Question

The cationic part of solid $\text{C}{\text{l}}_{\text{2}}{\text{O}}_{\text{6}}$ is having the “______________” shape.
(A) Linear
(B) Angular
(C) Tetrahedron
(D) Undefined

Answer 1

Hint: The shape and structure of a compound can be determined from the hybridization of the central atoms. We shall calculate the hybridisation using the formula and determine its shape.

Formula Used: ${\displaystyle \frac{\text{V + X - C + A}}{\text{2}}}$
where V is the number of valence shell electrons, X is the number of monovalent groups, C is the charge of the cation, and A is the charge on the anion.

Complete step by step answer:
In the solid state $\text{C}{\text{l}}_{\text{2}}{\text{O}}_{\text{6}}$ exists as an anion and a cation. The cationic part is $\text{Cl}{{\text{O}}_{\text{2}}}^{+}$ while the anionic part is $\text{Cl}{{\text{O}}_{\text{4}}}^{-}$ . In $\text{Cl}{{\text{O}}_{\text{2}}}^{+}$ the hybridization of the central atom, chlorine is:
   ${\displaystyle \frac{\text{7 + 0 - 1 + 0}}{\text{2}}}={\displaystyle \frac{6}{2}}=3$ .
As there are seven electrons in the valence shell of chlorine and one mono positive charge, Hence chlorine atom in $\text{Cl}{{\text{O}}_{\text{2}}}^{+}$ should be $\text{s}{\text{p}}^2$ hybridized with a trigonal planar structure.
As the third point of the trigonal planar structure is empty due to the absence of any atom, the structure should be angular in shape.
So, the correct answer of the question is option C, angular.

Note:
In the $\text{C}{\text{l}}_{\text{2}}{\text{O}}_{\text{6}}$ molecule, the chlorine atoms are the central atom and two $\text{Cl}{\text{O}}_{\text{3}}$ units form a dimer together. So considering only one part of the dimer $\text{Cl}{\text{O}}_{\text{3}}$
For chlorine the number of valence shell electrons is 7, and there are neither any monovalent groups nor any cationic or any anionic charge. Hence the hybridisation is $3.5$ . Which means that the hybridization is $\text{s}{\text{p}}^{\text{3}}$ . Hence the shape should be tetrahedral.