College Chemistry Quiz: Structure Of Ionic Solids
11 questions · exam conditions
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Structure Of Ionic SolidsQuestion 1 of 11

A binary ionic compound adopts the zinc blende structure when the radius ratio r(cation)/r(anion) falls within a specific range. Which radius ratio range favors the zinc blende structure over rock salt?

0.155 - 0.225
0.225 - 0.414
0.414 - 0.732
0.732 - 1.000
Above 1.000
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College Chemistry Quiz

College Chemistry Quiz: Structure Of Ionic Solids

Practice Structure Of Ionic Solids in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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 College Chemistry.

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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.

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Question 1

A binary ionic compound adopts the zinc blende structure when the radius ratio r(cation)/r(anion) falls within a specific range. Which radius ratio range favors the zinc blende structure over rock salt?

  1. 0.155 - 0.225
  2. 0.225 - 0.414 (correct answer)
  3. 0.414 - 0.732
  4. 0.732 - 1.000
  5. Above 1.000
Explanation: When you encounter questions about crystal structures and radius ratios, you're dealing with how geometric constraints determine which crystal lattice is most stable for a given ionic compound. The zinc blende structure is favored when the cation-to-anion radius ratio falls between 0.225 and 0.414. This range represents the geometric sweet spot where cations can fit efficiently into tetrahedral holes formed by the anion lattice. In zinc blende, each cation is surrounded by four anions in a tetrahedral arrangement, and this coordination requires the cation to be large enough to maintain stable bonding distances but small enough to fit comfortably in the tetrahedral site. Looking at the incorrect options: Choice A (0.155-0.225) represents ratios too small for zinc blende stability—cations this small relative to anions would rattle around in tetrahedral holes, making the structure unstable. Choice C (0.414-0.732) corresponds to the rock salt structure range, where cations are large enough to prefer octahedral coordination (surrounded by six anions) rather than tetrahedral. Choice D (0.732-1.000) represents ratios approaching or exceeding equal ionic sizes, which favor structures like cesium chloride where cations coordinate with eight anions. The key insight is that as radius ratios increase, preferred coordination numbers increase: tetrahedral (4) for zinc blende, octahedral (6) for rock salt, and cubic (8) for cesium chloride structure. Study tip: Memorize the critical radius ratio ranges: 0.225-0.414 for zinc blende, 0.414-0.732 for rock salt. Remember that smaller ratios mean smaller coordination numbers.

Question 2

The unit cell of an ionic compound contains 4 formula units of MX. If the structure has a coordination number of 6:6 (each ion surrounded by 6 of the opposite type), which structure type does this compound most likely adopt?

  1. Cesium chloride
  2. Zinc blende
  3. Rock salt (correct answer)
  4. Fluorite
  5. Rutile
Explanation: When analyzing crystal structures of ionic compounds, you need to match the given structural information—formula units per unit cell and coordination numbers—to known structure types. The key clues here are 4 formula units of MX per unit cell and 6:6 coordination (each ion surrounded by 6 of the opposite type). The rock salt structure perfectly matches these criteria. In rock salt (like NaCl), you have a face-centered cubic arrangement where each Na⁺ is octahedrally coordinated by 6 Cl⁻ ions, and vice versa. The unit cell contains exactly 4 formula units of NaCl. Let's examine why the other options don't fit: (A) Cesium chloride has 8:8 coordination, not 6:6. Each Cs⁺ is surrounded by 8 Cl⁻ ions in a cubic arrangement, and the unit cell contains only 1 formula unit. (B) Zinc blende has 4:4 coordination with a tetrahedral geometry. Each Zn²⁺ is surrounded by 4 S²⁻ ions, not 6, making this incompatible with the given coordination number. (D) Fluorite has 8:4 coordination—each Ca²⁺ is surrounded by 8 F⁻ ions, while each F⁻ is surrounded by 4 Ca²⁺ ions. This doesn't match the 6:6 requirement. Study tip: Memorize the coordination numbers for common ionic structures: cesium chloride (8:8), zinc blende (4:4), rock salt (6:6), and fluorite (8:4). When given coordination numbers in a problem, immediately eliminate structures that don't match, then verify with other structural details like formula units per unit cell.

Question 3

In the corundum structure of Al₂O₃, aluminum ions occupy 2/3 of the octahedral holes in a hexagonal close-packed array of oxide ions. What is the coordination number of Al³⁺?

