Chapter 8, Problem 121A

Interpretation Introduction

Interpretation:

The reason for the ${\text{CF}}_{\text{4}}$ molecule to be non-polar, even though it contains polar bonds, is to be explained.

Concept introduction:

The asymmetric structure and presence of polar covalent bonds confirm that the molecule is polar in nature. The molecule having some dipole moment is termed as a polar molecule and the molecule with no dipole moment or zero dipole moment is identified as a non-polar molecule.

Answer to Problem 121A

The charge distribution at any distance from its center is equal but opposite in direction. Therefore, ${\text{CF}}_{\text{4}}$ molecule is non-polar, even though it contains polar bonds.

Explanation of Solution

Let’s assume a bond is formed between two elements A and B. For a bond to be ionic, the electro negativity difference $\left(\Delta \epsilon \right)$ between A and B should be greater than 1.7. For a bond to be polar covalent, the electro negativity difference $\left(\Delta \epsilon \right)$ between A and B should be in between 0.4 and 1.7.

In ${\text{CF}}_{\text{4}}$, the electro negativity of carbon and fluorine is 2.55 and 3.98 respectively. The electro negativity difference $\left(\Delta \epsilon \right)$ between carbon and fluorine is calculated as shown below.

$\begin{array}{c}\Delta \epsilon =3.98-2.55\\ =1.43\end{array}$

The electro negativity difference $\left(\Delta \epsilon \right)$ is 1.43 which is in between 0.4 and 1.7. Therefore, the $\text{C}-\text{F}$ bond is polar covalent.

The given molecule ${\text{CF}}_{\text{4}}$ is tetrahedral and therefore symmetrical. The structure of ${\text{CF}}_{\text{4}}$ is shown below.

In ${\text{CF}}_{\text{4}}$, the charge distribution at any distance from its center is equal but opposite in direction. Therefore, the partial charges in ${\text{CF}}_{\text{4}}$ are balanced which makes ${\text{CF}}_{\text{4}}$ as a non-polar compound.

Conclusion

The charge distribution at any distance from its center is equal but opposite in direction. Therefore, ${\text{CF}}_{\text{4}}$ molecule is non-polar.