Chemical Bonding — Ionic, Covalent & Metallic Bonds Explained

Why do atoms bond at all? Because isolated atoms are generally not at their lowest energy state. Bonding releases energy and creates a more stable arrangement. Understanding why bonds form makes everything else — Lewis structures, shapes, hybridization — fall into place logically rather than feeling like a list to memorise.

Types of Chemical Bonds

There are three primary types of chemical bonds, each arising from a different mechanism:

Ionic Bonding

An ionic bond forms when one atom transfers electrons to another, creating oppositely charged ions that attract each other electrostatically.

When does ionic bonding occur? When the electronegativity difference between two atoms is large (generally > 1.7 on Pauling scale). Typically: a metal + a non-metal.

Formation of NaCl:

• Na (2,8,1) → Na⁺ (2,8) + e⁻ [IE₁ = 496 kJ/mol]
• Cl (2,8,7) + e⁻ → Cl⁻ (2,8,8) [EA = 349 kJ/mol released]
• Na⁺ and Cl⁻ attract → lattice energy released (−786 kJ/mol for NaCl)

The overall process is energetically favourable because the lattice energy released far exceeds the energy needed for ionization.

Properties of Ionic Compounds

• High melting/boiling points: Strong electrostatic forces in the crystal lattice
• Soluble in polar solvents (water), insoluble in non-polar solvents
• Conduct electricity in molten state or solution (free ions), but not in solid state (ions are fixed in lattice)
• Brittle: When stress is applied, like-charged layers align → repulsion → the crystal shatters
• No directional bonding: The electrostatic force acts equally in all directions

Covalent Bonding

A covalent bond forms when atoms share electron pairs. Both nuclei attract the shared electrons, holding the atoms together.

When does covalent bonding occur? When the electronegativity difference is small (< 1.7), usually between two non-metals.

Lewis Dot Structures

Lewis structures show all valence electrons explicitly: bonding pairs (as lines or dots between atoms) and lone pairs (dots on individual atoms).

Steps to draw a Lewis structure:

1. Count total valence electrons (add them all up; subtract 1 per positive charge; add 1 per negative charge)
2. Identify the central atom (usually the least electronegative, least hydrogen, most bonding sites)
3. Place single bonds between central atom and all outer atoms
4. Complete octets on outer atoms first using lone pairs
5. Give any remaining electrons to the central atom
6. If central atom has < 8 electrons, form double or triple bonds by converting lone pairs

Example — CO₂:

• Total valence electrons: C(4) + O(4+4) = 16
• Single bonds C–O use 4 electrons → 12 remaining
• Complete O octets: each O gets 3 lone pairs → 12 electrons used
• C has 0 electrons left — octet not satisfied
• Convert one lone pair from each O into a double bond → O=C=O
• Final: O=C=O, carbon has 8 electrons, each O has 4 lone pair electrons + 4 bonding = 8 ✓

Octet Rule and Its Exceptions

The octet rule states that atoms tend to form bonds until they are surrounded by 8 electrons. But there are three important exceptions:

1. Odd-electron molecules (radicals): NO, NO₂, ClO₂ — have an odd number of valence electrons, so one atom must have an unpaired electron. Full octet is impossible for all atoms.

2. Incomplete octet (electron-deficient): BF₃ (6 electrons around B), BeCl₂ (4 around Be), AlCl₃ (6 around Al) — these compounds exist as stable species with less than 8 electrons on the central atom. They are strong Lewis acids precisely because they can accept a lone pair.

3. Expanded octet (hypervalent compounds): PCl₅ (10 electrons around P), SF₆ (12 around S), ClF₃ (10 around Cl) — possible only for Period 3 and beyond because they have available d-orbitals to accommodate extra electrons.

Formal Charge

Formal charge = (Valence electrons of free atom) − (Lone pair electrons) − ½(Bonding electrons)

The structure with the smallest formal charges (closest to zero) on all atoms is the most stable Lewis structure. For resonance structures, prefer the one that puts negative formal charge on the more electronegative atom.

Resonance

When a single Lewis structure cannot adequately represent a molecule, we use resonance structures. The actual molecule is a resonance hybrid — a weighted average of all contributing structures.

Examples:

• Benzene (C₆H₆): Two Kekulé structures contribute equally. All C–C bonds are identical, with bond order 1.5.
• Ozone (O₃): Two equivalent structures; the central O has formal charge +1, terminal Os have −1 and 0 alternating. Bond order = 1.5.
• SO₄²⁻: Four equivalent resonance structures; all S–O bond lengths are equal.

VSEPR Theory

VSEPR (Valence Shell Electron Pair Repulsion) theory predicts molecular geometry based on one principle: electron pairs (bonding and lone pairs) around the central atom repel each other and arrange themselves to maximise separation.

Repulsion Order

Lone pair–Lone pair > Lone pair–Bond pair > Bond pair–Bond pair

This hierarchy is critical for understanding why bond angles in water (104.5°) and ammonia (107°) are less than the tetrahedral angle (109.5°).

