Question

Why does Cerium have $4$ oxidation states?

Answer 1

Hint: Cerium belongs to the lanthanide series. It is the second element of the lanthanide series. Its atomic number is $58$. It is a soft, white-silvery element. It is the most abundant member of the lanthanide series.

Complete answer:
The electronic configuration of cerium is $[Xe]4{f^1}5{d^1}6{s^2}$.In order to achieve stable electronic configuration, it can easily lose the outermost three or four electrons.
Lanthanides are very less electronegative. Hence, their electronegativity value is nearly equal to s-block elements. So, they easily lose electrons to form a cation.
Most lanthanoids can only use $3$ electrons as valence electrons. Cerium is an exception because of the stability of the empty $f - $ shell.
When Cerium loses $4$ electrons, it acquires a fully filled electronic configuration of Xenon. Therefore, it possesses extra stability.
Cerium often shows $ + 3$ oxidation state which is the characteristic oxidation state of lanthanide series. But Cerium also shows $ + 4$ oxidation state.
However,$ + 3$oxidation state of Cerium is more stable than the $ + 4$oxidation state due to greater stabilization of $4f$ orbital than the $5d$ and $6s$ orbitals. Elements in other states try to lose or gain electrons to get a $ + 3$ oxidation state.

Additional Information: Lanthanides are known as the rare earth elements of the modern periodic table as the occurrence of these elements is very small.

Note:
Lanthanides show variable oxidation states such as $ + 2, + 3, + 4$.$ + 3$is the most stable oxidation state. It is attained by removing two outermost electrons of $6s$ orbital and one electron from $4f$ orbital. This is due to the high energy difference in $4f$ and $6s$ orbital, hence it is difficult to remove an electron from a $4f$ subshell.