Question

Question: Match the xenon compounds in Column$- {\rm I}$with its structure in Column$- {\rm I}{\rm I}$ and assign the correct code:

Column$- {\rm I}$ Column$- {\rm I}{\rm I}$

(a)| ${\text{Xe}}{{\text{F}}_4}$ | $\left( {\rm I} \right)$ | Pyramidal
(b)| ${\text{Xe}}{{\text{F}}_6}$| $\left( {{\rm I}{\rm I}} \right)$| Square planar
(c)| ${\text{XeO}}{{\text{F}}_4}$| $\left( {{\rm I}{\rm I}{\rm I}} \right)$| Distorted octahedral
(d)| ${\text{Xe}}{{\text{O}}_3}$| $\left( {{\rm I}{\text{V}}} \right)$| Square pyramidal

A. $A - \left( {\rm I} \right),B - \left( {{\rm I}{\rm I}} \right),C - \left( {{\rm I}{\rm I}{\rm I}} \right),D - \left( {{\rm I}V} \right)$
B. $A - \left( {{\rm I}{\rm I}} \right),B - \left( {{\rm I}{\rm I}{\rm I}} \right),C - \left( {{\rm I}V} \right),D - \left( {\rm I} \right)$
C. $A - \left( {{\rm I}{\rm I}} \right),B - \left( {{\rm I}{\rm I}{\rm I}} \right),C - \left( {\rm I} \right),D - \left( {{\rm I}V} \right)$
D. $A - \left( {{\rm I}{\rm I}{\rm I}} \right),B - \left( {{\rm I}V} \right),C - \left( {\rm I} \right),D - \left( {{\rm I}{\rm I}} \right)$

Answer 1

${\text{VSEPR}}$ theory is used to predict geometry of individual molecules from the number of electron pairs surrounding the central atom. Full form of ${\text{VSEPR}}$ theory is valence shell electron pair repulsion theory. Different compounds have different geometry due to the different number of bond pairs and lone pairs.

Complete step by step answer: As we know ${\text{VSEPR}}$ theory is used to predict shape of compound. The electron pairs present around the central atom around repel each other and hence move as far apart as possible so that there are no further repulsions between them as a result molecule has minimum energy and maximum stability. If the central atom is surrounded by different types of atoms or is surrounded by both lone pairs and bond pairs geometry is decided. The order is given below
$\left( {lp - lp > lp - bp > bp - bp} \right)$
Where $lp$ is lone pair and $bp$ is bond pair repulsions respectively.
We will see one by one compound given in the option
${\text{Xe}}{{\text{F}}_4}$ Has $4$ bond pairs with ${\text{F}}$ fluorine and$2$lone pairs hence it exhibits square planar geometry
${\text{Xe}}{{\text{F}}_6}$ Has $6$ bond pairs with ${\text{F}}$ fluorine and $1$ lone pair hence it exhibits distorted octahedral geometry
${\text{XeO}}{{\text{F}}_4}$Has $5$ bond pairs with $4$ with ${\text{F}}$ fluorine and $1$ with oxygen $\left( {{O_2}} \right)$and 1 lone pair hence it exhibit square pyramidal geometry.
${\text{Xe}}{{\text{O}}_3}$ Has $3$ bond pairs with oxygen $\left( {{O_2}} \right)$ and $1$ lone pair hence it exhibits pyramidal geometry. Observing the option our answer to the given question is option B that is
$A - \left( {{\rm I}{\rm I}} \right),B - \left( {{\rm I}{\rm I}{\rm I}} \right),C - \left( {{\rm I}V} \right),D - \left( {\rm I} \right)$

Note: Remember the number of bond pairs and lone pairs for which different geometries of compounds are formed or you can just find them by knowing their electronic configuration. The Basis of the theory is that atoms are arranged in 3d in such a way that they experience least mutual repulsion so you can think in this way also, and predict the structure accordingly.