IB Chemistry Quiz: Apply The Ionic Model

What this quiz covers

This quiz focuses on Apply The Ionic Model, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. XY
2. X₂Y₃
3. X₃Y₂ (correct answer)
4. X₅Y₂

Question 2

1. Silicon dioxide (SiO₂)
2. Iodine (I₂)
3. Sodium chloride (NaCl) (correct answer)
4. Graphite (C)

Question 3

1. Each of three barium atoms transfers two electrons to each of two nitrogen atoms, forming Ba²⁺ and N³⁻ ions. (correct answer)
2. Each of two barium atoms transfers three electrons to each of three nitrogen atoms, forming Ba³⁺ and N²⁻ ions.
3. Each barium atom transfers one electron to a nitrogen atom, and the resulting ions share electrons to achieve octets.
4. Each of two nitrogen atoms accepts three electrons, one from each of six different barium atoms, forming a complex lattice.

Question 4

1. $\text{AlF}_3$ has the highest lattice energy because $\text{Al}^{3+}$ has the highest charge density, followed by $\text{MgF}_2$, then $\text{CaF}_2$ (correct answer)
2. $\text{CaF}_2$ has the highest lattice energy because $\text{Ca}^{2+}$ is the largest cation, creating stronger electrostatic attractions with $\text{F}^-$ ions
3. $\text{MgF}_2$ has the highest lattice energy because magnesium has the smallest atomic number among the three metals
4. All three compounds have similar lattice energies because they all contain fluoride ions with the same charge

Question 5

1. The electronegativity difference between the metal and chlorine is greater for cesium than sodium, requiring higher coordination numbers
2. The lattice energy is maximized differently for each compound based on the cation size, with larger cations preferring lower coordination geometries
3. Cesium has more electrons than sodium, creating stronger London dispersion forces that stabilize the higher coordination structure
4. The radius ratio $\frac{r_{cation}}{r_{anion}}$ is larger for $\text{CsCl}$ than $\text{NaCl}$, allowing the larger cesium ion to accommodate more chloride neighbors (correct answer)

Question 6

1. Each $\text{Ca}^{2+}$ substitution creates one cation vacancy to maintain charge neutrality, increasing ionic conductivity through vacancy migration (correct answer)
2. Each $\text{Ca}^{2+}$ substitution requires removal of one $\text{Cl}^-$ ion to maintain charge neutrality, creating anion vacancies throughout the crystal
3. The $\text{Ca}^{2+}$ ions cannot substitute for $\text{Na}^+$ because the size difference is too large for the rock salt structure to accommodate
4. Each $\text{Ca}^{2+}$ substitution creates local charge imbalances that are compensated by electron delocalization, making the crystal electrically conductive

Question 7

1. $\text{MgO}$ has higher thermal expansion because the smaller ions create more directional bonding that responds more sensitively to temperature changes
2. $\text{MgO}$ has higher thermal expansion because the higher lattice energy creates stronger temperature dependence of the interionic potential energy
3. Both compounds have identical thermal expansion coefficients because they have the same charges on their ions and identical crystal structures
4. $\text{BaO}$ has higher thermal expansion because the larger ions have weaker electrostatic attractions, making the lattice more responsive to thermal energy (correct answer)

Question 8

1. X⁻ is smaller than Y⁻ because smaller anions create higher lattice energy, leading to greater heat release when overcome by hydration
2. X⁻ is larger than Y⁻ because larger anions have lower lattice energy and higher hydration energy, favoring exothermic dissolution
3. X⁻ is smaller than Y⁻ because smaller anions have higher hydration energy that exceeds their higher lattice energy contribution (correct answer)
4. The anion sizes cannot be compared because the dissolution behavior depends primarily on the metal cation properties, not the anions

Question 9

1. $\text{LiF}$ will have higher conductivity because the smaller $\text{Li}^+$ and $\text{F}^-$ ions can move more rapidly through the molten state
2. $\text{CsI}$ will have higher conductivity because the larger $\text{Cs}^+$ and $\text{I}^-$ ions have weaker interionic forces and greater mobility in the melt (correct answer)
3. $\text{LiF}$ will have higher conductivity because the higher charge density creates stronger ion-dipole interactions that facilitate charge transport
4. Both compounds will have identical conductivity because they contain the same number of ions per formula unit and have the same crystal structure

