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IB Chemistry S3.1: The periodic table worked solutions

Worked solutions to every practice question in S3.1: The periodic table of Photon Chemistry: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question B1

[1 mark]
A main-group atom occupies period 4, group 2. Which statement follows directly?(A) It has two occupied shells and four valence electrons. (B) It has four occupied shells and two valence electrons. (C) Its atomic number is four. (D) Its outer electrons occupy a p sublevel.
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Period 4 means four occupied principal levels. Group 2 supplies two outer s electrons. Only option B satisfies both pieces of the address.

Final answer

B

Question B2

[1 mark]
A materials catalogue lists aluminium, silicon, nickel and neon. Which entry describes silicon correctly?(A) Metal, s-block (B) Metalloid, p-block (C) Non-metal, d-block (D) Transition element, p-block
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Silicon has an outer p electron and mixed metal/nonmetal properties. Its p-block position is independent of the property label metalloid.

Final answer

B

Question B3

[1 mark]
A neutral calcium atom forms \mathrm{Ca}^{2+}. Which change explains the smaller size of the ion?(A) The nucleus loses two protons. (B) Two neutrons are removed from the nucleus. (C) The atom gains two electrons and repulsion decreases. (D) Its outermost occupied shell is removed, leaving electrons closer to the same nucleus.
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Calcium loses its two 4s electrons. The remaining highest occupied shell is n=3, while the nucleus still contains 20 protons. Removing that outer shell explains the smaller cation.

Final answer

D

Question B4

[1 mark]
Which statement describes a negative first electron affinity?(A) A gaseous atom gains an electron and releases energy. (B) A gaseous atom loses an electron and releases energy. (C) A gaseous positive ion gains an electron and absorbs energy. (D) A bonded atom attracts a shared electron pair more strongly.
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The starting species must be a neutral gaseous atom, and the process must add an electron. A negative energy change denotes release. Option A meets all three conditions.

Final answer

A

Question B5

[1 mark]
Which prediction is correct for dilute aqueous iodine added to aqueous lithium chloride?(A) Chlorine forms because iodine has more occupied electron shells. (B) Lithium metal forms because lithium is in group 1. (C) No halogen displacement occurs because iodine gains electrons less readily than chlorine. (D) Hydrogen forms because every group 1 substance reacts with water.
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The oxidant here is iodine. It is weaker than chlorine and cannot take electrons from chloride in this aqueous comparison. Lithium is already present as ions; no lithium-metal water reaction is involved.

Final answer

C

Question B6

[1 mark]
A sulfur oxide has one sulfur atom for every three oxygen atoms. Which paired description is correct?(A) Sulfur state -6; acidic oxide. (B) Sulfur state +3; acidic oxide. (C) Sulfur state +6; basic oxide. (D) Sulfur state +6; acidic oxide.
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For neutral \mathrm{SO_3}, x+3(-2)=0, giving sulfur +6. The oxide forms an acid with water, so its behaviour is acidic.

Final answer

D

Question B7

HL[1 mark]
Which observation would best distinguish a typical first-row transition metal from a group 2 metal?(A) The solid conducts electricity. (B) The element forms at least one positive ion. (C) Its compounds exhibit several metal oxidation states. (D) The element can react with an acid.
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A typical first-row transition element can use more than one metal oxidation state; group 2 overwhelmingly uses +2. Electrical conduction or formation of a positive ion does not distinguish the groups.

Final answer

C

Question B8

HL[1 mark]
A complex solution absorbs mainly green light. Using the six-colour wheel in the data booklet, which colour is expected to dominate the transmitted light?(A) Green (B) Yellow (C) Orange (D) Red
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Remove the absorbed green band from the incident white light. The complementary dominant transmitted colour is red, giving option D.

Final answer

D

Question B9

[2 marks]
For a covalent bond between unlike atoms, define electronegativity and state whether it has an energy unit.
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Electronegativity describes how strongly a bonded atom attracts the shared electron pair. It is a relative, dimensionless quantity; it is not an electron-removal or attachment enthalpy.

Final answer

Attraction of an atom for the shared electron pair; dimensionless.

