IB Chemistry Quiz: Apply Models To Materials
20 questions · exam conditions
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Apply Models To MaterialsQuestion 1 of 20

The repeating unit of an addition polymer is shown as -[CH(CN)CH₂]-. Which molecule is the monomer used to form this polymer?

CH₃CH₂CN
CH₂(CN)CH₂OH
CH(CN)=CH₂
N≡C-C≡CH
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IB Chemistry Quiz

IB Chemistry Quiz: Apply Models To Materials

Practice Apply Models To Materials in IB 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 Apply Models To Materials, 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.

All questions

Question 1

The repeating unit of an addition polymer is shown as -[CH(CN)CH₂]-. Which molecule is the monomer used to form this polymer?

  1. CH₃CH₂CN
  2. CH₂(CN)CH₂OH
  3. CH(CN)=CH₂ (correct answer)
  4. N≡C-C≡CH
Explanation: Addition polymerization involves the breaking of a double bond in a monomer to form single bonds that link the monomer units together. To form the repeating unit -[CH(CN)CH₂]-, the single bond between the two carbon atoms in the backbone must have been a C=C double bond in the monomer. The monomer is therefore propenenitrile (acrylonitrile), which has the structure CH₂=CH(CN).

Question 2

An unknown solid compound is found to be brittle, have a melting point of 980 °C, and conduct electricity only when in a molten state. Based on these properties, the bonding in the compound is best described as:

  1. Metallic, with a lattice of cations and delocalized electrons.
  2. Polar covalent, with molecules held together by dipole-dipole forces.
  3. Ionic, with ions held in a fixed crystal lattice that are free to move when molten. (correct answer)
  4. Covalent network, with atoms joined by strong covalent bonds in a continuous lattice.
Explanation: The combination of properties—high melting point, brittleness, and electrical conductivity only in the molten state—is characteristic of an ionic compound. In the solid state, ions are held in fixed positions in a crystal lattice and cannot move to carry charge. When melted, the ions become mobile and can conduct electricity. Metallic solids conduct electricity in the solid state. Covalent molecular solids have low melting points. Covalent network solids have very high melting points but do not conduct electricity in any state (with exceptions like graphite).

Question 3

A new material is synthesized. It is a very hard, brittle solid with a melting point above 2500 °C. It does not conduct electricity under any conditions and is insoluble in both water and hexane. The bonding in this material is best described as:

  1. Ionic
  2. Metallic
  3. Polar molecular
  4. Covalent network (correct answer)
Explanation: The combination of very high melting point, extreme hardness, brittleness, and being an electrical insulator in all states is characteristic of a covalent network solid (or giant covalent structure). In these materials, atoms are held together by a vast, three-dimensional lattice of strong covalent bonds (e.g., diamond, silicon dioxide). Ionic compounds would conduct when molten, metallic compounds would conduct as solids, and molecular solids would have much lower melting points.

Question 4

What is the systematic IUPAC name for the monomer that polymerizes to form the repeating unit -[O-(CH₂)₄-CO]-? [AHL]

  1. 4-hydroxypentanoic acid
  2. 5-hydroxypentanoic acid (correct answer)
  3. Butane-1,4-diol
  4. Hexanedioic acid
Explanation: This polymer is a polyester. The repeating unit is -[O-(CH₂)₄-CO]-. This unit is formed via condensation polymerization of a single monomer containing both a hydroxyl (-OH) group and a carboxylic acid (-COOH) group. Let's reconstruct the monomer. The -CO- part comes from a -COOH group, and the -O- part comes from an -OH group. The monomer must be HO-(CH₂)₄-COOH. To name this, we number the carbon chain starting from the carboxylic acid carbon as C1. This makes the chain five carbons long (a pentanoic acid). The hydroxyl group is on the fifth carbon, so the name is 5-hydroxypentanoic acid.

Question 5

Which pair of molecules could undergo condensation polymerization to form a polyester? [AHL]

  1. HO-CH₂-CH₂-OH and HOOC-COOH (correct answer)
  2. CH₂=CH₂ and CH₂=CH-COOH
  3. HO-CH₂-CH₂-NH₂ and HOOC-CH₂-COOH
  4. H₂N-CH₂-CH₂-NH₂ and HOOC-COOH
Explanation: The formation of a polyester requires the reaction between a hydroxyl (-OH) functional group and a carboxylic acid (-COOH) functional group, forming an ester linkage (-COO-). To form a polymer, each monomer must have two functional groups (be difunctional). Choice A provides a diol (two hydroxyl groups) and a dicarboxylic acid (two carboxylic acid groups), which are the correct monomers for forming a polyester. Choice D would form a polyamide. Choice B monomers would undergo addition polymerization. Choice C would form a polymer with both ester and amide links.

