IB Chemistry Quiz: Apply Periodic Table Classification
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Apply Periodic Table ClassificationQuestion 1 of 20

Fluorine is a stronger oxidizing agent than chlorine. What is the main reason for fluorine's exceptional oxidizing strength?

Fluorine has a more exothermic electron affinity than chlorine.
The F-F bond is significantly weaker than the Cl-Cl bond due to electron repulsion.
Fluorine is the most electronegative element, which solely determines its oxidizing power.
The small size of the fluoride ion leads to a highly exothermic hydration enthalpy.
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IB Chemistry Quiz

IB Chemistry Quiz: Apply Periodic Table Classification

Practice Apply Periodic Table Classification 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 Periodic Table Classification, 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.

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Question 1

Fluorine is a stronger oxidizing agent than chlorine. What is the main reason for fluorine's exceptional oxidizing strength?

  1. Fluorine has a more exothermic electron affinity than chlorine.
  2. The F-F bond is significantly weaker than the Cl-Cl bond due to electron repulsion.
  3. Fluorine is the most electronegative element, which solely determines its oxidizing power.
  4. The small size of the fluoride ion leads to a highly exothermic hydration enthalpy. (correct answer)
Explanation: Oxidizing strength in aqueous solution depends on the overall enthalpy change of the process X₂(g) → 2X⁻(aq). This can be broken down into bond dissociation enthalpy, electron affinity, and hydration enthalpy using a Born-Haber cycle. While fluorine's electronegativity is high (C) and its bond is weak (B), the most significant contributing factor to its high oxidizing power is the extremely large exothermic hydration enthalpy of the small F⁻ ion. The high charge density of F⁻ attracts water molecules very strongly, releasing a large amount of energy. Note that fluorine's electron affinity is actually less exothermic than chlorine's (A), an exception due to electron repulsion in fluorine's small valence shell.

Question 2

Aqueous solutions containing the complex ions [Cu(H₂O)₆]²⁺ and [Zn(H₂O)₆]²⁺ are observed. The copper solution is blue, while the zinc solution is colorless. What is the best explanation for this difference?

  1. Copper is a more reactive metal than zinc.
  2. The Zn²⁺ ion is larger than the Cu²⁺ ion, which prevents the absorption of visible light.
  3. Water is a colored ligand that imparts its color to the copper complex but not the zinc complex.
  4. The Cu²⁺ ion has a partially filled d-subshell, while the Zn²⁺ ion has a full d-subshell. (correct answer)
Explanation: The color of transition metal complexes is due to d-d electronic transitions. When ligands approach the central metal ion, the degenerate d-orbitals split into two different energy levels. Electrons can be promoted from the lower level to the upper level by absorbing energy corresponding to a specific wavelength of visible light. The color we see is the complementary color of the light absorbed. For this to occur, the d-subshell must be partially filled. Cu²⁺ has an electron configuration of [Ar]3d⁹ (partially filled). Zn²⁺ has a configuration of [Ar]3d¹⁰ (completely filled). With a full d-subshell, no d-d electron transition is possible, so the [Zn(H₂O)₆]²⁺ complex does not absorb visible light and is colorless.

Question 3

The first six successive ionization energies (IE) for an element Q in Period 3 are given in kJ mol⁻¹: IE₁ = 578 IE₂ = 1817 IE₃ = 2745 IE₄ = 11577 IE₅ = 14842 IE₆ = 18379

Based on the provided ionization energy data, what is the identity of element Q?

  1. Magnesium (Mg)
  2. Aluminium (Al) (correct answer)
  3. Silicon (Si)
  4. Phosphorus (P)
Explanation: A large jump in successive ionization energies occurs when an electron is removed from a new, inner electron shell (a core electron). We analyze the ratios of successive IEs. The jump from IE₃ (2745) to IE₄ (11577) is significantly larger than the jumps between IE₁, IE₂, and IE₃. This indicates that the first three electrons are valence electrons and the fourth electron is a core electron. An element with three valence electrons is in Group 13. The element in Period 3 and Group 13 is Aluminium (Al).

Question 4

The first ionization energies (in kJ/mol) for consecutive elements in a period are: 578, 787, 1012, 1314, 1681. Based on these values and periodic trends, which classification pattern best represents these elements?

