Why does copper(II) carbonate decompose into copper(II) oxide rather than copper(I) oxide when heated? I recently did a lab at school surrounding the decomposition of copper(II) carbonate into copper(II) oxide, and subsequently into pure copper, when sufficient heat is added. What I don't understand is why the copper(II) carbonate can't decompose into copper(I) oxide, rather than copper(II) oxide.

For example, copper(II) carbonate decomposing into copper(II) oxide would be:

$$\ce{CuCO3 -&gt;CuO + CO2}$$

While copper(II) carbonate decomposing into copper(I) oxide would be:

$$\ce{4CuCO3 -&gt; 2Cu2O + 4CO2 + O2}$$

If anyone could help me understand why copper(II) carbonate decomposes into copper(II) oxide rather than copper(I) oxide, I would greatly appreciate it.

Edit: I've been nosing around Wikipedia a bit and discovered that copper(I) oxide reacts with atmospheric moisture to become copper(II) oxide. This would explain why copper (I) oxide is not present; it has degraded into copper(II) oxide. I'm still interested in how copper(I) oxide and water react with respect to reactants and products, so if anyone knows what the equation between the products and reactants is, I would like to know.
It comes to stability of products, in this case interaction of metal ion with ligands (here oxygen). In water solutions $\ce{Cu^2+}$ is usually more stable than $\ce{Cu^1+}$, it's quite different story in non-aqueous solvents like acetonitrile. In solid: $\ce{CuCl2}$ is more stable than $\ce{CuCl}$, but iodide $\ce{CuI}$ is more stable than $\ce{CuI2}$. 

The simplest explanation would be hard-soft acid-base theory: $\ce{Cu^2+}$ is a hard acid, and $\ce{Cu+}$ is a soft one.  Oxygen in copper oxide or $\ce{Cl-}$ in chloride are hard bases and $\ce{I-}$ is a soft one. The same to the same: soft acids goes to soft bases, and hard acids likes the hard bases.