Chapter 11, Problem 85AE

Interpretation Introduction

Interpretation:

The reason for which fluorine is given importance in terms of electronegativity and what combination of atoms on periodic table form the ionic bond have to be identified.

Concept Introduction:

Lewis structure:

The representation of valence shell electrons around the atom is known as Lewis structure or Lewis dot structure.  Electrons are represented as a dot in Lewis structures, a single dot represents unpaired electron and paired of dots represents paired electrons.

Explanation of Solution

Given,

The molar mass of the compound is $\text{166}\text{g/mol}$.

The percentage composition of carbon is $14.5\%$.

The mass of carbon can be calculated as shown below.

$\begin{array}{c}Massofcarbon=\frac{14.5}{100}\times 166\\ =24.07g\end{array}$

The percentage composition of chlorine is $85.5\%$.

$\begin{array}{c}Massofchlorine=\frac{85.5}{100}\times 166\\ =141.93g\end{array}$

The mass of carbon is $24.07g$.

The mass of chlorine is $141.93g$.

The empirical formula can be calculated as follows:

$\begin{array}{l}CCl\\ \text{Mass}24.07\text{g}141.93\text{g}\\ \text{Molar}\text{mass}12.011\text{g/mol}35.453\text{g/mol}\\ \text{Moles}\text{=}\frac{\text{mass}}{Molarmass}\frac{24.07\text{g}}{12.011\text{g/mol}}\frac{141.93\text{g}}{35.453\text{g/mol}}\\ 2.00mol4.00mol\\ Divideby2.00mol\frac{2.00mol}{2.00mol}\frac{4.00mol}{2.00mol}\\ 12\end{array}$

The empirical formula of the compound is $CC{l}_2$.

The empirical formula mass is sum of the mass of carbon and two chlorine atoms.

$\begin{array}{c}\text{Mass}\text{=}\text{12}\text{.011}\text{g/mol}\text{+2×35}\text{.453}\text{g/mol}\\ \text{=}82.917g/mol\end{array}$

The molar mass of the compound is $\text{166}\text{g/mol}$.

The molecular formula is calculated as follows:

$\begin{array}{c}\text{n}\text{=}\frac{\text{166}\text{g/mol}}{\text{82}\text{.917}\text{g/mol}}\\ =2\end{array}$

Here it is found that empirical formula is twice molecular formula. Therefore the molecular formula of the compound is $C_{2}C{l}_4$.

The Lewis structure of $C_{2}C{l}_4$ is drawn as follows:

The total number of valence electrons in $C_{2}C{l}_4$ is $36$. 4 valence electrons from each  carbon atom and 7 from each $Cl$ atom.

The four chlorine atoms are bonded to two carbon atoms. Write skeletal structure representing this and place two electrons between each carbon and chlorine atom.

Here $10$ valence electrons are used and 26 valence electrons are remaining. These electrons can be distributed over carbon and chlorine atoms to get an octet configuration.

Here all the atoms form octet. Therefore, Lewis structure of $C_{2}C{l}_4$ may be represented as follows: