AP EAPCET 22 May 2025 Shift 1 Paper

Among the given lanthanides, the elements that exhibit the $+4$ oxidation state are $\mathrm{Ce}$, $\mathrm{Pr}$, $\mathrm{Nd}$, $\mathrm{Tb}$, and $\mathrm{Dy}$.$\mathrm{Ce^{4+}}$ is well known, for example in $\mathrm{CeO_2}$ and $\mathrm{CeF_4}$.$\mathrm{Pr^{4+}}$ is known in compounds such as $\mathrm{PrO_2}$ and $\mathrm{PrF_4}$.$\mathrm{Nd^{4+}}$ is known in compounds such as $\mathrm{NdF_4}$.$\mathrm{Tb^{4+}}$ is known in compounds such as $\mathrm{TbO_2}$ and $\mathrm{TbF_4}$.$\mathrm{Dy^{4+}}$ is known in compounds such as $\mathrm{DyF_4}$.$\mathrm{Pm}$, $\mathrm{Sm}$, $\mathrm{Eu}$, and $\mathrm{Gd}$ do not show the $+4$ oxidation state here; $\mathrm{Sm}$ and $\mathrm{Eu}$ commonly show $+2$ along with $+3$, while $\mathrm{Gd}$ shows mainly $+3$.Thus, the number of given lanthanide elements exhibiting the $+4$ oxidation state is $5$.Hence, the correct option is $5$.