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Structure 2. Models of bonding and structure - The ionic model

Grade 12IBChemistry

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

🔑Concepts

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Ionic bonding is defined as the electrostatic attraction between oppositely charged ions (Mn+M^{n+} and Xm−X^{m-}).

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Ions are formed by the transfer of electrons from a metal atom (which forms a cation: M→Mn++ne−M \rightarrow M^{n+} + ne^-) to a non-metal atom (which forms an anion: X+ne−→Xn−X + ne^- \rightarrow X^{n-}).

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The stability of ions is often explained by the octet rule, where atoms gain or lose electrons to achieve a noble gas electron configuration (usually ns2np6ns^2 np^6).

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Ionic compounds do not exist as discrete molecules; instead, they form a giant ionic lattice, which is a regular 3D arrangement of alternating positive and negative ions.

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The strength of the ionic bond (lattice enthalpy) depends on the charge of the ions and their ionic radii. According to Coulomb's Law, the force of attraction is proportional to the product of charges (q1q2q_1 q_2) and inversely proportional to the square of the distance between them (r2r^2).

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Physical properties of ionic compounds include high melting and boiling points (due to strong electrostatic forces), brittleness (due to repulsion between like-charged ions when layers shift), and solubility in polar solvents like H2OH_2O.

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Ionic compounds conduct electricity only when molten or in aqueous solution because the ions are free to move (mobile charge carriers); they are insulators in the solid state as ions are fixed in the lattice.

📐Formulae

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

Elattice∝q1q2rcation+ranionE_{lattice} \propto \frac{q_1 q_2}{r_{cation} + r_{anion}}

Empirical Formula: MmXn where m×(charge of M)+n×(charge of X)=0\text{Empirical Formula: } M_m X_n \text{ where } m \times (\text{charge of } M) + n \times (\text{charge of } X) = 0

💡Examples

Problem 1:

Deduce the chemical formula for the ionic compound formed between Aluminum (AlAl, Group 13) and Oxygen (OO, Group 16).

Solution:

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

Explanation:

Aluminum is in Group 13 and loses 3 electrons to form the Al3+Al^{3+} cation. Oxygen is in Group 16 and gains 2 electrons to form the O2−O^{2-} anion. To achieve electrical neutrality, two Al3+Al^{3+} ions (total charge +6+6) must combine with three O2−O^{2-} ions (total charge −6-6).

Problem 2:

Compare the melting points of Sodium Chloride (NaClNaCl) and Magnesium Oxide (MgOMgO) and explain the difference.

Solution:

Tm(MgO)>Tm(NaCl)T_m(MgO) > T_m(NaCl)

Explanation:

The melting point of MgOMgO is significantly higher than that of NaClNaCl. This is because Mg2+Mg^{2+} and O2−O^{2-} have higher charges (+2+2 and −2-2) compared to Na+Na^+ and Cl−Cl^- (+1+1 and −1-1). Additionally, the ionic radii of Mg2+Mg^{2+} and O2−O^{2-} are smaller than Na+Na^+ and Cl−Cl^-. According to the lattice energy relationship E∝q1q2rE \propto \frac{q_1 q_2}{r}, the electrostatic attractions in MgOMgO are much stronger.