A-LevelChemistryChemical bonding9701/22 · May/June 2022 · Q19 Marks

Magnesium has a melting point of 650°C and high electrical conductivity. Explain these properties of magnesium by referring to its structure and bonding.

(a)Magnesium has a melting point of 650°C and high electrical conductivity. Explain these properties of magnesium by referring to its structure and bonding.[2]
(b)(i)When magnesium is heated in air, magnesium oxide, MgO, is the major product. Smaller amounts of magnesium nitride, Mg3N2, are also made. Calculate the oxidation number for magnesium and for the nitrogen species in Mg3N2 to complete Table 1.1. Table 1.1 species magnesium in Mg3N2 nitrogen in Mg3N2 oxidation number[1]
(b)(ii)Identify the type of reaction which takes place between magnesium and nitrogen. Explain your answer.[1]
(b)(iii)Define enthalpy change of formation.[2]
(b)(iv)When 3.645g of Mg(s) burns in excess N₂(g) to form Mg3N2(s), 23.05kJ of energy is released. Calculate the enthalpy change of formation, ∆H₁, of Mg3N2. Show your working.[3]

Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

[2] M1 comment explaining high melting point of Mg in terms of many strong metallic bonds OR many strong (electrostatic) attractions between cations and delocalised electrons OR strong bonds in giant metallic structure. M2 comment explaining electrical conductivity of Mg in terms of movement of delocalised electrons delocalised electrons can move through the structure
[1] species magnesium in Mg3N2 nitrogen in Mg3N2 oxidation number 2(+)/(+)II 3-/-III
[1] redox AND Mg loses electrons / Mg increases ox no / Mg is oxidised AND N gains electrons / N reduces ox no / N is reduced
[2] M1 (enthalpy change / energy change) when one mole of a compound / substance is formed M2 from its elements in their standard states
[3] M1 correct method to find amount Mg3N2 formed in reaction 3.645 ÷ 24.3 mol Mg produces 0.15 ÷ 3 mol of Mg3N2 OR 0.05 mol of Mg3N2 M2 use M1 in correct expression to find energy released by 1 mol Mg3N2 23.05 + M1 OR 23.05 ÷ mol Mg3N2 OR 23 050+ M1 OR 23 050 ÷ Mg3N2 M3 calculate value from M2 expression correctly ΔΗ = −(M2) = −461(.0) kJ mol-1 OR –461 000 J mol-1

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