Why does silver oxide form a coordination complex when treated with ammonia? Part of the preparation of Tollens' reagent occurs according to the formula: 

$$\ce{Ag2O (s) + 4 NH3 + 2 NaNO3 + H2O → 2 [Ag(NH3)2]NO3 + 2 NaOH}$$

I wonder how exactly the oxygen gets "unstuck" from the silver under the influence of ammonia. I've read about coordination complexes, so I understand what $\ce{[Ag(NH3)2]NO3}$ is, but I'm puzzled at what exactly happens to pull that oxygen apart from $\ce{Ag2O}$

I've done very little exploration of the issue, so this may be too basic a question - in which case just please give a hint on what to read up. 

I found this (now deleted) comment:


  $\ce{NH3}$ dissolves silver oxide because the nitrogen has a pair of electrons that it can share with the silver forming a diammine complex. It has nothing to do with pH adjustment. – A.K. Jan 20 '16 at 17:17


Chemguide states this: 


  This is made from silver(I) nitrate solution. You add a drop of sodium hydroxide solution to give a precipitate of silver(I) oxide, and then add just enough dilute ammonia solution to redissolve the precipitate.
It's quite elementary (perhaps). Silver (and gold) are known to show two main coordination numbers in its compounds, namely, $2$ and $4$. To start, we can consider $\ce{[Ag(OH)2]-}$ to be our initial complex as in aqueous solutions it can be thought of to be existing in an equilibrium with $\ce{Ag2O}$. See footnote [1].

A quick glance at the order for the strength of ligands tells us that $\ce{NH3}$ is a stronger ligand than $\ce{OH-}$. What I now intend to do is to compare the thermodynamic stability of $\ce{[Ag(OH)2]-}$ with that of the diammine complex $\ce{[Ag(NH3)2]+}$.

The strength of ligands is directly proportional to the thermal stability of a complex so yes, the diammine complex is more stable. 

A similar case is seen with copper(I). Cyanide complexes of copper(I) are known to be extremely stable and hence it prefers forming a complex with cyanide over other ligands.

[1] Nevertheless, saying this may be highly controversial because Ellingham diagrams tell us that $\ce{Ag(OH)}$ being unstable gets converted into $\ce{Ag2O}$ which is a much more stable compound for silver to be in. However in aqueous solutions, an equilibrium of such a kind can always exist.