Multiple choiceQuestion 1 · Recall
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By the end you can describe the reactions of the halogens with hydrogen, state the conditions each one needs, and explain the trend in reactivity using H–X bond strengths.
All four halogens react with hydrogen to make a hydrogen halide:
But the effort needed changes dramatically down the group.
| halogen | conditions | how it goes |
|---|---|---|
| fluorine | none: reacts in the cold, even in the dark | explosive |
| chlorine | ultraviolet light, or about 400 °C | reacts vigorously; explosive in sunlight |
| bromine | heat with a platinum catalyst | steady reaction |
| iodine | heat | reversible, low yield |
The hydrogen halides formed are colourless, acidic gases. In moist air they give steamy fumes because they dissolve in the water droplets to form an acid.
Reactivity with hydrogen decreases down Group 17. The key is the strength of the H–X bond that is formed.
| bond | bond enthalpy / kJ mol⁻¹ |
|---|---|
| – | 562 |
| – | 431 |
| – | 366 |
| – | 299 |
The reaction forms two H–X bonds per molecule of , and forming bonds releases energy.
So the trend in reactivity follows the trend in H–X bond enthalpy.
Think of the reaction as a trade: you break an H–H bond and an X–X bond, and you get two H–X bonds in return.
Fluorine gives you the best possible deal: two of the strongest H–X bonds on offer. Iodine gives the worst deal: two of the weakest.
Where the analogy stops: reactions also have to get over an activation energy barrier, which is why chlorine needs ultraviolet light even though the overall energy change is favourable. Bond strength explains the trend; the conditions also reflect how fast each reaction gets going.