Chapter 6, Problem 42QRT

(a)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{KrF}}_{\text{4}}$ and a formal charge of each atom has to be written.

Concept Introduction:

A Lewis structure is also known as Lewis dot diagrams or electron dot structures.  The bond between atoms and lone pairs of electrons that is present in the molecule.  Lewis structure represents each atom and their position in structure using the chemical symbol.  Excess electrons forms the lone pair are given by pair of dots, and are located next to the atom.

The formula for the formal charge can be written as

  $\text{Formal}\text{charge=}\text{Valence}\text{electrons-lone}\text{pair}\text{electrons}\text{-}\frac{\text{Bonding}\text{electrons}}{\text{2}}$

Explanation of Solution

The correct Lewis structure of the ${\text{KrF}}_{\text{4}}$ can be drawn as,

Draw a chart:

  $\overline{\begin{array}{l}\text{Kr}\text{F}\\ \overline{\underset{\_}{\begin{array}{l}\text{Valence}\text{electrons}87\\ \text{Lone}\text{pair}\text{electrons}46\\ \text{Bonding}\text{electrons}82\\ \text{Formal}\text{charge}8-(4+4)=07-(6+1)=0\end{array}}}\end{array}}$

The total number of valence electron is $36.$

(b)

Interpretation Introduction

Interpretation:

The Lewis structure for ${\text{ClO}}_{\text{3}}^{\text{-}}\text{ion}$ and a formal charge of each atom has to be written.

Concept Introduction:

Refer Part (a).

Explanation of Solution

The correct possible Lewis structure for ${\text{ClO}}_{\text{3}}^{\text{-}}\text{ion}$ can be drawn as,

First structure:

Draw a chart:

  $\overline{\begin{array}{l}\text{Cl}\text{O}\\ \overline{\underset{\_}{\begin{array}{l}\text{Valence}\text{electrons}76\\ \text{Lone}\text{pair}\text{electrons}26\\ \text{Bonding}\text{electrons}62\\ \text{Formal}\text{charge}7-(2+3)=+26-(6+1)=-1\end{array}}}\end{array}}$

The total number of valence electron is $26.$

To make double bond between chlorine and oxygen against the octet rule and then reduced the formal charge to zero.

Draw a chart:

    $\overline{\begin{array}{l}Cl=\text{O}\\ \overline{\underset{\_}{\begin{array}{l}\text{Valence}\text{electrons}76\\ \text{Lone}\text{pair}\text{electrons}24\\ \text{Bonding}\text{electrons}102\\ \text{Formal}\text{charge}7-(2+5)=06-(6+1)=-1\end{array}}}\end{array}}$

(c)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{SO}}_2{\text{Cl}}_{\text{2}}$ and a formal charge of each atom has to be written.

Concept Introduction:

Refer Part (a).

Explanation of Solution

The correct Lewis structure of the ${\text{SO}}_2{\text{Cl}}_{\text{2}}$ can be drawn as,

Draw a chart:

  $\overline{\begin{array}{l}SOCl\\ \overline{\underset{\_}{\begin{array}{l}\text{Valence}\text{electrons}667\\ \text{Lone}\text{pair}\text{electrons}066\\ \text{Bonding}\text{electrons}822\\ \text{Formal}\text{charge}6-(0+4)=+26-(6+1)=-17-(6+1)=\text{0}\end{array}}}\end{array}}$

The total number of valence electron is $32.$

To make double bond between sulfur and oxygen against the octet rule and then reduced the formal charge to zero.

  $\overline{\begin{array}{l}SOCl\\ \overline{\underset{\_}{\begin{array}{l}\text{Valence}\text{electrons}667\\ \text{Lone}\text{pair}\text{electrons}046\\ \text{Bonding}\text{electrons}1242\\ \text{Formal}\text{charge}6-(0+6)=06-(4+2)=07-(6+1)=\text{0}\end{array}}}\end{array}}$

Hence, the formal charge of the ${\text{SO}}_2{\text{Cl}}_{\text{2}}$ is zero.