Which is the right structure of ${\mathrm{XeF}}_4$?

Consider the following iodides:

$\begin{array}{ccc}{\mathrm{PI}}_3& {\mathrm{AsI}}_3& {\mathrm{SbI}}_3\\ 102°& 100.2°& 99°\end{array}$

The bond angle is maximum in ${\mathrm{PI}}_3$, which is:

${\mathrm{CO}}_2$ is isostructure with:

(I) ${\mathrm{HgCl}}_2$

(II) ${\mathrm{SnCl}}_2$

(III) ${\mathrm{NO}}_2$

(IV) ${\mathrm{C}}_2{\mathrm{H}}_2$

The correct answer is: