewis & actual geometry: H–C≡C–H (180°).
• Each C: 2 $\sigma$ + 2 $\pi$ → needs 2 hybrids; leave 2 $p$ orbitals unhybridized.
  • $1s + 1p \rightarrow 2\,sp$ hybrids (linear).
• Overlaps
  • C≡C $\sigma$: $sp<em>{\text{C1}}$ ∩ $sp</em>{\text{C2}}$
  • C–H $\sigma$ (×2): $sp<em>{\text{C}}$ ∩ $1s</em>{\text{H}}$
  • Two $\pi$ bonds: orthogonal $p_{\text{C}}$ pairs overlap sideways in two perpendicular planes.

Comprehensive Domain/Hybrid/Angle Table

Skills & Expectations

• From any Lewis structure you must be able to:
  • Label each bond as $\sigma$ or $\pi$.
  • Specify the orbital on each atom that participates in that bond (e.g.
    $sp^{3}$, $sp^{2}$, $sp$, $p$, $1s$).
• Drawing 3-D “balloon” hybrid-orbital pictures is not required.

Practice Problem Outline (Slide 25 Recap)

• Given a complex organic fragment containing F, C, N, O & H atoms, you filled a table:
  • For every bond you identified: type (σ/π), domain count → hybrid on first atom, domain count → hybrid on second atom.
  • Examples:
  • F₁–C₂ $\sigma$: F (4 domains → $sp^{3}$) vs C (3 domains → $sp^{2}$).
  • C₄≡N₅ triple bond contains 1 $\sigma$ (sp–sp) + 2 $\pi$ (p–p) overlaps.

Energetic Footnote – s→p Promotion

• Occasionally an $s$ electron is promoted to $p$ (e.g. C in CFH₃) to create enough unpaired electrons.
  • Energy cost < energy released by forming additional bonds, so overall thermodynamically favorable.

Real-World Implications & Connections

• Magnetic properties (paramagnetism/diamagnetism) are better rationalized by MO theory; VB theory remains preferred for routine shape and hybridization predictions.
• Bond strength hierarchy (triple > double > single) underpins reactivity trends in organic & inorganic chemistry.
• Hybridization provides a quantitative basis for explaining macroscopic molecular shape → critical for drug binding, material properties, enzymatic specificity.
• Understanding $\pi$ bonds clarifies conjugation & resonance (delocalized $\pi$ systems), foundational for UV–Vis absorption and color in organic dyes.

What Comes Next in the Course

• Unit 6 will focus on quantitative composition:
  • Percent mass of elements in compounds.
  • Empirical & molecular formula determination from experimental data.
  • Solution concentration calculations (molarity, mass %, etc.).