  1. Alkali metal → alkaline earth metal → transition metal → metalloid → nonmetal, with irregular increase due to d-orbital effects
  2. Alkaline earth metal → boron group → carbon group → nitrogen group → oxygen group, with steady increase across the period
  3. Transition metal → transition metal → post-transition metal → metalloid → halogen, with variable shielding effects from d electrons
  4. Alkali metal → alkaline earth metal → boron group → carbon group → nitrogen group, showing typical main group progression (correct answer)
Explanation: The steady increase in ionization energy (578→787→1012→1314→1681 kJ/mol) is characteristic of main group elements across a period. Starting from 578 kJ/mol suggests an alkali metal, followed by alkaline earth (787), then Groups 13-15 (1012, 1314, 1681). Choice A incorrectly includes transition metals. Choice B starts with alkaline earth but the first value (578) is too low. Choice C involves transition metals which would show more irregular patterns due to d-orbital shielding effects.

Question 5

Element Y has the following properties: it forms predominantly ionic compounds when bonding with nonmetals, its oxide is basic, and it has two electrons in its outermost s orbital. Element Z has completely filled d orbitals and forms colorless compounds in its most stable oxidation state. If Y and Z are in the same period, what is the most likely relationship between their positions?

  1. Y is an alkaline earth metal and Z is immediately to its right in the same period
  2. Y is an alkaline earth metal and Z is a post-transition metal with filled d¹⁰ configuration (correct answer)
  3. Y is an alkali metal and Z is a transition metal with partially filled d orbitals
  4. Y is a transition metal in Group 2 and Z is a noble gas with filled p orbitals
Explanation: Element Y has 2 electrons in its outermost s orbital and forms ionic compounds with basic oxides, indicating it's an alkaline earth metal (Group 2). Element Z has completely filled d orbitals and forms colorless compounds, suggesting d¹⁰ configuration (like Zn²⁺, Cd²⁺). These would be post-transition metals. Choice A is incorrect because Z wouldn't be immediately adjacent. Choice C misidentifies Y as alkali metal. Choice D incorrectly identifies Y as a transition metal and Z as a noble gas.

Question 6

Element X exhibits the following properties: it has a low first ionization energy, forms predominantly ionic compounds, conducts electricity in solid state, and its compounds often display characteristic flame colors. Element Y has high electronegativity, exists as diatomic molecules at room temperature, and readily accepts electrons to form stable anions. Based on these properties and typical periodic trends, which statement about their likely positions is most justified?

  1. X is likely a transition metal in the d-block while Y is a halogen, and they would form compounds with formula XY₂
  2. X is likely an alkali metal in Group 1 while Y is a halogen, and they would form ionic compounds with formula XY (correct answer)
  3. X is likely an alkaline earth metal while Y is a chalcogen, and their compounds would have covalent character
  4. X is likely a lanthanide element while Y is a noble gas, and they would not readily form stable compounds
Explanation: Element X shows classic alkali metal properties: low ionization energy, ionic compound formation, metallic conduction, and flame colors (due to easily excited valence electrons). Element Y shows halogen properties: high electronegativity, diatomic molecules (F₂, Cl₂, etc.), and stable anion formation. Group 1 metals form 1:1 ionic compounds with halogens (XY). Choice A suggests XY₂ which would require X to have +2 charge. Choice C suggests covalent character which contradicts the ionic nature described. Choice D incorrectly identifies Y as noble gas.

Question 7

A chemistry student analyzes three unknown elements from the same period and finds: Element A forms basic oxides and has low electronegativity; Element B forms amphoteric oxides and shows intermediate metallic character; Element C forms acidic oxides and has high electronegativity. Additionally, element B can form both ionic and covalent compounds depending on the reaction partner. Based on this periodic behavior, which classification sequence is most consistent with the data?

  1. A is an alkali metal, B is an alkaline earth metal, and C is a transition metal, representing early period elements
  2. A is an alkaline earth metal, B is a metalloid from Group 14, and C is a halogen, showing the metal-nonmetal transition
  3. A is a transition metal, B is a post-transition metal, and C is a metalloid, representing late transition series behavior
  4. A is an alkaline earth metal, B is a metalloid from Group 13 or 14, and C is from Group 16 or 17, demonstrating periodic trends (correct answer)
Explanation: The progression from basic oxides (A) to amphoteric oxides (B) to acidic oxides (C) with corresponding electronegativity changes represents the classic metal-to-nonmetal transition across a period. Element B's ability to form both ionic and covalent compounds is characteristic of metalloids in Groups 13-14. This sequence spans from Group 2 (alkaline earth) through metalloid region to Groups 16-17 (nonmetals). Choice A doesn't reach nonmetals. Choice B skips Groups 13-15. Choice C stays within metallic elements only.

