1. Electronic configuration of cerium
   Cerium, $Z=58$, has electronic configuration close to: $[Xe]4f^1 5d^1 6s^2$ or often written effectively as $[Xe]4f^1 5d^1 6s^2$.

Because lanthanoids commonly lose the two $6s$ electrons and sometimes one more electron, the $+3$ oxidation state is the most common. Cerium is special because it can also form a stable $+4$ oxidation state.

2. Check option A: "The common oxidation states of cerium are $+3$ and $+4$"
   This is true. Cerium commonly shows both $+3$ and $+4$ oxidation states.

3. Check option B: "Cerium (IV) acts as an oxidizing agent"
   This is true. Since $Ce^{4+}$ readily gets reduced to $Ce^{3+}$, it behaves as an oxidizing agent.

4. Check option C: "The $+4$ oxidation state of cerium is not known in solutions"
   This is false. Cerium(IV) does exist in solution, for example in acidic medium, and $Ce^{4+}/Ce^{3+}$ is a well-known redox couple used in volumetric analysis. Therefore this statement is incorrect.

5. Check option D: "The $+3$ oxidation state of cerium is more stable than the $+4$ oxidation state"
   This is generally true for lanthanoids, and for cerium the $+3$ state is common and stable, although $+4$ is also important due to extra stability in compounds like $CeO_2$.

6. Conclusion
   The incorrect statement is: $\boxed{\text{C}}$

7. Comparison with stored answer
   Stored correct answer is C, which matches the derived answer.