Question

In the following reaction sequence in aqueous solution, the species X, Y and Z respectively, are:
\[{s_2}{o_3}^{2 - }\xrightarrow{{A{g^ + }}}X\xrightarrow{{A{g^ + }}}Y\xrightarrow{{with\,time}}Z\]
          (Clear solution) (white precipitate) (black precipitate)

A. ${[Ag{({S_2}{O_3})_2}]^{3 - }},A{g_2}{S_2}{O_3},A{g_2}S$
B. ${[Ag{({S_2}{O_3})_2}]^{5 - }},A{g_2}{S_2}{O_3},A{g_2}S$
C. ${[Ag{(S{O_3})_2}]^{3 - }},A{g_2}{S_2}{O_3},Ag$
D. ${[Ag{(S{O_3})_3}]^{3 - }},A{g_2}S{O_4},Ag$

Answer 1

Hint:We know that thiosulphate ions when added to silver produce a clear solution. Also, we know that silver with thiosulphate produces a white precipitate of silver thiosulphate, $A{g_2}{S_2}{O_3}$

Complete step by step solution:
We are aware of the reaction that silver with thiosulphate ${S_2}{O_3}^{2 - }$ can give a white precipitate of silver thiosulphate ,$A{g_2}{S_2}{O_3}$. And thus, the white precipitate can possibly dissolve in an excess of thiosulphate.
1. While performing the experiments, we notice that the silver $A{g^ + }$ is added drop by drop to the thiosulphate solution. When we add drop by drop silver $A{g^ + }$ we get a complex. Since the coordination number of silver is 2, so we have two thiosulphate ions, and the charge of silver and thiosulphate add up to this complex to charge -3, thus the complex formed will be ${[Ag{({S_2}{O_3})_2}]^{3 - }}$. So, we got the X which is a clear solution of the formula ${[Ag{({S_2}{O_3})_2}]^{3 - }}$.
2. Now we have an excess of this complex and to this, we are adding the excess $A{g^ + }$, since the thiosulphate is limited, we get $A{g_2}{S_2}{O_3}$. This will be a white precipitate. So, Y which is a white precipitate is the $A{g_2}{S_2}{O_3}$.
3. So, after some time, the $A{g_2}{S_2}{O_3}$ formed will react with water ${H_2}O$, to give $A{g_2}S$
$A{g_2}{S_2}{O_3}\xrightarrow{{{H_2}O}}A{g_2}{S_{(s)}} + {H_2}S{O_4}$
This is a disproportionation reaction of thiosulphate to produce $A{g_2}S$ and ${H_2}S{O_4}$, since the $A{g^{2 + }}$ is water soluble it forms $A{g_2}S$. Therefore this is the black precipitate Z.

Therefore the correct option is A

Note:
The compound $A{g_2}{S_2}{O_3}$ will change its colour three or four times with time to finally give a black precipitate. Silver thiosulphate when it is fresh it is white. Slowly within seconds we can see there is a change to yellow, it then changes to orange, then brown and finally black