AP Chemistry Quiz: Structure Of Ionic Solids

What this quiz covers

This quiz focuses on Structure Of Ionic Solids, giving you a quick way to practice the rules, question types, and explanations that matter most for AP 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. Solid X typically has stronger attractions because the smaller anion allows ions to be closer together. (correct answer)
2. Solid X typically has weaker attractions because smaller ions have fewer electrons to share covalently.
3. Solid Y typically has stronger attractions because larger anions increase the number of ionion bonds.
4. Both solids have the same attractions because ionic bonding depends only on the ratio of ions, not their sizes.
5. Solid Y typically has stronger attractions because larger anions increase delocalized electron mobility.

Explanation: This question evaluates how ion size influences attraction strength in ionic lattices. In Solid X (NaF), the smaller F⁻ anion allows closer approach to Na⁺, strengthening electrostatic attractions compared to larger I⁻ in Solid Y (NaI). Smaller interionic distances increase lattice energy, leading to stronger bonding in similar crystal structures. This is based on Coulomb's law, where force is inversely proportional to distance squared. A tempting distractor is that Solid Y has stronger attractions due to larger anions increasing ion-ion bonds, which is incorrect because it misconceives bond strength as depending on ion count rather than distance. When comparing ionic solids with the same cation, consider anion size effects on interionic distance to predict attraction strength.

Question 2

1. The solid consists of discrete $\mathrm{CaF_2}$ molecules held together by intermolecular forces.
2. Each $\mathrm{Ca}^{2+}$ is surrounded by $\mathrm{F}^-$ ions, and each $\mathrm{F}^-$ is surrounded by $\mathrm{Ca}^{2+}$ ions in a 3D array. (correct answer)
3. The solid is a metallic lattice in which $\mathrm{F}^-$ ions contribute delocalized electrons.
4. The solid is an amorphous network in which ions are randomly distributed with no repeating pattern.
5. The solid contains alternating neutral $\mathrm{CaF}$ units that pack together without electrostatic attraction.

Explanation: This question assesses knowledge of the arrangement of ions in an ionic solid lattice. In the CaF₂ solid, each Ca²⁺ ion is surrounded by multiple F⁻ ions, and each F⁻ is surrounded by Ca²⁺ ions, forming a continuous 3D array held by electrostatic attractions. This structure ensures charge balance with a 1:2 ratio of Ca²⁺ to F⁻ ions, reflecting the +2 and -1 charges respectively. The repeating pattern maximizes attractions between opposite charges and minimizes repulsions, characteristic of ionic crystal structures. A tempting distractor is that the solid consists of discrete CaF₂ molecules held by intermolecular forces, which is wrong because it confuses ionic solids with molecular solids, ignoring the extended ionic lattice. When describing ionic solid structures, focus on the extended 3D arrangement of ions rather than discrete molecules.

Question 3

1. $1\ \mathrm{Al^{3+}}:1\ \mathrm{O^{2-}}$
2. $2\ \mathrm{Al^{3+}}:3\ \mathrm{O^{2-}}$ (correct answer)
3. $3\ \mathrm{Al^{3+}}:2\ \mathrm{O^{2-}}$
4. $1\ \mathrm{Al^{3+}}:2\ \mathrm{O^{2-}}$
5. $2\ \mathrm{Al^{3+}}:1\ \mathrm{O^{2-}}$

Explanation: This question tests the skill of calculating ion ratios for electrical neutrality in an ionic lattice. The Al³⁺ ion has a +3 charge and O²⁻ has a -2 charge, so to balance charges, two Al³⁺ provide +6 and three O²⁻ provide -6. This 2:3 ratio ensures the overall lattice is neutral, as required for stable ionic solids. The repeating pattern in the 3D structure reflects this ratio, maintaining electrostatic stability throughout the crystal. A tempting distractor is 3 Al³⁺:2 O²⁻, which is wrong because it results in a net +1 charge, stemming from the misconception of reversing the charge-based ratio. For neutrality in ionic compounds, use the least common multiple of charge magnitudes to find the balanced ion ratio.