Question B10

[2 marks]
A schematic first-ionization-energy trace follows three consecutive elements A, B and C. B is a noble-gas maximum, followed by a sharp fall at C. The horizontal axis is proton number.(a) Give the group of C and explain the fall. [2]
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The element after a noble gas is in group 1. Its removable electron occupies a new, more distant principal shell and is more shielded. It is therefore less strongly attracted, so its first ionization energy falls. These are first removals from different neutral atoms.

Final answer

Group 1; C starts a new occupied shell.

Question B11

HL[3 marks]
HL: a model complex absorbs at 620\,\mathrm{nm}. Use c=3.00\times10^8\,\mathrm{m\,s^{-1}}.(a) Calculate the absorbed frequency. [2] (b) Would absorption at 530 nm require a larger or smaller d-energy gap? [1]
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(a) Convert 620\,\mathrm{nm}=6.20\times10^{-7}\,\mathrm{m}. Then f=c/\lambda=(3.00\times10^8)/(6.20\times10^{-7})=4.838709677\times10^{14}\,\mathrm{Hz}, or 4.84\times10^{14}\,\mathrm{Hz}. (b) The shorter wavelength has higher frequency and photon energy, so the corresponding gap is larger.

Final answer

4.84\times10^{14}\,\mathrm{Hz}; larger at 530 nm.

Tutorial practice

Question U1

[5 marks]
Element X lies in period 4 and group 16. Only atoms with atomic numbers up to 36 are considered.(a) Identify X and write its condensed electron configuration. [2] (b) Explain how the configuration supports the period and group numbers. [2] (c) Name one other element expected to have similar chemical properties. [1]
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(a) A period-4, group-16 atom is selenium. Its configuration is [\mathrm{Ar}]3d^{10}4s^24p^4, totalling 18+10+2+4=34 electrons. (b) The highest occupied level is 4. The outer 4s^24p^4 pattern has six electrons and gives group 10+6=16. The filled 3d sublevel is internal for this count. (c) Sulfur is an acceptable family comparison; oxygen or tellurium also qualifies.

Final answer

Se; [\mathrm{Ar}]3d^{10}4s^24p^4; n=4, six outer electrons; e.g. S.

Question U2

[6 marks]
The following species are considered: \mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Na}^{+} and neutral \mathrm{Na}.(a) State the electron count for each of the three ions. [1] (b) Order the three ions by increasing radius and justify the order. [3] (c) Explain why the sodium atom is larger than its cation. [2]
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(a) Oxygen gains two electrons: 8+2=10; fluorine gains one: 9+1=10; sodium loses one: 11-1=10. (b) With the same electron count and occupied shells, increasing proton number contracts the cloud. Thus the increasing-radius order is \mathrm{Na^+}<\mathrm{F^-}<\mathrm{O^{2-}}. (c) Neutral Na occupies 3s, while its cation has only the first two shells. Loss of the outer electron also reduces repulsion, so the remaining cloud is more tightly held by the unchanged nucleus.

Final answer

10 each; \mathrm{Na^+}<\mathrm{F^-}<\mathrm{O^{2-}}; Na has an extra occupied shell.

Question U3

[5 marks]
An inventory contains \mathrm{CaH_2}, \mathrm{BaO_2}, \mathrm{H_2SO_4} and elemental \mathrm{S_8}.(a) Determine hydrogen in calcium hydride and oxygen in barium peroxide. [2] (b) Calculate the sulfur oxidation state in sulfuric acid. [1] (c) State the sulfur oxidation state in the elemental sample and explain its value. [2]
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(a) Calcium hydride gives +2+2x=0, hence H is -1. Barium peroxide similarly gives +2+2y=0, hence O is -1. (b) In sulfuric acid, 2(+1)+s+4(-2)=0, so S is +6. (c) S is zero in \mathrm{S_8}. Each bond joins identical atoms, so its electrons are formally allocated equally.

Final answer

H -1; O -1; S +6; elemental S 0.