Question 6

On a bonding triangle diagram, a substance is located in the region corresponding to a moderate electronegativity difference and a high average electronegativity. Which properties are most likely for this substance?

  1. A low melting point solid that is soft and does not conduct electricity.
  2. A liquid with a high boiling point that is volatile and miscible with water.
  3. A very high melting point solid that is an electrical insulator and insoluble in water. (correct answer)
  4. A malleable solid with a lustrous appearance that conducts electricity well.
Explanation: A moderate electronegativity difference and high average electronegativity are characteristic of polar covalent bonds, often found in covalent network solids (e.g., SiO₂, SiC). These materials have atoms held together by a vast network of strong covalent bonds, resulting in very high melting points, hardness, and general insolubility. They are typically electrical insulators because the electrons are localized in bonds. Choice C matches this description. Choice A describes a molecular solid, and choice D describes a metal.

Question 7

The monomer propene, CH₃CH=CH₂, can be polymerized. Depending on the catalyst, the resulting poly(propene) can be isotactic (all CH₃ groups on the same side of the chain) or atactic (CH₃ groups randomly oriented). Which prediction about their properties is most plausible?

  1. Atactic poly(propene) will be more crystalline and have a higher density than isotactic poly(propene).
  2. The C–H bonds in the methyl groups of isotactic poly(propene) are more polar than in the atactic form.
  3. Atactic poly(propene) will be harder and more rigid because the random groups prevent chain slippage.
  4. Isotactic poly(propene) allows for more efficient chain packing, leading to stronger intermolecular forces and a higher melting point. (correct answer)
Explanation: The regular, ordered structure of isotactic poly(propene), with all methyl groups on one side, allows the polymer chains to pack together closely and efficiently. This close packing increases the effectiveness of the London dispersion forces between the chains. Stronger intermolecular forces require more energy to overcome, resulting in a higher melting point, greater crystallinity, and higher density. The random orientation of methyl groups in the atactic form prevents efficient packing, leading to weaker intermolecular forces and a softer, amorphous material with a lower melting point.

Question 8

High-density poly(ethene) (HDPE) consists mainly of linear polymer chains, while low-density poly(ethene) (LDPE) has significant chain branching. Which statement correctly explains why HDPE has a higher melting point than LDPE?

  1. The C–C covalent bonds in the linear chains of HDPE are stronger than those in the branched chains of LDPE.
  2. The linear chains in HDPE can pack together more closely, leading to stronger London dispersion forces between them. (correct answer)
  3. Hydrogen bonding can occur between the linear chains of HDPE but not between the branched chains of LDPE.
  4. The monomers used to make HDPE have a higher molar mass than the monomers used to make LDPE.
Explanation: The melting of a polymer involves overcoming the intermolecular forces between polymer chains. HDPE has linear chains that can align and pack closely together in an ordered, crystalline fashion. This close packing maximizes the surface area contact between chains, resulting in stronger overall London dispersion forces (LDFs). In contrast, the branches on LDPE chains prevent them from packing efficiently, leading to weaker LDFs and a lower melting point. The monomer (ethene) and the covalent bond strengths are the same for both.

Question 9

Nitinol is a 'memory-metal' alloy of nickel and titanium. It can be deformed and will return to its original shape upon heating. This property depends on a reversible solid-state phase change. Which statement about the bonding in nitinol provides the best explanation for this ability?

  1. The alloy consists of alternating nickel and titanium ions held in a rigid, brittle lattice.
  2. Strong, directional covalent bonds between nickel and titanium atoms allow the material to 'remember' its shape.
  3. The non-directional metallic bonding allows atoms to rearrange into different crystal structures without fracturing. (correct answer)
  4. Weak intermolecular forces between nickel and titanium atoms are overcome at body temperature.
Explanation: Alloys are held together by metallic bonding, which involves a lattice of positive ions and a 'sea' of delocalized electrons. This bonding is non-directional, meaning the attraction is maintained even if the atoms shift positions. This allows the atoms in nitinol to rearrange from one solid crystal structure to another (a phase change) without breaking the material, which is essential for the shape-memory effect. Covalent bonds are directional and would lead to a brittle material, while ionic and intermolecular force models are incorrect for a metallic alloy.

Question 10

Two materials, X and Y, have similar molecular formulas but vastly different melting points (X: 1050°C, Y: 78°C). X is insoluble in water but conducts electricity when molten. Y is highly soluble in polar solvents and does not conduct electricity in any state. What structural difference most likely accounts for these contrasting properties?

  1. X has hydrogen bonding while Y has only dipole-dipole interactions between molecules
  2. X forms a covalent network structure while Y exists as discrete molecular units
  3. X has ionic bonding in a crystal lattice while Y has covalent bonding in discrete molecules (correct answer)
  4. X has metallic bonding with electron delocalization while Y has localized covalent bonds
Explanation: The high melting point and electrical conductivity when molten (but not when solid) indicates X is an ionic compound. The low melting point, solubility in polar solvents, and lack of conductivity indicates Y is a polar molecular compound. Option A doesn't explain the conductivity difference. Option B doesn't explain why X conducts when molten. Option D doesn't explain the solubility behavior of Y or why X only conducts when molten.

Question 11

A solid exhibits the following thermal expansion behavior: it expands significantly more in one direction than the other two when heated. Chemical analysis shows it contains only nonmetallic elements. X-ray crystallography reveals long chains of atoms connected by short bonds, with longer distances between chains. Which structural model best correlates with these observations?

  1. Three-dimensional covalent network with uniform bond lengths and isotropic thermal expansion
  2. Molecular crystal with discrete units held together by intermolecular forces in all directions
  3. Layered ionic structure with alternating cation and anion sheets and directional bonding
  4. One-dimensional polymer chains with strong intramolecular bonds and weak intermolecular interactions (correct answer)
Explanation: When analyzing crystal structures, you need to connect macroscopic properties (like thermal expansion) with microscopic arrangements of atoms and bonding patterns. The key insight here is recognizing how different bonding strengths in different directions create anisotropic (directional) behavior. The observations point to a structure with strong bonds in one direction and weak interactions in others. The significant expansion in one direction indicates weak intermolecular forces that easily stretch when heated, while minimal expansion in other directions suggests strong covalent bonds that resist thermal motion. Since the material contains only nonmetallic elements and shows long chains with short intramolecular bonds, this describes polymer chains held together by weak van der Waals forces or hydrogen bonds. Option D correctly identifies this as one-dimensional polymer chains. The strong covalent bonds along the chain resist expansion, while weak intermolecular forces between chains allow significant thermal expansion perpendicular to the chain direction. Option A is wrong because three-dimensional networks show isotropic (uniform) expansion in all directions, contradicting the observed anisotropy. Option B describes molecular crystals, but these typically don't form long chains and would show more uniform expansion. Option C suggests an ionic structure, but the material contains only nonmetallic elements, ruling out ionic bonding. Remember: anisotropic thermal expansion is a telltale sign of structural anisotropy. Look for materials where bonding strength varies significantly with direction—polymers, layered materials, and chain structures are prime examples on IB Chemistry exams.

Question 12

A crystalline solid has the following properties: melting point 2800°C, insoluble in all common solvents, extremely hard, and electrical resistivity of 101610^{16} Ω·cm. Based on these macroscopic properties, what can be concluded about its atomic-level structure and bonding?

  1. Close-packed metallic structure with partially filled d-orbitals and mobile electron sea
  2. Three-dimensional network of covalent bonds with complete electron localization between atoms (correct answer)
  3. Ionic crystal lattice with high charge density ions and strong electrostatic interactions
  4. Layered structure with strong intralayer covalent bonds and weak interlayer van der Waals forces
Explanation: The extremely high melting point, hardness, insolubility, and very high electrical resistivity all point to a covalent network structure like diamond or silicon carbide. The complete electron localization in covalent bonds explains the insulating behavior. Option A (metals) would be conductive. Option C (ionic) would have some solubility in polar solvents and would conduct when molten. Option D (layered) wouldn't explain the extreme hardness and would likely show some cleavage planes.

