Ionic Bonding

Atoms form ions to achieve a stable noble-gas electron configuration (full outer shell). Whether the atom loses or gains electrons depends on whether it is closer to the previous or next noble gas in the periodic table.

Metals (Groups 1-3): LOSE electrons → form POSITIVE ions (cations).

Group	Example	Outer e⁻	Loses	Ion
1	Na (2,8,1)	1	1 e⁻	Na⁺ (2,8)
1	K (2,8,8,1)	1	1 e⁻	K⁺
2	Mg (2,8,2)	2	2 e⁻	Mg²⁺
2	Ca (2,8,8,2)	2	2 e⁻	Ca²⁺
3	Al (2,8,3)	3	3 e⁻	Al³⁺

For example: $\text{Na} \rightarrow \text{Na}^+ + e^-$ — the single outer-shell electron is lost, leaving the previous shell (2, 8) — neon configuration.

Non-metals (Groups 5-7): GAIN electrons → form NEGATIVE ions (anions).

Group	Example	Outer e⁻	Gains	Ion
5	N (2,5)	5	3 e⁻	N³⁻ (2,8)
6	O (2,6)	6	2 e⁻	O²⁻ (2,8)
6	S (2,8,6)	6	2 e⁻	S²⁻ (2,8,8)
7	F (2,7)	7	1 e⁻	F⁻ (2,8)
7	Cl (2,8,7)	7	1 e⁻	Cl⁻ (2,8,8)

For example: $\text{Cl} + e^- \rightarrow \text{Cl}^-$ — the chlorine atom gains one electron, completing its outer shell to 8 electrons (argon configuration).

Quick rule. For metals, the charge magnitude equals the group number. For non-metals, the charge magnitude equals (8 minus the group number), with a negative sign. Group 4 elements usually share electrons (covalent bonding) rather than form ions, and Group 0 noble gases form no ions at all.

An ionic compound is always electrically neutral — total positive charge = total negative charge. Use this rule to deduce the chemical formula from the ion charges.

Procedure.

1. Write the cation and anion with their charges.
2. Find the lowest common multiple (LCM) of the charge magnitudes.
3. The number of each ion needed = LCM ÷ that ion's charge magnitude.
4. Write the formula with the cation first and the anion second.

Worked examples.

• Sodium chloride. Na⁺ + Cl⁻. LCM = 1. So 1 Na⁺ + 1 Cl⁻ → NaCl.
• Magnesium oxide. Mg²⁺ + O²⁻. LCM = 2. So 1 Mg²⁺ + 1 O²⁻ → MgO.
• Calcium chloride. Ca²⁺ + Cl⁻. LCM = 2. So 1 Ca²⁺ + 2 Cl⁻ → CaCl₂.
• Sodium oxide. Na⁺ + O²⁻. LCM = 2. So 2 Na⁺ + 1 O²⁻ → Na₂O.
• Aluminium chloride. Al³⁺ + Cl⁻. LCM = 3. So 1 Al³⁺ + 3 Cl⁻ → AlCl₃.
• Aluminium oxide. Al³⁺ + O²⁻. LCM = 6. So 2 Al³⁺ + 3 O²⁻ → Al₂O₃.
• Magnesium nitride. Mg²⁺ + N³⁻. LCM = 6. So 3 Mg²⁺ + 2 N³⁻ → Mg₃N₂.

Polyatomic ions require brackets when more than one is present:

• Calcium hydroxide: Ca²⁺ + 2 OH⁻ → Ca(OH)₂.
• Ammonium sulfate: 2 NH₄⁺ + SO₄²⁻ → (NH₄)₂SO₄.
• Aluminium nitrate: Al³⁺ + 3 NO₃⁻ → Al(NO₃)₃.
• Calcium carbonate: Ca²⁺ + CO₃²⁻ → CaCO₃ (no brackets, only one CO₃²⁻).

A dot-and-cross diagram shows the transfer of outer-shell electrons during ionic bond formation. Conventions:

• Draw only the outer-shell electrons (inner shells are not relevant for bonding).
• Use dots (•) for one element's electrons and crosses (×) for the other's. The choice is arbitrary but must be consistent.
• After electron transfer, show the resulting ions in square brackets with the charge as a superscript: [Na]⁺, [Cl]⁻, [Mg]²⁺, [O]²⁻.
• The transferred electrons appear inside the bracket of the anion (the species that GAINED them) — you can see they have been moved across.

Example 1 — sodium chloride (NaCl). Na atom (×) has 1 outer electron; Cl atom (•) has 7. The single × transfers to Cl. Result:

• [Na]⁺ — empty outer shell shown (the M shell is gone, the 2,8 configuration is now the outer).
• [Cl • • • • • • • ×]⁻ — 8 electrons in the outer shell (7 dots + 1 cross).

