Why AgI is insoluble in NH4OH solution while AgCl and AgBr is soluble in NH4OH???

We interrogate the equilibrium...

#AgX(s) + 2NH_3(aq) rightleftharpoons[Ag(NH_3)_2]^+ + X^-#

For #X=Cl, Br# the equilibrium LIES to the RIGHT as we face the page, and the silver ion goes up into solution as the SOLUBLE ammonia complex. For #X=I#, the equilibrium lies to the left as written. Likely, the #Ag-I# bond is stronger given that there is a better size match between silver cation and iodide anion, and silver iodide precipitates from solution as a canary-yellow powder.

Note that the #NH_4OH# term, i.e. #"ammonium hydroxide"# is a bit of a misnomer...ammonium hydroxide is concentrated aqueous ammonia, i.e. #NH_3*(H_2O)_n#....the given equilibrium LIES to the left as we face the page...

#NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^+ + HO^-#

This is misleading, because they all become more soluble in #"NH"_4"OH"(aq)#.

#"AgI"# seems to be the least soluble mainly because it has the most covalent character, and the two atoms have the most similar atomic radii in context, thereby forming the strongest bond in the lattice structure.

Therefore, its #K_(sp)# is the smallest by a factor of at least #10^4# and it is harder to make it soluble enough at ordinary concentrations to call it "soluble".

MECHANISM FOR BECOMING MORE SOLUBLE

They all become more soluble in #"NH"_4"OH"(aq)# by the same mechanism:

#2("NH"_4"OH"(aq) -> cancel("NH"_3(aq)) + "H"_2"O"(l))#
#"AgX"(s) rightleftharpoons cancel("Ag"^(+)(aq)) + "X"^(-)(aq)#
#ul(cancel("Ag"^(+)(aq)) + cancel(2"NH"_3(aq)) -> "Ag"("NH"_3)_2^(+)(aq))#
#"AgX"(s) + 2"NH"_4"OH"(aq) -> "Ag"("NH"_3)_2^(+)(aq) + "X"^(-)(aq) + 2"H"_2"O"(l)#

• The first reaction step has #K_1 -= K_b^(-2) = (1/(1.8 xx 10^(-5)))^2 = 3.086 xx 10^9#, and is heavily skewed towards aqueous ammonia.
• The second reaction step has #K_2 -= K_(sp) = 1.8 xx 10^(-10)# for #"AgCl"#, #7.7 xx 10^(-13)# for #"AgBr"#, and #8 xx 10^(-17)# for #"AgI"#, and is the only step that differs based on the anion.
• The third reaction step is has #K_3 -= K_f = 1.6 xx 10^7#, the formation constant of the silver ammonia complex.

This assumes that enough ammonia is used to completely consume the anion in #"AgX"#.

NEW EQUILIBRIUM CONSTANT

The overall reaction then has the composite equilibrium constant:

#beta = K_1K_2K_3#

#= {((3.086 xx 10^9)(1.8 xx 10^(-10))(1.6 xx 10^7),"AgCl"),((3.086 xx 10^9)(7.7 xx 10^(-13))(1.6 xx 10^7),"AgBr"),((3.086 xx 10^9)(8 xx 10^(-17))(1.6 xx 10^7),"AgI"):}#

#= color(green)({(8.89 xx 10^6,"AgCl"),(3.80 xx 10^4,"AgBr"),(3.95,"AgI"):})#

Out of all three, #"AgI"# is the only one that is considered a (somewhat) weak electrolyte upon full complexation with #"NH"_3(aq)#, because #beta# is on the order of #1-10#, as opposed to #1000# or more.

#"AgI"# was much more insoluble than #"AgCl"# or #"AgBr"# in the first place, so the complexation with ammonia did not increase its solubility enough for it to be reasonably soluble at ordinary concentrations.

EXPLAINING VIA KNOWN TRENDS

Here are some trends to observe that may OR MAY NOT explain why #"AgI"# is much more insoluble than #"AgCl"# and #"AgBr"#:

• In general, the more similar the ion OR atom size (whichever is MORE appropriate for the PARTICULAR compound), the stronger the bond formed and the less soluble in water. Atomic and ionic radii are listed below.

#r_("Ag"^(+)) = "115 pm"#, #r_("Cl"^(-)) = "184 pm"#, #r_("Br"^(-)) = "196 pm"#, #r_("I"^(-)) = "220 pm"#

#r_("Ag") = "144 pm"#, #r_("Cl") = "99 pm"#, #r_("Br") = "114 pm"#, #r_("I") = "133 pm"#

The ionic radii would not explain it, because it follows the opposite trend to the one expected.
The covalent radii would explain it, because it does follow the proper trend in bond strength.

• The extent of ionic character should be examined, to determine the validity of using covalent radii or ionic radii. Electronegativities are:

#"Ag": 1.9#, #"Cl": 3.0#, #"Br": 2.8#, #"I": 2.5#

We find that #"AgI"# contains the most covalent character because the electronegativities between those atoms are the most similar.

CONCLUSIONS

Therefore, we find #"AgI"# is the least soluble to the greatest extent because:

1. #"AgI"# has the most covalent character (#|Delta"EN"| = 0.6#, as opposed to #0.9# or #1.1#).

2. The covalent and metallic radii (#"144 pm"# vs. #"133 pm"#) should be used to justify this for #"AgI"#, instead of the ionic radii (which would have suggested the opposite trend), due to its having the highest covalent character.

Likely, the atomic radius matchup is the greater reason, based on the extent .