Question

Which of the following pairs does not have the same shape?

• A. $H_3{O}^{+},N{H}_3$
• B. $N{H}_4^{+},C{H}_4$
• C. $I_3^{-},Xe{F}_2$
• D. $Xe{F}_6,S{F}_6$

Answer 1

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 pyrimadal 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 pair and 2 bond pair 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.