The Periodic Table - Group VII properties (3)
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1.
Explain, in terms of the reactivity series of metals and non-metals, why chlorine is more reactive than iodine. Include a discussion of the relative ease with which each element can lose electrons.
The reactivity series of metals and non-metals is a scale that ranks elements in order of their tendency to lose electrons and form positive ions. Chlorine (Cl) is more reactive than iodine (I) because it is higher in the reactivity series. This indicates that chlorine has a greater tendency to lose electrons than iodine.
Losing Electrons: Reactivity in the reactivity series corresponds to the ease with which an element loses electrons to form a positive ion. A more reactive element readily loses electrons. Chlorine readily loses one electron to form a Cl+ ion. Iodine also loses one electron to form an I+ ion, but it does so with less tendency.
Reactivity Series and Electron Loss: The position of an element in the reactivity series directly reflects its ability to lose electrons. Elements higher in the series are more likely to lose electrons and form positive ions. Since chlorine is higher than iodine, it has a greater tendency to lose electrons and therefore displaces iodine from its compounds. The greater the tendency to lose electrons, the more reactive the element.
The reactivity series can be summarized as follows (simplified):
Reactivity | Chlorine (Cl) | Iodine (I) |
Reactivity Series Position | Higher | Lower |
Ease of Electron Loss | Easier | Harder |
2.
The reactivity of the halogens decreases down the group. Explain why the reactivity of fluorine is greater than that of chlorine, even though both are halogens. Describe one practical application that takes advantage of the high reactivity of fluorine.
Fluorine is the most reactive halogen due to its small atomic size and high electronegativity. Its small size allows it to approach and interact more readily with other atoms, while its high electronegativity means it has a strong tendency to attract electrons. This combination makes it much more likely to gain an electron and form a negative ion compared to chlorine, which is larger and less electronegative.
One practical application that takes advantage of fluorine's high reactivity is in the production of hydrofluoric acid (HF). HF is used as a cleaning agent for glass and also in the etching of silicon in the semiconductor industry. It reacts readily with glass (silicon dioxide) to dissolve it, due to the strong interaction between fluorine and silicon.
3.
Describe an experimental procedure to demonstrate the displacement reaction of bromine water with sodium thiosulfate. Include safety precautions and explain how you would identify the product formed.
Experimental Procedure:
- Setup: Place a small amount of bromine water in a conical flask. Add a few drops of sodium thiosulfate solution.
- Observation: Observe the reaction carefully. A cloudy solution will form initially.
- Heat: Gently warm the flask using a hot plate or by holding it in warm water.
- Reaction: As the solution warms, the cloudiness will gradually disappear.
Safety Precautions:
- Bromine Water: Bromine water is corrosive and can cause burns. Wear safety goggles and gloves when handling it. Avoid contact with skin and clothing.
- Sodium Thiosulfate: Although relatively safe, avoid inhaling the powder.
- Heating: Use caution when using a hot plate or hot water to avoid burns.
Identifying the Product:
The product formed is bromine gas (Br2). The disappearance of the cloudiness is due to the reaction of the bromine gas with the sodium thiosulfate (Na2S2O3) in the solution. The balanced equation for this reaction is:
Br2(aq) + 2Na2S2O3(aq) → 2NaBr(aq) + Na2S4O6(aq)
The disappearance of the cloudiness indicates that the bromine gas has reacted with the thiosulfate ions to form sodium bromide (NaBr) and sodium thiosulfate (Na2S4O6). The reaction is a chemical test for the presence of bromine.