IB Chemistry Quiz: Understand Models To Materials

19 questions · exam conditions

0:00

Understand Models To MaterialsQuestion 1 of 19

Kevlar is a condensation polymer known for its exceptional strength. It is formed from 1,4-phenylenediamine and terephthaloyl chloride (a derivative of a dicarboxylic acid). What type of linkage connects the monomer units in Kevlar and what small molecule is eliminated during its formation? [AHL Content]

Ester linkage; H₂O is eliminated

Amide linkage; HCl is eliminated

Ether linkage; H₂O is eliminated

Amide linkage; H₂ is eliminated

← Back to quizzes

IB Chemistry Quiz

IB Chemistry Quiz: Understand Models To Materials

Practice Understand 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 Understand 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

Ester linkage; H₂O is eliminated
Amide linkage; HCl is eliminated (correct answer)
Ether linkage; H₂O is eliminated
Amide linkage; H₂ is eliminated

Explanation: The reaction is between an amine group (-NH₂) from 1,4-phenylenediamine and an acyl chloride group (-COCl) from terephthaloyl chloride. This is a condensation reaction that forms an amide linkage (-CONH-). In this specific reaction, the hydrogen from the amine group and the chlorine from the acyl chloride group are eliminated, forming a molecule of hydrogen chloride (HCl) as the small molecule by-product. Therefore, the polymer contains amide linkages and HCl is eliminated.

Question 2

The electronegativity values for Be, Cl, and Cs are 1.6, 3.2, and 0.8 respectively. Based on these values, how should the bonding in BeCl₂ and CsCl be characterized?

BeCl₂ is ionic and CsCl is polar covalent.
Both BeCl₂ and CsCl are predominantly ionic.
BeCl₂ is polar covalent and CsCl is predominantly ionic. (correct answer)
Both BeCl₂ and CsCl are polar covalent.

Explanation: The character of a bond is determined by the difference in electronegativity (ΔEN) between the two atoms. For BeCl₂, ΔEN = 3.2 - 1.6 = 1.6. This value falls in the range typically classified as polar covalent. For CsCl, ΔEN = 3.2 - 0.8 = 2.4. This large difference indicates a predominantly ionic bond. Therefore, BeCl₂ has polar covalent bonds and CsCl has ionic bonding. The distractors misinterpret these ΔEN values: A reverses the classifications, B incorrectly classifies BeCl₂ as ionic, and D incorrectly classifies CsCl as polar covalent.

Question 3

A material is synthesized that has a very high melting point, is extremely hard, and does not conduct electricity under any conditions. It is insoluble in common solvents. The electronegativity difference between its constituent elements is 1.7. What is the most likely classification for this material?

Ionic solid
Metallic solid
Covalent network solid (correct answer)
Molecular solid

Explanation: The combination of very high melting point, extreme hardness, and being an electrical insulator under all conditions are hallmark properties of a covalent network solid (like diamond or silicon dioxide). The atoms are held in a rigid 3D lattice by strong covalent bonds. An electronegativity difference of 1.7 is on the borderline between polar covalent and ionic, but the other properties, especially the lack of conductivity when molten, strongly rule out an ionic solid (A). Metallic solids (B) are conductors. Molecular solids (D) have low melting points.

Question 4

The monomer for the polymer neoprene is 2-chloro-1,3-butadiene (CH₂=C(Cl)-CH=CH₂). What would be the most likely repeating unit of the polymer formed under conditions that favor 1,4-addition polymerization?

-[CH₂-C(Cl)=CH-CH₂]- (correct answer)
-[CH₂-C(Cl)(CH₂)-CH₃]-
-[CH₂=C(Cl)-CH-CH₂]-
-[CH₂-CHCl-CH=CH]-

Explanation: The monomer is CH₂=C(Cl)-CH=CH₂. In 1,4-addition polymerization of a conjugated diene, the π bonds at positions 1 and 4 are used to form the polymer backbone, while a new π bond forms between carbons 2 and 3. The repeating unit becomes -[CH₂-C(Cl)=CH-CH₂]- where the chlorine substituent remains on carbon 2. Distractor B shows incorrect saturation. Distractor C shows an unreacted monomer structure. Distractor D shows incorrect connectivity.

