Statement-I: ${\mathrm{PbF}}_4$ is ionic

Statement-II: ${\mathrm{PbI}}_4$ is highly stable

I) ${\mathrm{SnF}}_4 \mathrm{and} {\mathrm{PbF}}_4$ are ionic

II) ${\mathrm{PbI}}_4$ cannot exist. Initially formed two Pb-I bonds are not sufficiently energetic to excite the electrons in lead;

${\mathrm{Pb}}^{+4}$ is an oxidizing agent and ${\mathrm{I}}^{-}$ is a reducing agent