Question 8

The first ionization energy of sulfur is less than that of phosphorus. What is the best explanation for this exception to the general periodic trend?

  1. Sulfur has a larger atomic radius, so the outermost electron is further from the nucleus and more easily removed.
  2. Phosphorus has a stable, half-filled 3p sub-level, making it more difficult to remove an electron compared to sulfur.
  3. The effective nuclear charge experienced by the valence electrons in sulfur is significantly lower than in phosphorus.
  4. Sulfur has one doubly-occupied 3p orbital, and repulsion between the paired electrons makes one easier to remove. (correct answer)
Explanation: Phosphorus has the electron configuration [Ne] 3s²3p³, with three unpaired electrons in separate p-orbitals (a half-filled subshell). Sulfur has the configuration [Ne] 3s²3p⁴, with one p-orbital containing a pair of electrons. The mutual repulsion between these two paired electrons in the same orbital reduces the energy required to remove one of them, making sulfur's first ionization energy lower than that of phosphorus, despite sulfur's higher nuclear charge. This effect outweighs the stability of the half-filled subshell in phosphorus.

Question 9

A solution of potassium bromide is mixed with chlorine water. Which statement describes the expected observation and the correct justification?

  1. No reaction occurs because bromine is more reactive than chlorine.
  2. The solution turns orange-brown as chlorine displaces bromide ions, forming bromine. (correct answer)
  3. The solution remains colorless but a precipitate of potassium chloride forms.
  4. The solution turns pale green as bromide ions displace chlorine, forming chloride ions.
Explanation: In Group 17, the halogens, reactivity (oxidizing ability) decreases down the group. This means a more reactive halogen can displace a less reactive halide ion from its salt solution. Chlorine is above bromine in Group 17, so it is more reactive. Chlorine will oxidize bromide ions to bromine. The equation for the reaction is Cl₂(aq) + 2KBr(aq) → 2KCl(aq) + Br₂(aq). Aqueous bromine (Br₂) is orange-brown, so this color will appear in the solution. Distractor A incorrectly states the reactivity trend. Distractor D reverses the roles of the halogens. Distractor C is incorrect because potassium chloride is soluble in water and does not precipitate.

Question 10

An unknown element from Period 3 forms an oxide that dissolves in both 1.0 mol dm⁻³ hydrochloric acid and 1.0 mol dm⁻³ sodium hydroxide. What is the identity of the element?

  1. Sodium (Na)
  2. Magnesium (Mg)
  3. Aluminium (Al) (correct answer)
  4. Sulfur (S)
Explanation: The ability of an oxide to react with both an acid (HCl) and a base (NaOH) defines it as amphoteric. Across Period 3, the oxides trend from basic (Na₂O, MgO) to amphoteric (Al₂O₃) to acidic (SiO₂, P₄O₁₀, SO₃, Cl₂O₇). Sodium oxide and magnesium oxide are basic and would only react with the acid. Sulfur oxides are acidic and would only react with the base. Aluminium oxide (Al₂O₃) is the classic example of an amphoteric oxide in Period 3. Therefore, the element is aluminium.

Question 11

An element X forms a stable chloride with the formula XCl₃ and a stable oxide with the formula X₂O₃. Based on this information, in which group of the periodic table is element X most likely found?

  1. Group 1
  2. Group 2
  3. Group 13 (correct answer)
  4. Group 15
Explanation: The formulas of the compounds indicate the oxidation state of element X. In XCl₃, since chloride has an oxidation state of -1, X must have an oxidation state of +3. In X₂O₃, since oxide has an oxidation state of -2, X must also have an oxidation state of +3 (2x + 3(-2) = 0, so 2x = 6, x = +3). An element that commonly forms a stable +3 ion is found in Group 13 (e.g., Al³⁺, Ga³⁺). Group 1 elements form +1 ions. Group 2 elements form +2 ions. Group 15 elements can have a +3 oxidation state, but their most common negative oxidation state is -3, and they also commonly show a +5 state (e.g., PCl₅, N₂O₅). The most straightforward conclusion is Group 13.

