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Salt Analysis question

2025 · 2 Apr · Shift 2 · Q13
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Salt Analysis question

2025 · 2 Apr · Shift 2 · Q13

JEE MainChemistrySalt AnalysisMCQ+4 / −1
Formation of Na4[Fe(CN)5NOS]\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]Na4​[Fe(CN)5​NOS], a purple coloured complex formed by addition of sodium nitroprusside in sodium carbonate extract of salt indicates the presence of :
  1. A
    Sulphide ion
  2. B
    Sulphite ion
  3. C
    Sulphate ion
  4. D
    Sodium ion
View written solutionFree

Correct answer: A

  1. Identify the test mentioned

    The question refers to the formation of a purple coloured complex on adding sodium nitroprusside to the sodium carbonate extract of a salt.

    The complex given is: Na4[Fe(CN)5NOS]\mathrm{Na}_4[\mathrm{Fe}(\mathrm{CN})_5\mathrm{NOS}]Na4​[Fe(CN)5​NOS]

  2. Recall the salt analysis test

    In qualitative inorganic analysis, sodium nitroprusside is a confirmatory reagent for sulphide ion (S2−)\left(\mathrm{S}^{2-}\right)(S2−).

    When sulphide ion is present, it reacts with sodium nitroprusside to give a violet/purple complex.

  3. Match with the options

    • A: Sulphide ion →\rightarrow→ gives purple colour with sodium nitroprusside ✅
    • B: Sulphite ion →\rightarrow→ does not give this characteristic purple nitroprusside complex ❌
    • C: Sulphate ion →\rightarrow→ does not respond in this way ❌
    • D: Sodium ion →\rightarrow→ unrelated to this test ❌
  4. Conclusion

    Therefore, the presence of the purple complex Na4[Fe(CN)5NOS]\mathrm{Na}_4[\mathrm{Fe}(\mathrm{CN})_5\mathrm{NOS}]Na4​[Fe(CN)5​NOS] indicates the presence of: Sulphide ion\boxed{\text{Sulphide ion}}Sulphide ion​

  5. Comparison with stored answer

    Stored correct answer: A

    Derived answer: A

    Hence, they agree.

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