Q10.64CP - Chemistry: Molecular Nature Of Matter And Change

Question

Consider the following reaction of silicon tetrafluoride: ${\mathbf{SiF}}_4\text{ + }{\mathbf{F}}^{-}\stackrel{}{\to }{\mathbf{SiF}}_5^{-}$

1. Which depiction below best illustrates the change in molecular shape around Si?
2. Give the name and ${\mathbf{AX}}_m{\mathbf{E}}_n$ designation of each shape in the depiction chosen in part (a).

Step-by-Step Solution

Answer

(a) The depiction “B” best illustrates the change in molecular shape around Si. 

(b) The tetrahedral shape of ${\mathrm{SiF}}_4$ with ${\mathrm{AX}}_4$ designation changes to trigonal bipyramidal shape of ${\mathrm{SiF}}_4$ with ${\mathrm{AX}}_4$ designation on reaction with F.

Step 1: The depiction which best illustrates the change in molecular shape around Si.

We know that the molecular shapes of the molecules from the Lewis structure are determined by employing Valence-Shell Electron-Pair Repulsion (VSEPR) theory. The molecular shapes are assigned specific ${\mathrm{AX}}_{\mathrm{m}}{\mathrm{E}}_{\mathrm{n}}$ designations.

Where m and n are integers

A - central atom

X - Surrounding atom E - Nonbonding valence-electron group (usually a lone pair).

From the given reaction in the question, ${\mathbf{SiF}}_{\mathbf{4}}\mathbf{+}{\mathbf{F}}^{\mathbf{-}}\stackrel{}{\mathbf{\to }}{\mathbf{SiF}}_{\mathbf{5}}^{\mathbf{-}}$, the depiction “B” best illustrates the change in molecular shape around Si. As ${\mathrm{SiF}}_4$ has a tetrahedral shape and ${\mathrm{SiF}}_5$ has trigonal bipyramidal shape depiction of “B” alone will show the change in molecular shape around Si. While the rest of the other depiction does not match to the change in molecular shape around Si.

Step 2: Name and AX m E n Designation of each Molecule

We can explain the change in molecular shape around Si by considering ${\mathrm{SiF}}_4$ Lewis's structure and molecular shape. In ${\mathrm{SiF}}_4$ the central atom is silicon, which is surrounded by four fluorine atoms. It contains 4 bonding pairs and 0 lone pairs with ${\mathrm{AX}}_4$ designation having a tetrahedral shape.

Valence electron of $\mathrm{Si}=4\mathrm{e}$; 

Valence electrons of $\mathrm{F}=7\mathrm{e}$;

Total no. of valence electrons$=4+4\mathrm{x}7=32\mathrm{e}$.

Lewis structure of ${\mathbf{SiF}}_{\mathbf{4}}$ and its tetrahedral molecular shape

When ${\mathbf{SiF}}_{\mathbf{4}}$ reacts with ${\mathbf{F}}_{\mathbf{5}}^{\mathbf{-}}$ to form  ${\mathbf{SiF}}_{\mathbf{5}}^{\mathbf{-}}$. In ${\mathbf{SiF}}_{\mathbf{5}}^{\mathbf{-}}$, the central atom is silicon, which is surrounded by five fluorine atoms with a -1 ionic charge. It contains 5 bonding pairs and 0 lone pairs with ${\mathrm{AX}}_5$ designation having trigonal bipyramidal shape.

Valence electron of $\mathrm{Si}=4\mathrm{e}$; 

Valence electrons of $\mathrm{F}=7\mathrm{e}$;

We have to add 1e to the total number of valence electrons due to the -1 ionic charge.

Total no. of valence electrons $=4+5\mathrm{x}7+1=40\mathrm{e}$.  

Lewis structure of ${\mathbf{SiF}}_5^{-}$ and its trigonal bipyramidal molecular shape

Thus, depiction “B” correctly shows the change in molecular shape around Si. The shape and designation of ${\mathrm{SiF}}_4$ is a tetrahedral shape with ${\mathrm{AX}}_4$ designation which changes to ${\mathrm{SiF}}_5^{-}$ with trigonal bipyramidal shape with ${\mathrm{AX}}_5$ designation on reaction with ${\mathrm{F}}_{}^{-}$.