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The Periodic Table

Group 7: the halogens — displacement reactions and the trend

  • 3 things to know
  • AS Unit 1

Overview

Recall questions on the halogens — displacement reactions and the trend.

What you need to know

Answer each one before you reveal it.

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Key term

Both sodium chloride and sodium iodide can be made by direct combination of the elements. Which forms more readily, and why?

Process

Rank the halide ions in order of reducing power and explain the order.

Key term

Why is fluorine the most powerful oxidising agent of all the elements?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Both sodium chloride and sodium iodide can be made by direct combination of the elements. Which forms more readily, and why?

    Model answer

    Sodium chloride. Chlorine is a much stronger oxidising agent than iodine because its atom is smaller and less shielded, so it takes an electron from sodium far more readily. The reaction with chlorine is fast and highly exothermic, whereas sodium and iodine react far more sluggishly.

  2. 2

    Why does the formula of a metal halide depend on the group the metal is in?

    Model answer

    The halogen atom always gains one electron to form a 1− ion, so the number of halide ions needed is set by the charge on the metal ion. A Group 1 metal forms M⁺ and gives MX, a Group 2 metal forms M²⁺ and gives MX₂, and aluminium forms Al³⁺ and gives AlX₃, so that the compound is electrically neutral overall.

  3. 3

    Define an oxidising agent and explain what makes the halogens act as oxidising agents.

    Model answer

    An oxidising agent removes electrons from another species and is itself reduced. Each halogen atom is one electron short of a full outer shell and is highly electronegative, so it readily takes an electron from another species according to X₂ + 2e⁻ → 2X-.

  4. 4

    State and explain the trend in oxidising power down Group 7.

    Model answer

    Oxidising power decreases down the group, so fluorine is the strongest oxidising agent and iodine the weakest. Down the group the atomic radius increases and the outer shell is shielded from the nucleus by more inner electrons, so the attraction of the nucleus for an incoming electron falls and the halogen becomes less able to take an electron from another species.

  5. 5

    Give the standard electrode potentials for the halogens and use them to rank oxidising power.

    Model answer

    The values for the X₂ to 2X- systems are about +2.87 V for fluorine, +1.36 V for chlorine, +1.07 V for bromine and +0.54 V for iodine. The more positive the value, the greater the tendency to be reduced and so to act as an oxidising agent, which confirms the order F₂ greater than Cl₂ greater than Br₂ greater than I₂.

  6. 6

    Rank the halide ions in order of reducing power and explain the order.

    Model answer

    Iodide is the strongest reducing agent, then bromide, then chloride, with fluoride the weakest. A reducing agent must give up an electron, and the outer electrons of the large iodide ion are far from the nucleus and heavily shielded, so they are lost most easily. The order of reducing power for the ions is the exact reverse of the order of oxidising power for the elements.

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