Chemical Bonding
Chemical bonding explains how atoms combine to make stable molecules. The PMDC MDCAT 2026 syllabus expects you to predict molecular geometry using VSEPR theory, identify hybridisation states, distinguish sigma from pi bonds, calculate dipole moments, and rank bond energies. This is one of the most heavily tested chapters of inorganic chemistry — expect 3–4 MCQs.
VSEPR Theory
The Valence Shell Electron-Pair Repulsion theory (Sidgwick–Powell, refined by Gillespie–Nyholm) predicts the geometry of a molecule from the number of electron pairs around the central atom. Pairs — bonding and lone — arrange themselves to minimise repulsion: lone-pair–lone-pair > lone-pair–bond-pair > bond-pair–bond-pair.
VSEPR geometry — full reference
| Total e-pairs | BP | LP | Geometry / shape | Bond angle | Hybridisation | Example |
|---|---|---|---|---|---|---|
| 2 | 2 | 0 | Linear | 180° | sp | BeCl2, CO2 |
| 3 | 3 | 0 | Trigonal planar | 120° | sp2 | BF3, BCl3 |
| 3 | 2 | 1 | Bent / V-shape | ~118° | sp2 | SO2, O3 |
| 4 | 4 | 0 | Tetrahedral | 109.5° | sp3 | CH4, NH4+ |
| 4 | 3 | 1 | Trigonal pyramidal | 107° | sp3 | NH3, PH3 |
| 4 | 2 | 2 | Bent / V-shape | 104.5° | sp3 | H2O, H2S |
| 5 | 5 | 0 | Trigonal bipyramidal | 120°/90° | sp3d | PCl5 |
| 5 | 4 | 1 | See-saw | ~90°/120° | sp3d | SF4 |
| 5 | 3 | 2 | T-shape | ~90° | sp3d | ClF3, BrF3 |
| 5 | 2 | 3 | Linear | 180° | sp3d | XeF2, I3− |
| 6 | 6 | 0 | Octahedral | 90° | sp3d2 | SF6 |
| 6 | 5 | 1 | Square pyramidal | ~90° | sp3d2 | BrF5 |
| 6 | 4 | 2 | Square planar | 90° | sp3d2 | XeF4 |
Why H2O bond angle < NH3 < CH4
All three central atoms have approximately tetrahedral electron-pair geometry, but lone pairs occupy more space than bond pairs and squeeze the bonding pairs together. CH4 (no LP) = 109.5°; NH3 (1 LP) = 107°; H2O (2 LP) = 104.5°.
Hybridization
Atomic orbitals on the central atom mix to form an equal number of equivalent hybrid orbitals oriented toward the bonded atoms. Hybridisation lets us account for observed bond angles and the equivalence of bonds (e.g. all four C–H bonds in methane).
1 s + 1 p → 2 sp orbitals. Two unhybridised p orbitals remain, perpendicular to each other and to the sp axis. Examples: BeCl2, C in HC≡CH (acetylene), C in CO2. The pi bonds in alkynes use the unhybridised p orbitals.
1 s + 2 p → 3 sp2 orbitals lying in one plane. One unhybridised p orbital remains, perpendicular to the plane. Examples: BF3, C in C2H4 (ethene), aromatic ring carbons.
1 s + 3 p → 4 sp3 orbitals pointing to the corners of a tetrahedron. No unhybridised p orbitals. Examples: CH4, NH3, H2O, sp3 carbons of all alkanes.
sp3d → trigonal bipyramidal (PCl5); sp3d2 → octahedral (SF6). Only available to elements with empty d orbitals (period 3 and below).
Quick formula for hybridisation
H = ½ (V + M − C + A) where V = valence electrons of central atom, M = monovalent atoms attached, C = positive charge, A = negative charge. H = 2 → sp; 3 → sp2; 4 → sp3; 5 → sp3d; 6 → sp3d2.
Sigma and Pi Bond
Two flavours of covalent bond, distinguished by the way the orbitals overlap.
Formed by head-on / axial overlap of orbitals along the inter-nuclear axis. Maximum electron density between the two nuclei. Stronger than a pi bond. Free rotation is possible around a sigma bond. Every covalent bond contains exactly one sigma bond.
Formed by sideways / lateral overlap of two parallel unhybridised p orbitals. Electron density is above and below the inter-nuclear axis — not between the nuclei. Weaker than a sigma bond and prevents free rotation (gives rise to cis/trans isomerism). Found in double and triple bonds in addition to one sigma.
| Property | Sigma (σ) bond | Pi (π) bond |
|---|---|---|
| Mode of overlap | Head-on / axial | Sideways / lateral |
| Electron density | Between the two nuclei (along axis) | Above and below the inter-nuclear axis |
| Strength | Stronger | Weaker |
| Free rotation around bond? | Yes | No (locks geometry → cis/trans) |
| Orbitals used | s-s, s-p, p-p (axial) or hybrid orbitals | Only unhybridised parallel p orbitals |
| Found in | Every covalent bond | Multiple bonds (alongside one σ) |
Counting in multiple bonds
- Single bond = 1 σ.
