Chapter 11, Problem 11.100QE

Interpretation Introduction

Interpretation:

Lewis structure for ${\text{CF}}_{\text{4}}$ and ${\text{CF}}_{\text{2}}{\text{CCl}}_{\text{2}}$ has to be drawn, hybridization of the carbon atoms has to be given and also whether they are polar or nonpolar has to be predicted.  The molecule that is expected to have high boiling point has to be given and also the type of intermolecular force present in both compounds has to be predicted.

Explanation of Solution

Given compounds are ${\text{CF}}_{\text{4}}$ and ${\text{CF}}_{\text{2}}{\text{CCl}}_{\text{2}}$.  Lewis structures can be drawn considering the number of valence electrons.

Lewis Structures:

Lewis structure for ${\text{CF}}_{\text{4}}$:

The total number of valence electron in carbon is four and that of fluorine is seven.  Therefore, the total number of valence electrons in ${\text{CF}}_{\text{4}}$ is calculated as shown below;

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=1\times 4+4\times 7\\ =4+28\\ =32\end{array}$

Skeletal structure for ${\text{CF}}_{\text{4}}$ is drawn considering the carbon atom as the central atom as shown below;

Eight electrons are utilized for the skeletal structure of ${\text{CF}}_{\text{4}}$.  Remaining electrons are placed on fluorine atoms as lone pair of electrons.  Therefore, the Lewis structure of ${\text{CF}}_{\text{4}}$ is drawn as shown below;

Lewis structure for ${\text{CF}}_{\text{2}}{\text{CCl}}_{\text{2}}$:

The total number of valence electron in carbon is four and that of fluorine, chlorine is seven.  Therefore, the total number of valence electrons in ${\text{CF}}_{\text{2}}{\text{CCl}}_{\text{2}}$ is calculated as shown below;

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{valence}\text{electrons}=2\times 4+4\times 7\\ =8+28\\ =36\end{array}$

Skeletal structure for ${\text{CF}}_{\text{2}}{\text{CCl}}_{\text{2}}$ is drawn considering the carbon atom as the central atom as shown below;

Ten electrons are utilized for the skeletal structure of ${\text{CF}}_{\text{2}}{\text{CCl}}_{\text{2}}$.  Remaining electrons are placed on fluorine atoms and chlorine atoms as lone pair of electrons until octet is achieved.  After that the remaining electrons are placed on carbon atom.  Therefore, the Lewis structure of ${\text{CF}}_{\text{2}}{\text{CCl}}_{\text{2}}$ is drawn as shown below;

A double bond is made between the two carbon atoms.  Therefore, the Lewis structure of ${\text{CF}}_{\text{2}}{\text{CCl}}_{\text{2}}$ is drawn as follows;

Hybridization:

Hybridization of the carbon atom can be determined by the steric number.  Steric number is the total number of lone pair of electrons present on the central atom and the number of atoms that is bonded to the central atom.

Hybridization of carbon atom in ${\text{CF}}_{\text{4}}$:

In ${\text{CF}}_{\text{4}}$, there is no lone pair of electrons on central carbon atom.  The total number of atoms that is bonded to the carbon atom is four.  Therefore, the steric number is four.  The hybridization of carbon atom that has steric number of four is $s{p}^3$.

Hybridization of carbon atoms in ${\text{CF}}_{\text{2}}{\text{CCl}}_{\text{2}}$:

In ${\text{CF}}_{\text{2}}{\text{CCl}}_{\text{2}}$, there is no lone pair of electrons on central carbon atom.  The total number of atoms that is bonded to the carbon atom is three.  Therefore, the steric number is three.  The hybridization of carbon atoms that has steric number of three is $s{p}^2$.

Polar and nonpolar:

The molecule ${\text{CF}}_{\text{4}}$ has a tetrahedral structure.  As all the terminal in the tetrahedron is fluorine atom, there will not be any resultant dipole leading to the compound ${\text{CF}}_{\text{4}}$ to be nonpolar.

The molecule ${\text{CF}}_{\text{2}}{\text{CCl}}_{\text{2}}$ is also nonpolar because this molecule has a symmetric planar structure.

Intermolecular forces:

The molecules ${\text{CF}}_{\text{4}}$ and ${\text{CF}}_{\text{2}}{\text{CCl}}_{\text{2}}$ are nonpolar molecules.  London dispersion force is the only force that is present in nonpolar molecules.

Boiling point:

Strength of the intermolecular forces that is present in ${\text{CF}}_{\text{4}}$ and ${\text{CF}}_{\text{2}}{\text{CCl}}_{\text{2}}$ are responsible for the boiling point of the molecules.  London dispersion forces increases with the increase in number of electrons.  Considering this, the number of electrons present in ${\text{CF}}_{\text{4}}$ is less than that of ${\text{CF}}_{\text{2}}{\text{CCl}}_{\text{2}}$.  This means that ${\text{CF}}_{\text{2}}{\text{CCl}}_{\text{2}}$ will have higher boiling point than ${\text{CF}}_{\text{4}}$.