question_answer

Addition of sodium thiosulphate solution to a solution of silver nitrate gives X as white precipitate, insoluble in water but soluble in excess thiosulphate solution to give Y.. On boiling in water, Y gives Z. X, Y and Z respectively are

A) \[A{{g}_{2}}{{S}_{2}}{{O}_{3}},N{{a}_{3}}[Ag{{({{S}_{2}}{{O}_{3}})}_{2}}],A{{g}_{2}}S\]

B)  \[A{{g}_{2}}S{{O}_{4}},Na[Ag{{({{S}_{2}}{{O}_{3}})}_{2}}],A{{g}_{2}}{{S}_{2}}\]

C)  \[A{{g}_{2}}{{S}_{2}}{{O}_{3}},N{{a}_{5}}[Ag{{({{S}_{2}}{{O}_{3}})}_{3}}],AgS\]

D)  \[A{{g}_{2}}S{{O}_{3}},N{{a}_{3}}[Ag{{({{S}_{2}}{{O}_{3}})}_{2}}],A{{g}_{2}}O\]

Correct Answer: A

Solution :

 The sequence of reactions are \[\underset{\begin{smallmatrix}  Silver \\  nitrate \end{smallmatrix}}{\mathop{2AgN{{O}_{3}}}}\,+\underset{\begin{smallmatrix}  Sodium \\  thiosulphate \end{smallmatrix}}{\mathop{N{{a}_{2}}{{S}_{2}}{{O}_{3}}}}\,\xrightarrow{{}}\underset{\begin{smallmatrix}  (x) \\  (White\,ppt.) \end{smallmatrix}}{\mathop{A{{g}_{2}}{{S}_{2}}{{O}_{3}}}}\,+2NaN{{O}_{3}}\]   Hence, \[X=A{{g}_{2}}{{S}_{2}}{{O}_{3}}\] \[Y=N{{a}_{3}}[Ag{{({{S}_{2}}{{O}_{3}})}_{2}}]Z=A{{g}_{2}}S\]