Explanation

(d)
The general electronic configuration for lanthanides is
[Xe]($n - 2)f^{1 - 14}\ (n - 1)d^{1}ns^{2}.$
$\because$ After the loss of both of the $6s$-electrons and also the solitary $d$-electrons, the lanthanoids gain stable configurations.
$\therefore$ (+3) oxidation state is most common among lanthanides.