  1. 3
  2. 4
  3. 6 (correct answer)
  4. 8
  5. 9
Explanation: When you encounter questions about crystal structures and coordination numbers, focus on the specific geometric environment around each ion. The coordination number tells you how many nearest neighbors surround a central atom or ion. In the corundum structure, aluminum ions (Al³⁺) occupy octahedral holes within a hexagonal close-packed array of oxide ions (O²⁻). The key insight is understanding what "octahedral holes" means geometrically. An octahedral hole is a space surrounded by six atoms arranged at the vertices of an octahedron. Therefore, any ion occupying an octahedral hole will be surrounded by exactly six nearest neighbors. Since Al³⁺ ions occupy these octahedral sites, each aluminum ion is coordinated by six oxide ions, giving Al³⁺ a coordination number of 6. This makes C the correct answer. Let's examine why the other options are incorrect: A) 3 represents trigonal planar coordination, which doesn't match octahedral geometry. B) 4 corresponds to tetrahedral coordination, where ions would occupy tetrahedral holes, not octahedral ones. D) 8 represents cubic coordination, which would occur if aluminum occupied cubic holes rather than octahedral sites. The fact that only 2/3 of the octahedral holes are occupied is important for maintaining charge neutrality in Al₂O₃, but it doesn't change the coordination environment of the aluminum ions that are present. Study tip: Always connect hole types to coordination numbers: tetrahedral holes = 4 coordination, octahedral holes = 6 coordination, cubic holes = 8 coordination. The geometry of the hole determines the coordination number.

Question 4

In the nickel arsenide (NiAs) structure, arsenic atoms form a hexagonal close-packed lattice with nickel atoms occupying all octahedral holes. What is the stoichiometry of this compound?

  1. NiAs₂
  2. Ni₂As
  3. NiAs (correct answer)
  4. Ni₃As₂
  5. Ni₂As₃
Explanation: When you encounter crystal structure problems, you need to determine the ratio of atoms by counting how many of each type fit in the unit cell based on the packing arrangement and hole occupancy. In the NiAs structure, arsenic atoms form a hexagonal close-packed (hcp) lattice. In any close-packed structure, there are exactly as many octahedral holes as there are atoms forming the lattice. Since all octahedral holes are occupied by nickel atoms, this means there's a 1:1 ratio of nickel to arsenic atoms, giving us the formula NiAs. Here's why each wrong answer fails: Answer A (NiAs₂) would require twice as many arsenic atoms as nickel atoms, which would happen if nickel atoms formed the lattice and arsenic occupied both octahedral and tetrahedral holes. Answer B (Ni₂As) would require twice as many nickel atoms, which could occur if arsenic formed the lattice but only half the octahedral holes were filled by nickel. Answer D (Ni₃As₂) represents a more complex ratio that doesn't match the described structure at all. The key insight is recognizing that "all octahedral holes occupied" in a close-packed lattice always gives a 1:1 ratio. Since there are exactly as many octahedral holes as lattice atoms in any close-packed structure, complete occupancy yields equal numbers of both atom types. Study tip: Memorize that close-packed structures contain equal numbers of octahedral holes and lattice positions. This relationship appears frequently in crystal structure problems and makes stoichiometry calculations straightforward.

Question 5

The lattice energy of an ionic solid MX is proportional to (q₁ × q₂)/r₀, where q₁ and q₂ are ionic charges and r₀ is the interionic distance. If the lattice energy of NaCl (r₀ = 2.81 Å) is 786 kJ/mol, what is the approximate lattice energy of MgO (r₀ = 2.10 Å)?

  1. 1050 kJ/mol
  2. 2100 kJ/mol
  3. 3200 kJ/mol
  4. 4200 kJ/mol (correct answer)
  5. 5400 kJ/mol
Explanation: When you encounter lattice energy problems, you're dealing with the electrostatic forces holding ionic crystals together. The key relationship is that lattice energy is proportional to q1×q2r0\frac{q_1 \times q_2}{r_0}, where the charges are in the numerator and distance is in the denominator. To solve this, you need to set up a ratio comparing NaCl to MgO. For NaCl: Na⁺ has charge +1, Cl⁻ has charge -1, so q1×q2=(+1)(−1)=−1q_1 \times q_2 = (+1)(-1) = -1 (we use absolute value, so 1). For MgO: Mg²⁺ has charge +2, O²⁻ has charge -2, so q1×q2=(+2)(−2)=4q_1 \times q_2 = (+2)(-2) = 4 in absolute value. Setting up the proportion: UMgOUNaCl=qMg×qOqNa×qCl×rNaClrMgO\frac{U_{MgO}}{U_{NaCl}} = \frac{q_{Mg} \times q_O}{q_{Na} \times q_{Cl}} \times \frac{r_{NaCl}}{r_{MgO}} UMgO786=41×2.812.10=4×1.34=5.36\frac{U_{MgO}}{786} = \frac{4}{1} \times \frac{2.81}{2.10} = 4 \times 1.34 = 5.36 Therefore: UMgO=786×5.36=4213 kJ/molU_{MgO} = 786 \times 5.36 = 4213 \text{ kJ/mol} This matches answer D) 4200 kJ/mol. Answer A) 1050 kJ/mol only accounts for the distance ratio, ignoring the charge effect. Answer B) 2100 kJ/mol considers the 4× charge increase but ignores the distance difference. Answer C) 3200 kJ/mol uses incorrect charge calculations or ratios. Study tip: Always remember that lattice energy increases dramatically with higher charges (squared effect) and decreases with larger ionic radii. The charge effect typically dominates over size effects in these calculations.

