Chemistry 9701/22 — May/June 2022

Cambridge AS Level · AS Level Structured Questions · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Chemical Bonding · Introduction to Organic Chemistry · Electrochemistry · Chemical Energetics · Atoms, Molecules and Stoichiometry · Atomic Structure · +14 more

Q1Chemical BondingElectrochemistryChemical EnergeticsAtoms, Molecules and StoichiometryFree sample

(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.

DifficultyMedium-Easy

Worked solution

Answer

Magnesium has a giant metallic structure: positive $\text{Mg}^{2+}$ ions are held in a lattice by a 'sea' of delocalised electrons.

There are many strong metallic bonds (strong electrostatic attractions between the cations and the delocalised electrons). A large amount of energy is needed to break these bonds, giving a high melting point.
The delocalised electrons are free to move throughout the structure and can carry charge, so magnesium conducts electricity.

Final answer

Giant metallic structure; strong metallic bonds explain high melting point; mobile delocalised electrons explain conductivity.

Detailed explanation

Background Concept

Metallic bonding is the electrostatic attraction between a lattice of positive metal cations and a mobile 'sea' of delocalised valence electrons. In a giant metallic structure, every atom contributes one or more outer electrons to this shared electron cloud, so the positive ions are held strongly in position. This model explains typical metallic properties: high melting and boiling points, malleability, and electrical conductivity.

Understanding the Question

The question asks for two properties of magnesium — a high melting point and high electrical conductivity — to be explained in terms of its structure and bonding. The word 'explain' means each property needs a reason, not just a statement. Melting point depends on how much energy is needed to break the metallic lattice; conductivity depends on whether charged particles are free to move.

Approach

First state the structure of magnesium. Then, for each property, link it to a feature of that structure:

high melting point → strong metallic bonds need much energy to break;
electrical conductivity → delocalised electrons can move through the lattice.

Keep the two explanations separate so each mark-scheme point is clear.

Step-by-Step Reasoning

Magnesium atoms lose their two outer electrons, forming $\text{Mg}^{2+}$ ions. These ions form a regular giant lattice surrounded by delocalised electrons.
To melt magnesium, the metallic bonds must be broken. Because there are many strong electrostatic attractions between cations and the electron sea, a large quantity of energy is required. This gives magnesium a high melting point.
In the solid metal, the delocalised electrons are not attached to any individual ion. When a voltage is applied, these electrons drift through the lattice and carry charge, so magnesium conducts electricity.

Both mark-scheme points are therefore covered: many strong metallic bonds for the melting point, and mobile delocalised electrons for conductivity.

Key Takeaways

Metallic properties follow directly from the free-electron model. Bond strength controls melting point; mobility of charge carriers controls electrical conductivity.

Common Mistakes

Saying 'magnesium has strong intermolecular forces' — metallic bonding is not an intermolecular force.
Saying 'the metal ions move through the lattice to carry the current' — in a solid metal, it is the delocalised electrons that move; the cations are fixed.
Writing only 'strong bonds' without linking them to the large energy needed to break them.
Omitting the word 'delocalised', which is essential for the conductivity explanation.

Things to Be Careful About

The mark scheme awards one mark for the melting-point reason and one for the conductivity reason. State the structure once, then explicitly connect 'strong metallic bonds' to melting point and 'mobile/delocalised electrons' to conductivity.

Techniques used

describe the giant metallic structure of magnesiumrelate melting point to the strength of metallic bondsexplain electrical conductivity by the movement of delocalised electrons

(b)

When magnesium is heated in air, magnesium oxide, MgO, is the major product. Smaller amounts of magnesium nitride, $\text{Mg}_3\text{N}_2$ , are also made.

(i)

Calculate the oxidation number for magnesium and for the nitrogen species in $\text{Mg}_3\text{N}_2$ to complete Table 1.1.

Table 1.1

species	magnesium in $\text{Mg}_3\text{N}_2$	nitrogen in $\text{Mg}_3\text{N}_2$
oxidation number

DifficultyEasy

Worked solution

Answer

species	magnesium in $\text{Mg}_3\text{N}_2$	nitrogen in $\text{Mg}_3\text{N}_2$
oxidation number	+2	-3

Final answer

Mg: +2; N: -3

Detailed explanation

Background Concept

Oxidation number is a bookkeeping charge assigned to an atom in a compound. Key rules: an element in its elemental state has oxidation number 0; a Group 2 metal normally has oxidation number +2 in its compounds; the sum of all oxidation numbers in a neutral compound is zero.

Understanding the Question

The question asks for the oxidation numbers of both magnesium and nitrogen in $\text{Mg}_3\text{N}_2$ . This is a direct application of the rule that the oxidation numbers in a neutral formula must add to zero.

Approach

Assign magnesium its well-known group oxidation number, then use the condition that the total must be zero to find the oxidation number of nitrogen.

