Question

When $AgN{O}_3$ is treated with hypo solution, a white ppt of A is formed. This gets dissolved when excess of hypo solution is added due to the formation of compound B. Precipitate A on boiling with water turns black due to the formation of C. Find the incorrect option:

• A. White precipitate is due to $N{a}_3\left[Ag\right(S_2{O}_3{)}_2]$
• B. Black precipitate is due to $A{g}_{2}S$
• C. Along with $C$, $H_{2}S{O}_4$ is also formed
• D. Average oxidation state of $S$ in hypo solution is $+2$

Answer 1

The correct option is A White precipitate is due to $N{a}_3\left[Ag\right(S_2{O}_3{)}_2]$

$2AgN{O}_3+N{a}_2{S}_2{O}_3\to \underset{whiteppt\left(A\right)}{A{g}_2{S}_2{O}_3}\downarrow +2NaN{O}_3$
$A{g}_2{S}_2{O}_3+3N{a}_2{S}_2{O}_3\to \underset{B}{2N{a}_3\left[Ag\right(S_2{O}_3{)}_2]}$
$A{g}_2{S}_2{O}_3+H_{2}O\stackrel{boil}{\to }\underset{C}{A{g}_{2}S}\downarrow +S{O}_4^{2-}+2H^{+}$
So,
a) White precipitate is due to $A{g}_2{S}_2{O}_3$
b) Black precipitate is due to $A{g}_{2}S$
c) Along with $C$, $H_{2}S{O}_4$ is also formed
d) Average oxidation state of $S$ in $N{a}_2{S}_2{O}_3$ is $+2$.