Chapter 18, Problem 88E

Interpretation Introduction

Interpretation: The bond angle and hybridization of the given compounds needs to be determined.

Concept Introduction:

Lewis dot structure is the representation which shows the bonding between atoms present in a molecule. It shows lone pairs and bond pairs that exist on each bonded atom. Lewis dot structure is also known as Lewis dot formula or electron dot structure.

The shape or geometry of molecule can be predicted with the help of hybridization and VSEPR theory. It can be checked with the below formula:

  $\text{Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms}\text{.}$

Answer to Problem 88E

• $\text{XeF}$ = ${\text{sp}}^{\text{3}}\text{d= linear; 180}°$
• ${\text{XeO}}_{\text{2}}\text{F}$ = ${\text{sp}}^{\text{3}}\text{d= see saw; 90}°,\text{180}°$
• ${\text{XeO}}_{\text{3}}$ = ${\text{sp}}^{\text{3}}\text{ = pyramide ; 103}°$
• ${\text{XeO}}_4$ = ${\text{sp}}^{\text{3}}\text{ = tetrahedral ; 109}°$
• ${\text{XeF}}_4$ = ${\text{sp}}^{\text{3}}{\text{d}}^{\text{2 }}\text{= square planer ; 90}°$
• ${\text{XeO}}_{\text{3}}{\text{F}}_2$ = ${\text{sp}}^{\text{3}}\text{d= trigonal bipyramide; 90}°,180°$
• ${\text{XeO}}_{\text{2}}{\text{F}}_4$ = ${\text{sp}}^{\text{3}}{\text{d}}^{\text{2}}\text{= octahedral ; 90}°,180°$

Explanation of Solution

Given information:

  

The hybridization can be checked with the help of Lewis structure and VSEPR theory which provide the geometry of molecules with lone pairs.

• 

  $\begin{array}{l}\text{Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms}\text{.}\\ \text{Hybridization = 2 + 3}\\ {\text{Hybridization = sp}}^{\text{3}}\text{d= linear; 180}°\end{array}$

• 

  $\begin{array}{l}\text{Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms}\text{.}\\ \text{Hybridization = 4 + 1}\\ {\text{Hybridization = sp}}^{\text{3}}\text{d= see saw; 90}°,\text{180}°\end{array}$

  

  $\begin{array}{l}\text{Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms}\text{.}\\ \text{Hybridization = 3 + 1=4}\\ {\text{Hybridization = sp}}^{\text{3}}\text{ = pyramide ; 103}°\end{array}$

  

  $\begin{array}{l}\text{Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms}\text{.}\\ \text{Hybridization = 4 + 0 =4}\\ {\text{Hybridization = sp}}^{\text{3}}\text{ = tetrahedral ; 109}°\end{array}$

  

  $\begin{array}{l}\text{Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms}\text{.}\\ \text{Hybridization = 4 + 2 = 6}\\ {\text{Hybridization =sp}}^{\text{3}}{\text{d}}^{\text{2 }}\text{= square planer ; 90}°\end{array}$

  

  $\begin{array}{l}\text{Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms}\text{.}\\ \text{Hybridization = 5 + 0 =5}\\ {\text{Hybridization = sp}}^{\text{3}}\text{d= trigonal bipyramide; 90}°,180°\end{array}$

  

  $\begin{array}{l}\text{Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms}\text{.}\\ \text{Hybridization = 6 + 0 = 6}\\ {\text{Hybridization = sp}}^{\text{3}}{\text{d}}^{\text{2}}\text{= octahedral ; 90}°,180°\end{array}$

Conclusion

Thus,

• $\text{XeF}$ = ${\text{sp}}^{\text{3}}\text{d= linear; 180}°$
• ${\text{XeO}}_{\text{2}}\text{F}$ = ${\text{sp}}^{\text{3}}\text{d= see saw; 90}°,\text{180}°$
• ${\text{XeO}}_{\text{3}}$ = ${\text{sp}}^{\text{3}}\text{ = pyramide ; 103}°$
• ${\text{XeO}}_4$ = ${\text{sp}}^{\text{3}}\text{ = tetrahedral ; 109}°$
• ${\text{XeF}}_4$ = ${\text{sp}}^{\text{3}}{\text{d}}^{\text{2 }}\text{= square planer ; 90}°$
• ${\text{XeO}}_{\text{3}}{\text{F}}_2$ = ${\text{sp}}^{\text{3}}\text{d= trigonal bipyramide; 90}°,180°$
• ${\text{XeO}}_{\text{2}}{\text{F}}_4$ = ${\text{sp}}^{\text{3}}{\text{d}}^{\text{2}}\text{= octahedral ; 90}°,180°$