Question 4

1. AlO
2. AlO$_2$
3. Al$_2$O$_3$ (correct answer)
4. Al$_3$O$_2$
5. Al$_2$O

Explanation: This question tests the skill of deriving the empirical formula for an ionic solid based on ion charges and charge neutrality. The correct answer, choice C, gives Al₂O₃ as the simplest whole-number ratio, where two Al³⁺ ions (total +6 charge) balance three O²⁻ ions (total -6 charge) in the repeating lattice. This formula reflects the need for electrical neutrality in the crystal, achieved by finding the least common multiple of the charges (6) and adjusting the ion counts accordingly. The principle ensures the lattice is stable with no net charge per formula unit. A tempting distractor is choice B, AlO₂, which arises from the misconception of directly using the charge values as subscripts without balancing the total charges. To find ionic empirical formulas, always use the crisscross method or least common multiple to ensure charge balance in the ratio.

Question 5

1. The crystal consists of a repeating lattice of ions held together by electrostatic attractions between oppositely charged ions. (correct answer)
2. The crystal consists of discrete MgF$_2$ molecules held together by hydrogen bonding between molecules.
3. The crystal consists of neutral Mg and F atoms in a sea of mobile electrons that allows malleability.
4. The crystal consists of alternating ions, but the attractions are primarily due to shared electron pairs in covalent bonds.
5. The crystal consists of ions that are randomly arranged, so there is no repeating pattern in the solid.

Explanation: This question tests the understanding of the basic structure of ionic solids, focusing on the arrangement and bonding of ions in a crystal lattice. The correct answer, choice A, accurately describes magnesium fluoride as a repeating lattice of Mg²⁺ and F⁻ ions held together by strong electrostatic attractions between oppositely charged ions, which is consistent with the extended 3D pattern mentioned in the question. This structure arises because ionic solids form from the transfer of electrons, resulting in cations and anions that attract each other in a way that maximizes opposite-charge interactions and minimizes like-charge repulsions. The alternating pattern ensures charge balance and stability throughout the crystal. A tempting distractor is choice B, which incorrectly suggests discrete MgF₂ molecules held by hydrogen bonding, reflecting the misconception of treating ionic compounds as molecular rather than as extended ionic lattices. To analyze ionic solid structures, always recall that they consist of infinite arrays of ions bound by ionic bonds, not discrete molecules with intermolecular forces.

Question 6

1. Low melting point because only weak intermolecular forces must be overcome.
2. Formation of a molecular crystal because NaO units act as discrete molecules.
3. High electrical conductivity in the solid state due to delocalized electrons.
4. Malleability because layers of atoms can slide while bonding is maintained.
5. Brittleness because shifting ions can bring like charges into contact. (correct answer)

Explanation: This question tests the identification of properties unique to ionic solids compared to others like metals. The correct answer, choice C, notes brittleness due to like-charge repulsion when ions shift, typical for the rigid lattice of Na⁺ and O²⁻ ions. This contrasts with metals' malleability from delocalized electrons. The property arises from the alternating ion arrangement that fractures under deformation. A tempting distractor is choice B, malleability, which applies to metals and misapplies metallic bonding to ionic structures, a common confusion. To compare solid types, link properties directly to their bonding and particle mobility.

Question 7

1. The ions are held together by weak intermolecular forces that break easily upon impact.
2. The ions can rearrange without changing attractions, allowing layers to slide past one another.
3. When the lattice shifts, like charges can become adjacent, increasing repulsion and causing fracture. (correct answer)
4. The solid contains mobile electrons that cause the lattice to crack under stress.
5. Covalent bonds between K and Br atoms break, producing brittle fragments.

