Q11.55CP - Chemistry: Molecular Nature Of Matter And Change

Question

Silicon tetrafluoride reacts with to produce the hexafluorosilicate ion, ${{\mathbf{SiF}}_6}^{2-}$ ;  ${\mathbf{GeF}}_4$behaves similarly, but ${\mathbf{CF}}_{\mathbf{4}}$ does not.

1. Draw Lewis structures for ${\mathbf{SiF}}_4$ , ${{\mathbf{GeF}}_6}^{2-}$ , and  ${\mathbf{CF}}_4$.
2. What is the hybridization of the central atom in each species?
3. Why doesn’t  ${\mathbf{CF}}_4$ react with $\mathbf{F}$ to form ${{\mathbf{CF}}_6}^{2-}$ ?

Step-by-Step Solution

Answer

The process of formation of a new hybrid orbital with equal energy by mixing atomic orbital is known as hybridization.

The answer for

a.

b. ${\mathbf{sp}}^3$

c. Does not have vacant d-orbitals

Step 1: Lewis structure of SiF 4 , GeF 6 2 - and CF 4

The structure in which atoms share electrons based on the octet rule so that each atom has eight electrons in its outer shell is known as the Lewis structure.

From the Lewis structure, it is concluded that  ${\mathbf{SiF}}_4$, ${{\mathbf{GeF}}_6}^{2-}$and ${{\mathbf{CF}}_4}^{}$ contains four, six and four single bonds.

Step 2: Hybridization of SiF 4 , GeF 6 2 - and CF 4

${\mathbf{SiF}}_4$

The central silicon atom is surrounded by four fluorine atoms which form 4 hybrid orbitals. Hence, its hybridization is ${\mathbf{sp}}^3$  with tetrahedral geometry.

${{\mathbf{GeF}}_6}^{2-}$

The central germanium atom is surrounded by six fluorine atoms which form 6 hybrid orbitals. Hence, its hybridization is ${\mathbf{sp}}^3{\mathbf{d}}^2$ with octahedral geometry.

 

${\mathbf{CF}}_4$

 

The central carbon atom is surrounded by four fluorine atoms which form 4 hybrid orbitals. Hence, its hybridization is ${\mathbf{sp}}^3$  with tetrahedral geometry.

 

Step 3: Formation of CF 6 2 -

Silicon possesses vacant d-orbitals in its valence shell so it can extend its coordination number beyond four. Hence, ${{\mathbf{SiF}}_6}^{2-}$ exists. On the other hand, Carbon does not have vacant d-orbitals in its valence shell so it cannot extend its coordination number beyond four. Hence,${{\mathbf{CF}}_6}^{2-}$  does not exist.