Identify the element and its series

Cerium, $\mathrm{Ce}$, is a lanthanoid element.

Write its electronic configuration

The atomic number of cerium is $58$.

Its electronic configuration is approximately: $\mathrm{Ce}: [\mathrm{Xe}]\,4f^1 5d^1 6s^2$

Determine the common oxidation states

Lanthanoids most commonly show the oxidation state $+3$ due to loss of two $6s$ electrons and one $5d/4f$ electron.

For cerium, an additional electron can also be lost to give the stable $+4$ oxidation state.

Thus, cerium commonly shows: $+3 \text{ and } +4$

Check the options

• A: $+3, +4$ ✓ Correct
• B: $+2, +3$ ✗ $+2$ is not a common oxidation state of cerium
• C: $+2, +4$ ✗ $+2$ is not common
• D: $+3, +5$ ✗ $+5$ is not a common oxidation state of cerium

Final answer

The most common oxidation states of cerium are: $\boxed{+3, +4}$