Question U4

HL[6 marks]
The data booklet lists I_1(\mathrm{Be})=900, I_1(\mathrm{B})=801, I_1(\mathrm{N})=1402 and I_1(\mathrm{O})=1314\,\mathrm{kJ\,mol^{-1}}.(a) Explain the boron–beryllium reversal using the electron removed from each atom. [2] (b) Explain the oxygen–nitrogen reversal in terms of orbital occupancy. [2] (c) State the general first-ionization trend across period 2 and the reason these results do not invalidate it. [2]
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(a) Be loses a 2s electron. B loses 2p, which is higher in energy and more shielded/less penetrating, so it requires less energy. (b) N has three separate 2p electrons. O has a paired 2p orbital, and pair repulsion makes removal easier. (c) Rising nuclear charge with relatively little core-shielding change generally raises first IE across period 2. The two local changes in removable-electron environment explain departures from a strictly increasing sequence.

Final answer

B removes 2p rather than 2s; O removes a paired 2p electron; the overall trend rises.

Question U5

[6 marks]
Compare neutral Li/Na down group 1 and Na/Cl across period 3.(a) Explain which of Li and Na has the larger atomic radius. [2] (b) Explain which of Na and Cl has the higher first ionization energy. [2] (c) State the direction of change in atomic radius and electronegativity from Na towards Cl. [2]
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(a) Na has an extra occupied principal shell. Greater distance and core shielding outweigh its greater nuclear charge, so Na is larger. (b) Na and Cl lose electrons from the third shell; Cl has greater nuclear charge with comparatively little change in core shielding. Stronger attraction requires more removal energy. (c) Across this period the electron cloud generally contracts and attraction for a shared bond pair increases: radius decreases, electronegativity increases.

Final answer

Na larger; Cl higher IE; radius falls and electronegativity rises.

Question U6

[5 marks]
Separate water samples receive sodium metal or magnesium oxide in a demonstration.(a) Write the sodium reaction, including states, and name the gas. [2] (b) Give one visible observation from the sodium demonstration. [1] (c) Write the magnesium oxide hydration equation and explain why complete dissolution should not be assumed. [2]
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(a) \ce{2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g)}; the gas is hydrogen. (b) Bubbles or movement/disappearance of the metal are acceptable visible observations. (c) \ce{MgO(s) + H2O(l) -> Mg(OH)2(s)}. Hydration can be slow and magnesium hydroxide is sparingly soluble. The small dissolved fraction makes the liquid alkaline; a solid suspension can remain.

Final answer

2\mathrm{Na}+2\mathrm{H_2O}\to2\mathrm{NaOH}+\mathrm{H_2}; gas/moving metal; \mathrm{MgO}+\mathrm{H_2O}\to\mathrm{Mg(OH)_2}, sparingly soluble.

Question U7

HL[4 marks]
HL: use Zn [\mathrm{Ar}]3d^{10}4s^2, Cu [\mathrm{Ar}]3d^{10}4s^1 and Ni [\mathrm{Ar}]3d^84s^2. Restrict colour predictions to the simple d–d model.(a) Explain why Cu qualifies as a transition element while Zn does not. [2] (b) Compare the d occupancies of \mathrm{Cu^+} and \mathrm{Ni^{2+}}, and identify the ion that can show visible d–d absorption in a suitable ligand field. [2]
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(a) Cu can form Cu^{2+} with an incomplete 3d^9 sublevel, satisfying the transition-element criterion. Zn and its normal Zn^{2+} ion both have a full 3d^{10} sublevel. (b) Removing the outer s electrons gives Cu^+ d^{10} and Ni^{2+} d^8. The latter has electrons and vacancies in split d levels and can absorb visible photons if the gap is suitable. Cu^+ lacks the ordinary d–d route; other absorption mechanisms are outside this model.

Final answer

Cu forms d^9 Cu^{2+}; Zn/Zn^{2+} are d^{10}. Cu^+ d^{10}; Ni^{2+} d^8 can absorb.

Question U8

[5 marks]
Dilute aqueous bromine is added separately to potassium iodide and potassium chloride. Compare group trends.(a) Identify the halide solution in which displacement occurs. [1] (b) Write the net ionic displacement equation and identify the oxidizing species. [2] (c) Explain why the reactivity directions down groups 1 and 17 differ. [2]
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(a) Bromine displaces iodide, but not chloride. (b) \ce{Br2(aq) + 2I^-(aq) -> 2Br^-(aq) + I2(aq)}. Bromine accepts electrons and is the oxidant. (c) Extra shells and shielding make electron loss easier down group 1, increasing metal reactivity. They weaken attraction for an incoming electron down group 17, decreasing halogen oxidizing strength.

