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Questions on Group 2: trends down the group and reactions with water
The Group 2 elements beryllium to barium, their electron configurations and the +2 oxidation state, the trends in atomic radius, first ionisation energy and melting point with the magnesium anomaly, reactivity down the group, the reactions of magnesium, calcium, strontium and barium with water, magnesium with steam, the reactions with oxygen, chlorine and dilute hydrochloric and sulfuric acids, the oxides with water and the rising pH of the solutions, the oxides as bases, and the extraction of titanium from TiCl₄ with magnesium as the reducing agent.
21 original questions · 69 marks · the Group 2: trends down the group and reactions with water notes · Group 2 and Group 7 · Flashcards for this unit
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Explain the trends in atomic radius, first ionisation energy and reactivity with water from magnesium to barium. Include an equation in your answer.
Mark scheme
Indicative content
- atomic radius increases down the group because each element has one more occupied shell of electrons than the element above it
- first ionisation energy decreases down the group although the nuclear charge increases, so less energy is needed to remove an outer electron
- the outer electrons are further from the nucleus and more shielded by inner shells, so they are less strongly attracted, which outweighs the greater nuclear charge
- reactivity with water increases down the group, from a very slow reaction for magnesium with cold water to a vigorous one for barium
- because the sum of the first two ionisation energies falls, the two outer electrons are lost more easily and the metal is oxidised to M²⁺ more readily
- a correct balanced equation with state symbols, for example Ba(s) + 2H₂O(l) → Ba(OH)₂(aq) + H₂(g), or the steam reaction of magnesium
Levels
- Level 3 (5–6 marks) all three trends are stated and each is explained: radius through the added shell, ionisation energy through distance and shielding outweighing nuclear charge, and reactivity through the easier loss of both outer electrons, with a correct balanced equation with state symbols
- Level 2 (3–4 marks) the three trends are stated and two are explained, or all three are explained and the equation is unbalanced or absent
- Level 1 (1–2 marks) one or two trends are stated, typically with the greater nuclear charge given as the reason the outer electrons are lost more easily; no correct equation
Give the full electron configurations of a calcium atom and of a Ca²⁺ ion. Then use the ionisation energies in the table to explain why calcium is +2 in all its compounds and never +3.
Ionisation of calcium First Second Third Energy / kJ mol⁻¹ 590 1145 4912 Mark scheme
- B1 calcium atom: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
- B1 Ca²⁺ ion: 1s² 2s² 2p⁶ 3s² 3p⁶, the two 4s electrons having been removed
- B1 the first two electrons come from the 4s subshell, but a third would come from the full 3p subshell, closer to the nucleus and less shielded
- B1 the third ionisation energy is more than four times the second, more energy than any reaction supplies, so Ca³⁺ never forms
Explain why the melting point decreases from calcium to barium, and identify the element from magnesium to barium whose melting point does not fit the trend.
Mark scheme
- B1 the metals have metallic bonding: a lattice of 2+ ions attracted to delocalised electrons
- B1 down the group the ions are larger with the same 2+ charge, so the delocalised electrons are further from the centre of each ion
- B1 the attraction is weaker, so less energy is needed to break up the lattice; reject any reference to intermolecular forces or molecules
- B1 magnesium does not fit: at 650 °C it melts below calcium, 842 °C, although its ions are smaller; ignore any attempted explanation
Explain the roles of magnesium and of argon in the reduction TiCl₄(g) + 2Mg(l) → Ti(s) + 2MgCl₂(l) at about 1000 °C, identifying the species oxidised and the species reduced.
Mark scheme
- B1 magnesium is oxidised, from 0 in Mg to +2 in MgCl₂
- B1 titanium is reduced, from +4 in TiCl₄ to 0 in Ti, so magnesium is the reducing agent
- B1 at about 1000 °C titanium and magnesium would react with the oxygen and nitrogen in air
- B1 argon is unreactive and excludes air, so the titanium is not contaminated with oxide or nitride, which make it brittle; 'argon is inert' alone is not enough
Compare the reactions of magnesium and barium with cold water.