Question 10

1. $\text{NaCl} < \text{MgCl}_2 < \text{AlCl}_3$ because higher cation charge creates more ions per formula unit and stronger ion-dipole interactions
2. $\text{AlCl}_3 < \text{MgCl}_2 < \text{NaCl}$ because higher charged cations create more ion pairing, reducing the effective number of charge carriers
3. $\text{NaCl} < \text{MgCl}_2 < \text{AlCl}_3$ because the total number of ions increases with cation charge: 2, 3, and 4 ions respectively (correct answer)
4. All three solutions have equal conductivity because they have the same molar concentration and contain the same anion type

Question 11

1. In $\text{X}_2\text{Y}$, X has charge +1 and Y has charge -2; in $\text{XY}_2$, X has charge +2 and Y has charge -1
2. In $\text{X}_2\text{Y}$, X has charge +2 and Y has charge -4; in $\text{XY}_2$, X has charge +4 and Y has charge -2
3. In $\text{X}_2\text{Y}$, X has charge +1 and Y has charge -2; in $\text{XY}_2$, X has charge +4 and Y has charge -2 (correct answer)
4. The charges cannot be determined from melting point data alone because ionic size effects dominate over charge effects

Question 12

1. $\text{BaSO}_4$ because both $\text{Ba}^{2+}$ and $\text{SO}_4^{2-}$ are large ions that create weak ion-dipole interactions with water
2. $\text{Al}_2\text{O}_3$ because the high charge density of $\text{Al}^{3+}$ and $\text{O}^{2-}$ ions creates very strong lattice energy that water cannot overcome (correct answer)
3. $\text{AgCl}$ because silver has d-electrons that interfere with the ionic bonding model and reduce water solubility
4. $\text{CaCO}_3$ because carbonate ions are too large to fit between water molecules in the hydration sphere

Question 13

1. 1:1
2. 2:3 (correct answer)
3. 3:2
4. 1:3

Question 14

1. K⁺(g) and Br⁻(g)
2. Ca²⁺(g) and S²⁻(g)
3. Al³⁺(g) and N³⁻(g) (correct answer)
4. Na⁺(g) and Cl⁻(g)

Question 15

1. It increases, because the increasing size of the cation leads to greater electron cloud polarization.
2. It decreases, because the increasing radius of the cation weakens the electrostatic attraction to the chloride ion. (correct answer)
3. It remains relatively constant, because the charge on the cation (+1) and anion (-1) does not change.
4. It decreases, because the electronegativity of the cation decreases, making the bond less ionic in character.

Question 16

1. Al(HCO₃)₃ (correct answer)
2. AlHCO₃
3. Al₂(HCO₃)₃
4. Al₃HCO₃

Question 17

1. The S²⁻ ion is smaller than the S atom because the added electrons are pulled closer by the nucleus.
2. The S²⁻ ion is larger than the S atom because the same nuclear charge is attracting more electrons, increasing electron-electron repulsion. (correct answer)
3. They have the same radius because the number of protons and the energy levels occupied remain unchanged.
4. The S²⁻ ion is smaller than the S atom because gaining electrons increases the effective nuclear charge.

Question 18

1. +2
2. +3 (correct answer)
3. +1
4. +6

Question 19

1. The model is fully accurate; attraction between discrete positive and negative ions is the only interaction present.
2. The model is incomplete because it ignores the covalent bonds that exist between all ions in the lattice structure.
3. The model is an oversimplification, as polarization of the anion by the cation introduces some covalent character. (correct answer)
4. The model is flawed because ionic bonding also involves repulsive forces between the nuclei of adjacent ions.

Question 20

1. It is ionically bonded to a single sodium ion (Na⁺) forming a discrete molecule.
2. It is surrounded by four sodium ions (Na⁺) in a tetrahedral geometry.
3. It is surrounded by six sodium ions (Na⁺) in an octahedral geometry. (correct answer)
4. It is surrounded by eight sodium ions (Na⁺) in a cubic geometry.