Question 12

Which statement best distinguishes between electronegativity and electron affinity?

  1. Electronegativity is the energy released when an atom gains an electron, while electron affinity describes the attraction of an atom for electrons in a covalent bond.
  2. Electronegativity is a calculated value on a relative scale, while electron affinity is a measurable energy change for an isolated gaseous atom. (correct answer)
  3. Both properties measure the same phenomenon, but electronegativity is used for metals and electron affinity is used for non-metals.
  4. High electronegativity always corresponds to a highly exothermic electron affinity, with no exceptions.
Explanation: Electronegativity (e.g., the Pauling scale) is a relative measure of the ability of an atom to attract a bonding pair of electrons in a covalent bond. It is a calculated, dimensionless quantity. Electron affinity is the energy change (often released, so exothermic) that occurs when a neutral atom in the gaseous state gains an electron to form a negative ion. It is an experimentally measurable quantity with units of energy (e.g., kJ mol⁻¹). Choice A reverses the definitions. Choice C is incorrect. Choice D is not strictly true; for example, noble gases have some electronegativity value but have positive (endothermic) electron affinities.

Question 13

Of the elements Na, Si, S, and Cl, which has the highest first ionization energy and which has the largest atomic radius?

  1. Highest IE: Cl; Largest radius: Na (correct answer)
  2. Highest IE: Cl; Largest radius: Si
  3. Highest IE: Na; Largest radius: Cl
  4. Highest IE: S; Largest radius: Na
Explanation: All four elements are in Period 3. First ionization energy generally increases across a period from left to right due to increasing effective nuclear charge. Therefore, Cl will have the highest first ionization energy. Atomic radius generally decreases across a period from left to right, also due to increasing effective nuclear charge pulling the electron shells closer. Therefore, Na, being the leftmost element, will have the largest atomic radius. So, the correct combination is Highest IE: Cl and Largest radius: Na.

Question 14

Which sequence correctly arranges the halide ions F⁻, Cl⁻, and Br⁻ in order of increasing ionic radius?

  1. F⁻ < Cl⁻ < Br⁻ (correct answer)
  2. Br⁻ < Cl⁻ < F⁻
  3. Cl⁻ < F⁻ < Br⁻
  4. F⁻ < Br⁻ < Cl⁻
Explanation: Fluorine, chlorine, and bromine are all in Group 17. When comparing ions from the same group, the primary factor determining size is the principal quantum number (n) of the outermost electron shell. F⁻ has its valence electrons in the n=2 shell, Cl⁻ in the n=3 shell, and Br⁻ in the n=4 shell. As the number of electron shells increases down the group, the ionic radius also increases. Therefore, the correct order of increasing radius is F⁻ < Cl⁻ < Br⁻.

Question 15

Solid potassium oxide (K₂O) is added to a beaker of pure water. Which of the following correctly identifies the major species present (other than water) and the resulting nature of the solution?

  1. K⁺(aq) and O²⁻(aq); neutral
  2. K(s) and O₂(g); neutral
  3. K₂O(aq); acidic
  4. K⁺(aq) and OH⁻(aq); basic (correct answer)
Explanation: Potassium is a Group 1 metal, so its oxide, K₂O, is a basic oxide. When a soluble metal oxide dissolves in water, it reacts to form the corresponding metal hydroxide. The oxide ion (O²⁻) is a very strong base and reacts completely with water: K₂O(s) + H₂O(l) → 2KOH(aq). Potassium hydroxide (KOH) is a strong base and fully dissociates in water into potassium ions (K⁺) and hydroxide ions (OH⁻). The presence of excess hydroxide ions makes the solution basic (alkaline).

Question 16

Manganese (Z=25) can exhibit a wide range of oxidation states, including +2, +3, +4, +6, and +7. What is the reason for this property?