Step-by-Step Reasoning

Magnesium is in Group 2, so its oxidation number in compounds is +2.
There are three magnesium atoms, so the total contribution from magnesium is $3 \times (+2) = +6$ .
The compound is neutral, so the two nitrogen atoms must contribute $-6$ in total.
Therefore each nitrogen atom has oxidation number $-6/2 = -3$ .

This is consistent with magnesium nitride being an ionic compound containing the nitride ion, $\text{N}^{3-}$ .

Key Takeaways

For a neutral formula, the sum of oxidation numbers is always zero. Known group oxidation numbers can be used to deduce the oxidation number of an unknown element.

Common Mistakes

Writing $-6$ for each nitrogen atom instead of dividing by 2 because there are two N atoms.
Forgetting that magnesium is +2 in its compounds.
Confusing oxidation number with charge: for simple ions in ionic compounds they are equal, but it is still safer to use the sum rule.

Things to Be Careful About

Use the correct sign and arithmetic: $3(+2) + 2x = 0$ , so $x = -3$ . Write +2 and -3 rather than 2 and 3.

Techniques used

assign oxidation numbers using group and electronegativity rulesuse the neutrality of the formula to deduce an unknown oxidation number

(ii)

Identify the type of reaction which takes place between magnesium and nitrogen.

Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

Redox.

Magnesium is oxidised: its oxidation number increases from 0 in Mg to +2 in $\text{Mg}_3\text{N}_2$ , so it loses electrons.

Nitrogen is reduced: its oxidation number decreases from 0 in $\text{N}_2$ to -3 in $\text{Mg}_3\text{N}_2$ , so it gains electrons.

Final answer

Redox; Mg is oxidised and N2 is reduced.

Detailed explanation

Background Concept

A redox reaction is one in which oxidation and reduction happen together. Oxidation is loss of electrons or an increase in oxidation number; reduction is gain of electrons or a decrease in oxidation number. An element in its elemental state has oxidation number 0.

Understanding the Question

The question asks you to identify the type of reaction between magnesium and nitrogen and to justify your answer. The justification must come from showing that oxidation numbers change, or from describing electron transfer.

Approach

Compare the oxidation numbers of magnesium and nitrogen in the reactants and in the product. If one element increases and another decreases, the reaction is redox.

Step-by-Step Reasoning

Magnesium starts as Mg(s), an element, so its oxidation number is 0. In $\text{Mg}_3\text{N}_2$ it is +2. The increase from 0 to +2 shows magnesium loses electrons and is oxidised.
Nitrogen starts as $\text{N}_2$ (g), an element, so each N atom has oxidation number 0. In $\text{Mg}_3\text{N}_2$ each N is -3. The decrease shows nitrogen gains electrons and is reduced.
Because both oxidation and reduction occur in the same reaction, the reaction is redox.

Key Takeaways

A combination reaction can be redox when elements combine to form a compound with different oxidation states. Always identify both the oxidised species and the reduced species.

Common Mistakes

Saying 'magnesium gains electrons' — magnesium loses electrons.
Calling the reaction only 'reduction' because nitrogen is reduced.
Forgetting that elements in their standard state have oxidation number 0.

Things to Be Careful About

Use the oxidation numbers from the table or write them in your answer. The mark scheme accepts either oxidation-number reasoning or electron-transfer reasoning; the safest answer includes both.

Techniques used

compare oxidation numbers of reactants and productsidentify oxidation and reduction by electron transfer

(iii)

Define enthalpy change of formation.

DifficultyEasy

Worked solution

Answer

The enthalpy change when one mole of a compound is formed from its elements in their standard states.

Final answer

The enthalpy change when one mole of a compound is formed from its elements in their standard states.

Detailed explanation

Background Concept

Enthalpy change of formation is a defined standard enthalpy change. It is always quoted per mole of product formed, and the reactants must be the elements in their standard states under standard conditions. For example, the formation reaction of $\text{Mg}_3\text{N}_2$ is from solid Mg and gaseous N2.

Understanding the Question

The question asks for a definition of enthalpy change of formation. It is a definition mark, so the wording must contain both required ideas: 'one mole of compound' and 'from its elements in their standard states'.

Approach

Recall the standard definition and make sure both halves are present. Do not mention combustion or neutralisation, and do not add extra quantities that are not part of the definition.

Step-by-Step Reasoning

State that the change refers to the formation of one mole of a compound/substance.
State that the substance is formed from its elements, and that those elements must be in their standard states.

'Standard states' means the physical state and form in which the element is most stable under standard conditions, e.g. graphite for carbon and N2(g) for nitrogen.

Key Takeaways

Definition questions in energetics are awarded largely for precise wording. Including 'one mole' and 'elements in their standard states' is essential.

Common Mistakes

Writing 'energy change when a compound is formed' without saying 'one mole'.
Saying 'from its atoms' instead of 'from its elements'.
Omitting 'standard states'.

