Question

$Na_2S_2O_3(solution) \xrightarrow{'X'/H^+} coloured \ solution$ Where 'X' is/are -

• A. $CuSO_4(aq.)$
• B. $FeCl_3(aq.)$
• C. $Cl_2$ water
• D. $Cr_2O_7^{-2}$

Answer 1

Answer: (B), (D)

B, D

Answer 2

The reaction is between $Na_2S_2O_3$ solution and reagent 'X' in the presence of acid ($H^+$), producing a coloured solution. We need to identify which of the given options for 'X' results in a coloured solution under these conditions.

Let's examine each option:

(A) $CuSO_4(aq.)$: Contains $Cu^{2+}$ ions (blue in solution). Thiosulphate ($S_2O_3^{2-}$) reacts with $Cu^{2+}$. In acidic conditions, $S_2O_3^{2-}$ decomposes to sulfur (yellow precipitate) and $SO_2$. Also, $Cu^{2+}$ can be reduced by $S_2O_3^{2-}$ to $Cu^{+}$, which often forms insoluble compounds like $Cu_2S_2O_3$, which then decomposes to black $Cu_2S$ and yellow $S$. While transient coloured complexes might form, the predominant outcome in acidic conditions involves the formation of precipitates ($S$, $Cu_2S$), resulting in a turbid mixture rather than a clear coloured solution.

(B) $FeCl_3(aq.)$: Contains $Fe^{3+}$ ions (typically yellow-brown in solution). Thiosulphate ($S_2O_3^{2-}$) reduces $Fe^{3+}$ to $Fe^{2+}$. The reaction is $2Fe^{3+} + 2S_2O_3^{2-} \rightarrow 2Fe^{2+} + S_4O_6^{2-}$. $Fe^{3+}$ is yellow/brown, and $Fe^{2+}$ is pale green in aqueous solution. Both are coloured ions in solution. There is also a known reaction where $Fe^{3+}$ reacts with $S_2O_3^{2-}$ to form a transient deep violet complex, $[Fe(S_2O_3)]^{+}$. This violet complex is a coloured species in solution, although it is unstable and quickly decomposes as $Fe^{3+}$ is reduced to $Fe^{2+}$. In acidic conditions, acid decomposition of $S_2O_3^{2-}$ to sulfur may also occur, leading to turbidity. However, the formation of the transient violet complex and the final pale green $Fe^{2+}$ solution represent the presence of coloured species in solution.

(C) $Cl_2$ water: Contains $Cl_2$ (pale greenish-yellow). $Cl_2$ is a strong oxidizing agent and oxidizes $S_2O_3^{2-}$ to $SO_4^{2-}$ in acidic conditions: $S_2O_3^{2-} + 4Cl_2 + 5H_2O \rightarrow 2SO_4^{2-} + 8Cl^- + 10H^+$. The reactants ($S_2O_3^{2-}$, $Cl_2$ which is consumed, $H_2O$, $H^+$) and the products ($SO_4^{2-}$, $Cl^-$, $H^+$) are all colourless species in solution. The initial colour of $Cl_2$ water disappears as it reacts. This reaction does not produce a coloured solution.

(D) $Cr_2O_7^{-2}$: Contains dichromate ions ($Cr_2O_7^{2-}$, orange). Dichromate is a strong oxidizing agent and oxidizes $S_2O_3^{2-}$ in acidic conditions. The primary reaction is $Cr_2O_7^{2-} + 3S_2O_3^{2-} + 8H^+ \rightarrow 2Cr^{3+} + 3S_4O_6^{2-} + 4H_2O$. Dichromate ($Cr_2O_7^{2-}$, orange) is reduced to chromium(III) ions ($Cr^{3+}$, green). Tetrathionate ($S_4O_6^{2-}$) is colourless. Thus, the reaction results in a solution containing green $Cr^{3+}$ ions, which is a coloured solution. If dichromate is in excess or acidity is high, oxidation might proceed to sulfate, $SO_4^{2-}$, which is also colourless, still resulting in green $Cr^{3+}$. This reaction forms a stable coloured solution (green).

Comparing options (B) and (D), both produce coloured solutions. (B) gives a transient violet colour followed by a pale green solution (with potential sulfur precipitate). (D) gives a stable green solution. Both fit the description "coloured solution". Therefore, both (B) and (D) are correct options.

Final Answer Check:

• $Na_2S_2O_3 + CuSO_4/H^+$: Forms precipitates ($S$, $Cu_2S$), not a clear coloured solution.
• $Na_2S_2O_3 + FeCl_3/H^+$: Forms transient violet complex, then pale green $Fe^{2+}$ solution (with potential $S$ precipitate). Coloured solution formed.
• $Na_2S_2O_3 + Cl_2/H^+$: Reactants and products are colourless (except initial $Cl_2$ colour which is consumed). No coloured solution formed.
• $Na_2S_2O_3 + Cr_2O_7^{2-}/H^+$: Forms green $Cr^{3+}$ solution. Coloured solution formed.

Both (B) and (D) yield coloured solutions. The question asks for 'X' is/are, implying multiple correct options are possible.