Question 13

A researcher observes that a crystalline material becomes an excellent electrical conductor when heated above 800°C, but reverts to insulating behavior when cooled. The material maintains its crystal structure throughout this process, and no chemical decomposition occurs. Which bonding model and structural feature combination best explains this reversible conductivity transition?

  1. Ionic bonding where thermal energy provides sufficient activation energy for ion migration through the crystal lattice
  2. Covalent bonding where thermal expansion creates conduction pathways between previously isolated molecular units
  3. Metallic bonding where temperature changes the degree of orbital overlap and electron delocalization
  4. Semiconductor behavior where thermal energy promotes electrons from valence band to conduction band across a moderate band gap (correct answer)
Explanation: The reversible transition from insulator to conductor with temperature, while maintaining crystal structure, is characteristic of semiconductor behavior. At high temperatures, thermal energy excites electrons across the band gap into the conduction band. Upon cooling, electrons return to the valence band, restoring insulating behavior. Option A would involve ion migration which typically causes permanent structural changes. Option B doesn't explain the sharp transition temperature. Option C describes materials that are always conductive, just with temperature-dependent resistance.

Question 14

A material scientist observes that compound Z can be cleaved along specific crystallographic planes with relatively little force, yet individual layers are extremely difficult to compress. The material is a good conductor parallel to the layers but an insulator perpendicular to them. What structural model best explains this anisotropic behavior?

  1. Metallic bonding with preferential electron flow in crystallographic directions due to orbital overlap
  2. Ionic bonding with alternating charge layers creating directional electrostatic field gradients
  3. Layered structure with strong intralayer covalent bonds and weak interlayer van der Waals interactions (correct answer)
  4. Network covalent structure with systematic defects creating preferential fracture planes
Explanation: The combination of easy cleavage along planes, high resistance to compression within layers, and anisotropic conductivity strongly suggests a layered structure like graphite. Strong covalent bonds within layers explain the compression resistance and in-plane conductivity, while weak van der Waals forces between layers explain easy cleavage and poor perpendicular conductivity. Option A doesn't explain the easy cleavage. Option B doesn't typically show such pronounced anisotropy. Option D doesn't explain the conductivity anisotropy.

Question 15

Two compounds with identical chemical formulas AB2AB_2 exhibit dramatically different properties. Compound I melts at 1600°C and conducts electricity only when molten. Compound II melts at -25°C and is completely miscible with nonpolar solvents. What fundamental difference in bonding and molecular architecture accounts for these contrasting behaviors?

  1. Compound I has coordinate covalent bonding while Compound II has pure covalent bonding
  2. Compound I forms a discrete molecular structure while Compound II forms an extended network
  3. Compound I has ionic bonding in an extended lattice while Compound II has covalent bonding in discrete molecules (correct answer)
  4. Compound I has hydrogen bonding interactions while Compound II has only London dispersion forces
Explanation: The high melting point and conduction only when molten indicates Compound I is ionic with an extended crystal lattice structure. The low melting point and miscibility with nonpolar solvents indicates Compound II consists of discrete nonpolar covalent molecules. Option A doesn't explain the conductivity or solubility differences. Option B reverses the structural descriptions. Option D doesn't account for the electrical conductivity of molten Compound I.

Question 16

A material exhibits excellent thermal and electrical conductivity, high tensile strength, and can be deformed without breaking. When analyzed by X-ray diffraction, it shows a face-centered cubic (fcc) structure. Which combination of bonding model and structural feature best explains these macroscopic properties?

  1. Metallic bonding with delocalized electrons and close-packed arrangement allowing slip planes (correct answer)
  2. Covalent network bonding with directional bonds and rigid three-dimensional framework
  3. Ionic bonding with electrostatic attractions and layered crystalline structure
  4. Van der Waals forces with molecular interactions and hexagonal close-packed structure
Explanation: The combination of excellent conductivity (thermal and electrical), high strength, malleability, and fcc structure is characteristic of metallic bonding. The delocalized electron sea explains conductivity, while the close-packed fcc structure with slip planes explains the ability to deform without breaking. Option B describes covalent networks (like diamond) which are hard but brittle. Option C describes ionic compounds which are generally insulators and brittle. Option D describes molecular solids which have poor conductivity and low strength.

