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Atoms, Elements and Compounds - Ions and ionic bonding

Grade 11IGCSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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An ion is an atom or group of atoms that has become electrically charged by losing or gaining one or more electrons. Metals lose electrons to form positive ions called cations (e.g., Na+Na^+, Mg2+Mg^{2+}), while non-metals gain electrons to form negative ions called anions (e.g., Cl−Cl^-, O2−O^{2-}).

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Ionic bonding is the strong electrostatic force of attraction between oppositely charged ions. This occurs typically between a metal and a non-metal.

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The electronic configuration of ions usually achieves a stable 'noble gas' structure (a full outer shell). For example, NaNa (2,8,12,8,1) becomes Na+Na^+ (2,82,8).

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Ionic compounds form a giant ionic lattice, which is a regular 3D3D arrangement of alternating positive and negative ions.

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Physical properties of ionic compounds include high melting and boiling points (due to strong electrostatic forces), solubility in water (usually), and electrical conductivity only when molten or in aqueous solution (as ions are free to move).

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Valency and Charge: Group I elements form 1+1+ ions, Group II form 2+2+ ions, Group III form 3+3+ ions, Group VI form 2−2- ions, and Group VII form 1−1- ions.

📐Formulae

M→Mn++ne− (Oxidation/Cation formation)M \rightarrow M^{n+} + ne^- \text{ (Oxidation/Cation formation)}

X+ne−→Xn− (Reduction/Anion formation)X + ne^- \rightarrow X^{n-} \text{ (Reduction/Anion formation)}

Total positive charge+Total negative charge=0\text{Total positive charge} + \text{Total negative charge} = 0

F=kq1q2r2 (Electrostatic force between ions)F = k \frac{q_1 q_2}{r^2} \text{ (Electrostatic force between ions)}

💡Examples

Problem 1:

Describe the formation of Magnesium Oxide from its atoms and deduce its chemical formula.

Solution:

Mg+12O2→MgOMg + \frac{1}{2}O_2 \rightarrow MgO

Explanation:

Magnesium (Group II) has the electronic configuration 2,8,22,8,2. It loses 22 electrons to form Mg2+Mg^{2+}. Oxygen (Group VI) has the configuration 2,62,6 and gains 22 electrons to form O2−O^{2-}. The 2+2+ and 2−2- charges balance perfectly in a 1:11:1 ratio, resulting in the formula MgOMgO.

Problem 2:

Deduce the formula for Aluminium Sulfate.

Solution:

Al2(SO4)3Al_2(SO_4)_3

Explanation:

Aluminium forms a Al3+Al^{3+} ion. The sulfate polyatomic ion is SO42−SO_4^{2-}. To balance the charges, we need two Al3+Al^{3+} ions (total charge +6+6) and three SO42−SO_4^{2-} ions (total charge −6-6), ensuring the compound is electrically neutral.

Problem 3:

Explain why NaClNaCl does not conduct electricity as a solid but does conduct when dissolved in water.

Solution:

Presence of mobile ions.

Explanation:

In the solid state, ions in NaClNaCl are held in fixed positions within the giant lattice by strong electrostatic forces and cannot move. When dissolved in water (H2OH_2O), the lattice breaks down and the ions (Na+Na^+ and Cl−Cl^-) become mobile, allowing them to carry an electric current.

Ions and ionic bonding - Revision Notes & Key Formulas | IGCSE Grade 11 Chemistry