Answer:
Option A,C,D
Explanation:
Solubilities of silver halides in water decrease from fluoride (AgF) to iodide (AgI). Silver fluoride is readily soluble in water hence when AgNO3 solution is added to HF solution (HF being weak acid, its solution maintain a 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 Agl when AgNO3 solution is added to their aqueous solution.
$HCl(aq)+AgNO_{3}(aq) \rightarrow AgCl(s)$(Curdy white)+ $HNO_{3}(aq)$
$HBr (aq)+AgNO_{3}(aq) \rightarrow AgBr(s)$(pale yellow)
$HI(aq)+AgNO_{3}(aq) \rightarrow AgI(s)$(yellow)+$HNO_{3}(aq)$
The solubilities decreases from AgCl to AgI, AgCl dissolves in aqueous ammonia, AgBr dissolves only slightly in concentrated ammonia while AgI does not dissolve in ammonia solution $Na_{2}S_{2}O_{3}$ solution dissolve all toi AgCl,AgBr,AgI by forming compound $[Ag(S_{2}O_{3})_{2}]^{3-}$ as $S_{2}O_{3}^{2-}$ is a strong complexing agent than ammonia