VSEPR Shapes — Complete Table

Hybridization

Hybridization is the concept of mixing atomic orbitals of similar energies to form new hybrid orbitals with equal energy and different spatial orientation. It explains observed molecular geometries that pure s and p orbitals can't account for.

sp Hybridization

One s + one p → two sp hybrid orbitals (180° apart, linear geometry)

Remaining: 2 pure p orbitals (available for π bonds)

Examples: BeCl₂, C₂H₂ (acetylene), CO₂

In C₂H₂: Each carbon is sp hybridised. The two sp orbitals on each C form the C–H σ bond and the C–C σ bond. The two remaining p orbitals on each C form two π bonds → triple bond (1σ + 2π).

sp² Hybridization

One s + two p → three sp² hybrid orbitals (120° apart, trigonal planar)

Remaining: 1 pure p orbital (for one π bond)

Examples: BF₃, C₂H₄ (ethylene), SO₃, benzene, NO₃⁻, CO₃²⁻

In C₂H₄: Each carbon has three sp² orbitals (2 for C–H bonds, 1 for C–C σ bond) and one leftover p orbital. The two p orbitals side-overlap to form the π bond → double bond = 1σ + 1π.

sp³ Hybridization

One s + three p → four sp³ hybrid orbitals (109.5° ideal, tetrahedral)

Examples: CH₄, NH₃, H₂O, CCl₄, SiH₄

In CH₄: Carbon mixes 2s and all three 2p → four equivalent sp³ orbitals, each forming a σ bond with H. Bond angle = 109.5°, perfectly tetrahedral.

In NH₃: N has four sp³ orbitals — three used for N–H bonds, one holds a lone pair. Lone pair repulsion compresses bond angle to 107°.

In H₂O: O has four sp³ orbitals — two for O–H bonds, two hold lone pairs. Greater lone pair–lone pair repulsion reduces bond angle to 104.5°.

sp³d Hybridization

One s + three p + one d → five sp³d hybrid orbitals (trigonal bipyramidal)

Examples: PCl₅, SF₄, ClF₃, XeF₂

PCl₅ has five sp³d orbitals in a trigonal bipyramidal arrangement: three equatorial bonds at 120° and two axial bonds at 90° to the equatorial plane.

sp³d² Hybridization

One s + three p + two d → six sp³d² hybrid orbitals (octahedral, 90°)

Examples: SF₆, PCl₆⁻, XeF₄, [Co(NH₃)₆]³⁺

SF₆: S uses one 3s, three 3p, and two 3d orbitals → six equivalent sp³d² orbitals pointing toward the corners of a regular octahedron. All S–F bonds identical; all angles 90°.

Metallic Bonding

In metals, valence electrons are delocalised — they don't belong to any specific atom but roam freely through the entire metal lattice. This "sea of electrons" holds positively charged metal cores in place.

Properties explained by metallic bonding:

• High electrical conductivity: Free electrons carry charge
• High thermal conductivity: Free electrons transfer kinetic energy
• Malleability and ductility: Metal layers can slide over each other; electron sea adjusts and maintains bonding
• Metallic lustre: Free electrons absorb and re-emit light across all visible frequencies
• High melting points (generally): Strong attraction between electron sea and metal cations; stronger for smaller, higher-charged cations

Molecular Orbital Theory (MOT) — Basics

MOT treats electrons as delocalised over the entire molecule. Atomic orbitals combine to form molecular orbitals (MOs): bonding MOs (lower energy, electron-rich between nuclei) and antibonding MOs (higher energy, node between nuclei, denoted with *).

MO Filling Order for Period 2 Diatomics

For Li₂ to N₂ (before O₂): σ1s < σ1s < σ2s < σ2s < π2p = π2p < σ2p < π2p = π2p < σ*2p

For O₂ and beyond: σ1s < σ1s < σ2s < σ2s < σ2p < π2p = π2p < π2p = π2p < σ*2p

The key change: σ(2p) drops below the two π(2p) orbitals for O₂ and F₂.

Key MOT Examples

O₂ (16 electrons): Bond order = (10 − 6)/2 = 2 → double bond ✓ Two electrons enter degenerate π*2p orbitals with parallel spins → O₂ is paramagnetic (has unpaired electrons). This is one of MOT's greatest triumphs — it explains O₂'s paramagnetism, which Lewis structure cannot.

N₂ (14 electrons): Bond order = (10 − 4)/2 = 3 → triple bond ✓ No unpaired electrons → N₂ is diamagnetic ✓

NO (15 electrons): Bond order = (10 − 5)/2 = 2.5 One unpaired electron in π* → paramagnetic NO⁺ (loses the π* electron): bond order = 3 → stronger and shorter bond than NO itself

He₂ (4 electrons): Bond order = (2 − 2)/2 = 0 → He₂ does not exist ✓

Hydrogen Bonding

Hydrogen bonding is a special type of dipole–dipole interaction that forms when H is bonded to a highly electronegative atom (F, O, or N) with lone pairs. The partially positive H is attracted to the lone pair on an electronegative atom of another molecule.

Conditions: H must be bonded to F, O, or N; the electronegative atom of the acceptor must have a lone pair.

