IB Chemistry Quiz: Apply The Metallic Model
20 questions · exam conditions
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Apply The Metallic ModelQuestion 1 of 20

A hypothetical element 'Y' has a much higher melting point and is a better electrical conductor than potassium (K). Element 'Y' is in the same period as potassium. What is the most likely identity of Y?

Calcium (Ca)
Argon (Ar)
Bromine (Br)
Rubidium (Rb)
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IB Chemistry Quiz

IB Chemistry Quiz: Apply The Metallic Model

Practice Apply The Metallic Model 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 The Metallic Model, giving you a quick way to practice the rules, question types, and explanations that matter most for IB 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 hypothetical element 'Y' has a much higher melting point and is a better electrical conductor than potassium (K). Element 'Y' is in the same period as potassium. What is the most likely identity of Y?

  1. Calcium (Ca) (correct answer)
  2. Argon (Ar)
  3. Bromine (Br)
  4. Rubidium (Rb)
Explanation: Potassium (K) is in Period 4. Rubidium (Rb) is below K in Group 1 and would have a lower melting point. Argon (Ar) and Bromine (Br) are non-metals and poor conductors. Calcium (Ca) is in the same period as K. It has two valence electrons to contribute to the delocalized sea and forms a +2 cation, compared to K's one electron and +1 cation. This results in much stronger metallic bonding (higher melting point) and a higher density of charge carriers (better conductivity).

Question 2

In a Daniel cell, a zinc strip is placed in a ZnSO₄ solution and a copper strip in a CuSO₄ solution. When connected by a wire, electrons flow from zinc to copper. Which statement best describes the function of the metallic bonding within the copper wire?

  1. The positive copper cations in the wire move from the anode side to the cathode side, carrying the charge.
  2. The copper atoms in the wire are oxidized to release electrons that then travel to the cathode.
  3. The metallic bonds are broken and reformed along the length of the wire, passing the electrons along in a relay.
  4. The wire provides a medium through which mobile delocalized electrons can drift from a region of high potential energy to one of lower potential energy. (correct answer)
Explanation: The wire acts as an external circuit. The metallic bonding within the copper wire means there is a pre-existing 'sea' of mobile, delocalized electrons. When the cell is active, this sea of electrons experiences an electromotive force, causing a net drift of electrons from the anode (Zn) to the cathode (Cu). The bonding structure of the wire facilitates this electron flow; it does not involve the movement of cations or the breaking of bonds.

Question 3

Which of the following statements about alloys is correct?

  1. Alloys are chemical compounds formed between two or more metals in a fixed stoichiometric ratio.
  2. The properties of an alloy, such as hardness and conductivity, are always an average of the properties of its constituent metals.
  3. Alloys can be formed because the non-directional nature of metallic bonding can accommodate atoms of different sizes in the lattice. (correct answer)
  4. Adding a non-metal like carbon to a metal like iron to make steel converts the bonding from metallic to covalent network.
Explanation: Alloys are solid solutions (mixtures), not compounds with fixed ratios. Their properties are often very different from, and not a simple average of, their components (e.g., they are usually harder and less conductive). The bonding in steel is still fundamentally metallic, although the carbon atoms add complexity. The key reason alloys can form is that the delocalized electron sea can surround and bind together cations of different sizes and charges, a direct result of the non-directional nature of the metallic bond.

Question 4

An alloy is made from 90% Metal X (atomic radius 135 pm) and 10% Metal Y (atomic radius 155 pm). A second alloy is made from 90% Metal X and 10% Metal Z (atomic radius 138 pm). How will the properties of the two alloys likely compare?

  1. The X-Y alloy will be harder and a poorer electrical conductor than the X-Z alloy. (correct answer)
  2. The X-Z alloy will be harder and a poorer electrical conductor than the X-Y alloy.
  3. Both alloys will be softer than pure Metal X because the impurities weaken the lattice.
  4. The hardness of the two alloys will be nearly identical, but the X-Y alloy will be a better conductor.
Explanation: The hardness of an alloy is related to the degree of disruption of the metallic lattice. A greater difference in atomic radii causes more disruption, making it harder for atomic layers to slide. The radius difference between X and Y (20 pm) is much larger than between X and Z (3 pm). Therefore, the X-Y alloy will be significantly harder. This disruption also impedes the flow of delocalized electrons, so the more distorted lattice (X-Y) will also be a poorer electrical conductor.