Question 5

A material is formed from two elements, Q and R. It is a solid at room temperature, conducts electricity only in the molten state, and is brittle. The electronegativity of Q is 3.4 and R is 1.2. Which region of the bonding triangle would this material most likely occupy?

The metallic vertex, due to its conductivity.
The covalent vertex, due to its brittle nature.
The ionic vertex, based on its properties and the electronegativity difference. (correct answer)
The region between the ionic and metallic vertices, as it conducts when molten.

Explanation: The properties described (brittle solid, conducts only when molten) are characteristic of an ionic compound. The large electronegativity difference (ΔEN = 3.4 - 1.2 = 2.2) further supports the formation of an ionic bond. Therefore, the material would be located near the ionic vertex of the bonding triangle. Distractor A is incorrect because metals conduct electricity in the solid state. Distractor B is incorrect because while some covalent network solids are brittle, they do not conduct electricity when molten. Distractor D is incorrect because the combination of properties, especially non-conductivity in the solid state, points strongly to ionic, not a metallic-ionic intermediate.

Question 6

Nitinol is an alloy of nickel and titanium known for its shape-memory properties. It is a substitutional alloy. What can be deduced about the nickel and titanium atoms that allows for the formation of this type of alloy?

Their atomic radii are of a similar magnitude. (correct answer)
Their electronegativity values are significantly different.
One element has a much smaller atomic radius than the other.
They belong to the same group in the periodic table.

Explanation: Substitutional alloys are formed when atoms of one metallic element replace atoms of another in the metallic lattice. This is most readily achieved when the atoms of the different elements have similar atomic radii (typically within about 15% of each other), allowing them to fit into the lattice without causing excessive strain. Nickel and titanium are both transition metals in the same period with similar atomic radii, which facilitates the formation of a substitutional alloy. A large electronegativity difference (B) would lead to ionic or intermetallic compound formation. A large difference in atomic radii (C) would lead to the formation of an interstitial alloy. Belonging to the same group (D) is not a requirement; Ni is in Group 10 and Ti is in Group 4.

Question 7

An unknown substance is a brittle solid with a high melting point. It does not conduct electricity as a solid but does conduct when molten or dissolved in water. The substance is formed from a Group 2 element and a Group 16 element. Which statement is consistent with this information?

The substance is a covalent network solid held together by strong directional bonds.
The substance has metallic bonding with delocalized electrons that become mobile only at high temperatures.
The substance is an ionic compound composed of a lattice of cations and anions. (correct answer)
The substance is a molecular solid with strong covalent bonds and weak intermolecular forces.

Explanation: The described properties—brittle solid, high melting point, conducts electricity only when molten or aqueous—are the classic characteristics of an ionic compound. In the solid state, the ions are held in fixed positions within the crystal lattice and cannot move to carry charge. When molten or dissolved, the ions are free to move and can conduct electricity. A compound formed between a Group 2 metal (which forms a 2+ ion) and a Group 16 non-metal (which forms a 2- ion) will have a large electronegativity difference, favoring the formation of an ionic lattice. The other options describe other material types with inconsistent properties.

Question 8

Consider three substances: SiO₂, MgF₂, and PCl₃. Based on the general principles of electronegativity and bonding models, which sequence correctly arranges these substances from most ionic to most covalent character?

MgF₂ > PCl₃ > SiO₂
MgF₂ > SiO₂ > PCl₃ (correct answer)
SiO₂ > PCl₃ > MgF₂
PCl₃ > SiO₂ > MgF₂

Explanation: The ionic/covalent character is determined by the electronegativity difference (ΔEN). We can predict this using the periodic table trends. MgF₂ involves a metal from group 2 and the most electronegative non-metal (F), resulting in a very large ΔEN and ionic character. SiO₂ involves a metalloid (Si) and a highly electronegative non-metal (O); this will have a large ΔEN but is known to be a covalent network solid, indicating highly polar covalent character. PCl₃ involves two non-metals (P and Cl) that are relatively close in the periodic table, so their ΔEN will be the smallest, indicating polar covalent character. The order of decreasing ΔEN, and thus decreasing ionic character, is Mg-F > Si-O > P-Cl. Therefore, the sequence from most ionic to most covalent is MgF₂ > SiO₂ > PCl₃.

Question 9

A molecular crystal of iodine ( $\text{I}_2$ ) is compared to a covalent network solid of silicon carbide (SiC). Both contain covalent bonds, yet their mechanical and thermal properties differ dramatically. What is the primary structural reason for these differences?

The iodine crystal has stronger individual covalent bonds but weaker overall structure due to molecular packing arrangement limitations.
Silicon carbide contains continuous covalent bonding throughout the structure while iodine has strong intramolecular bonds but weak intermolecular forces. (correct answer)
The difference in electronegativity between Si-C bonds versus I-I bonds results in fundamentally different mechanical response mechanisms.
Iodine molecules can rotate freely within the crystal lattice while silicon carbide has fixed atomic positions that prevent deformation.

Explanation: SiC is a covalent network where every atom is covalently bonded to neighbors, creating a 3D network of strong bonds. I₂ is a molecular crystal where strong covalent bonds exist within molecules but only weak van der Waals forces exist between molecules. This explains the dramatic property differences. Option A incorrectly suggests I-I bonds are stronger. Option C focuses on electronegativity rather than the key structural difference. Option D oversimplifies the structural differences.

Question 10

The repeating unit of an addition polymer is -[CF₂-CCl₂]-. What is the formula of the monomer used to create this polymer?

F₂C=CCl₂ (correct answer)
FClC=CFCl
F₂C=CF₂
Cl₂C=CH₂

Explanation: Addition polymerization involves the breaking of a pi bond in an alkene monomer to form a long saturated chain. To find the monomer, identify the repeating two-carbon unit in the polymer backbone and re-form the double bond between those two carbon atoms. The repeating unit is -CF₂-CCl₂-. The two carbons in the backbone are a CF₂ group and a CCl₂ group. Re-forming the double bond between them gives the monomer F₂C=CCl₂ (1,1-dichloro-2,2-difluoroethene).

Question 11

Polyvinyl chloride (PVC) is formed from the monomer chloroethene (CH₂=CHCl). Which statement correctly compares the carbon-carbon bonding in the monomer and the polymer?

The monomer contains only sigma bonds, while the polymer contains both sigma and pi bonds.
The average C-C bond length in the polymer is shorter than the C=C bond length in the monomer.
The monomer has sp² and sp³ hybridized carbons, while the polymer has only sp³ hybridized carbons.
The C=C double bond in the monomer is broken and replaced by C-C single bonds forming the polymer backbone. (correct answer)

Explanation: In addition polymerization, the pi bond of the C=C double bond in the alkene monomer (chloroethene) breaks. The electrons from this pi bond are then used to form new C-C single (sigma) bonds that link the monomer units together into a long chain. Therefore, the polymer backbone consists entirely of C-C single bonds. Distractor A is incorrect; the monomer has one sigma and one pi bond between the carbons. Distractor B is incorrect; C-C single bonds are longer than C=C double bonds. Distractor C is incorrect; all carbons in the monomer that are part of the double bond are sp² hybridized, while all carbons in the polymer backbone are sp³ hybridized.

Question 12

The average electronegativity of two elements in a compound is very low, while the difference in their electronegativity is also very low. Where in the bonding triangle would this compound be located?

Near the metallic corner. (correct answer)
Near the ionic corner.
Near the covalent corner.
In the center of the triangle.

