Question

When excess of $N{a}_2{S}_2{O}_3$ is added to dil.$AgN{O}_3$, a soluble compound ‘X’ is formed. However, when dil. $N{a}_2{S}_2{O}_3$ is added to conc. $AgN{O}_3$ solution, a white precipitate turns yellow and finally black. Precipitate ‘Y’ is formed here. Which is the correct pair?

• A. ‘X’ is $A{g}_{2}S$ and ‘Y’ is $N{a}_3\left[Ag\right(S_2{O}_3{)}_2]$
• B. ‘X’ is $N{a}_3\left[Ag\right(S_2{O}_3{)}_2]$ and ‘Y’ is $A{g}_{2}S$
• C. ‘X’ is $A{g}_2{S}_2{O}_3$ and ‘Y’ is $A{g}_{2}S$
• D. ‘X’ is $A{g}_2{S}_2{O}_3$ and ‘Y’ is $N{a}_3\left[Ag\right(S_2{O}_3{)}_2]$

Answer 1

The correct option is B ‘X’ is $N{a}_3\left[Ag\right(S_2{O}_3{)}_2]$ and ‘Y’ is $A{g}_{2}S$

In excess thiosulphate, dil.$AgN{O}_3$ forms a soluble complex.
$N{a}_2{S}_2{O}_3\left(excess\right)+dil.AgN{O}_3\to N{a}_3\left[Ag\right(S_2{O}_3{)}_2]+NaN{O}_3$
But when conc.$AgN{O}_3$ reacts with dilute thiosulphate, it gives a white precipitate of silver thiosulphate which changes to yellow and finally gives a black precipitate of $A{g}_{2}S$.
$dil.N{a}_2{S}_2{O}_3+2AgN{O}_3\to A{g}_2{S}_2{O}_3\left(whiteppt\right)+2NaN{O}_3$
$A{g}_2{S}_2{O}_3\to A{g}_{2}S+H_{2}S{O}_4$
So $X=N{a}_3\left[Ag\right(S_2{O}_3{)}_2],Y=A{g}_{2}S$