Explanation: This question tests understanding of the brittleness of ionic solids and how their structure relates to mechanical properties. Ionic solids are brittle because when stress is applied, the ordered lattice can shift, causing ions of like charge to become adjacent to each other, resulting in strong electrostatic repulsion that causes the crystal to fracture along cleavage planes. In the KBr lattice, K⁺ and Br⁻ ions alternate in a regular pattern, but mechanical stress can displace layers so that K⁺ ions align with K⁺ ions and Br⁻ with Br⁻, creating repulsion. Choice B incorrectly suggests that layers can slide without changing attractions, which describes metallic solids but not ionic ones - in ionic solids, shifting disrupts the attractive pattern. To predict mechanical properties, consider how structural changes affect the balance of attractive and repulsive forces.

Question 8

1. Ti$_2$O
2. TiO
3. Ti$_4$O$_2$
4. TiO$_2$ (correct answer)
5. Ti$_2$O$_3$

Explanation: This question tests determining an empirical formula for charge neutrality in a packed ionic crystal. The correct answer, choice B, is TiO₂, with one Ti⁴⁺ (+4) balanced by two O²⁻ (total -4), consistent with hard-sphere packing in a regular pattern. This simplest ratio ensures lattice stability. The formula reflects ion proportions for neutrality. A tempting distractor is choice A, TiO, assuming 1:1 despite charges, a misconception from equating ratios without balancing. For packed ion formulas, prioritize charge balance over packing details.

Question 9

1. Hydrogen bonding between sulfate ions holds the solid together in layers.
2. Metallic bonding from delocalized electrons shared among Ba and sulfate units holds the solid together.
3. London dispersion forces between neutral formula units are the primary attraction in the solid.
4. Covalent bonds between Ba and S atoms hold discrete molecules together in the solid.
5. Attractive forces between $\text{Ba}^{2+}$ and $\text{SO}_4^{2-}$ ions throughout the lattice hold the solid together. (correct answer)

Explanation: This question tests understanding of the forces holding ionic solids together. In an ionic solid containing Ba²⁺ and SO₄²⁻, the attractive electrostatic forces between the positively charged barium ions and negatively charged sulfate ions throughout the three-dimensional lattice hold the solid together. These ionic attractions extend in all directions, creating a stable crystal structure. Choice B incorrectly suggests covalent bonds between Ba and S atoms, which misunderstands that BaSO₄ is an ionic compound where the sulfate ion maintains its polyatomic structure - the Ba-SO₄ interaction is ionic, not covalent. To identify bonding in ionic solids, recognize that metal cations and polyatomic anions interact through electrostatic attractions, not covalent bonds.

Question 10

1. Two $\text{Na}^+$ ions are needed to balance the charge of one $\text{S}^{2-}$ ion in the lattice. (correct answer)
2. NaS would be unstable because ionic compounds must contain equal numbers of cations and anions.
3. NaS would form molecules with covalent Na–S bonds instead of ions.
4. NaS would conduct electricity as a solid due to delocalized electrons, so it cannot form.
5. NaS would require sulfur to have a $-1$ oxidation state, which is not allowed in solids.

Explanation: This question tests understanding of charge balance in ionic compounds and formula unit determination. In ionic solids, the formula unit represents the simplest whole number ratio of ions that achieves electrical neutrality. Since Na⁺ has a +1 charge and S²⁻ has a -2 charge, two sodium ions are needed to balance the charge of one sulfide ion, resulting in Na₂S. Choice B incorrectly suggests that ionic compounds must have equal numbers of cations and anions, which is only true when the charges have equal magnitude - this misconception ignores the importance of charge balance over particle count. To determine ionic formula units, multiply the number of each ion by its charge and ensure the sum equals zero.

Question 11

1. Low melting point because only weak intermolecular forces hold particles together.
2. High melting point because strong electrostatic attractions exist throughout a 3D array of ions. (correct answer)
3. High electrical conductivity as a solid because electrons are delocalized across the lattice.
4. Ability to form long covalent networks because ions share electrons in all directions.
5. Malleability because layers of ions can slide without changing electrostatic forces.

