Why Cu+ is unstable in aqueous medium? This was one of the question that was given in the school exam.

But what I am expecting is that $\ce{Cu+}$  attains a $d^{10}$ configuration by losing one electron from s-subshell. Since it has fully filled d-orbital, it should be stable. But it is found that it is unstable and tends to disproportionate in aqueous medium or with moist air to form $\ce{Cu^2+}$ where  $\ce{Cu^2+}$ has incompletely filled d-orbital ($\ce{d^9}$ configuration) . For example:

$$\ce{Cu2O(s) + H2SO4(aq) -&gt; Cu(s) + CuSO4(aq) + H2O(l)}$$

$$\ce{Cu2O(s) + 2H+(aq) -&gt; Cu(s) + Cu^2+(aq) + H2O(l)}$$

Oxidation number changes: $\ce{2Cu(I) -&gt; Cu(0) + Cu(II)}$

(source)

Why is this so?
You are quite correct in that it appears at first sight that $\ce{Cu+}$ should be more stable than $\ce{Cu^2+}$, but in aqueous media it isn’t. 

Stability in aqueous conditions depends on the hydration energy of the ions when they bond to the water molecules (an exothermic process). The $\ce{Cu^2+}$ ion has a greater charge density than the $\ce{Cu+}$ ion and so forms much stronger bonds releasing more energy. 

The extra energy needed for the second ionisation of the copper is more than compensated for by the hydration, so much so that the $\ce{Cu+}$ ion loses an electron to become $\ce{Cu^2+}$ which can then release this hydration energy.  A nearby $\ce{Cu+}$ ion is the most facile reduction target for the removed electron, which is why $\ce{Cu(s)}$ is also formed.

Hence, $\ce{Cu^2+}$ is more stable than $\ce{Cu+}$ in aqueous medium.