Question

Addition of Sodium thiosulphate solution to a solution of a silver nitrate given ‘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. $A{g_2}{S_2}{O_3},N{a_3}\left[ {Ag{{({S_2}{O_3})}_2}} \right],A{g_2}S$
B. $A{g_2}S{O_4},Na\left[ {Ag{{({S_2}{O_3})}_2}} \right],A{g_2}{S_2}$
C. $A{g_2}{S_2}{O_3},N{a_5}\left[ {Ag{{({S_2}{O_3})}_3}} \right],AgS$
D. $A{g_2}S{O_3},N{a_3}\left[ {Ag{{({S_2}{O_3})}_2}} \right],A{g_2}O$

Answer 1

Hint: Precipitation reaction is a reaction that yields an insoluble product when two solutions are mixed. We described a precipitation reaction in which a colorless solution of silver nitrate was mixed with a yellow orange solution. The most important step in analyzing an unknown reaction is to write down all the species whether molecules or dissociate ions that are actually present in the solution.

Complete step by step answer:
Here we see the reaction of silver nitrate which given compound contains white precipitate and that product reacts with excess thiosulphate and gives us ‘Y’ compound. The boiling of that compound gives us the ‘Z’ compound.
The reaction for ‘X’, ‘Y’ and ‘Z’ respectively are given below,
 $N{a_2}{S_2}{O_3}\xrightarrow{{2AgN{O_3}}}A{g_2}{S_2}{O_3}$
                                                    X
 \[A{g_2}{S_2}{O_3}\xrightarrow{{N{a_2}{S_2}{O_3}}}N{a_3}\left[ {Ag{{({S_2}{O_3})}_2}} \right]\]
                                                       Y
 \[N{a_3}\left[ {Ag{{({S_2}{O_3})}_2}} \right]\xrightarrow{{water}}A{g_2}S\]
                                                       Z
Here we get the products of $A{g_2}{S_2}{O_3},N{a_3}\left[ {Ag{{({S_2}{O_3})}_2}} \right],A{g_2}S$.

So, option (A) is correct answer.

Note: Reaction with sodium thiosulphate solution to a solution of silver nitrate given white precipitate, insoluble in water but soluble in excess thiosulphate solution to give another product. Silver formed by solution physical development tends not to form filaments, so the original images tended to be brownish rather than black this is because the small, fairly compact Ag particles scattered blue light preferentially. The various processes occurring during development must be properly balanced for the system work.