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Chemical Bonding - Ionic Bonding

Grade 9IB

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

🔑Concepts

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Ionic bonding involves the electrostatic attraction between oppositely charged ions, formed by the complete transfer of one or more electrons from a metal atom to a non-metal atom.

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Metal atoms lose valence electrons to form positively charged ions called cations (e.g., Na+Na^+, Mg2+Mg^{2+}).

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Non-metal atoms gain electrons to form negatively charged ions called anions (e.g., Cl−Cl^-, O2−O^{2-}).

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The 'Octet Rule' states that atoms tend to gain, lose, or share electrons to achieve a stable electron configuration similar to a noble gas, typically with 88 electrons in the outermost shell.

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Ionic compounds do not exist as discrete molecules but form a 'Giant Ionic Lattice', a regular 3D3D arrangement of alternating positive and negative ions.

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Properties of ionic compounds include high melting and boiling points due to strong electrostatic forces, brittleness, and electrical conductivity only when molten or in aqueous solution (free ionsfree\ ions).

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The chemical formula of an ionic compound represents the simplest whole-number ratio of ions (the empirical formula) such that the total positive charge equals the total negative charge.

📐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 (Coulombic attraction force between ions)F = k \frac{q_1 q_2}{r^2} \text{ (Coulombic attraction force between ions)}

💡Examples

Problem 1:

Explain the formation of the ionic bond in Sodium Chloride (NaClNaCl) using electron configurations.

Solution:

Sodium (NaNa) has the configuration [2,8,1][2, 8, 1]. It loses 1e−1e^- to become Na+Na^+ [2,8][2, 8]. Chlorine (ClCl) has the configuration [2,8,7][2, 8, 7]. It gains 1e−1e^- to become Cl−Cl^- [2,8,8][2, 8, 8].

Explanation:

The transfer of one electron from NaNa to ClCl allows both atoms to achieve a stable octet. The resulting Na+Na^+ and Cl−Cl^- ions are held together by strong electrostatic forces in a lattice.

Problem 2:

Determine the chemical formula for the compound formed between Aluminium (AlAl) and Oxygen (OO).

Solution:

Al3++O2−→Al2O3Al^{3+} + O^{2-} \rightarrow Al_2O_3

Explanation:

Aluminium is in Group 13 and forms a 3+3+ ion (Al3+Al^{3+}). Oxygen is in Group 16 and forms a 2−2- ion (O2−O^{2-}). To balance the charges, we find the least common multiple of 33 and 22, which is 66. Thus, 2×(+3)=+62 \times (+3) = +6 and 3×(−2)=−63 \times (-2) = -6. The formula is Al2O3Al_2O_3.

Problem 3:

Why does Magnesium Oxide (MgOMgO) have a significantly higher melting point than Sodium Chloride (NaClNaCl)?

Solution:

The ions in MgOMgO (Mg2+Mg^{2+} and O2−O^{2-}) have higher charges than the ions in NaClNaCl (Na+Na^+ and Cl−Cl^-).

Explanation:

According to the relationship F∝q1q2F \propto q_1q_2, the electrostatic attraction between doubly charged ions (+2+2 and −2-2) is much stronger than between singly charged ions (+1+1 and −1-1). More energy is required to overcome these stronger forces in the MgOMgO lattice.