Question 6

The density of an ionic solid MX with rock salt structure is 4.25 g/cm³. If the unit cell edge length is 5.85 Å and the molar mass of MX is 85.5 g/mol, how many formula units are contained in the unit cell?

  1. 2
  2. 4 (correct answer)
  3. 6
  4. 8
  5. 12
Explanation: When you encounter questions about unit cell calculations, you're connecting macroscopic properties (density, molar mass) with microscopic crystal structure. The key is using the relationship between these quantities to find how many formula units fit in one unit cell. To solve this, you'll use the density formula rearranged for the number of formula units: Z=ρ×Vcell×NAMZ = \frac{\rho \times V_{cell} \times N_A}{M}, where Z is the number of formula units, ρ is density, V_cell is unit cell volume, N_A is Avogadro's number, and M is molar mass. First, calculate the unit cell volume: Vcell=(5.85×10−8 cm)3=2.00×10−22 cm3V_{cell} = (5.85 \times 10^{-8} \text{ cm})^3 = 2.00 \times 10^{-22} \text{ cm}^3 Then substitute: Z=4.25 g/cm3×2.00×10−22 cm3×6.022×1023 mol−185.5 g/mol=4Z = \frac{4.25 \text{ g/cm}^3 \times 2.00 \times 10^{-22} \text{ cm}^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}}{85.5 \text{ g/mol}} = 4 Choice A (2) would correspond to a simple cubic structure, which is rare for ionic compounds. Choice C (6) isn't a typical value for common crystal structures. Choice D (8) would suggest a more complex structure like fluorite, but rock salt structure specifically contains 4 formula units per unit cell. The rock salt structure (like NaCl) always has 4 formula units per unit cell - this is a fundamental characteristic of this crystal type. Remember this structural relationship: when you see "rock salt structure" mentioned, expect Z = 4 as a key checkpoint for your calculations.

Question 7

In the inverse spinel structure of Fe₃O₄, Fe³⁺ ions occupy both tetrahedral and octahedral sites while Fe²⁺ ions occupy only octahedral sites. If the normal spinel formula is AB₂O₄, what is the cation distribution in Fe₃O₄?

  1. Tetrahedral: Fe²⁺; Octahedral: Fe³⁺
  2. Tetrahedral: Fe³⁺; Octahedral: Fe²⁺
  3. Tetrahedral: Fe³⁺; Octahedral: Fe³⁺ + Fe²⁺ (correct answer)
  4. Tetrahedral: Fe²⁺ + Fe³⁺; Octahedral: Fe³⁺
  5. Tetrahedral: Empty; Octahedral: Fe²⁺ + Fe³⁺
Explanation: When you encounter spinel structure questions, focus on understanding how cations distribute between tetrahedral and octahedral sites. The normal spinel formula AB₂O₄ has A cations in tetrahedral sites and B cations in octahedral sites, while inverse spinels mix the cations differently. For Fe₃O₄, you need to account for both Fe²⁺ and Fe³⁺ ions. The question states that Fe³⁺ ions occupy both tetrahedral and octahedral sites, while Fe²⁺ ions occupy only octahedral sites. This immediately tells you the distribution: tetrahedral sites contain only Fe³⁺, while octahedral sites contain both Fe³⁺ and Fe²⁺ ions. Answer A incorrectly places Fe²⁺ in tetrahedral sites, directly contradicting the given information that Fe²⁺ ions occupy only octahedral sites. Answer B suggests Fe²⁺ ions are exclusively in octahedral sites while Fe³⁺ are exclusively in tetrahedral sites, missing that Fe³⁺ ions occupy both types of sites. Answer D incorrectly places Fe²⁺ in tetrahedral sites and suggests Fe³⁺ ions are only in octahedral sites, contradicting both parts of the given information. Answer C correctly captures the distribution: Fe³⁺ ions in tetrahedral sites, and both Fe³⁺ and Fe²⁺ ions sharing the octahedral sites. Study tip: For spinel structure problems, always carefully track which oxidation states can occupy which sites. The term "inverse spinel" signals that the cation distribution differs from the simple AB₂O₄ pattern, so expect mixed occupancy in at least one type of site.

Question 8

In the layered structure of CdI₂, Cd²⁺ ions occupy octahedral holes between close-packed layers of I⁻ ions. What fraction of the octahedral holes are filled by Cd²⁺ ions?

