Identify from the following pairs which do not have same shape?

The correct option is D $Xe{F}_6,S{F}_6$
$N{H}_3$ has a hybridisation of $s{p}^3$. Nitrogen has a lone pair of electrons and so it has a pyramidal shape and a slightly less bond angle due to lone pair-bond pair repulsion.


$H_3{O}^{+}$ has a steric number of $4$ giving a hybridisation of $s{p}^3$. The three $H$- atoms and a lone pair of electrons tend to a shape of trigonal pyramidal to minimise repulsion.


$N{H}_4^{+}$ and $C{H}_4$ have $s{p}^3$ hybridisation and both have tetrahedral geometry.

$I_3^{-}$ and $Xe{F}_2$ have a steric number of $5$ and have $s{p}^{3}d$ hybridisation. They have $3$ lone pairs and $2$ bond pairs of electrons and they have a geometry of trigonal bypiramidal and linear shape.



$Xe{F}_6$ has a steric number of $7$. So, it will have a hybridisation of $s{p}^3{d}^3$ and a geometry of distorted octahedral or square pyramidal due to the presence of lone pair on the central atom.

$S{F}_6$ has a steric number of $6$ and has a hybridisation of $s{p}^3{d}^2$. So, it has an octahedral geometry.