Question 5

The enthalpy of atomization is the energy required to convert one mole of an element in its standard state into gaseous atoms. For sodium, this value is +107 kJ mol⁻¹, and for magnesium, it is +148 kJ mol⁻¹. What can be deduced from this data?

  1. Sodium is more reactive than magnesium.
  2. The metallic bonding in magnesium is stronger than in sodium. (correct answer)
  3. The first ionization energy of sodium is lower than that of magnesium.
  4. Magnesium atoms are smaller than sodium atoms.
Explanation: The enthalpy of atomization is a direct measure of the strength of the forces holding the atoms together in the solid state. For a metal, this corresponds to the strength of the metallic bond. Since more energy is required to separate magnesium atoms (148 kJ mol⁻¹) than sodium atoms (107 kJ mol⁻¹), it can be deduced that the metallic bonding in magnesium is stronger. While options C and D are true statements, they are not direct deductions from the enthalpy of atomization data.

Question 6

When a piece of aluminum foil is hammered into a thinner sheet, it maintains its metallic properties and doesn't fracture like an ionic crystal would. This malleability can be explained by the metallic bonding model because:

  1. The delocalized electrons can redistribute themselves to maintain bonding as the metal atoms slide past each other into new positions without breaking bonds. (correct answer)
  2. The directional nature of metallic bonds allows them to bend and flex without breaking, similar to how covalent bonds behave in organic polymers.
  3. The strong electrostatic attractions between metal cations and anions can withstand the mechanical stress applied during the hammering process.
  4. The metallic lattice structure becomes more stable when compressed, causing the atoms to form stronger bonds in the thinner configuration.
Explanation: The metallic bonding model explains malleability through the non-directional nature of the electron sea. When metal atoms are displaced, the delocalized electrons can adjust their distribution to maintain bonding in new positions. B is incorrect because metallic bonds are non-directional, unlike covalent bonds. C is wrong because metals don't contain anions - they have cations in an electron sea. D is incorrect because the question asks about maintaining properties during deformation, not about increased stability.

Question 7

The hardness of metal alloys often exceeds that of pure metals, even though both are held together by metallic bonding. Which explanation best accounts for this observation using the metallic bonding model?

  1. Different sized atoms in alloys create a more uniform electron distribution, strengthening the overall metallic bonding throughout the crystal structure.
  2. The presence of different atoms disrupts the regular arrangement of metal cations, making it more difficult for layers of atoms to slide past each other under stress. (correct answer)
  3. Alloy formation increases the total number of delocalized electrons available, creating stronger electrostatic attractions between the cations and the electron sea.
  4. The mixing of different metals creates hybrid orbitals that form stronger metallic bonds compared to the pure metallic bonds in single-element metals.
Explanation: In the metallic bonding model, hardness relates to resistance to deformation. When different-sized atoms are present in an alloy, they create irregularities that impede the sliding of atomic layers, increasing hardness while maintaining the electron sea structure. A is incorrect because uniform distribution wouldn't necessarily strengthen bonding. C is wrong because the total electrons don't necessarily increase in alloys. D incorrectly applies molecular orbital concepts inappropriately to metallic bonding.

Question 8

When sodium metal is cut with a knife, it reveals a shiny surface that quickly tarnishes in air. The initial shininess can be explained by the metallic bonding model, but why does the surface lose its metallic luster so rapidly?

  1. The delocalized electrons at the surface gradually escape into the atmosphere, leaving behind a layer of positively charged sodium atoms that appear dull.
  2. Exposure to air causes the surface electrons to become localized on individual sodium atoms, disrupting the electron sea and eliminating the reflective properties.
  3. The metallic bonding at the surface is replaced by ionic bonding as sodium reacts with oxygen and moisture, forming compounds that lack delocalized electrons. (correct answer)
  4. Air molecules physically block the movement of delocalized electrons at the surface, preventing them from responding to light and creating the characteristic metallic reflection.
Explanation: Fresh sodium surfaces are shiny because delocalized electrons can interact with light. Tarnishing occurs when sodium reacts with oxygen and water vapor to form ionic compounds (Na₂O, NaOH) that lack the electron sea structure. These ionic products don't have the mobile electrons needed for metallic luster. A is incorrect because electrons don't simply escape. B wrongly suggests localization without chemical reaction. D incorrectly describes a physical blocking mechanism rather than chemical transformation.

Question 9

A student observes that a copper wire can be bent repeatedly without breaking, but eventually fails after many bend cycles. Using the metallic bonding model, which explanation best accounts for both the initial flexibility and eventual failure?

  1. The delocalized electrons initially accommodate bending by redistributing, but repeated stress gradually depletes the electron sea until insufficient electrons remain to maintain bonding.
  2. Metallic bonding allows initial deformation through electron mobility, but repeated bending creates permanent structural defects that accumulate until they compromise the material's integrity. (correct answer)
  3. The non-directional nature of metallic bonds permits bending, but cyclic stress converts some metallic bonds to directional covalent bonds, which are more brittle and prone to failure.
  4. Initial flexibility results from electron delocalization, but repeated mechanical stress causes some metal atoms to lose electrons permanently, creating weak points in the structure.
Explanation: The metallic bonding model explains flexibility through the non-directional electron sea that can maintain bonding during deformation. However, repeated bending creates cumulative structural damage (grain boundaries, dislocations, micro-cracks) that eventually overwhelm the material's ability to maintain integrity, regardless of the bonding type. A incorrectly suggests electron depletion. C wrongly describes conversion to covalent bonds. D incorrectly describes permanent electron loss from atoms.

Question 10

A student measures the electrical conductivity of several Group 1 metals and finds that conductivity decreases down the group (Li > Na > K > Rb > Cs). Using the metallic bonding model, what factor most likely accounts for this trend?

  1. Smaller atoms have more tightly held valence electrons, which increases their mobility and enhances electrical conductivity throughout the metallic structure.
  2. As atomic size increases down the group, the delocalized electrons become more spread out, reducing the density of the electron sea and decreasing conductivity.
  3. Larger atoms form weaker metallic bonds due to reduced orbital overlap, which decreases the stability of the delocalized electron system and limits conductivity.
  4. The increasing atomic radius down the group creates larger distances between metal cations, making it more difficult for delocalized electrons to move efficiently between atoms. (correct answer)
Explanation: As atomic size increases down Group 1, the distance between metal cations increases, creating longer pathways for electron movement and reducing conductivity. The metallic bonding model explains conductivity through electron mobility, which decreases with larger inter-atomic distances. A is incorrect because tightly held electrons are less mobile. B focuses on electron density rather than mobility pathways. C incorrectly emphasizes bond strength over electron transport efficiency.

Question 11

A student compares the electrical conductivity of copper wire at different temperatures and observes that conductivity decreases as temperature increases. Which explanation best accounts for this observation using the metallic bonding model?

  1. Higher temperatures cause the delocalized electrons to gain kinetic energy, making them move faster and conduct electricity more efficiently through the metal lattice.
  2. Increased thermal motion of metal cations disrupts the regular lattice structure, causing more frequent collisions that impede the flow of delocalized electrons. (correct answer)
  3. At higher temperatures, the metallic bonds become stronger due to increased orbital overlap, which restricts the movement of electrons through the structure.
  4. Elevated temperatures cause some delocalized electrons to become localized on specific metal atoms, reducing the overall number of charge carriers available.
Explanation: According to the metallic bonding model, electrical conductivity depends on the mobility of delocalized electrons through the 'sea' of electrons. As temperature increases, metal cations vibrate more vigorously, creating more obstacles for electron flow and reducing conductivity. A is incorrect because faster electrons don't necessarily conduct better if they encounter more obstacles. C is wrong because metallic bonds typically weaken at higher temperatures. D is incorrect because the metallic model doesn't predict electron localization due to temperature increases.

Question 12

The thermal conductivity of metals generally follows the same trend as electrical conductivity, with good electrical conductors also being good thermal conductors. Using the metallic bonding model, what is the most likely explanation for this correlation?

  1. Both properties depend on the same mechanism: the movement of delocalized electrons, which can transfer both electrical charge and kinetic energy throughout the structure. (correct answer)
  2. The strong metallic bonds that facilitate electrical conduction also create rigid pathways that allow heat to travel efficiently through direct atom-to-atom contact.
  3. Metals with high electrical conductivity have more valence electrons, which increases both the charge-carrying capacity and the heat storage capacity of the material.
  4. The metallic lattice structures that optimize electrical conductivity also minimize the distance between atoms, allowing for more efficient heat transfer through lattice vibrations.
Explanation: Both electrical and thermal conductivity in metals primarily depend on the mobility of delocalized electrons. Electrons can carry both electrical charge and thermal energy (kinetic energy) through the metal structure. B is incorrect because thermal conductivity in metals is mainly due to electron movement, not direct atomic contact. C confuses the number of valence electrons with mobility and heat storage. D focuses on lattice vibrations, which play a secondary role compared to electron transport in metals.

Question 13

The melting points of transition metals are generally much higher than those of Group 1 metals, despite both groups exhibiting metallic bonding. Which factor best explains this difference according to the metallic bonding model?

  1. Transition metals have partially filled d-orbitals that create stronger directional bonding compared to the purely non-directional bonding in Group 1 metals.
  2. The smaller ionic radii of transition metal cations allow for more efficient packing in the crystal lattice, creating stronger overall metallic bonding.
  3. Transition metals contribute more electrons to the delocalized electron sea and have higher charge densities, resulting in stronger electrostatic attractions within the metallic structure. (correct answer)
  4. The d-electrons in transition metals are more tightly bound to the nucleus, creating more stable metallic bonds that require more energy to break during melting.
Explanation: Transition metals typically contribute more valence electrons (s + d electrons) to the electron sea than Group 1 metals (only s electrons), and their cations have higher charges. This creates stronger electrostatic attractions between the cations and the more dense electron sea, requiring more energy to disrupt during melting. A is incorrect because metallic bonding remains non-directional. B focuses on packing rather than electron contribution. D wrongly suggests tightly bound electrons contribute to metallic bonding strength.

Question 14

Mercury is unique among metals because it exists as a liquid at room temperature, yet it still exhibits typical metallic properties like electrical conductivity and metallic luster. How does the metallic bonding model account for mercury's unusual physical state while maintaining metallic characteristics?

  1. Mercury has weaker metallic bonding due to relativistic effects on its electrons, resulting in insufficient attractive forces to maintain a solid structure while preserving the delocalized electron system. (correct answer)
  2. The large atomic size of mercury creates greater distances between atoms, weakening the metallic bonds enough to allow fluidity but maintaining enough electron delocalization for metallic properties.
  3. Mercury's electron configuration produces a less dense electron sea compared to other metals, reducing the strength of metallic bonding while still providing mobile electrons for conduction.
  4. The metallic bonding in mercury is temperature-sensitive and transitions between strong (solid-like) and weak (liquid-like) states rapidly, creating an apparent liquid that retains metallic electron behavior.
Explanation: Mercury's liquid state results from relativistic effects that influence its electron behavior, leading to weaker metallic bonding than expected for its position in the periodic table. However, the metallic bonding model still applies - the electron sea remains intact but with weaker attractive forces, allowing fluidity while maintaining conductivity and luster. B oversimplifies atomic size effects. C incorrectly describes electron density rather than relativistic effects. D incorrectly describes rapid bonding transitions rather than consistently weaker bonding.

Question 15

A chemistry student observes that when magnesium ribbon is heated, it maintains its metallic properties until it suddenly ignites and forms a white ionic compound (MgO). Which statement best explains this transition using bonding models?

  1. The metallic bonding model applies until ignition, when sufficient thermal energy causes the delocalized electrons to become localized, converting metallic bonds directly into ionic bonds.
  2. Heating increases the mobility of electrons in the metallic structure until they gain enough energy to escape entirely, leaving behind positively charged metal atoms.
  3. The metallic bonding model explains the initial heating behavior, but ignition involves a chemical reaction with oxygen that replaces metallic bonding with ionic bonding between Mg²⁺ and O²⁻. (correct answer)
  4. The transition occurs because thermal energy converts the non-directional metallic bonds into directional covalent bonds, which then rearrange into the ionic structure of magnesium oxide.
Explanation: The observation describes two distinct phases: heating (physical change) and ignition (chemical reaction). During heating, metallic bonding explains the retained properties. Ignition represents a chemical reaction with oxygen, forming new ionic bonds between Mg²⁺ and O²⁻ ions. A is incorrect because metallic electrons don't simply become localized to form ionic bonds. B describes electron emission, not oxide formation. D incorrectly suggests metallic bonds convert to covalent then ionic bonds.