Explanation: The bonding triangle plots compounds based on the average electronegativity of the constituent elements (x-axis) and the difference in their electronegativity (y-axis). The metallic corner is characterized by elements with low average electronegativity (as metals have low EN values) and a very low difference in electronegativity (as it's often an element bonding with itself or another metal with a similar EN). The ionic corner has a large difference and intermediate average. The covalent corner has a high average (non-metals have high EN) and a low difference. Therefore, low average and low difference corresponds to the metallic corner.

Question 13

A polymer is synthesized via a condensation reaction between ethane-1,2-diol and propanedioic acid. What is the identity of the repeating linkage and the small molecule eliminated? [AHL Content]

Amide linkage, HCl eliminated
Ester linkage, H₂ eliminated
Ether linkage, H₂O eliminated
Ester linkage, H₂O eliminated (correct answer)

Explanation: The reaction occurs between a diol (contains two -OH groups) and a dicarboxylic acid (contains two -COOH groups). The reaction between an alcohol functional group and a carboxylic acid functional group is esterification, which produces an ester linkage (-COO-). This is a condensation reaction because a small molecule is eliminated. For each ester link formed, a hydrogen atom from the alcohol and a hydroxyl group from the carboxylic acid are removed to form a molecule of water (H₂O). Therefore, the polymer is a polyester, and water is the eliminated molecule.

Question 14

Bronze is a substitutional alloy of copper and tin. Which statement best explains why bronze is significantly harder and less malleable than pure copper?

The tin atoms form strong ionic bonds with the copper atoms, holding the lattice in a more rigid structure.
The different sizes of the tin and copper atoms disrupt the regular arrangement of the metallic lattice, making it more difficult for layers of atoms to slide over one another. (correct answer)
The tin atoms increase the number of delocalized electrons, which strengthens the metallic bonding throughout the structure.
The tin atoms are much smaller and fit into the spaces in the copper lattice, preventing dislocation movement.

Explanation: In a substitutional alloy like bronze, tin atoms (which are larger than copper atoms) replace copper atoms in the lattice. This difference in atomic size disrupts the regular, orderly layers of atoms found in pure copper. This disruption makes it more difficult for the layers to slide past each other when a force is applied, which is the mechanism for malleability. Consequently, the alloy is harder and less malleable. Distractor A is incorrect as bonding remains metallic, not ionic. Distractor C is incorrect; alloying generally decreases electrical conductivity because the lattice disruption hinders electron flow. Distractor D describes an interstitial alloy, which bronze is not; tin atoms are too large to fit into the interstices of the copper lattice.

Question 15

Which statement best describes the primary structural reason for the enhanced properties of many alloys compared to their constituent pure metals?

Alloying introduces covalent bonds into the metallic lattice, which are stronger and more directional.
The presence of different atoms creates a more ordered and dense crystal lattice, increasing its strength.
The disruption of the regular metallic lattice by atoms of a different size impedes the movement of dislocations. (correct answer)
Alloys are chemical compounds with fixed stoichiometry, leading to superior and more predictable properties.

Explanation: The properties of metals, such as malleability and ductility, are due to the ability of layers of atoms to slide over one another, a process facilitated by the movement of dislocations in the crystal lattice. In an alloy, atoms of a different element (often of a different size) disrupt the regular, repeating arrangement of the lattice. This disruption makes it more difficult for dislocations to propagate through the material, thus impeding the sliding of atomic layers. This results in the alloy being harder, stronger, and less malleable than the pure parent metals. Distractor A is incorrect as the bonding remains metallic. Distractor B is incorrect because alloying creates a less ordered, not more ordered, lattice. Distractor D is incorrect as alloys are mixtures (solid solutions), not compounds with fixed stoichiometry.

Question 16

What is a key difference between addition polymerization and condensation polymerization? [AHL Content]

Only addition polymerization involves the formation of a small molecule by-product such as water.
The empirical formula of an addition polymer is identical to that of its monomer, which is not true for a condensation polymer. (correct answer)
Condensation polymerization requires unsaturated monomers, while addition polymerization requires monomers with two different functional groups.
Addition polymerization results in polymers with lower average molar mass than condensation polymerization.

