Detection Of Halogens
The sodium fusion extract of an organic compound is acidified with dilute nitric acid and treated with silver nitrate, giving a pale yellow precipitate insoluble in ammonia solution. Which halogen is present?
Select the correct option:
Solution
Iodine
Halogens in organic compounds are converted by sodium fusion into sodium halides (NaX). The extract is first boiled with dilute nitric acid to decompose any NaCN or Na_2S (which would otherwise precipitate as silver cyanide or sulphide), and then silver nitrate is added. The colour and ammonia-solubility of the silver halide identify the halogen: AgCl is white and fully soluble in ammonia; AgBr is pale yellow and only sparingly soluble in ammonia; AgI is deep yellow and completely insoluble in ammonia. A pale yellow precipitate insoluble in ammonia is therefore silver iodide, confirming iodine. Option Chlorine is wrong because AgCl is white and dissolves readily in ammonia. Option Fluorine is incorrect because silver fluoride is soluble and gives no precipitate. The bromine option is wrong because AgBr is partially soluble in ammonia and is cream rather than fully insoluble yellow. This solubility-difference logic, together with the prior removal of cyanide and sulphide, is the NCERT basis for distinguishing the halides reliably. Plausibility check: ammonia insolubility increases steadily from AgCl to AgI as the silver halide solubility product decreases down the group, so the most insoluble, deepest-coloured precipitate corresponds to iodide, consistent with the observed pale yellow ammonia-insoluble solid.
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About This Question
- Subject
- chemistry
- Chapter
- principles related to practical chemistry
- Topic
- detection of halogens
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
Iodine
Halogens in organic compounds are converted by sodium fusion into sodium halides (NaX). The extract is first boiled with dilute nitric acid to decompose any NaCN or Na_2S (which would otherwise precipitate as silver cyanide or sulphide), and then silver nitrate is added. The colour and ammonia-solubility of the silver halide identify the halogen: AgCl is white and fully soluble in ammonia; AgBr is pale yellow and only sparingly soluble in ammonia; AgI is deep yellow and completely insoluble in ammonia. A pale yellow precipitate insoluble in ammonia is therefore silver iodide, confirming iodine. Option Chlorine is wrong because AgCl is white and dissolves readily in ammonia. Option Fluorine is incorrect because silver fluoride is soluble and gives no precipitate. The bromine option is wrong because AgBr is partially soluble in ammonia and is cream rather than fully insoluble yellow. This solubility-difference logic, together with the prior removal of cyanide and sulphide, is the NCERT basis for distinguishing the halides reliably. Plausibility check: ammonia insolubility increases steadily from AgCl to AgI as the silver halide solubility product decreases down the group, so the most insoluble, deepest-coloured precipitate corresponds to iodide, consistent with the observed pale yellow ammonia-insoluble solid.
This medium difficulty chemistry question is from the chapter principles related to practical chemistry, covering the topic of detection of halogens. It appeared in the 2025 exam.
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