Write a note on oxidation states shown by lanthanoids.

(N/A) In lanthanoids,$Ln(II)$ and $Ln(III)$ compounds are dominant species. However,occasionally $(+2)$ and $(+4)$ ions in solution or in solid compounds are also obtained. This irregularity is because of extra stability of $f^{0}, f^{7}$ and $f^{14}$ orbitals.
In aqueous solution,$Sm^{2+}, Yb^{2+}$ and $Eu^{2+}$ act as strong reducing agents as these ions readily get oxidized to $(+3)$ state because of higher stability in $(+3)$ oxidation state.
Elements $Pr, Nd, Tb$ and $Dy$ show $(+4)$ oxidation state but only in oxides $MO_{2}$. $Tb(IV)$ has half-filled $f$-orbitals and it is an oxidant.
$Ce(IV)$ is a very good analytical reagent. The $E^{\circ}$ value of $Ce^{4+}/Ce^{3+}$ is $+1.74 \ V$,which suggests that $Ce^{4+}$ is a very good oxidizing agent and easily gets reduced to $Ce^{3+}$. It can even oxidize water. However,the reaction rate is very slow.