Explanation: This question tests understanding of how ionic lattice structure determines physical properties. Solid X (CsI) has an ionic lattice structure with strong electrostatic attractions between Cs⁺ and I⁻ ions extending throughout a three-dimensional array, resulting in a high melting point because significant energy is required to overcome these attractions. In contrast, Solid Y (CO₂) consists of discrete molecules held together by weak intermolecular forces, giving it a much lower melting point. Choice A incorrectly attributes a low melting point to ionic solids, confusing them with molecular solids - this misconception fails to recognize the strength of ionic bonding. To predict properties of ionic solids, consider that the extensive network of strong electrostatic attractions leads to high melting points, brittleness, and electrical conductivity when melted or dissolved.

Question 12

1. The crystal contains equal numbers of $\text{Li}^+$ and $\text{N}^{3-}$ ions because the lattice alternates charges.
2. The crystal contains three $\text{Li}^+$ ions for every one $\text{N}^{3-}$ ion in a repeating ionic lattice. (correct answer)
3. The crystal contains LiN molecules that are held together by hydrogen bonding between units.
4. The crystal contains neutral Li and N atoms, since ions cannot exist in the solid state.
5. The crystal contains delocalized electrons that allow it to be malleable like a metal.

Explanation: This question tests understanding of charge balance and ion ratios in ionic crystals. Since Li⁺ has a +1 charge and N³⁻ has a -3 charge, three lithium ions are needed to balance the charge of one nitride ion, resulting in Li₃N as the formula unit and a 3:1 ratio of Li⁺ to N³⁻ throughout the crystal lattice. This ratio ensures electrical neutrality: 3(+1) + 1(-3) = 0. Choice A incorrectly suggests equal numbers of each ion, which would result in a net negative charge - this misconception confuses alternating positions in the lattice with the overall stoichiometric ratio. To determine ion ratios in ionic solids, use the charges to find the smallest whole numbers that sum to zero total charge.

Question 13

1. The ratio is $1:1$, implying equal numbers of cations and anions in alternating positions.
2. The ratio is $2:1$, implying two $\text{Sr}^{2+}$ ions for each $\text{Cl}^-$ ion in the lattice.
3. The ratio is $1:2$, implying one $\text{Sr}^{2+}$ for every two $\text{Cl}^-$ ions in a repeating ionic array. (correct answer)
4. The ratio is $2:2$, implying the compound must form discrete $\text{Sr}_2\text{Cl}_2$ molecules.
5. The ratio is variable, implying the lattice adjusts composition based on temperature and pressure.

Explanation: This question tests understanding of predicting ion ratios in ionic compounds based on charge balance. Since Sr²⁺ has a +2 charge and Cl⁻ has a -1 charge, two chloride ions are needed to balance each strontium ion, giving a 1:2 ratio of Sr²⁺ to Cl⁻ in the ionic solid (SrCl₂). This ratio ensures electrical neutrality: 1(+2) + 2(-1) = 0, and implies a repeating ionic array where each strontium ion is surrounded by chloride ions in a three-dimensional lattice. Choice B incorrectly reverses the ratio to 2:1, which would create an unbalanced positive charge - this error often occurs when students confuse which ion needs more to balance the charge. To predict ion ratios, remember that the ion with the smaller charge magnitude will have the larger subscript in the formula.

Question 14

1. Each particle of the solid is a discrete $\text{Fe}_2\text{O}_3$ molecule with covalent bonds.
2. The solid contains a repeating arrangement with a $2:3$ ratio of $\text{Fe}^{3+}$ to $\text{O}^{2-}$ ions. (correct answer)
3. The solid contains equal numbers of $\text{Fe}^{3+}$ and $\text{O}^{2-}$ ions, but the subscripts indicate coordination number.
4. The solid contains $\text{Fe}^{2+}$ and $\text{O}^-$ ions, which combine to give the written formula.
5. The solid contains Fe atoms in a sea of electrons and O atoms in interstitial holes.