Mark scheme
- B1 magnesium reacts very slowly, giving only a few bubbles, whereas barium reacts vigorously with rapid fizzing
- B1 both give hydrogen and the metal hydroxide, with the metal oxidised from 0 to +2
- B1 magnesium hydroxide is sparingly soluble and stays as a white solid, whereas barium hydroxide dissolves
- B1 so the barium solution is much more alkaline, with a higher pH, than the water around the magnesium
Evaluate the claim that titanium is expensive because its ore is rare.
Mark scheme
- B1 the claim is not supported: titanium is one of the most abundant elements in the Earth's crust, and rutile is plentiful
- B1 the cost comes from the extraction, which is a batch process held at high temperatures for long periods
- B1 it needs chlorine, argon and magnesium, and the magnesium is itself extracted by electrolysis, which uses a great deal of electricity
- B1 carbon cannot be used as the reducing agent because it forms brittle titanium carbide, so the cheaper route used for iron is not available
Explain why the first ionisation energy decreases from magnesium to barium.
Mark scheme
- B1 the outer electron is in a shell further from the nucleus, so atomic radius increases; allow 'more energy levels'
- B1 there are more filled inner shells, so the outer electron is more shielded from the nucleus
- B1 the attraction between the nucleus and the outer electron is weaker, outweighing the increase in nuclear charge, so less energy is needed to remove it
Describe what is seen when magnesium is heated in steam, and write an equation with state symbols for the reaction.
Mark scheme
- B1 the magnesium burns with a bright white light
- B1 a white solid (powder) of magnesium oxide is left, and the hydrogen can be lit at the end of the tube
- B1 Mg(s) + H₂O(g) → MgO(s) + H₂(g); reject Mg(OH)₂ as the product with steam
Complete the equation for the reaction of barium with cold water, including state symbols, Ba(s) + 2H₂O(l) →, and state what happens to the oxidation state of barium.
Mark scheme
- B1 the products are Ba(OH)₂ and H₂, in a 1 : 1 ratio
- B1 the state symbols are Ba(OH)₂(aq) and H₂(g); allow Ba²⁺(aq) + 2OH⁻(aq) for the hydroxide
- B1 barium goes from 0 to +2, so it is oxidised and acts as the reducing agent
Explain why the reactivity of the Group 2 metals with water increases down the group.
Mark scheme
- B1 when the metal reacts each atom loses its two outer electrons, so it is oxidised
- B1 down the group the outer electrons are further from the nucleus and more shielded, so they are less strongly attracted
- B1 the sum of the first and second ionisation energies falls, so the electrons are lost more easily and the metal reacts faster
Calculate the mass of magnesium, in kg, needed to react with 500 kg of titanium(IV) chloride. Ar values: Ti 47.9, Cl 35.5, Mg 24.3.
Mark scheme
- M1 Mr(TiCl₄) = 47.9 + 4 × 35.5 = 189.9, so amount of TiCl₄ = 5.00 × 10⁵ g ÷ 189.9 = 2633 mol
- M1 TiCl₄ + 2Mg → Ti + 2MgCl₂, so amount of Mg = 2 × 2633 = 5266 mol
- A1 mass of Mg = 5266 × 24.3 = 1.28 × 10⁵ g = 128 kg, to three significant figures; error carried forward from M2
Calculate the volume of hydrogen, in cm³ at room temperature and pressure, given off when 0.400 g of calcium reacts completely with water. Ar of Ca = 40.1; molar gas volume = 24.0 dm³ mol⁻¹.
Mark scheme
- M1 amount of Ca = 0.400 g ÷ 40.1 g mol⁻¹ = 9.975 × 10⁻³ mol
- M1 Ca(s) + 2H₂O(l) → Ca(OH)₂(aq) + H₂(g) is 1 : 1, so 9.975 × 10⁻³ mol of H₂ forms
- A1 volume = 9.975 × 10⁻³ × 24.0 = 0.239 dm³ = 239 cm³, to three significant figures
Suggest why sodium can be used in place of magnesium to reduce titanium(IV) chloride, write the equation with state symbols for a reaction at 850 °C, and compare the number of moles of each metal needed per mole of titanium.
