Question

Which of the following compounds is formed when Na2SO3 is boiled with sulphur?
A) \[\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\]
B)\[\text{N}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{5}}}\]
C) \[\text{N}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{6}}}\]
D) \[\text{N}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{3}}\]

Answer 1

Hint: The sodium sulphate $\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{3}}}$in presence of sulphur undergoes a disproportionate reaction. It is an oxidation-reduction reaction in which the solid sulphur undergoes the oxidation and the sulphur in the sodium sulphate undergoes the reduction resulting in the formation of thiosulphate.

Complete step by step answer:
The sulphites are converted into the thiosulphates in presence of sulphur. the reaction is as follows:
\[\begin{align}
  & \text{SO}_{\text{3}}^{\text{2-}}\text{ + S }\to \text{ }{{\text{S}}_{\text{2}}}\text{O}_{\text{3}}^{\text{2-}} \\
 & \text{(sulphite) (sulphur) (thiosulphate)} \\
\end{align}\]
This is an oxidation-reduction reaction. Sulphur S acts as the reducing agent and itself undergoes the oxidation. The sodium sulphite $\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{3}}}$ acts as the oxidizing agent which itself undergoes the reduction.
The reaction of the oxidation and reduction of the sulphur and sodium sulphite as follows:
\[\text{N}{{\text{a}}_{\text{2}}}{{\text{S}}^{\text{IV}}}{{\text{O}}_{\text{3(aq)}}}\text{+}{{\text{S}}^{\text{0}}}_{\text{(S)}}\to \text{N}{{\text{a}}_{\text{2}}}\text{S}_{\text{2}}^{\text{II}}{{\text{O}}_{\text{3}}}\]
In this reaction the oxidation state of sulphur changes from $\text{0}\to \text{(II)}$ in solid sulphur. The oxidation state of sulphur in the sulphite changes from $\text{(IV)}$ to $\text{(II)}$ .
\[\begin{align}
  & {{\text{S}}^{\text{0}}}\text{-2}{{\text{e}}^{\text{-}}}\to {{\text{S}}^{\text{II}}}\text{ (Oxidaton reaction)} \\
 & {{\text{S}}^{\text{IV}}}+\text{2}{{\text{e}}^{\text{-}}}\to {{\text{S}}^{\text{II}}}\text{ (Reduction reaction)} \\
\end{align}\]
Thus, when sodium sulphite $\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{3}}}$is boiled with the sulphur it produces sodium thiosulphate.

Hence, (D) is the correct option.

Additional information:
A)$\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{3}}}$:
\[\text{N}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}\] Is an important industrial chemical, and world production exceeds one million tonnes per year. It is made by passing the SO2 into an aqueous solution \[\text{N}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}\] to give aqueous NaHSO3, then treating the solution with more \[\text{N}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}\]
The main use of Na2SO3 is as a bleach for wood pulp in the paper industry. It is also used to treat the boiler feed water. Small amount is used in the photographic developer.
This sulphite liberates the SO2 on treatment with dilute acid.
B)\[\text{N}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}\]:
Thiosulphate can be synthesized by oxidizing the polysulphides with air.
$\text{2N}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{3}}}\text{+3}{{\text{O}}_{\text{2}}}\xrightarrow{\text{Heat in air }}\text{2N}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}\text{+2S}$
In analytical chemistry, the salt of thiosulphate or the thiosulfate anion in the solution reacts with the iodine stoichiometrically. In the reaction of thiosulphate with the iodine causes the oxidation of thiosulphate to the tetrathionate. The reaction is widely used as the titrant in iodometry for the estimation of the metal ion in the solution.
Iodine very readily and rapidly oxidizes the thiosulphate ions \[{{\text{S}}_{\text{2}}}\text{O}_{\text{3}}^{\text{2-}}\] to form tetrathionate ions ${{\text{S}}_{\text{4}}}\text{O}_{\text{6}}^{\text{2-}}$ and the \[{{\text{I}}_{\text{2}}}\] is reduced to ${{\text{I}}^{\text{-}}}$ ions.
\[\begin{align}
  & \text{2N}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}\text{ + }{{\text{I}}_{\text{2}}}\to \text{ N}{{\text{a}}_{\text{2}}}{{\text{S}}_{4}}{{\text{O}}_{6}}\text{ }+\text{ 2NaI} \\
 & \text{ (Sodium tetrathionate)} \\
\end{align}\]

Note: This reaction of the conversion of the oxidation state of product into the same oxidation state is called as the disproportionation reaction. Here both the sulphur in the product are in the same oxidation state of$\text{II or +2}$.