Explanation: This question tests understanding of how chemical formulas represent ionic solids and their ion ratios. The formula Fe₂O₃ indicates that the ionic solid contains Fe³⁺ and O²⁻ ions in a 2:3 ratio throughout the repeating lattice structure to maintain electrical neutrality (2 × (+3) + 3 × (-2) = 0). This formula represents the simplest whole number ratio of ions in the extended three-dimensional array, not discrete molecules or coordination numbers. Choice A incorrectly suggests Fe₂O₃ exists as discrete molecules with covalent bonds, which is a common misconception - transition metal oxides typically form ionic lattices, not molecular compounds. To interpret ionic formulas, recognize that subscripts indicate the ratio of ions needed for charge balance in the crystal structure.

Question 15

1. It contains one $\text{Sr}^{2+}$ ion and one $\text{Br}^-$ ion, giving a net charge of $+1$.
2. It contains one $\text{Sr}^{2+}$ ion and two $\text{Br}^-$ ions, giving a net charge of $0$. (correct answer)
3. It contains two $\text{Sr}^{2+}$ ions and one $\text{Br}^-$ ion, giving a net charge of $+3$.
4. It contains two $\text{Sr}^{2+}$ ions and two $\text{Br}^-$ ions, giving a net charge of $+2$.
5. It contains one $\text{Sr}^{2+}$ ion and three $\text{Br}^-$ ions, giving a net charge of $-1$.

Explanation: This question tests understanding of formula units in ionic compounds. Sr²⁺ has a +2 charge and Br⁻ has a -1 charge, so a neutral formula unit requires one Sr²⁺ ion and two Br⁻ ions, giving SrBr₂ with a net charge of (+2) + 2(-1) = 0. The formula unit represents the simplest whole-number ratio of ions that gives electrical neutrality. Choice A incorrectly uses only one Br⁻ ion, which would leave a net positive charge, showing the error of not properly balancing ionic charges. When determining formula units, ensure the sum of all positive and negative charges equals zero.

Question 16

1. It has stronger electrostatic attractions because the ions have larger charge magnitudes. (correct answer)
2. It has weaker attractions because the ions are held by dipoledipole forces.
3. It forms more covalent bonds per formula unit, increasing the melting point.
4. It contains more delocalized electrons, strengthening metallic bonding.
5. It melts at a higher temperature because it consists of larger molecules.

Explanation: This question evaluates the ability to compare melting points of ionic solids based on ion charges and electrostatic attraction strength. The Mg²⁺/O²⁻ solid has ions with +2 and -2 charges, leading to stronger Coulombic attractions compared to the +1/-1 charges in Li⁺/Br⁻. Higher charge magnitudes increase lattice energy, requiring more thermal energy to disrupt the lattice, thus raising the melting point. While ion sizes also influence attraction strength, the charge difference is the primary factor here without lattice energy calculations. A tempting distractor is that it forms more covalent bonds per formula unit, which is incorrect because it misapplies covalent bonding concepts to predominantly ionic compounds. To predict relative melting points of ionic solids, prioritize comparing ion charge magnitudes before considering sizes.

Question 17

1. Covalent bonds between discrete NaCl molecules break, producing fragments of intact molecules.
2. Delocalized electrons move to absorb the impact, then the lattice springs back.
3. Ions melt locally on impact because ionic bonds are weak at room temperature.
4. Hydrogen bonds between formula units rupture, allowing the solid to cleave easily.
5. Layers can shift so like charges align, causing strong repulsion that fractures the crystal. (correct answer)

Explanation: This question tests the understanding of brittleness in ionic solids due to their lattice structure and electrostatic interactions. In the NaCl lattice, alternating $\mathrm{Na}^{+}$ and $\mathrm{Cl}^{-}$ ions are held rigidly by strong attractions between opposite charges, but applying force can cause layers to shift. When layers slip, like-charged ions align, leading to strong repulsions that fracture the crystal, explaining the brittleness. This behavior arises from the ionic bonding model, where the lattice lacks the ductility seen in metals with delocalized electrons. A tempting distractor is that covalent bonds between discrete NaCl molecules break, which is incorrect because it misconceives ionic solids as molecular rather than extended lattices of ions. To analyze properties of ionic solids, consider how disruptions to the lattice affect electrostatic forces between ions.

