The elements of Group 17 are called halogens.
- (a)Complete Table 1.1. Table 1.1 halogen | colour at 293 K chlorine | bromine | iodine | [1]
- (b)State the trend in volatility of the halogens chlorine, bromine and iodine. Explain your answer.[3]
- (c)(i)Iodine is made by reacting bromine with sodium iodide. Construct an ionic equation for the reaction of bromine with sodium iodide.[1]
- (c)(ii)State the role of bromine in the reaction. Explain your answer.[1]
- (d)Concentrated sulfuric acid is added to separate samples containing equal amounts of NaCl, NaBr and NaI. All three samples initially react to produce the hydrogen halide.
- (d)(i)Write an equation to describe the acid-base reaction that occurs when concentrated sulfuric acid reacts with NaBr.[1]
- (d)(ii)Deduce which sodium halide, NaCl, NaBr or NaI, produces the largest percentage yield of hydrogen halide when concentrated sulfuric acid is added. Explain your answer by considering the relative reactivity of the halide ions as reducing agents. identity of sodium halide explanation[3]
Mark scheme
Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.
- [1] All three correct for one mark
- chlorine green
- bromine brown
- iodine (dark) grey
- [3] M1 (it/volatility) decreases
- M2 more electrons in molecules
- M3 (so) increased strength of instantaneous dipole – induced dipole forces (of attraction)
- [1] Br2 + 2I → 2Br¯ + I2
- [1] (Br2/bromine is an) oxidising agent
- AND
- removes electron(s) from iodide (ions) I- OR increases the oxidation number of I-/ iodide (ion)
- [1] NaBr + H2SO4 → HBr + NaHSO4
- OR
- 2NaBr + H2SO4 → 2HBr + Na2SO4
- [3] M1 sodium chloride
- M2 Cl- is not a strong enough reducing agent to reduce H2SO4
- M3 Br- & I- are strong enough reducing agents to react (further) with (the S in) H2SO4
- Alternative M2 and M3
- M2 ability to behave as reducing agents increases down the group
- M3 EITHER Cl- is not strong enough to reduce H2SO4 OR Br- and I react (further) with (the S in) H2SO4
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