Final answer

Iodide; \mathrm{Br_2}+2\mathrm{I^-}\to2\mathrm{Br^-}+\mathrm{I_2}; Br_2 oxidant. Metal loss easier down group; halogen gain less favourable.

Advanced practice

Question A1

[7 marks]
A fuel sample contains sulfur as well as carbon. A treatment plant oxidises sulfur completely to sulfur dioxide. Its exhaust then contacts calcium oxide, which captures carbon dioxide as a solid carbonate. Treat the capture reaction as complete for the calculation.(a) Write the sulfur oxidation equation. [1] (b)(i) State an environmental consequence of sulfur dioxide emission. [1] (b)(ii) Write a balanced model equation for sulfur dioxide reacting with water to form sulfurous acid, \mathrm{H_2SO_3}. [1] (c) Write the reaction between calcium oxide and carbon dioxide. [1] (d) Determine the mass of pure calcium oxide needed to capture 22.0\,\mathrm{kg} of carbon dioxide. Use molar masses 56.08 and 44.01\,\mathrm{g\,mol^{-1}}, respectively. [2] (e) Explain why carbon dioxide capture alone does not remove the sulfur-related concern. [1]
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(a) \ce{S(s) + O2(g) -> SO2(g)}. (b)(i) Acid deposition can damage lake organisms or vegetation. (ii) In the stipulated aqueous model, \ce{SO2 + H2O <=> H2SO3}. (c) \ce{CaO(s) + CO2(g) -> CaCO3(s)}. (d) Convert 22.0\,\mathrm{kg} to 22000\,\mathrm{g}. Then n=22000/44.01=499.886389\ldots\,\mathrm{mol}. The ratio is 1:1, so m= (22000/44.01)(56.08)=28033.6287\ldots\,\mathrm{g}=28.0\,\mathrm{kg} to three significant figures. Keep the unrounded amount in the multiplication. (e) The calculation establishes carbon dioxide capture only; sulfur dioxide is a separate pollutant whose removal is not established.

Final answer

\mathrm{SO_2}; acid deposition; \mathrm{CaCO_3}; 28.0\,\mathrm{kg} CaO; sulfur pollution still needs attention.

Question A2

HL[6 marks]
An illustrative transition-element record gives the energy cost of each successive removal from gaseous species. Each entry is the cost of that removal alone, per mole, not a cumulative total. These values are not an actual element identification dataset.
Electron removed: 2 / 3 / 4; cost / \mathrm{kJ\,mol^{-1}}: 2110 / 3620 / 9870. (a) Calculate the increase in energy cost from the second to the third removal, and from the third to the fourth removal. [2] (b) Explain how the record can support variable oxidation states +2 and +3, and why it suggests a greater energy cost for reaching +4. [2] (c) Evaluate the claim that these costs alone prove that every compound with the element in state +2 is more stable than one with it in state +3. [2]
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(a) The differences are 3620-2110=1510\,\mathrm{kJ\,mol^{-1}} and 9870-3620=6250\,\mathrm{kJ\,mol^{-1}}. These compare individual removal costs, not cumulative energies. (b) The second and third costs are relatively close, supporting access to different electron counts and hence +2 and +3. Fourth removal costs substantially more. (c) The claim is unsupported. Gaseous ionization is only one contribution: bond formation, lattice and hydration energies can change the total balance. The arbitrary model does not identify an element or prove any universal compound stability order.

Final answer

1510, 6250\,\mathrm{kJ\,mol^{-1}}; +2/+3 plausible; total formation energies are needed.

