Question

The decreasing order of solubility of silver halides is:

• A. $AgI>AgBr>AgCl>AgF$
• B. $AgF>AgCl>AgBr>AgI$
• C. $AgCl>AgF>AgBr>AgI$
• D. $AgBr>AgF>AgI>AgCl$

Answer 1

The correct option is B $AgF>AgCl>AgBr>AgI$

$AgI$ has the maximum covalent character (minimum ionic character) among the given compounds because of the large size of $I^{-}$. Due to the larger size of $I^{-}$ there is more polarisation in $AgI$ whilst, $AgF$ has the least covalent character due to smallest size of the flouride ion.
If nothing regarding the solvent is mentioned, by default, the solvent is considered to be water, which is highly polar. So, the more ionic compound will be more soluble.