Types:

• Intermolecular H-bonding: Between different molecules (H₂O, HF, NH₃, alcohols, carboxylic acids)
• Intramolecular H-bonding: Within the same molecule (o-nitrophenol, salicylaldehyde)

Why does H-bonding matter for exam questions?

• H₂O has an anomalously high boiling point (100°C) compared to H₂S (−60°C) — explained by extensive H-bonding network
• HF has a higher boiling point than HCl despite lower molecular mass
• Intramolecular H-bonded compounds (o-nitrophenol) have lower boiling points and are more soluble in organic solvents than their intermolecular H-bonded isomers (p-nitrophenol)

5 Common Mistakes Students Make

Mistake 1: Counting only bonding pairs for VSEPR, ignoring lone pairs. Lone pairs absolutely affect molecular shape — they compress bond angles. Water has 4 electron pairs (2 bonding, 2 lone), and it's the lone pairs that push the bond angle from 109.5° to 104.5°.

Mistake 2: Saying NH₃ and H₂O have different hybridization. Both N in NH₃ and O in H₂O are sp³ hybridised. The difference is in how many sp³ orbitals hold lone pairs vs bonds, which affects geometry but not hybridization type.

Mistake 3: Forgetting that BF₃ violates the octet rule. BF₃ has only 6 electrons around boron. It is electron-deficient, hence a Lewis acid. Many students wrongly assume all stable molecules obey the octet rule.

Mistake 4: Confusing bond order with number of bonds drawn. Bond order from MOT accounts for both bonding and antibonding electrons. Bond order of O₂⁻ is 1.5 (not 2), because the extra electron goes into an antibonding orbital.

Mistake 5: Writing formal charges incorrectly in resonance structures. Formal charge = Valence electrons − Lone pair electrons − ½(Bonding electrons). A common error is dividing the wrong quantity by 2, or forgetting to count lone pairs. Practice on CO₂, SO₃, and NO₃⁻ to get fluent.

Practice Questions

Q1. Draw the Lewis structure of SO₂ and identify the formal charges.

Q2. Predict the shape and bond angle of NH₃ using VSEPR theory.

Q3. Why is the bond angle in H₂O (104.5°) less than in NH₃ (107°)?

Q4. What is the hybridization of the central atom in PCl₅? Describe its shape.

Q5. Calculate the bond order of O₂ and explain why it is paramagnetic.

Q6. Compare the melting points of NaCl and CCl₄. Explain in terms of bonding.

Q7. Explain why HF has a higher boiling point than HCl, despite having a lower molecular mass.

Q8. XeF₄ is square planar while SF₄ is see-saw shaped. Why?

Frequently Asked Questions

What is the difference between σ and π bonds?

A σ (sigma) bond forms by head-on (axial) overlap of orbitals. Electron density is concentrated along the bond axis. Every single, double, and triple bond contains exactly one σ bond. A π (pi) bond forms by side-on (lateral) overlap of p orbitals. Electron density is above and below the bond axis. Double bonds have 1σ + 1π; triple bonds have 1σ + 2π. π bonds are weaker than σ bonds and prevent rotation about the bond axis (hence cis-trans isomerism in alkenes).

How do I know whether to use sp, sp², or sp³ hybridization?

Count the steric number (SN) around the central atom: SN = number of atoms bonded to central atom + number of lone pairs on central atom. SN = 2 → sp; SN = 3 → sp²; SN = 4 → sp³; SN = 5 → sp³d; SN = 6 → sp³d². This formula works for all main-group compounds without needing to analyse orbital mixing from scratch.

Why does CO₂ have no net dipole but H₂O does?

Both CO₂ and H₂O have polar bonds. CO₂ is linear — the two C=O bond dipoles point in exactly opposite directions and cancel. H₂O is bent (104.5°) — the two O–H bond dipoles do not cancel. The resultant dipole points from the midpoint of the H–H line toward O. Molecular geometry determines whether bond dipoles cancel. Always check shape before predicting polarity.

What makes a compound a Lewis acid vs Lewis base?

A Lewis acid is an electron pair acceptor (electron-deficient species: BF₃, AlCl₃, Fe³⁺). A Lewis base is an electron pair donor (has lone pairs: NH₃, H₂O, Cl⁻). This concept is broader than the Brønsted–Lowry definition (which involves H⁺ transfer). All Brønsted bases are Lewis bases, but not vice versa.

Is water truly sp³ hybridised? Teachers debate this.

The sp³ hybridisation of O in water is the standard NCERT/JEE treatment and is correct for exam purposes. Formal MO calculations show some deviation, but for predicting shape, bond angles, and exam questions, sp³ hybridisation for O in H₂O is the accepted framework. Stick with it.

What is the significance of bond order in MOT?

Bond order tells you the net number of bonding electron pairs. Higher bond order = shorter bond length = stronger bond = higher bond dissociation energy. For example: N₂ (bond order 3, bond length 110 pm, very stable), O₂ (bond order 2, 121 pm), F₂ (bond order 1, 141 pm). Bond order can be fractional — NO has bond order 2.5, meaning the bond is between a double and triple bond in character.