Home / AS & A Level Chemistry 11.3 Some reactions of the halide ions Exam Style Practice Questions Paper 1
X and Y are sodium salts of Group 17 elements.
When X reacts with concentrated sulfuric acid, hydrogen sulfide, \(\mathrm{H_2S}\), is produced.
When Y reacts with concentrated sulfuric acid, there is no change in the oxidation number of the sulfur.
Which statement is correct?
Correct Answer: \( \boxed{\mathrm{A}} \)
X is sodium iodide because iodide ions reduce concentrated sulfuric acid to hydrogen sulfide.
Y is sodium chloride because chloride reacts with concentrated sulfuric acid in an acid-base reaction only, with no change in sulfur oxidation number.
Iodide ions are strong reducing agents and reduce aqueous bromine to bromide ions.
Therefore, the correct answer is (A).
Mixing aqueous silver nitrate and aqueous sodium chloride produces a precipitate.
Addition of which reagent to the mixture gives a colourless solution?
Correct Answer: \( \boxed{\mathrm{A}} \)
Mixing aqueous silver nitrate and sodium chloride forms a white precipitate of silver chloride:
\( \mathrm{AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)} \)
Silver chloride dissolves in aqueous ammonia because it forms the soluble diamminesilver(I) complex:
\( \mathrm{AgCl(s) + 2NH_3(aq) \rightleftharpoons [Ag(NH_3)_2]^+(aq) + Cl^-(aq)} \)
The other reagents do not dissolve the silver chloride precipitate.
Therefore, the correct answer is (A).
Sodium bromide is warmed with concentrated sulfuric acid.
Which row describes the change in the oxidation number of the sulfur and the role of the bromide ions in the reaction?
| Change in oxidation number of sulfur | Role of bromide ions | |
|---|---|---|
| A | \(+6\) to \(+4\) | Oxidising agent |
| B | \(+6\) to \(+4\) | Reducing agent |
| C | \(+4\) to \(0\) | Oxidising agent |
| D | \(+4\) to \(0\) | Reducing agent |
Correct Answer: \( \boxed{\mathrm{B}} \)
Concentrated sulfuric acid acts as an oxidising agent and oxidises bromide ions to bromine.
The sulfur in sulfuric acid is reduced from oxidation number \(+6\) to \(+4\), forming sulfur dioxide.
Since the bromide ions lose electrons, they act as the reducing agent.
Therefore, the correct answer is (B).