[gertrudetrumpet]

Hi, I was wondering why silver only forms ag(nh3)2+ instead of tetrahedral or octahedral. Correct me if I am wrong, but I don't think that steric has anything to do with why. Couldn't the 5d orbitals, which are empty, also be used in the coordinate covalent bonds?

[dipesh747]

It doesn't only co-ordinate by 2. e.g. http://pubs.acs.org/doi/abs/10.1021/ic00219a015

[gertrudetrumpet]

Ah I see, but uh so why is the coordination number of 2 so stable in silver? Is the stability of the complex influenced by the charge on the ion? Like copper ammonium complex stops at 4 since the equilibrium constant becomes negative after that right?

[dipesh747]

Is the stability of the complex influenced by the charge on the ion?

If you are not sure about the answer to that question then you really have no hope of understanding. Read things around ligand and crystal field theory and work from there. There is a lot of reading for you to do to understand the answer to your question, I spent a whole module a university learning about D-block chemistry. (and much more after that lol)

[gertrudetrumpet]

Sorry for my ignorance, but I have not learned inorganic chemistry yet. I do know a bit about crystal and ligand field theory, but I would really appreciate an answer. I am a high school student and just am curious about why silver is so stable when 2 of its water ligands are replaced.

[dipesh747]

Ag will be oxidation number + or 2+, so it will be d10 or d9. If it d10 its crystal field stabilisation energy will be 0. This is a very stable species so it is preferred.

[gertrudetrumpet]

Right, but I don't see how that has to do with silver only accepting 2 ammines in the ligand exchange with h20...aren't both ligands neutral?

[dipesh747]

You only asked about coordination stability not the stability of one complex over another.

Well look at it this way; you have Ag(OH2)2 + 2NH3 --> Ag(NH3)2 + 2H2O

Now the gibbs free energy will be favourable, which is what drives the reaction (thermodynamically driven).

Now delta G = delta H - T delta S

Now entropy isn't driving the reaction so enthalpy will be. Now you can think about why enthalpy is driving the reaction and you have your answer.

[gertrudetrumpet]

Ok so is it that the dative bond dissociation enthalpy for nh3 is more than h2o, making the compound more thermodynamically stable, with a negative delta h value? If that is so, why doesn't silver get more substituted with nh3 and form tetraamminesilver? Steric or electronic effects?

[dipesh747]

its due to salvation effects, water prefers being in the solution compared to NH3.

[gertrudetrumpet]

Then why doesn't silver form hexaamminesilver?

[AWK]

Silver(I) forms Ag(NH₃)₄⁺ but only in liquid ammonia








printing error corrected

[dipesh747]

Silver doesn't form octahedral complexes.