Question 17

Material P has a bulk modulus of 180 GPa and shows plastic deformation under stress. Material Q has a bulk modulus of 442 GPa but fractures brittlely under the same stress conditions. Both materials have similar atomic masses and densities. What difference in atomic-level bonding and structure most likely explains this mechanical behavior contrast?

  1. Material P has metallic bonding with slip planes while Material Q has covalent bonding with rigid directional bonds (correct answer)
  2. Material P has ionic bonding with mobile charge carriers while Material Q has metallic bonding with localized electrons
  3. Material P has van der Waals bonding with weak interactions while Material Q has hydrogen bonding with directional attractions
  4. Material P has covalent network structure with defects while Material Q has perfect ionic crystal arrangement
Explanation: The plastic deformation of Material P indicates the ability for atoms to move past each other without breaking bonds, characteristic of metallic bonding with slip planes. The brittle fracture of Material Q despite high bulk modulus indicates strong but directional bonds that cannot accommodate deformation, characteristic of covalent bonding. Option B incorrectly describes ionic materials as having mobile charge carriers in the solid state. Option C describes bonding too weak to account for the high bulk moduli. Option D reverses the expected behavior of these bond types.

Question 18

Analysis of a material's structure reveals that it has a coordination number of 12 for each atom, with electrons that can move freely throughout the structure. However, unlike typical materials with these characteristics, this substance is relatively soft and has a low melting point. Which factor most likely accounts for this unexpected combination of properties?

  1. The presence of d-electrons creates antibonding interactions that weaken the overall structure
  2. Large atomic size results in diffuse electron clouds and weaker electrostatic attractions (correct answer)
  3. Partial covalent character in the bonding reduces the metallic bond strength significantly
  4. Defects in the crystal structure create stress concentration points that facilitate deformation
Explanation: Coordination number 12 and freely moving electrons indicate metallic bonding in a close-packed structure. The unexpected softness and low melting point suggest weak metallic bonding, which occurs with large atoms (like alkali metals) where the valence electrons are far from the nucleus and the metallic bonds are weak due to large atomic radii. Option A doesn't typically apply to simple metals. Option C doesn't explain the structural features. Option D would affect mechanical properties but not necessarily the melting point.

Question 19

Bronze is a substitutional alloy where some copper atoms in the lattice are replaced by larger tin atoms. How does this substitution primarily affect the physical properties of the metal?

  1. It increases the electrical conductivity by providing more mobile charge carriers from the tin atoms.
  2. It decreases malleability because the irregular atom sizes disrupt the slip planes in the metallic lattice. (correct answer)
  3. It converts the non-directional metallic bonding into directional covalent bonds, making the structure brittle.
  4. It lowers the melting point by weakening the electrostatic attraction between cations and delocalized electrons.
Explanation: The key feature of alloys that makes them harder and less malleable than pure metals is the disruption of the regular crystal lattice. In bronze, the larger tin atoms prevent the layers of copper atoms from sliding past one another easily. This resistance to slipping of atomic layers is observed as decreased malleability and increased hardness. The bonding remains non-directional and metallic. While alloying does affect conductivity (usually decreasing it) and melting point, the primary effect on malleability is due to lattice disruption.

Question 20

Consider the oxides MgO, Al₂O₃, and SiO₂. Using electronegativity values from the data booklet, which statement correctly describes the trend in bonding character across these period 3 oxides?

  1. The bonding becomes progressively more metallic due to the increase in valence electrons from Mg to Si.
  2. The bonding character transitions from predominantly ionic to polar covalent as the electronegativity difference decreases. (correct answer)
  3. All three compounds exhibit pure covalent bonding as oxygen is a non-metal.
  4. The bonding character transitions from covalent to ionic as the metallic character of the period 3 element increases.
Explanation: From the data booklet, EN values are: Mg=1.31, Al=1.61, Si=1.90, O=3.44. The electronegativity difference (ΔEN) for each oxide is: MgO: 3.44-1.31=2.13; Al₂O₃: 3.44-1.61=1.83; SiO₂: 3.44-1.90=1.54. As we move from Mg to Si, the electronegativity of the element increases, so its ΔEN with oxygen decreases. This corresponds to a trend from predominantly ionic bonding (MgO) towards polar covalent bonding (SiO₂), with Al₂O₃ being intermediate (amphoteric).