Things to Be Careful About

Use the word 'compound' or 'substance' as allowed by the mark scheme. Ensure 'from its elements in their standard states' appears, since this is the second mark.

Techniques used

state the formal definition of enthalpy change of formationspecify standard states

(iv)

When 3.645g of Mg(s) burns in excess $\text{N}_2(\text{g})$ to form $\text{Mg}_3\text{N}_2(\text{s})$ , 23.05kJ of energy is released.

Calculate the enthalpy change of formation, $\Delta H_f$ , of $\text{Mg}_3\text{N}_2$ . Show your working.

\Delta H_f (\text{Mg}_3\text{N}_2) =

DifficultyMedium

Worked solution

Working

$M_r(\text{Mg}) = 24.3$

n(\text{Mg}) = \frac{3.645}{24.3} = 0.150\ \text{mol}

Balanced equation:

3\text{Mg}(\text{s}) + \text{N}_2(\text{g}) \rightarrow \text{Mg}_3\text{N}_2(\text{s})

n(\text{Mg}_3\text{N}_2) = \frac{0.150}{3} = 0.0500\ \text{mol}

Energy released per mole of $\text{Mg}_3\text{N}_2$ :

\frac{23.05\ \text{kJ}}{0.0500\ \text{mol}} = 461\ \text{kJ mol}^{-1}

Since the reaction is exothermic, the enthalpy change is negative:

\Delta H_f(\text{Mg}_3\text{N}_2) = -461\ \text{kJ mol}^{-1}

Answer

$-461\ \text{kJ mol}^{-1}$

Final answer

-461 kJ mol^-1

Detailed explanation

Background Concept

The enthalpy change of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states. Because this question gives the energy released when a particular mass of magnesium reacts, we must scale that energy to one mole of the product $\text{Mg}_3\text{N}_2$ . The balanced formation equation is essential because the stoichiometric ratio between Mg and $\text{Mg}_3\text{N}_2$ is 3 : 1.

Understanding the Question

3.645 g of magnesium burns in excess nitrogen, releasing 23.05 kJ, to form $\text{Mg}_3\text{N}_2$ . 'Excess N2' means all the magnesium is used up, so magnesium is the limiting reactant. We need the enthalpy change per mole of $\text{Mg}_3\text{N}_2$ , not per mole of magnesium.

Approach

Convert the mass of magnesium to moles using $n = m/M_r$ .
Write the balanced equation and use the 3 : 1 ratio to find the moles of $\text{Mg}_3\text{N}_2$ formed.
Divide the energy released by the moles of $\text{Mg}_3\text{N}_2$ to get the enthalpy change per mole.
Add a negative sign because energy is released (exothermic).

Step-by-Step Reasoning

$M_r(\text{Mg}) = 24.3$ .
$n(\text{Mg}) = 3.645 / 24.3 = 0.150$ mol.
Balanced equation:

$3\text{Mg}(\text{s}) + \text{N}_2(\text{g}) \rightarrow \text{Mg}_3\text{N}_2(\text{s})$

From the equation, 3 mol Mg form 1 mol $\text{Mg}_3\text{N}_2$ , so:

$n(\text{Mg}_3\text{N}_2) = \frac{0.150}{3} = 0.0500\ \text{mol}$

The heat released by 0.0500 mol of product is 23.05 kJ, so the heat released per mole is:

$\frac{23.05}{0.0500} = 461\ \text{kJ mol}^{-1}$

The reaction is exothermic, so the system loses enthalpy; therefore $\Delta H_f = -461$ kJ mol $^{-1}$ . If working in joules, $23.05$ kJ $= 23{,}050$ J and the answer is $-461{,}000$ J mol $^{-1}$ .

Key Takeaways

Formation enthalpy is always per mole of product formed.
The stoichiometric ratio from the balanced equation must be used to convert moles of reactant to moles of product.
An exothermic reaction gives a negative $\Delta H$ .

Common Mistakes

Using 0.150 mol as the amount of $\text{Mg}_3\text{N}_2$ instead of dividing by 3.
Forgetting the negative sign because the reaction releases heat.
Writing the answer as kJ without per mole.
Mixing up kJ and J in the final unit.
Reporting the enthalpy change per mole of magnesium rather than per mole of $\text{Mg}_3\text{N}_2$ .

Things to Be Careful About

Include the balanced equation and state symbols where required. The energy released must be divided by the moles of product, not reactant. Keep the unit kJ mol $^{-1}$ (or J mol $^{-1}$ ) and remember that 'released energy' corresponds to a negative $\Delta H$ .

Techniques used

write the balanced equation for formation of Mg3N2calculate moles of Mg from mass and molar massuse the stoichiometric ratio to find moles of Mg3N2divide the energy released by the moles of product and assign the sign

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