Question 18

1. $\mathrm{Ba(PO_4)_2}$
2. $\mathrm{BaPO_4}$
3. $\mathrm{Ba_2(PO_4)_3}$
4. $\mathrm{Ba_3PO_4}$
5. $\mathrm{Ba_3(PO_4)_2}$ (correct answer)

Explanation: This question assesses determining the empirical formula for an ionic compound with polyatomic ions. The Ba²⁺ has a +2 charge and PO₄³⁻ has a -3 charge, so three Ba²⁺ provide +6 and two PO₄³⁻ provide -6 for balance. This results in the formula Ba₃(PO₄)₂, representing the repeating ratio in the lattice. The structure treats PO₄³⁻ as a single unit, with the lattice arranged to maximize attractions between Ba²⁺ and the polyatomic anions. A tempting distractor is Ba(PO₄)₂, which is incorrect because it yields a net -4 charge, misunderstanding the charge balance for polyatomic ions. When writing formulas with polyatomic ions, ensure the subscripts balance the total charges while keeping the polyatomic unit intact.

Question 19

1. $\mathrm{MgN}$
2. $\mathrm{Mg_2N_3}$
3. $\mathrm{Mg_3N_2}$ (correct answer)
4. $\mathrm{Mg_6N_6}$
5. $\mathrm{MgN_2}$

Explanation: This question tests the skill of determining the empirical formula for an ionic solid based on ion charges to ensure charge neutrality. The ions involved are Mg²⁺ with a +2 charge and N³⁻ with a -3 charge, requiring a ratio that balances the total positive and negative charges in the lattice. To achieve neutrality, three Mg²⁺ ions provide +6 charge, and two N³⁻ ions provide -6 charge, resulting in the formula Mg₃N₂. This ratio represents the simplest whole-number repeating unit in the 3D ionic lattice, where opposite charges attract to hold the structure together. A tempting distractor is Mg₂N₃, which is incorrect because it would result in a net charge of +1 per formula unit, misunderstanding the need for exact charge balance in ionic compounds. When determining empirical formulas for ionic solids, always find the lowest whole-number ratio of ions that results in zero net charge.

Question 20

1. It is a lattice of $\mathrm{Sr^{2+}}$ and $\mathrm{Cl^-}$ ions; the formula gives the simplest whole-number ratio in the crystal. (correct answer)
2. It is composed of $\mathrm{SrCl_2}$ molecules with covalent bonds that remain intact in the crystal.
3. It is a metallic lattice in which $\mathrm{Cl^-}$ provides a sea of delocalized electrons.
4. It is a network covalent solid in which all atoms are connected by covalent bonds throughout the crystal.
5. It is held together primarily by London dispersion forces between neutral formula units.

Explanation: This question tests understanding the distinction between formula units and molecular structures in ionic solids. SrCl₂(s) is an ionic lattice of Sr²⁺ and Cl⁻ ions in a 1:2 ratio, with the formula indicating the simplest charge-balanced unit, not discrete molecules. The solid is held by electrostatic attractions in a 3D array, without covalent bonds within 'molecules.' This reflects the ionic model, where the lattice extends indefinitely rather than consisting of packed molecules. A tempting distractor is that it is composed of SrCl₂ molecules with covalent bonds, which is wrong because it misclassifies ionic compounds as molecular, ignoring the ionic lattice nature. To clarify ionic solid structures, emphasize that formula units represent ratios in the lattice, not individual molecules.