Question A3

HL[6 marks]
Two hypothetical complexes of the same metal ion at the same oxidation state, L and M, are measured at equal concentration in the same cuvette. Their spectra each have one dominant visible band. The table lists absorbances at selected wavelengths.
\begin{array}{c|ccc}\lambda/\mathrm{nm}&450&580&660\\\hline A_L&0.06&0.72&0.09\\A_M&0.81&0.08&0.04\end{array}(a) Select the listed wavelength that provides the greatest sensitivity for measuring the concentration of L. [1] (b) Deduce the main observed colours of L and M using the booklet wheel. [2] (c) Explain why ligand identity can account for different absorption bands even though the metal and oxidation state are unchanged. [2] (d) Name one measurement condition that must remain constant when comparing the two absorbances. [1]
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(a) Of the listed samples, L has its greatest absorbance at 580\,\mathrm{nm}, giving the greatest concentration response under the stated conditions. (b) In the dominant-band model, L removes yellow and appears violet. M removes blue and appears orange. Three sampled wavelengths do not fix an exact real perceived colour. (c) A different ligand environment can alter the separation of d energies even at the same metal oxidation state. The changed gap then absorbs a different photon energy and wavelength. (d) Keep cuvette path length constant; concentration, solvent or temperature are also valid specified controls.

Final answer

580 nm; L violet, M orange; different ligand splitting; e.g. fixed path length.

Question A4

[6 marks]
A hypothetical pair of metallic solids X and Y has the same packing geometry. Supplied values: X, molar mass 24.0\,\mathrm{g\,mol^{-1}} and density 1.50\,\mathrm{g\,cm^{-3}}; Y, 60.0\,\mathrm{g\,mol^{-1}} and 3.00\,\mathrm{g\,cm^{-3}}. Use V_m=M/\rho and, only under this packing assumption, r_Y/r_X=(V_{m,Y}/V_{m,X})^{1/3}.(a) Calculate both molar volumes. [2] (b) Estimate the radius ratio and the percentage by which Y is larger. [2] (c) A student claims that the denser solid must have smaller atoms. Evaluate the claim using the calculation and identify one limitation of the radius model. [2]
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(a) V_{m,X}=24.0/1.50=16.0\,\mathrm{cm^3\,mol^{-1}} and V_{m,Y}=60.0/3.00=20.0\,\mathrm{cm^3\,mol^{-1}}. (b) r_Y/r_X=(20/16)^{1/3}=1.077217345\ldots, or 1.08. The increase is 100(1.077217345-1)=7.72\%. Use the unrounded ratio. (c) Y is twice as dense but has the larger estimated radius, so density alone does not rank atomic size: mass per atom also changes. Equal packing geometry is a model assumption; different packing or bonding could change the relation between bulk molar volume and an assigned atomic radius.

Final answer

16.0, 20.0\,\mathrm{cm^3\,mol^{-1}}; 1.08; 7.72% larger; density alone is insufficient; packing may differ.

Connected questions

Question X1

Connected questions[6 marks]
S3.1 + S1.4 + S1.5. A 0.460\,\mathrm{g} sodium sample reacts completely with excess water. Use M(\mathrm{Na})=22.99\,\mathrm{g\,mol^{-1}} and a supplied gas molar volume 24.0\,\mathrm{dm^3\,mol^{-1}} at the collection conditions.(a) Write the balanced equation with states and predict the solution character. [2] (b) Calculate the volume of dry hydrogen collected, in \mathrm{cm^3}. [3] (c) Explain why equal moles of potassium would usually react more vigorously under comparable conditions. [1]
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(a) \ce{2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g)}. Dissolved hydroxide makes the solution alkaline. (b) n(\mathrm{Na})=0.460/22.99=0.0200086994\,\mathrm{mol}. The 2:1 ratio gives n(\mathrm{H_2})=0.0100043497\,\mathrm{mol}. Thus V=0.0100043497(24.0)=0.240104393\,\mathrm{dm^3}=240\,\mathrm{cm^3} to three significant figures. (c) K has an extra occupied shell with greater shielding; weaker attraction makes electron loss easier.

Final answer

Alkaline; 240\,\mathrm{cm^3} H_2; K loses its more distant, shielded electron more easily.

