Topic 8 · page 24 of 29

Colours of the halogens and displacement reactions

A more reactive halogen oxidises the halide ion of a less reactive one and displaces the halogen. Learn the colour of each halogen in water and in a non-polar solvent, so you can read the result of each test.

13 min read8.25

Read first: Trends in Group 7

Colours of the halogens 8.25

The specification expects you to know the colours of the elements in three settings: as the element at standard conditions, in aqueous solution, and in a non-polar organic solvent such as hexane or cyclohexane.

Chlorine Bromine Iodine
Element, standard conditions pale green gas red-brown liquid grey-black solid
In water very pale green (almost colourless) orange (yellow when dilute) brown
In a non-polar solvent pale green orange-red violet (purple)
Three pairs of test tubes, one pair each for chlorine, bromine and iodine. In each pair the left tube shows the halogen in water as a single coloured solution, and the right tube shows two layers: the lower water layer is almost colourless and the upper non-polar solvent layer is coloured. Colours: chlorine very pale green in water and pale green in the solvent; bromine orange in water and orange-red in the solvent; iodine brown in water and violet in the solvent.
Halogens in water and in a non-polar solvent. The solvent layer floats on top of the water.

The halogens are more soluble in a non-polar solvent than in water, because they are non-polar molecules, so the colour is stronger in the solvent. The solvent is less dense than water, so it forms the upper layer. When you shake a halogen solution with a solvent, the coloured halogen moves into the upper layer.

Displacement reactions 8.25

A halogen that is a stronger oxidising agent takes electrons from the halide ion of a weaker one. The weaker halogen is displaced.

ClX2(aq)+2 BrX−(aq)→2 ClX−(aq)+BrX2(aq) \ce{Cl2(aq) + 2Br-(aq) -> 2Cl-(aq) + Br2(aq)}

Chlorine goes 0 → −1 (reduced). Bromide goes −1 → 0 (oxidised). Chlorine is the oxidising agent; bromide is the reducing agent. The half-equations are:

ClX2+2 eX−→2 ClX−2 BrX−→BrX2+2 eX− \ce{Cl2 + 2e- -> 2Cl-} \qquad \ce{2Br- -> Br2 + 2e-}
Chloride, Cl⁻(aq) Bromide, Br⁻(aq) Iodide, I⁻(aq)
Chlorine water, Cl₂(aq) no reaction orange: Br₂ formed brown: I₂ formed
Bromine water, Br₂(aq) no reaction no reaction brown: I₂ formed
Iodine, I₂(aq) no reaction no reaction no reaction

The ionic equations are:

ClX2(aq)+2 IX−(aq)→2 ClX−(aq)+IX2(aq)BrX2(aq)+2 IX−(aq)→2 BrX−(aq)+IX2(aq) \ce{Cl2(aq) + 2I-(aq) -> 2Cl-(aq) + I2(aq)} \qquad \ce{Br2(aq) + 2I-(aq) -> 2Br-(aq) + I2(aq)}

PracticalDisplacement reactions

  1. Put about 1 cm³ of each potassium halide solution (KCl, KBr, KI) into separate test tubes.
  2. Add about 1 cm³ of chlorine water to each. Record the colour of each solution.
  3. Add about 1 cm³ of cyclohexane to each, put the stopper in, shake, and let the layers separate. Record the colour of the upper layer.
  4. Repeat with bromine water in place of chlorine water, and with each halide.

The organic layer makes the result clearer. With chlorine and bromide, the upper layer is orange-red. With chlorine or bromine and iodide, the upper layer is violet. No halogen forms with a halide it cannot displace: the upper layer is the colour of the halogen that was added.

Worked examplePredicting a displacement

Predict what you would see when bromine water is added to potassium chloride, then to potassium iodide, followed by cyclohexane.

  1. KCl: Br₂ is a weaker oxidising agent than Cl₂, so it cannot take electrons from Cl⁻. No reaction. The upper layer is orange-red (the bromine).
  2. KI: Br₂ is a stronger oxidising agent than I₂, so it oxidises I⁻: BrX2+2 IX−→2 BrX−+IX2\ce{Br2 + 2I- -> 2Br- + I2}. The aqueous layer turns brown and the upper layer is violet.

Check yourself

  1. State the colours of chlorine, bromine and iodine dissolved in water.
  2. State the colour of iodine in cyclohexane, and say which layer it is in.
  3. Chlorine water is added to potassium iodide solution. State the observation and write the ionic equation.
  4. Explain, in terms of electrons, why bromine displaces iodide ions from solution but not chloride ions.
  5. Bromine water is added to potassium chloride solution, and then cyclohexane is added and shaken. Describe what you see in each tube.
  6. State the changes in oxidation number in ClX2+2 BrX−→2 ClX−+BrX2\ce{Cl2 + 2Br- -> 2Cl- + Br2} and say which species is the reducing agent.

Answers cover these until you have tried the questions

  1. Chlorine: very pale green. Bromine: orange. Iodine: brown.
  2. Violet (purple), in the upper layer, because cyclohexane is less dense than water and does not mix with it.
  3. The solution turns brown (iodine is made). ClX2(aq)+2 IX−(aq)→2 ClX−(aq)+IX2(aq)\ce{Cl2(aq) + 2I-(aq) -> 2Cl-(aq) + I2(aq)}
  4. Bromine is a stronger oxidising agent than iodine, so it can take electrons from I⁻. It is a weaker oxidising agent than chlorine, so it cannot take electrons from Cl⁻.
  5. The solution stays orange (no change): there is no reaction. With cyclohexane, the upper layer is orange-red (bromine in cyclohexane); the lower layer is almost colourless.
  6. Cl: 0 → −1 (reduced). Br: −1 → 0 (oxidised). The reducing agent is Br⁻ (it gives electrons and is oxidised).