Solveeit Logo

Question

Question: In the following reaction sequence in aqueous solution, the species X, Y and Z respectively, are: ...

In the following reaction sequence in aqueous solution, the species X, Y and Z respectively, are:
s2o32Ag+XAg+YwithtimeZ{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(S2O3)2]3,Ag2S2O3,Ag2S{[Ag{({S_2}{O_3})_2}]^{3 - }},A{g_2}{S_2}{O_3},A{g_2}S
B. [Ag(S2O3)2]5,Ag2S2O3,Ag2S{[Ag{({S_2}{O_3})_2}]^{5 - }},A{g_2}{S_2}{O_3},A{g_2}S
C. [Ag(SO3)2]3,Ag2S2O3,Ag{[Ag{(S{O_3})_2}]^{3 - }},A{g_2}{S_2}{O_3},Ag
D. [Ag(SO3)3]3,Ag2SO4,Ag{[Ag{(S{O_3})_3}]^{3 - }},A{g_2}S{O_4},Ag

Explanation

Solution

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, Ag2S2O3A{g_2}{S_2}{O_3}

Complete step by step solution:
We are aware of the reaction that silver with thiosulphate S2O32{S_2}{O_3}^{2 - } can give a white precipitate of silver thiosulphate ,Ag2S2O3A{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 Ag+A{g^ + } is added drop by drop to the thiosulphate solution. When we add drop by drop silver Ag+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(S2O3)2]3{[Ag{({S_2}{O_3})_2}]^{3 - }}. So, we got the X which is a clear solution of the formula [Ag(S2O3)2]3{[Ag{({S_2}{O_3})_2}]^{3 - }}.
2. Now we have an excess of this complex and to this, we are adding the excess Ag+A{g^ + }, since the thiosulphate is limited, we get Ag2S2O3A{g_2}{S_2}{O_3}. This will be a white precipitate. So, Y which is a white precipitate is the Ag2S2O3A{g_2}{S_2}{O_3}.
3. So, after some time, the Ag2S2O3A{g_2}{S_2}{O_3} formed will react with water H2O{H_2}O, to give Ag2SA{g_2}S
Ag2S2O3H2OAg2S(s)+H2SO4A{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 Ag2SA{g_2}S and H2SO4{H_2}S{O_4}, since the Ag2+A{g^{2 + }} is water soluble it forms Ag2SA{g_2}S. Therefore this is the black precipitate Z.

Therefore the correct option is A

Note:
The compound Ag2S2O3A{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