Question X2

Connected questions[6 marks]
S3.1 + R3.1 + S1.4. A pure 0.800\,\mathrm{g} MgO sample treats 50.0\,\mathrm{cm^3} of 0.600\,\mathrm{mol\,dm^{-3}} HCl. Use M(\mathrm{MgO})=40.31\,\mathrm{g\,mol^{-1}} and assume complete reaction wherever acid remains.(a) Classify MgO and write its net ionic reaction with acid. [2] (b) Identify the limiting reagent and calculate the mass of MgO left. [3] (c) Explain why a low solubility in water would not make MgO a neutral oxide. [1]
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(a) MgO is basic. \ce{MgO(s) + 2H^+(aq) -> Mg^{2+}(aq) + H2O(l)}. (b) n(\mathrm{H^+})=0.600(0.0500)=0.0300\,\mathrm{mol}, sufficient for 0.0150\,\mathrm{mol} MgO. Initially 0.800/40.31=0.0198461920\,\mathrm{mol} MgO is present, so acid limits. The reacted mass is 0.0150(40.31)=0.60465\,\mathrm{g} and the remaining mass is 0.19535\,\mathrm{g}=0.195\,\mathrm{g}. (c) Acid–base character is established by reaction with acid or base, independently of whether much oxide dissolves in pure water.

Final answer

Basic; \mathrm{MgO}+2\mathrm{H^+}\to\mathrm{Mg^{2+}}+\mathrm{H_2O}; acid limiting; 0.195 g MgO remains.

Question X3

Connected questions · HL[6 marks]
HL: S3.1 + S1.3 + R3.4. A simplified ligand replacement is \ce{[Cu(H2O)6]^{2+} + 4Cl^- <=> [CuCl4]^{2-} + 6H2O}. A model sample has dominant absorption at 600 nm. Use h=6.63\times10^{-34}\,\mathrm{J\,s} and c=3.00\times10^8\,\mathrm{m\,s^{-1}}.(a) Deduce both copper oxidation states and decide whether replacement is redox. [2] (b) Calculate the absorbed photon energy. [2] (c) Explain how ligand replacement can shift the absorption wavelength without changing the metal oxidation state. [2]
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(a) Water is neutral, so the hydrated metal is +2. For the chloride complex, x+4(-1)=-2 gives x=+2. No oxidation-state change occurs, so this replacement is not redox. (b) 600\,\mathrm{nm}=6.00\times10^{-7}\,\mathrm{m}. Using E=hc/\lambda, E=(6.63\times10^{-34})(3.00\times10^8)/(6.00\times10^{-7})=3.315\times10^{-19}\,\mathrm{J}=3.32\times10^{-19}\,\mathrm{J} to three significant figures. (c) Different ligand interactions can change the separation of the d energies. An absorbed photon must match the new gap, so its frequency/wavelength can change even though Cu remains +2.

Final answer

+2 in both; no redox; 3.32\times10^{-19}\,\mathrm{J}; ligand environment changes d-level splitting.

Question X4

Connected questions[6 marks]
S3.1 + R3.2 + S1.4. A 10.0\,\mathrm{cm^3} bromine-water sample is mixed with excess aqueous iodide. The liberated iodine consumes 15.0\,\mathrm{cm^3} of 0.0200\,\mathrm{mol\,dm^{-3}} thiosulfate. Use the supplied reaction \ce{I2 + 2S2O3^{2-} -> 2I^- + S4O6^{2-}}. Assume complete reactions.(a) Write the net ionic reaction that liberates iodine and identify the oxidant. [2] (b) Calculate the original bromine concentration in \mathrm{mol\,dm^{-3}}. [3] (c) Explain why iodine added to aqueous chloride would not liberate chlorine. [1]
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(a) \ce{Br2(aq) + 2I^-(aq) -> 2Br^-(aq) + I2(aq)}. Bromine gains electrons and is the oxidant. (b) n(\mathrm{S_2O_3^{2-}})=0.0200(0.0150)=3.00\times10^{-4}\,\mathrm{mol}. The supplied equation gives half as much iodine: 1.50\times10^{-4}\,\mathrm{mol}. Displacement produces one iodine molecule per bromine molecule, so n(\mathrm{Br_2})=1.50\times10^{-4}\,\mathrm{mol}. Divide by the original sample volume, 0.0100\,\mathrm{dm^3}, giving 0.0150\,\mathrm{mol\,dm^{-3}}. (c) Iodine is a weaker oxidant than chlorine in this aqueous comparison, so it cannot accept electrons from chloride to form chlorine.

Final answer

Br_2 oxidant; 0.0150\,\mathrm{mol\,dm^{-3}}; iodine is weaker than chlorine as an oxidant.

From the book

Photon Chemistry: Study Guide for the IB Diploma (SL & HL)

The periodic table from the core ideas to full exam questions, with question types, worked examples and graded practice.

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