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Chemistry - Ions and Ionic Bonds

Grade 9IGCSE

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

🔑Concepts

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Ions are charged particles formed when atoms lose or gain electrons to achieve a stable noble gas electronic configuration (a full outer shell). Metals typically lose electrons to form positive ions called cations (e.g., Na→Na++e−Na \rightarrow Na^+ + e^-), while non-metals gain electrons to form negative ions called anions (e.g., Cl+e−→Cl−Cl + e^- \rightarrow Cl^-).

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An ionic bond is the strong electrostatic force of attraction between oppositely charged ions. This bond typically forms between metallic and non-metallic elements.

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Ionic compounds do not exist as isolated molecules; they form a giant ionic lattice. This is a regular, repeating three-dimensional arrangement where each positive ion is surrounded by negative ions and vice versa.

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Due to the strength of the electrostatic forces within the lattice, ionic compounds have high melting and boiling points. They require significant thermal energy to overcome the forces holding the ions together.

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Ionic compounds are generally brittle because a small displacement in the lattice brings ions of the same charge next to each other, causing repulsion that shatters the crystal.

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Electrical conductivity: Ionic solids do not conduct electricity because the ions are fixed in position. However, they conduct electricity when molten or aqueous because the ions are free to move and carry the charge.

📐Formulae

Total Positive Charge+Total Negative Charge=0Total\ Positive\ Charge + Total\ Negative\ Charge = 0

Mn++Xm−→MmXnM^{n+} + X^{m-} \rightarrow M_m X_n

Energy∝q1×q2r2Energy \propto \frac{q_1 \times q_2}{r^2}

Na→Na++e−Na \rightarrow Na^+ + e^-

O+2e−→O2−O + 2e^- \rightarrow O^{2-}

💡Examples

Problem 1:

Determine the chemical formula for the ionic compound formed between Magnesium (MgMg, Group 2) and Chlorine (ClCl, Group 7).

Solution:

The formula is MgCl2MgCl_2.

Explanation:

Magnesium is in Group 2, so it loses 2 electrons to form Mg2+Mg^{2+}. Chlorine is in Group 7, so it gains 1 electron to form Cl−Cl^-. To balance the charges so the overall charge is zero: (1×+2)+(2×−1)=0(1 \times +2) + (2 \times -1) = 0 Thus, one Mg2+Mg^{2+} ion bonds with two Cl−Cl^- ions.

Problem 2:

Explain why Sodium Chloride (NaClNaCl) has a high melting point (801∘C801^\circ C).

Solution:

The high melting point is due to the strong electrostatic attractions in the giant ionic lattice.

Explanation:

In NaClNaCl, the Na+Na^+ and Cl−Cl^- ions are held together by strong ionic bonds in a giant lattice. Breaking these bonds to melt the solid requires a large amount of thermal energy. The strength of the bond is represented by the attraction between the opposite charges: F=kq1q2r2F = k \frac{q_1 q_2}{r^2}.

Problem 3:

Write the formula for Aluminium Oxide using the crossover method.

Solution:

The formula is Al2O3Al_2O_3.

Explanation:

Aluminium (AlAl) forms a 3+3+ ion (Al3+Al^{3+}) and Oxygen (OO) forms a 2−2- ion (O2−O^{2-}). By crossing over the numerical values of the charges to serve as subscripts: Al charge is 3→Subscript for O is 3Al \text{ charge is } 3 \rightarrow \text{Subscript for } O \text{ is } 3 O charge is 2→Subscript for Al is 2O \text{ charge is } 2 \rightarrow \text{Subscript for } Al \text{ is } 2 Result: Al2O3Al_2O_3.