Reactivity Trends and Displacement Reactions

As you go down Group 7, the relative molecular mass increases. This leads to stronger intermolecular forces, meaning higher melting and boiling points. At AQA standard, you must know their physical states and appearances at room temperature:

• Fluorine: Pale yellow gas.
• Chlorine: Pale green gas (forms a pale green aqueous solution in water).
• Bromine: Dark red-brown liquid (forms an orange or yellow-orange aqueous solution).
• Iodine: Dark grey/grey-black solid (forms a brown aqueous solution, but sublimates into a purple vapor).

If you need to test for chlorine gas, place damp blue litmus paper in the gas. The paper will turn red because the gas is acidic, and it will then bleach white.

Explaining the Trend in Reactivity

• Why does fluorine react aggressively with almost everything, while iodine barely reacts at all?
• In Group 7, reactivity decreases as you move down the group from fluorine to astatine.
• To achieve a full outer shell and become stable, halogens need to gain one electron to form a negative halide ion with a -1 charge (e.g., $Cl^-$).

To secure full marks when explaining this trend, you must describe the causal mechanism in step-by-step order. As you go down the group, the atoms have more electron shells, which increases the atomic radius (the distance from the nucleus to the outer shell). These extra inner shells also increase electron shielding.

The combination of increased distance and increased shielding results in a weaker nuclear attraction between the positive nucleus and an incoming electron. Because this electrostatic attraction is weaker, it becomes harder for the atom to gain the one electron it needs, making the element less reactive. Even though astatine is radioactive and rarely tested in school labs, we can use this trend to predict it is the least reactive element in the group.

Halogen Displacement Reactions

• In chemistry, an aggressive element can "steal" a place in a compound, forcing a weaker element out.
• This process is called a displacement reaction, where a more reactive halogen displaces a less reactive halogen from an aqueous solution of its metal halide salt.
• Metal halide salts (like potassium bromide, $KBr$) do NOT have a colour in aqueous solution; they are perfectly colourless.

To perform this reaction, follow these steps: First, add an aqueous solution of a halogen to a colourless metal halide solution. Then, observe the mixture. If the added halogen is higher up Group 7 (more reactive), it displaces the less reactive halide ion from the salt. Finally, the newly formed halogen molecule causes the solution to change colour. If the added halogen is lower in the group (less reactive), there is no reaction and no colour change.

Typical AQA displacement observations include:

• Chlorine water + KBr: Solution turns orange (bromine is formed).
• Chlorine water + KI: Solution turns brown (iodine is formed).
• Bromine water + KI: Solution turns brown (iodine is formed).

For Higher Tier students, an organic solvent like cyclohexane can be added to clearly distinguish the products. Bromine in cyclohexane appears orange/red-brown, whereas iodine in cyclohexane appears purple/violet/lilac.

Redox and Ionic Half-Equations (Higher Tier)

• Tracking the movement of electrons reveals the true mechanism behind displacement reactions.
• Displacement reactions are redox reactions because both oxidation and reduction happen simultaneously.
• The halogen molecule acts as an oxidising agent because it gains electrons (it is reduced), while the halide ion acts as a reducing agent because it loses electrons (it is oxidised).

During these reactions, the metal ion (like potassium, $K^+$) acts as a spectator ion. This means it does NOT change its state or its electronic configuration during the reaction, so we can ignore it when writing ionic equations.

Worked Example: Chlorine Displacing Bromine

Step 1: Write the full balanced symbol equation. Remember that halogens are diatomic.

$$Cl_2(aq) + 2KBr(aq) \rightarrow 2KCl(aq) + Br_2(aq)$$

Step 2: Write the ionic equation. Remove the potassium spectator ions.

$$Cl_2(aq) + 2Br^-(aq) \rightarrow 2Cl^-(aq) + Br_2(aq)$$

Step 3: Split into oxidation and reduction half-equations.

• Reduction half-equation: $Cl_2(aq) + 2e^- \rightarrow 2Cl^-(aq)$ (Chlorine gains electrons).
• Oxidation half-equation: $2Br^-(aq) \rightarrow Br_2(aq) + 2e^-$ (Bromide ions lose electrons).