Example 2 — magnesium oxide (MgO). Mg (×) has 2 outer electrons; O (•) has 6. Both × electrons transfer to O. Result:

• [Mg]²⁺
• [O • • • • • • ××]²⁻

Example 3 — calcium chloride (CaCl₂). Ca (×) loses 2 electrons; each Cl (•) gains 1. Result:

• [Ca]²⁺ in the middle.
• TWO [Cl • • • • • • • ×]⁻ ions, one on each side.

Example 4 — sodium oxide (Na₂O). Each Na has 1 outer electron; O needs 2. So TWO Na atoms (each × shown) donate 1 electron each to ONE O atom. Result:

• TWO [Na]⁺ ions.
• ONE [O • • • • • • ××]²⁻ ion.

Example 5 — aluminium oxide (Al₂O₃). Each Al loses 3 electrons; each O gains 2. LCM = 6: 2 Al and 3 O are needed. The 6 transferred electrons fill the three oxide ions. Result:

• 2 × [Al]³⁺.
• 3 × [O]²⁻ (each with 8 outer electrons).

Mark-scheme tip. The square brackets and charges are MARK-BEARING. Don't draw "atoms with arrows" instead of "ions in brackets" — Edexcel mark schemes specifically require the brackets and labels.

An ionic compound does NOT exist as isolated "molecules" of one cation + one anion (despite what a formula like NaCl might suggest). Instead, the ions arrange in a giant ionic lattice — a regular 3D pattern with billions of alternating + and − ions held together by electrostatic forces in every direction.

Sodium chloride lattice. Each Na⁺ ion is surrounded by 6 Cl⁻ neighbours (one above, one below, four in the same plane). Each Cl⁻ is surrounded by 6 Na⁺ neighbours in the same way. This 6:6 arrangement extends indefinitely in all three dimensions.

Why the lattice is so stable. Each ion experiences ELECTROSTATIC ATTRACTION not just to its 6 nearest neighbours but, more weakly, to every other oppositely-charged ion in the crystal. The net effect is an extremely strong "glue" holding the entire crystal together — far stronger than any individual ion-ion attraction would suggest.

Energy to overcome the lattice. To melt an ionic compound, you must supply enough thermal energy to overcome all these electrostatic forces simultaneously. This requires a lot of energy → high melting and boiling points.

Factors affecting lattice strength.

1. Charge magnitude. Larger charges → stronger forces (force ∝ product of charges). Mg²⁺O²⁻ is held by forces ~4× stronger than Na⁺Cl⁻ (2×2 vs 1×1).
2. Ionic radius / distance. Smaller ions can pack closer → stronger forces (force ∝ 1/distance²). Mg²⁺ < Na⁺ and O²⁻ < Cl⁻, so MgO ions are closer than NaCl ions.

Result: MgO mp (2852 °C) ≫ NaCl mp (801 °C) — both factors push in the same direction.

Four characteristic properties of ionic compounds, each explained by the giant ionic lattice + electrostatic forces story:

1. High melting and boiling points. The lattice contains strong electrostatic forces of attraction between oppositely-charged ions, acting in every direction throughout the crystal. A large amount of energy is needed to overcome these forces and break the lattice apart. Hence ionic compounds are typically SOLID at room temperature with high mp / bp (e.g. NaCl mp 801 °C).

2. Hard but brittle. The lattice is held tightly by electrostatic attraction, so the solid resists scratching and indentation — it is hard. However, when a force is applied that causes layers of the lattice to slide past one another, ions of LIKE CHARGE may end up adjacent. The resulting like-charge repulsion causes the crystal to split / shatter along that plane — the solid is brittle. (Compare with metals, which are also hard but DUCTILE because metallic bonds don't depend on a fixed arrangement.)

3. Soluble in water; usually insoluble in organic solvents. Water is a POLAR molecule — its oxygen is slightly negative and its hydrogens slightly positive. When NaCl is placed in water:

• Slightly-positive H atoms attract Cl⁻ ions.
• Slightly-negative O atoms attract Na⁺ ions.
• These attractions pull ions out of the lattice, surrounding each ion with a "shell" of water molecules → the lattice dissolves.

Non-polar solvents (petrol, oil) lack a strong dipole and cannot break the lattice — so ionic compounds are insoluble in them.

4. Electrical conductivity depends on state.

State	Conducts?	Reason
Solid	NO	Ions are fixed in lattice positions and cannot move to carry charge.
Molten	YES	Heat overcomes electrostatic forces; ions become free to move.
Aqueous (dissolved)	YES	Lattice broken by water; ions free to move through the solution.

In ionic compounds, the charge carriers are IONS (not electrons — that's metals). Always use the phrase "ions free to move" in your answer; the mark-scheme phrase "ions free to carry charge" is also credited.

Common Edexcel exam style. A table or figure of conductivity data; you must (a) identify the substance as ionic, (b) explain WHY the solid does not conduct and the molten / aqueous form does. This is worth 4 marks and is very high-frequency.