Mark scheme
- B1 sodium loses its outer electron very easily, so it is a strong reducing agent, like magnesium
- B1 TiCl₄(g) + 4Na(l) → Ti(s) + 4NaCl(l)
- B1 four moles of sodium are needed for each mole of titanium, against two of magnesium, because each sodium atom gives one electron and each magnesium atom two
Explain why calcium reacting with water gives a cloudy mixture, while barium reacting with water gives a clear solution.
Mark scheme
- B1 both metals form their hydroxide and hydrogen
- B1 calcium hydroxide is only sparingly soluble, so some of it comes out of solution as a white solid
- B1 barium hydroxide is much more soluble, because hydroxide solubility increases down the group, so it stays dissolved
Explain why calcium reacts with dilute sulfuric acid for only a short time, while magnesium goes on reacting until the metal or the acid is used up.
Mark scheme
- B1 calcium sulfate is only sparingly soluble, so it forms as a solid layer on the surface of the calcium
- B1 the layer keeps the acid away from the metal, so the reaction slows and stops
- B1 magnesium sulfate is soluble, so it dissolves and the magnesium surface stays exposed to the acid
Calculate the volume of hydrogen, in cm³ at room temperature and pressure, given off when 0.250 g of strontium reacts with excess dilute hydrochloric acid. Ar of Sr = 87.6; molar gas volume = 24.0 dm³ mol⁻¹.
Mark scheme
- M1 amount of Sr = 0.250 g ÷ 87.6 g mol⁻¹ = 2.854 × 10⁻³ mol
- M1 Sr(s) + 2HCl(aq) → SrCl₂(aq) + H₂(g) is 1 : 1 for strontium and hydrogen, so 2.854 × 10⁻³ mol of H₂ forms
- A1 volume = 2.854 × 10⁻³ × 24.0 = 6.85 × 10⁻² dm³ = 68.5 cm³, to three significant figures
Suggest why an open bag of quicklime, calcium oxide, gets warm and its lumps crumble to powder after rain gets into it. Give an equation with state symbols, and suggest the pH of the mixture when the powder is shaken with more water.
Mark scheme
- B1 calcium oxide reacts with the water in an exothermic reaction, forming powdery calcium hydroxide
- B1 CaO(s) + H₂O(l) → Ca(OH)₂(s)
- B1 about pH 12, because calcium hydroxide is sparingly soluble and releases some hydroxide ions; allow 11 to 12.5
Describe what is seen when hot magnesium ribbon is lowered into a gas jar of chlorine, and give the equation, with state symbols, and the change in oxidation state of each element.
Mark scheme
- B1 the magnesium glows or burns with a white light and a white solid forms on the walls of the jar
- B1 Mg(s) + Cl₂(g) → MgCl₂(s)
- B1 magnesium goes from 0 to +2 and is oxidised; chlorine goes from 0 to −1 and is reduced
Calculate the volume of carbon dioxide, in cm³ at room temperature and pressure, given off when 0.750 g of calcium carbonate reacts with excess dilute hydrochloric acid. Mr of CaCO₃ = 100.1; molar gas volume = 24.0 dm³ mol⁻¹.
Mark scheme
- M1 amount of CaCO₃ = 0.750 g ÷ 100.1 g mol⁻¹ = 7.493 × 10⁻³ mol
- M1 CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l) is 1 : 1 for carbonate and carbon dioxide
- A1 volume = 7.493 × 10⁻³ × 24.0 = 0.180 dm³ = 180 cm³, to three significant figures
Suggest why beryllium shows no visible reaction with water, even as steam, although it is in Group 2.
Mark scheme
- B1 beryllium has the smallest atoms in the group, so its outer electrons are closest to the nucleus and least shielded
- B1 it has the highest ionisation energies, so it loses electrons least readily; a surface oxide layer also protects the metal
Suggest the approximate pH of the solution formed when strontium oxide is shaken with water, and explain your answer.
Mark scheme
- B1 about pH 12 to 13, between the values for calcium oxide and barium oxide
- B1 the oxide forms strontium hydroxide, which is more soluble than calcium hydroxide and less soluble than barium hydroxide, so it gives an intermediate hydroxide ion concentration
Practise Group 2: trends down the group and reactions with water one question at a time