Question

$As{}_{2}{{S}_{3}}$ is soluble in ${{(N{{H}_{4}})}_{2}}{{S}_{2}}$ (yellow ammonium sulphide) due to the formation of${{(N{{H}_{4}})}_{3}}As{{S}_{4}}$.
If the given statement is true enter 1else enter 0.

Answer 1

Generally yellow ammonium sulphide is used as a laboratory reagent in qualitative analysis to precipitate metal sulphides. Yellow ammonium sulphide is a mixture of ammonium polysulphides and it can be prepared by dissolving sulphur in ammonia solution.

Complete step by step answer:
-The chemical Arsenic sulphide reacts with the ammonium sulphide and forms a thiosalt as the product.
-The above statement can be represented as a chemical reaction as below.
$\underset{\text{Arsenic sulphide}}{\mathop{As{}_{2}{{S}_{3}}}}\,+\underset{\text{ammonium sulphide}}{\mathop{3{{(N{{H}_{4}})}_{2}}{{S}_{2}}}}\,\to \underset{\text{Thiosalt}}{\mathop{2{{(N{{H}_{4}})}_{3}}As{{S}_{4}}}}\,$
-In the above reaction one mole of Arsenic sulphide reacts with three moles of ammonium sulphide and forms two moles of thiosulfate as the product.
-Arsenic sulphide is soluble in ammonium sulphide then only thiosalt is going to form.
-Therefore the given statement is correct then we have to enter 1.
Additional information:
-Commonly sulphides of Group IIB elements react with ammonium sulphide and form their respective thiosalts as the product. The formed thiosalt reacts with hydrochloric acid and forms colored precipitate.

& \underset{\text{Arsenic sulphide}}{\mathop{As{}_{2}{{S}_{3}}}}\,+\underset{\text{ammonium sulphide}}{\mathop{3{{(N{{H}_{4}})}_{2}}{{S}_{2}}}}\,\to \underset{\text{Thiosalt}}{\mathop{2{{(N{{H}_{4}})}_{3}}As{{S}_{4}}}}\, \\\ & {{(N{{H}_{4}})}_{3}}As{{S}_{4}}+6HCl\to A{{s}_{2}}{{S}_{6}}\downarrow +6N{{H}_{4}}Cl+3{{H}_{2}}S \\\ \end{aligned}$$ **Note:** The oxidation number of arsenic in arsenic sulphide is +3. While the oxidation number of arsenic in its thiosalt is +4. Means arsenic is going to lose one electron, and increases its oxidation number from +3 to +4, and then this reaction is called oxidation. Oxidation means loss of electron, if an electron is coming out from an element then the oxidation number of the particular element increases. In case of reduction it is vice versa.