Q10.72CP - Chemistry: Molecular Nature Of Matter And Change

Question

Dichlorine heptaoxide, ${\mathbf{Cl}}_2{\mathbf{O}}_7$ , can be viewed as two ${\mathbf{ClO}}_4$  groups sharing an O atom. Draw a Lewis structure for ${\mathbf{Cl}}_2{\mathbf{O}}_7$  with the lowest formal charges, and predict any deviation from the ideal for the Cl-O-Cl bond angle.

Step-by-Step Solution

Answer

The Lewis structure for  ${\mathbf{Cl}}_2{\mathbf{O}}_7$.

 

 

The deviation in Cl-O-Cl bond angle is  ${\mathbf{\text{118.6}}}^{\mathbf{\text{o}}}$.

Step 1: The Lewis Structure of Cl 2 O 7 with the lowest formal charges.

In dichlorine heptaoxide, each chlorine atom has seven valence electrons and each oxygen atom has six valence electrons. So, the total number of valence electrons in the molecule is 54.

Out of seven oxygen, three oxygen atoms are attached to one chlorine atom through double bonds and three oxygen atoms are attached to another chlorine atom through double bonds. Rest one oxygen atom presents at a bridge position between two chlorine atoms. This oxygen atom is attached through single bonds and two lone pairs are present on it.

The hybridization of Chlorine is ${\mathbf{sp}}^3$  and, the geometry will be tetrahedral.

Step 2: The Formal charges.

The formal charge on chlorine  $\mathbf{=}\mathbf{7}\mathbf{-}\mathbf{(}\mathbf{0}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\times }\mathbf{8}}\mathbf{)}\mathbf{=}\mathbf{+}\mathbf{3}$

The formal charge on oxygen (bridging)  $\mathbf{=}\mathbf{6}\mathbf{-}\mathbf{(}\mathbf{4}\mathbf{+}\frac{1}{2\times 4}\mathbf{)}\mathbf{=}\mathbf{0}$

The formal charge on oxygen (terminal)  $\mathbf{=}\mathbf{6}\mathbf{-}\mathbf{(}\mathbf{6}\mathbf{-}\frac{1}{2\times 2}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{1}$

Step 3: The Deviations

The geometry of ${\mathbf{Cl}}_2{\mathbf{O}}_7$  is tetrahedral but due to the presence of two lone pairs of electrons around the bridging oxygen, the expected ideal bond angle between Cl-O-Cl reduces from $\mathbf{109}\mathbf{.}\mathbf{5}\mathbf{°}$  to $\mathbf{118}\mathbf{.}\mathbf{6}\mathbf{°}$ .