Question

Precipitate 'A' from group II analysis is dissolved in ammonium sulphide solution to give a yellow solution 'B'. Dil. HCl is added to solution 'B' to get a yellow precipitate 'C'. 'C' dissolves in concentrated $HN{O}_3$ to form a acidic solution which gives a canary yellow precipitate on heating with ammonium molybdate. Precipitate 'C' is

• A. $A{s}_2{S}_3$
• B. $A{s}_2{S}_5$
• C. $\left(N{H}_4{)}_3\right[As\left(M{o}_3{O}_{10}{)}_4\right]$
• D. None of the above

Answer 1

The correct option is B $A{s}_2{S}_5$

If group- II precipitate dissolves in yellow ammonium sulphide and the colour of the solution is yellow, this indicates the presence of $A{s}^{3+}$ ions. Ammonium thioarsenide (B) formed on dissolution of $A{s}_2{S}_3$(A) decomposes with dil. HCl, and a yellow precipitate of arsenic (V) sulphide (C) is formed which dissolves in concentrated nitric acid on heating due to the formation of arsenic acid. On adding ammonium
molybdate solution to the reaction mixture and heating, a canary yellow precipitate (D) is formed. This confirms the presence of $A{s}^{3+}$ ions.

$\underset{A}{A{s}_2{S}_3}+3(N{H}_4{)}_2{S}_2\to \underset{B}{2(N{H}_4{)}_{3}As{S}_4}+S$
$2(N{H}_4{)}_{3}As{S}_4+6HCl\to \underset{C}{A{s}_2{S}_5}+3H_{2}S+6N{H}_{4}Cl$
$A{s}_2{S}_5+10HN{O}_3+4H_{2}O\to 6H_{3}As{O}_4+10NO+5S$
$H_{3}As{O}_4+(N{H}_4{)}_{2}Mo{O}_4+21HN{O}_3\to \underset{D}{\left(NH4\right)\left[As\right(M{o}_3{O}_{10}{)}_4]}+21N{H}_{4}N{O}_3+12H_{2}O$