Explanation: In addition polymerization, monomers (typically alkenes) add to one another in such a way that the polymer contains all the atoms of the monomer unit. Therefore, the empirical formula of the repeating unit in the polymer is identical to the empirical formula of the monomer. In condensation polymerization, a small molecule (like H₂O or HCl) is eliminated for each bond formed between monomer units. As a result, the repeating unit in a condensation polymer has fewer atoms than the monomer(s) it is formed from, and thus a different empirical formula.

Question 17

Which pair of monomers could react to form a polyester? [AHL Content]

A diamine and a dicarboxylic acid
A diol and a dicarboxylic acid (correct answer)
Two different alkenes
A diol and a diamine

Explanation: A polyester is a condensation polymer characterized by ester linkages (-COO-) in its backbone. An ester linkage is formed in an esterification reaction between a hydroxyl group (-OH) and a carboxylic acid group (-COOH), with the elimination of a water molecule. To form a long polymer chain, each monomer must be difunctional. Therefore, a polyester is formed from the reaction of a monomer with two hydroxyl groups (a diol) and a monomer with two carboxylic acid groups (a dicarboxylic acid). A diamine and a dicarboxylic acid (A) form a polyamide (like nylon). Two alkenes (C) would undergo addition polymerization. A diol and a diamine (D) would not readily form a polymer chain via condensation.

Question 18

The polymerization of propene (CH₃CH=CH₂) produces a polymer with the common name polypropylene. Which statement correctly describes the structure of this polymer?

The polymer backbone consists of alternating C=C double bonds and C-C single bonds.
A methyl group is attached to every carbon atom in the polymer backbone.
The repeating unit has the formula C₃H₆ and contains only sp³ hybridized carbon atoms. (correct answer)
The polymer is formed via a condensation reaction with the elimination of H₂.

Explanation: The addition polymerization of propene (CH₃CH=CH₂) breaks the C=C double bond. The repeating unit is -[CH(CH₃)-CH₂]-. The empirical formula of this repeating unit is C₃H₆, the same as the monomer. Within this polymer structure, the carbon from the original CH₃ group and the carbon it is attached to in the backbone are sp³, and the carbon from the original CH₂ group is also sp³. Therefore, all carbons in the polymer are sp³ hybridized. Distractor A is incorrect; the backbone is composed entirely of C-C single bonds. Distractor B is incorrect; the methyl group is attached to every second carbon atom in the backbone. Distractor D is incorrect; this is an addition polymerization, not condensation, and nothing is eliminated.

Question 19

Steel is an interstitial alloy of iron and carbon. Which statement correctly explains why steel is stronger than pure iron?

The carbon atoms replace iron atoms of a similar size, strengthening the lattice through increased mass.
The formation of iron carbide (Fe₃C) creates ionic bonds within the metallic structure, making it more rigid.
The carbon atoms donate their valence electrons to the delocalized sea, significantly increasing the strength of the metallic bond.
The small carbon atoms occupy the spaces between iron atoms, restricting the movement of layers and dislocations within the metallic lattice. (correct answer)

Explanation: In an interstitial alloy like steel, small non-metal atoms (carbon) occupy the interstitial spaces (holes) between the larger metal atoms (iron) in the crystal lattice. The presence of these atoms pins the layers of iron atoms in place, making it much more difficult for them to slide past one another. This restricts the movement of dislocations and is the primary reason for the increased hardness and strength of steel compared to pure iron. Distractor A incorrectly describes a substitutional alloy. Distractor C is an oversimplification; while carbon's electrons contribute, the primary strengthening mechanism is physical disruption. Distractor D is partially true in some steels (cementite is a phase), but the general explanation for strengthening in all steels is the interstitial model, and the bonding in iron carbide is complex, not purely ionic.