Chemical Bonding – Study Notes
(Optometry – Chemistry)
1. Simple Explanation
Chemical bonds are the forces that hold atoms together in molecules and solids.
- Ionic bonds: transfer of electrons from a metal to a non‑metal, creating oppositely charged ions that attract.
- Covalent bonds: sharing of one or more electron pairs between non‑metal atoms.
- Metallic bonds: delocalised “sea” of electrons shared by a lattice of metal atoms, giving metals their characteristic properties (conductivity, malleability).
In optometry, bonding determines the structure and properties of ocular tissues, drug molecules, and lens materials.
2. Key Points
| Concept | What It Means | Why It Matters in Optometry |
|---|---|---|
| Electronegativity (χ) | Ability of an atom to attract shared electrons. | Determines bond polarity → influences solubility of eye medications. |
| Electronegativity Difference (Δχ) | χ₂ – χ₁. | Δχ > 1.7 → ionic; 0.4 < Δχ < 1.7 → polar covalent; Δχ < 0.4 → non‑polar covalent. |
| Octet Rule | Atoms tend to attain 8 valence electrons (or 2 for H). | Helps draw Lewis structures of pharmaceutical molecules. |
| Lewis Structures | Visual representation of bonding and lone pairs. | Used to predict reactivity of ocular drugs. |
| Bond Order | Single = 1, double = 2, triple = 3. | Higher order → shorter, stronger bonds → affects material durability. |
| VSEPR Theory | Predicts molecular geometry from electron‑pair repulsion. | Geometry affects how molecules fit into ocular structures (e.g., drug binding). |
| Resonance | Delocalisation of electrons over multiple structures. | Stabilises molecules like amino acids in the eye. |
| Ionic vs Covalent vs Metallic | Different bonding types. | Determines physical state, solubility, optical properties of lens materials. |
Bond Strength & Length
| Bond | Typical Bond Energy (kJ mol⁻¹) | Typical Bond Length (Å) |
|---|---|---|
| C–H | 413 | 1.09 |
| C–O | 358 | 1.43 |
| O–O | 146 | 1.48 |
| Na⁺–Cl⁻ (ionic lattice) | 595 | 2.36 |
(Values approximate; used for comparison.)
3. Worked Example
Determine the nature of bonding in NaCl, H₂O, and CO₂
| Molecule | Atoms Involved | χ (approx.) | Δχ | Bond Type | Polarity |
|---|---|---|---|---|---|
| NaCl | Na (0.93) – Cl (3.16) | 2.23 | Ionic | Polar | |
| H₂O | H (2.20) – O (3.44) | 1.24 | Covalent | Polar | |
| CO₂ | C (2.55) – O (3.44) | 0.89 | Covalent | Non‑polar |
Step‑by‑Step
-
Find electronegativities
Na = 0.93, Cl = 3.16, H = 2.20, O = 3.44, C = 2.55. -
Compute Δχ
Na–Cl: 3.16 – 0.93 = 2.23
*H–O: 3.44 – 2.20 =