  1. Manganese is a metalloid and can therefore gain or lose many electrons.
  2. The successive ionization energies of manganese are all very low and similar in value.
  3. The 4s and 3d sub-levels are very close in energy, allowing for the loss of a variable number of electrons. (correct answer)
  4. Manganese has a large atomic radius which reduces the nucleus's hold on all valence electrons.
Explanation: This is a characteristic property of transition metals. The electron configuration of Mn is [Ar] 4s²3d⁵. The 4s and 3d orbitals are very close in energy. Consequently, manganese can lose its two 4s electrons to form the stable Mn²⁺ ion, but it can also lose some or all of its five 3d electrons in chemical reactions, leading to higher oxidation states up to +7 (loss of all 4s and 3d electrons). This small energy difference between the s and d sub-levels of the valence shell is the key reason for variable oxidation states in transition metals.

Question 17

Which statement correctly compares the radius of a sodium atom (Na) with a sodium ion (Na⁺) and a chlorine atom (Cl) with a chloride ion (Cl⁻)?

  1. Na is larger than Na⁺, and Cl is larger than Cl⁻.
  2. Na is smaller than Na⁺, and Cl is smaller than Cl⁻.
  3. Na is larger than Na⁺, and Cl is smaller than Cl⁻. (correct answer)
  4. Na is smaller than Na⁺, and Cl is larger than Cl⁻.
Explanation: When a metal atom like sodium (Na) loses an electron to form a cation (Na⁺), it loses its entire outermost electron shell (n=3). The remaining electrons are pulled more strongly by the unchanged nuclear charge, so the cation is significantly smaller than the parent atom. When a non-metal atom like chlorine (Cl) gains an electron to form an anion (Cl⁻), the nuclear charge remains the same but must attract an additional electron. This increases electron-electron repulsion and causes the electron cloud to expand, so the anion is larger than the parent atom. Therefore, Na > Na⁺ and Cl < Cl⁻.

Question 18

Element Z has a high melting point, is a semiconductor, and forms an acidic oxide with the formula ZO₂. In which section of the periodic table is Z located?

  1. Group 1, metals
  2. Group 2, metals
  3. Group 14, metalloids (correct answer)
  4. Group 17, non-metals
Explanation: The properties described are characteristic of a metalloid. A high melting point is typical of elements with covalent network structures. Being a semiconductor is the defining electrical property of a metalloid. An acidic oxide is typical of non-metals, but metalloids often form weakly acidic or amphoteric oxides. In Group 14, carbon (non-metal) forms acidic CO₂, silicon and germanium (metalloids) form SiO₂ and GeO₂, and tin and lead (metals) form amphoteric oxides. Silicon (Si) fits all the descriptions perfectly: high melting point, semiconductor, and forms an acidic oxide (SiO₂). Thus, element Z is located among the Group 14 metalloids.

Question 19

Consider the isoelectronic species S²⁻, Cl⁻, K⁺, and Ca²⁺. Which sequence correctly arranges these species in order of decreasing radius?

  1. Ca²⁺ > K⁺ > Cl⁻ > S²⁻
  2. S²⁻ > Cl⁻ > K⁺ > Ca²⁺ (correct answer)
  3. K⁺ > Ca²⁺ > S²⁻ > Cl⁻
  4. Cl⁻ > S²⁻ > K⁺ > Ca²⁺
Explanation: All four species are isoelectronic, meaning they have the same number of electrons (18, the configuration of Argon). For isoelectronic species, the radius is determined by the nuclear charge (number of protons). A higher nuclear charge pulls the electrons more strongly, resulting in a smaller radius. The number of protons are S (16), Cl (17), K (19), and Ca (20). Therefore, the species with the fewest protons (S²⁻) will be the largest, and the one with the most protons (Ca²⁺) will be the smallest. The correct order of decreasing radius is S²⁻ > Cl⁻ > K⁺ > Ca²⁺.

Question 20

Which property is NOT characteristic of most d-block transition elements?

  1. Formation of colored complex ions in solution.
  2. Existence of variable oxidation states in their compounds.
  3. Formation of basic oxides only. (correct answer)
  4. Exhibiting catalytic activity in many chemical reactions.
Explanation: Transition elements are well-known for forming colored complexes (A), having variable oxidation states (B), and acting as catalysts (D). For example, copper(II) sulfate solution is blue, manganese shows oxidation states from +2 to +7, and iron is used in the Haber process. However, their oxides are not exclusively basic. While they can form basic oxides in lower oxidation states (e.g., FeO), they can form amphoteric (e.g., Cr₂O₃) or acidic oxides (e.g., CrO₃, Mn₂O₇) in higher oxidation states. Therefore, forming basic oxides only is not a characteristic property.