Chapter 3.4, Problem 3.9Q

(a)

Interpretation Introduction

Interpretation:

The molecule given is polar or not has to be predicted, if it is polar by using a vector the direction in the electrons are pulled has to be indicated.

Explanation of Solution

The given molecule is ${\text{CS}}_2$.

The geometry of the molecule has to be determined first.  To determine the polarity, draw the Lewis structure.

  $\begin{array}{l}\text{Skeletal}\text{Structure}\text{Count}\text{Valence}\text{Electrons}\\ \text{1}\text{C}\text{×}{\text{4e}}^{\text{-}}\text{=}{\text{4e}}^{\text{-}}\\ \text{S C}\text{S}\text{2}\text{S}\text{×}{\text{6e}}^{\text{-}}\text{=}{\text{12e}}^{\text{-}}\\ \text{------------------}\\ {\text{16e}}^{\text{-}}\end{array}$

Distribute valence electron:

    

Vectors can be used to determine the polarity of this compound.  The vectors should point in the direction of each electronegative sulfur atom.  Since the dipole moment cancels each other, the molecule is not polar.

    

(b)

Interpretation Introduction

Interpretation:

The molecule ${\text{NF}}_{\text{3}}$ is polar or not has to be predicted, if it is polar by using a vector the direction in the electrons are pulled has to be indicated.

Explanation of Solution

The given molecule is ${\text{NF}}_{\text{3}}$.

The geometry of the molecule has to be determined first.  To determine the polarity, draw the Lewis structure.

  $\begin{array}{l}\text{Skeletal}\text{Structure}\text{Count}\text{Valence}\text{Electrons}\\ \text{1}\text{N}\text{×}{\text{5e}}^{\text{-}}\text{=}{\text{5e}}^{\text{-}}\\ \text{F}\text{N}\text{F}\text{3}\text{F}\text{×}{\text{7e}}^{\text{-}}\text{=}{\text{21e}}^{\text{-}}\\ \text{F}\text{------------------}\\ {\text{26e}}^{\text{-}}\end{array}$

Distribute valence electron:

    

Vectors can be used to determine the polarity of this compound.  The vectors should point in the direction of electronegative fluorine atom.  The given compound is polar.

    

(c)

Interpretation Introduction

Interpretation:

The molecule given is polar or not has to be predicted, if it is polar by using a vector the direction in the electrons are pulled has to be indicated.

Explanation of Solution

The given molecule is $\text{HCl}$.

The geometry of the molecule has to be determined first.  To determine the polarity, draw the Lewis structure.

  $\begin{array}{l}\text{Skeletal}\text{Structure}\text{Count}\text{Valence}\text{Electrons}\\ \text{1}\text{H}\text{×}{\text{1e}}^{\text{-}}\text{=}{\text{1e}}^{\text{-}}\\ \text{HCl}\text{1Cl}\text{×}{\text{7e}}^{\text{-}}\text{=}{\text{7e}}^{\text{-}}\\ \text{------------------}\\ 8{\text{e}}^{\text{-}}\end{array}$

Distribute valence electron:

    

The given molecule is polar because there is an electronegativity difference between two atoms.

(d)

Interpretation Introduction

Interpretation:

The molecule given is polar or not has to be predicted, if it is polar by using a vector the direction in the electrons are pulled has to be indicated.

Explanation of Solution

The given molecule is ${\text{SiCl}}_{\text{4}}$.

The geometry of the molecule has to be determined first.  To determine the polarity, draw the Lewis structure.

  $\begin{array}{l}\text{Skeletal}\text{Structure}\text{Count}\text{Valence}\text{Electrons}\\ \text{Cl}\text{1}\text{S}\text{×}{\text{6e}}^{\text{-}}\text{=}{\text{6e}}^{\text{-}}\\ \text{Cl}\text{Si Cl}\text{4Cl}\text{×}{\text{7e}}^{\text{-}}\text{=}{\text{28e}}^{\text{-}}\\ \text{Cl}\text{------------------}\\ 34{\text{e}}^{\text{-}}\end{array}$

Distribute valence electron:

    

Vectors can be used to determine the polarity of this compound.  The vectors should point in the direction of each electronegative chlorine atom.  Since the dipole moment cancels each other, the molecule is non-polar.