  1. 1/6
  2. 1/3
  3. 1/2 (correct answer)
  4. 2/3
  5. All
Explanation: When analyzing ionic crystal structures, you need to understand the relationship between the chemical formula and how ions fill available holes in the crystal lattice. In CdI₂, the iodide ions (I⁻) form close-packed layers with octahedral holes between them. To determine what fraction of these holes are filled, use the stoichiometry from the chemical formula. The formula CdI₂ tells you there's 1 Cd²⁺ ion for every 2 I⁻ ions. In close-packed structures, there's exactly 1 octahedral hole per anion. Since you have 2 I⁻ ions in the formula unit, there are 2 octahedral holes available. With only 1 Cd²⁺ ion to fill these holes, the fraction filled is 1 Cd2+2 octahedral holes=12\frac{1 \text{ Cd}^{2+}}{2 \text{ octahedral holes}} = \frac{1}{2}. Looking at the wrong answers: Choice (A) 1/6 would correspond to a formula like CdI₆, which doesn't exist for this structure type. Choice (B) 1/3 might come from incorrectly thinking about tetrahedral holes or miscounting the available sites. Choice (D) 2/3 reverses the relationship—this would require more cations than the formula allows. The key insight is that the chemical formula directly tells you the cation-to-anion ratio, and knowing there's one octahedral hole per anion in close-packed structures lets you calculate the fraction immediately. Always start with the formula stoichiometry when analyzing hole-filling in ionic crystals—it's your roadmap to the answer.

Question 9

Which statement correctly explains the relationship between lattice energy and solubility for ionic compounds?

  1. Higher lattice energy usually lowers solubility unless hydration compensates (correct answer)
  2. Higher lattice energy always increases solubility in water
  3. Solubility depends only on molar mass, not lattice energy
  4. Lattice energy affects melting point but never solubility
Explanation: This question tests college-level chemistry understanding of the structure of ionic solids, focusing on lattice formations and their properties. Ionic solids are characterized by a lattice structure where ions are held together by electrostatic forces, influencing properties like melting point and solubility. The solubility of ionic compounds in water depends on the balance between lattice energy and hydration energy, where high lattice energy can hinder dissolution if not overcome by ion-water interactions. Choice A is correct because it accurately describes how higher lattice energy usually lowers solubility unless compensated by strong hydration. Choice B is incorrect as it wrongly states that higher lattice energy always increases solubility, ignoring the opposing effect on lattice stability. Instruct students to differentiate between lattice energy and hydration energy by considering ion size and charge effects. Encourage practice with solubility predictions for compounds like NaCl and CaO to grasp these competing factors.

Question 10

What impact do crystal defects have on the conductivity of ionic solids?

  1. Defects always make ionic solids conduct well as solids
  2. Defects can increase ion mobility, raising solid-state conductivity (correct answer)
  3. Defects eliminate ions, so conductivity drops to zero
  4. Defects convert ionic solids into metals with free electrons
Explanation: This question tests college-level chemistry understanding of the structure of ionic solids, focusing on lattice formations and their properties. Ionic solids are characterized by a lattice structure where ions are held together by electrostatic forces, influencing properties like melting point and solubility. Crystal defects, such as vacancies or interstitials, can disrupt the perfect lattice, allowing for ion movement even in the solid state. Choice B is correct because it explains how defects increase ion mobility, thereby enhancing solid-state ionic conductivity. Choice A is incorrect as it overstates that defects always make solids conduct well, ignoring that pure lattices are insulators in solid form. Instruct students to differentiate between perfect and defective lattices by studying defect types like Schottky and Frenkel. Encourage practice with examples of doped ionic conductors to understand conductivity mechanisms.

Question 11

How does the lattice energy of MgO compare to NaCl, and how does it affect melting point?

  1. MgO has lower lattice energy, so it melts higher
  2. MgO has higher lattice energy, so it melts higher (correct answer)
  3. NaCl has higher lattice energy, so MgO melts lower
  4. They have equal lattice energy, so melting points match
Explanation: This question tests college-level chemistry understanding of the structure of ionic solids, focusing on lattice formations and their properties. Ionic solids are characterized by a lattice structure where ions are held together by electrostatic forces, influencing properties like melting point and solubility. In comparing MgO and NaCl, MgO has higher lattice energy due to the greater charges on its ions (Mg²⁺ and O²⁻) compared to Na⁺ and Cl⁻, and smaller ionic radii, leading to stronger attractions. Choice B is correct because it accurately states that MgO has higher lattice energy, resulting in a higher melting point as more energy is required to disrupt the lattice. Choice A is incorrect as it mistakenly claims MgO has lower lattice energy, which would imply a lower melting point, confusing the relationship between charge and lattice strength. Instruct students to differentiate between ionic compounds by calculating or comparing lattice energies using factors like ion charge and size. Encourage practice with examples like MgO versus NaCl to understand how these factors impact physical properties such as melting points.