Question 16

Which property of a metal is most directly related to the average kinetic energy of its vibrating cations within the lattice?

  1. Electrical conductivity
  2. Malleability
  3. Temperature (correct answer)
  4. Lustre
Explanation: Temperature is a measure of the average kinetic energy of the particles in a substance. In a solid metal, the particles are the cations vibrating about their fixed positions in the lattice and the mobile electrons. The kinetic energy of these vibrating cations is directly proportional to the absolute temperature of the metal. The other properties relate to electron mobility, non-directional bonding, and electron-photon interactions.

Question 17

Brass is an alloy of copper and zinc. Which statement best explains why brass is harder than pure copper?

  1. Zinc has more delocalized electrons than copper, strengthening the overall metallic bond.
  2. Zinc atoms form strong, directional covalent bonds with copper atoms, locking the lattice in place.
  3. The differently sized zinc atoms disrupt the regular lattice structure, making it more difficult for layers of atoms to slide over one another. (correct answer)
  4. The presence of zinc reduces the density of the delocalized electron sea, which paradoxically increases the rigidity of the structure.
Explanation: Alloys are generally harder than their constituent pure metals because the different sizes of the atoms disrupt the orderly layers of the crystal lattice. This disruption makes it more difficult for the layers to slide past one another when a force is applied, increasing the hardness and strength of the material.

Question 18

Which statement correctly compares the charge carriers responsible for electrical conductivity in solid magnesium and molten magnesium chloride?

  1. Both conduct electricity via mobile electrons, but they are more mobile in the molten salt.
  2. Solid magnesium conducts via mobile electrons, while molten magnesium chloride conducts via mobile Mg²⁺ and Cl⁻ ions. (correct answer)
  3. Solid magnesium is an insulator, while molten magnesium chloride conducts via mobile electrons.
  4. Both conduct electricity via mobile Mg²⁺ cations.
Explanation: Metallic solids like magnesium conduct electricity because they have a lattice of positive ions surrounded by a 'sea' of mobile delocalized electrons. These electrons are the charge carriers. Ionic compounds like magnesium chloride do not conduct when solid because their ions are in fixed positions. When molten, the ions (Mg²⁺ and Cl⁻) are free to move and carry charge, allowing the substance to conduct electricity.

Question 19

The strength of the metallic bond in Group 1 metals decreases from lithium to caesium. What is the primary reason for this trend?

  1. The number of delocalized electrons per atom decreases down the group.
  2. The increasing atomic radius leads to a weaker attraction between the nucleus of each cation and the delocalized electrons. (correct answer)
  3. Electronegativity decreases down the group, indicating a reduced ability to form strong bonds.
  4. The packing efficiency of the metal atoms in the crystal lattice decreases significantly down the group.
Explanation: All Group 1 metals have one valence electron and form a +1 cation. The primary factor changing down the group is the atomic (and ionic) radius. As the radius increases from Li to Cs, the delocalized electrons are, on average, further from the positive nuclei. This increased distance results in a weaker electrostatic attraction, thus weakening the metallic bond and leading to lower melting points and enthalpies of atomization.

Question 20

Both magnesium and silicon dioxide have high melting points. Why is magnesium a good electrical conductor while silicon dioxide is an insulator?

  1. Magnesium has delocalized electrons that are free to move, whereas silicon dioxide has its valence electrons localized in strong covalent bonds. (correct answer)
  2. The bonds in magnesium are metallic and therefore weaker than the covalent bonds in silicon dioxide, allowing electrons to flow freely.
  3. Magnesium is a metal and all metals are conductors, whereas silicon dioxide is a non-metal compound and therefore an insulator.
  4. The atoms in magnesium are less tightly packed than in silicon dioxide, leaving channels for electrons to move through the structure.
Explanation: The high melting point of both substances is due to strong bonding that requires much energy to overcome. However, the nature of the bonding dictates conductivity. Magnesium has metallic bonding with delocalized, mobile electrons that can carry a current. Silicon dioxide is a covalent network solid where electrons are held tightly in localized sigma bonds between silicon and oxygen atoms and are not free to move.