- Double bond = 1 σ + 1 π (e.g. C=C in ethene).
- Triple bond = 1 σ + 2 π (e.g. C≡C in ethyne).
Dipole Moment
A polar bond produces a separation of partial charges, generating a bond dipole. The molecular dipole is the vector sum of all bond dipoles.
μ = q × d, where q is the magnitude of the partial charge and d is the bond length. Unit: Debye (D); 1 D = 3.336 × 10−30 C·m. Direction: from δ+ to δ−.
Examples
- HCl: μ = 1.08 D (polar covalent bond).
- H2O: μ = 1.85 D (bent shape; bond dipoles do not cancel).
- NH3: μ = 1.47 D (pyramidal; lone pair adds to dipole).
- CO2: μ = 0 D (linear; equal and opposite C=O dipoles cancel).
- CH4: μ = 0 D (tetrahedral; symmetric).
- BF3: μ = 0 D (trigonal planar; symmetric).
Trend: Polarity follows electronegativity difference. Electronegativity (Pauling): F (4.0) > O (3.5) > N (3.0) > Cl (3.0) > C (2.5) > H (2.1). ΔEN of 0–0.4 → mostly non-polar covalent; 0.4–1.7 → polar covalent; > 1.7 → mostly ionic.
Bond Energy
Bond energy (or bond dissociation enthalpy) is the energy required to break one mole of a particular covalent bond in the gas phase. Units: kJ/mol. Larger bond energy means stronger and (usually) shorter bond.
Trends in bond energy
- Within the same group, bond energy decreases down the column — longer bond, weaker overlap. H–F (568 kJ/mol) > H–Cl (432) > H–Br (366) > H–I (298).
- For the same two atoms: triple > double > single. C≡C (839 kJ/mol) > C=C (614) > C–C (347).
- Bond length is inversely related to bond energy — the shorter the bond, the stronger.
- N≡N is exceptionally strong (945 kJ/mol) — explains the inertness of dinitrogen.
Why σ is stronger than π
The axial (head-on) overlap that forms a sigma bond places maximum electron density along the line joining the nuclei — the most stabilising arrangement. Sideways (lateral) overlap of p orbitals is less efficient because the orbitals are parallel rather than pointing at each other, so the pi bond contains less electron density between the nuclei and is therefore weaker (~ 60–65 percent of a sigma bond).
Worked MCQs
Five MCQs that capture the high-yield testing patterns for this chapter. Read the explanation even when you get the answer right — that's where the deeper concept lives.
Q1. The shape of NH3 according to VSEPR theory is:
Nitrogen has four electron pairs (3 BP + 1 LP) → tetrahedral electron-pair geometry, but the molecular shape (which ignores lone pairs) is trigonal pyramidal. The lone pair compresses the H–N–H angle from 109.5° to 107°.
Q2. The hybridisation of carbon atoms in ethyne (HC≡CH) is:
Each carbon in ethyne forms two sigma bonds (to H and to the other C). Two sigma frameworks → 2 hybrid orbitals → sp. The two remaining unhybridised p orbitals on each C overlap sideways to form the two pi bonds of the triple bond.
Q3. Which of the following has zero dipole moment?
CO2 is linear (O=C=O); the two equal but oppositely directed C=O bond dipoles cancel exactly. H2O and NH3 are bent and pyramidal respectively, so their bond dipoles do not cancel; HCl has a single polar bond.
Q4. A C=C double bond consists of:
Every multiple bond contains exactly one sigma (head-on overlap) plus one or more pi bonds (sideways overlap of unhybridised p orbitals). For a double bond there is one of each; the pi bond is the one broken in addition reactions.
Q5. Among H–F, H–Cl, H–Br and H–I, the bond energy is highest for:
H–F has the shortest bond and the largest electronegativity difference, giving it the strongest overlap and hence the largest bond energy (568 kJ/mol). Bond strength falls steadily down the halogen group as atomic radius grows and overlap weakens.
Quick Recap
- VSEPR shapes: linear (2), trigonal planar (3), tetrahedral (4), trigonal bipyramidal (5), octahedral (6).
- Lone pairs squeeze bond angles: CH4 109.5° > NH3 107° > H2O 104.5°.
- Hybridisation: sp = linear (180°), sp2 = trigonal planar (120°), sp3 = tetrahedral (109.5°).
- Sigma = axial overlap, stronger; pi = lateral overlap, weaker; double = 1σ + 1π; triple = 1σ + 2π.
- Dipole moment μ = q × d (Debye); zero for symmetric molecules (CO2, CH4, BF3).
- Bond energy unit kJ/mol; H–F highest; bond energy ↑ with bond order, ↓ down a group.