Thank you for reporting, we will resolve it shortly
Q.
Which of the following halides react(s) with $AgNO_3$ (aq) to give a precipitate that dissolves in $Na_2 S_2O_3$(aq)?
The d-and f-Block Elements
Solution:
Solubilities of silver halides in water decreases from fluoride $(AgF)$ to iodide $(AgI)$. Silver fluoride is readialy soluble in water, hence when $AgNO_3$ solution is added to $HF$ solution ($HF$ being weak acid, its solution maintain very low concentration of $F^-$) no precipitate of $AgF$ is formed.
$HCl, HBr$ and $HI$ being all strong acid, forms precipitates of $AgCl, AgBr$ and $AgI$ when $AgNO_3$ solution is added to their aqueous solution
$HCl(aq)+AgNO_3(aq) \rightarrow \underset{\text{Curdy white}}{AgCl(s)}+HNO_3(aq)$
$HBr(aq)+AgNO_3(aq) \rightarrow \underset{\text{Pale yellow}}{AgBr(s)} +HNO_3(aq)$
$HI(aq)+AgNO_3(aq) \rightarrow \underset{\text{yellow}}{AgI(s)}+HNO_3(aq)$
The solubilities decreases from $AgCl$ to $Agl, AgCl$ dissolves in aqueous ammonia, $AgBr$ dissolves only slightly in concentrated ammonia while $AgI$ does not dissolve in ammonia solution.
$Na_2 S_2O_3$ solution dissolve all three, $AgCl, AgBr, Agl$ by forming complex $[Ag(S_2 O_3)_2]^{3-}$ as $S_2O_3^{2-}$ is a stronger complexing agent than ammonia.