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Chemical Bonding and Structure - Ionic bonding and structure

Grade 11IBChemistry

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

🔑Concepts

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Ionic bonding is the result of the electrostatic attraction between oppositely charged ions (cationscations and anionsanions).

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It typically occurs between a metal (low electronegativity) and a non-metal (high electronegativity), involving the transfer of electrons to achieve a stable octet (noble gasnoble \ gas configuration).

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Ionic compounds form a giant ionic lattice structure, which is a regular 3D arrangement of ions where each ion is surrounded by ions of the opposite charge.

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The strength of the ionic bond is influenced by the lattice enthalpy: it increases with higher ionic charges and smaller ionic radii, as described by Coulomb's Law.

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Physical properties include: high melting and boiling points (due to strong electrostatic forces), brittleness (sliding layers cause like-charges to align and repel), and solubility in polar solvents like H2OH_2O.

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Electrical conductivity: Ionic compounds do not conduct electricity in the solid state because ions are in fixed positions, but they do conduct when molten or in aqueous solution as ions are free to move.

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Common polyatomic ions include Nitrate (NO3−NO_3^-), Carbonate (CO32−CO_3^{2-}), Sulfate (SO42−SO_4^{2-}), Phosphate (PO43−PO_4^{3-}), Ammonium (NH4+NH_4^+), and Hydroxide (OH−OH^-).

📐Formulae

F=kq1q2r2F = k \frac{q_1 q_2}{r^2}

ΔHlat∝Qcation⋅Qanionrcation+ranion\Delta H_{lat} \propto \frac{Q_{cation} \cdot Q_{anion}}{r_{cation} + r_{anion}}

M(s)+12X2(g)→MX(s)M(s) + \frac{1}{2}X_2(g) \rightarrow MX(s)

💡Examples

Problem 1:

Deduce the chemical formula of the compound formed between Aluminum (Z=13Z=13) and Oxygen (Z=8Z=8).

Solution:

Al2O3Al_2O_3

Explanation:

Aluminum has the electron configuration [Ne]3s23p1[Ne] 3s^2 3p^1 and loses 3 electrons to form Al3+Al^{3+}. Oxygen has the configuration [He]2s22p4[He] 2s^2 2p^4 and gains 2 electrons to form O2−O^{2-}. To achieve electrical neutrality, the charges must balance: 2×(+3)+3×(−2)=02 \times (+3) + 3 \times (-2) = 0. Thus, the ratio is 2:32:3.

Problem 2:

Explain why MgOMgO has a higher melting point (2852∘C2852^\circ C) than NaClNaCl (801∘C801^\circ C).

Solution:

The electrostatic attraction in MgOMgO is significantly stronger than in NaClNaCl.

Explanation:

According to the lattice enthalpy relationship ΔHlat∝q1q2r\Delta H_{lat} \propto \frac{q_1 q_2}{r}, Mg2+Mg^{2+} and O2−O^{2-} have charges of magnitude 22, whereas Na+Na^+ and Cl−Cl^- have charges of magnitude 11. The product of charges for MgOMgO (2×2=42 \times 2 = 4) is four times greater than for NaClNaCl (1×1=11 \times 1 = 1). Additionally, Mg2+Mg^{2+} is smaller than Na+Na^+, allowing for a smaller inter-ionic distance rr, further increasing the attractive force.

Problem 3:

Predict whether the compound NH4NO3NH_4NO_3 contains ionic or covalent bonds.

Solution:

It contains both ionic and covalent bonds.

Explanation:

The bond between the ammonium cation (NH4+NH_4^+) and the nitrate anion (NO3−NO_3^-) is ionic. However, the atoms within the polyatomic ions themselves (N−HN-H and N−ON-O bonds) are held together by covalent bonds.

Ionic bonding and structure - Revision Notes & Key Formulas | IB Grade 11 Chemistry