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QJEEM23Ch9591: JEE Main 2023 Chemistry - Class 11 JEEM23Ch9591 Solution: Match List I with List II .tg…

Match List I with List II

LIST I
Element detected
LIST II
Reagent used / Product formed
A. Nitrogen I. Na2[Fe(CN)5NO]\mathrm{Na_2[Fe(CN)_5NO]}
B. Sulphur II. AgNO3\mathrm{AgNO_3}
C. Phosphorous III. Fe4[Fe(CN)6]3\mathrm{Fe_4[Fe(CN)_6]_3}
D. Halogen IV. (NH4)2MoO4\mathrm{(NH_4)_2MoO_4}

Choose the correct answer from the options given below:

Multiple Choice Options

  • A A-III, B-I, C-IV, D-II
  • B A-II, B-I, C-IV, D-III
  • C A-IV, B-II, C-I, D-III
  • D A-II, B-IV, C-I, D-III

Step-by-Step Verified Solution

Nitrogen Detection (Lassaigne's Test)

Nitrogen in organic compounds is detected by converting it into sodium cyanide, which further reacts with iron(II) sulfate to form sodium hexacyanoferrate(II). By reacting this with iron(III) ions, Prussian blue is formed:

amp;Na+C+N→NaCNamp;6NaCN+FeSO4→Na4[Fe(CN)6]+Na2SO4amp;Na4[Fe(CN)6]+Fe3+→Fe4[Fe(CN)6]3 (Prussian blue)\begin{aligned}&\mathrm{Na}+\mathrm{C}+\mathrm{N} \rightarrow \mathrm{NaCN} \\\\&6 \mathrm{NaCN}+\mathrm{FeSO}_4 \rightarrow \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]+\mathrm{Na}_2 \mathrm{SO}_4 \\\\&\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]+\mathrm{Fe}^{3+} \rightarrow \mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \text{ (Prussian blue)}\end{aligned}

Sulphur Detection

Sulphur can be detected using Sodium nitroprusside:

Na2[Fe(CN)5NO]+Na2S→Na4[Fe(CN)5NOS] [Purple]\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]+\mathrm{Na}_2 \mathrm{S} \rightarrow \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right] \text{ [Purple]}

Phosphorous Detection

Phosphorus detection involves reaction with ammonium molybdate to form a canary yellow compound:

amp;Na3PO4+3HNO3→H3PO4+3NaNO3amp;H3PO4+12(NH4)2MoO4+21HNO3→amp;(NH4)3PO4⋅12MoO3+21NH4NO3+12H2O (canary yellow)\begin{aligned}&\mathrm{Na}_3 \mathrm{PO}_4+3 \mathrm{HNO}_3 \rightarrow \mathrm{H}_3 \mathrm{PO}_4+3 \mathrm{NaNO}_3 \\\\&\mathrm{H}_3 \mathrm{PO}_4+12\left(\mathrm{NH}_4\right)_2 \mathrm{MoO}_4+21 \mathrm{HNO}_3 \rightarrow\\\\& \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3+21 \mathrm{NH}_4 \mathrm{NO}_3+12 \mathrm{H}_2 \mathrm{O} \text{ (canary yellow)}\end{aligned}

Halogen Detection

Halogens react with silver nitrate to form specific colored precipitates:

amp;NaCl+AgNO3(aq)→AgCl+NaNO3 (White)amp;NaBr+AgNO3(aq)→AgBr+NaNO3 (Pale yellow)amp;NaI+AgNO3(aq)→AgI+NaNO3 (Yellow)\begin{aligned}&\mathrm{NaCl}+\mathrm{AgNO}_3(\mathrm{aq}) \rightarrow \mathrm{AgCl}+\mathrm{NaNO}_3 \text{ (White)} \\\\&\mathrm{NaBr}+\mathrm{AgNO}_3(\mathrm{aq}) \rightarrow \mathrm{AgBr}+\mathrm{NaNO}_3 \text{ (Pale yellow)} \\\\&\mathrm{NaI}+\mathrm{AgNO}_3(\mathrm{aq}) \rightarrow \mathrm{AgI}+\mathrm{NaNO}_3 \text{ (Yellow)}\end{aligned}

Based on the above reactions, the correct matching is:

Option A

A-III (Nitrogen), B-I (Sulphur), C-IV (Phosphorous), D-II (Halogen).

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