Chapter 7, Problem 7.28P

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{XeF}}_{\text{2}}{\text{,XeF}}_{\text{4}}{\text{,XeF}}_{\text{6}}{\text{,XeOF}}_{\text{4}}$ and ${\text{XeO}}_{\text{2}}{\text{F}}_{\text{2}}$ has to be written.

Concept Introduction:

Lewis Structure:  A Lewis structure shows a covalent bond as pair of electrons shared between two atoms.

Procedure to write Lewis formulas:

1. 1) The symbols of the atoms that are bonded together in the molecule next to one another are arranged.
2. 2) The total number of valence electrons in the molecule is calculated by adding the number of valence electrons for all the atoms in the molecules.  If the species is an ion, then the charge of ion into account by adding electrons, if it is a negative ion or subtracting electrons if it is a positive ion.
3. 3) A two-electron covalent bond is represented by placing a line between the atoms, which are assumed to be bonded to each other.
4. 4) The remaining valence electrons as lone pairs about each atom are arranged so that the octet rule is satisfied for each other.

Formal charge (F.C):  The charges that assigned to each atom in a molecule or ion by a set of arbitrary rules and don not actually represent the actual charges on the atoms are called as formal charges.

The formal charge is calculated using the formula,

  $\text{F}\text{.C=Valence}\text{electrons-No}\text{of}\text{non-bonding}\text{electrons-}\frac{\text{No}\text{of}\text{bonding}\text{electrons}}{\text{2}}$

The Lewis structure with zero formal charge or least separated formal charges is the preferred structure of the molecule.

Explanation of Solution

$Xe{F}_2:$

The total number of valence electrons in ${\text{XeF}}_{\text{2}}$ is,

Number of valence electrons in xenon=$\left(\text{1}\right)\left(\text{8}\right)\text{=8}$ electrons

Number of valence electrons in fluorine=$\left(\text{2}\right)\left(\text{7}\right)\text{=14}$ electrons

The total number of valence electrons is 22 electrons.

One xenon atom forms two bonds with fluorine that means six electrons form that bonds and the remaining sixteen are used to satisfy the octet rule of fluorine atoms.

The Lewis formula of ${\text{XeF}}_{\text{2}}$ is,

There would be presence of extra two electrons, since only twenty valence electrons are present.

As Xenon belongs to the fifth period, and it can expand the octet rule.  The formal charge for each atom is calculated as,

Formal charge on xenon=$\text{8-6-}\frac{4}{\text{2}}\text{=0}$

Formal charge on fluorine=$\text{7-6-}\frac{\text{2}}{\text{2}}\text{=0}$

$Xe{F}_4:$

The total number of valence electrons in ${\text{XeF}}_4$ is,

Number of valence electrons in xenon=$\left(\text{1}\right)\left(\text{8}\right)\text{=8}$ electrons

Number of valence electrons in fluorine=$\left(4\right)\left(\text{7}\right)\text{=28}$ electrons

The total number of valence electrons is 22 electrons.

One xenon atom forms four bonds with fluorine that means eight electrons form that bonds and the remaining twenty-eight are used to satisfy the octet rule of fluorine atoms.

The Lewis formula of ${\text{XeF}}_4$ is,

There would be presence of extra four electrons, since only thirty two valence electrons are present.

As Xenon belongs to the fifth period, and it can expand the octet rule.  The formal charge for each atom is calculated as,

Formal charge on xenon=$\text{8-6-}\frac{4}{\text{2}}\text{=0}$

Formal charge on fluorine=$\text{7-6-}\frac{\text{2}}{\text{2}}\text{=0}$

$Xe{F}_6:$

The total number of valence electrons in ${\text{XeF}}_{\text{6}}$ is,

Number of valence electrons in xenon=$\left(\text{1}\right)\left(\text{8}\right)\text{=8}$ electrons

Number of valence electrons in fluorine=$\left(6\right)\left(\text{7}\right)\text{=42}$ electrons

The total number of valence electrons is 50 electrons.

One xenon atom forms six bonds with fluorine that means twelve electrons form that bonds and the remaining thirty eight are used to satisfy the octet rule of fluorine atoms.

The Lewis formula of ${\text{XeF}}_{\text{6}}$ is,

There would be presence of extra two electrons, since only forty four valence electrons are present.

As Xenon belongs to the fifth period, and it can expand the octet rule.

The formal charge for each atom is calculated as,

Formal charge on xenon=$\text{8-6-}\frac{4}{\text{2}}\text{=0}$

Formal charge on fluorine=$\text{7-6-}\frac{\text{2}}{\text{2}}\text{=0}$

$XeO{F}_4:$

The total number of valence electrons in ${\text{XeOF}}_{\text{4}}$ is,

Number of valence electrons in xenon=$\left(\text{1}\right)\left(\text{8}\right)\text{=8}$ electrons

Number of valence electrons in fluorine=$\left(4\right)\left(\text{7}\right)\text{=28}$ electrons

Number of valence electrons in oxygen=$\left(\text{1}\right)\left(\text{6}\right)\text{=6}$ electrons

The total number of valence electrons is 42 electrons.

One xenon atom forms four bonds with fluorine and a pi bond with oxygen that means twelve electrons form that bonds and the remaining thirty are used to satisfy the octet rule of fluorine atoms.  A pi bond is used because of extra unassigned electrons.

The Lewis formula of ${\text{XeOF}}_{\text{4}}$ is,

There would be presence of extra two electrons, since only forty valence electrons are present.

As Xenon belongs to the fifth period, and it can expand the octet rule.  The formal charge for each atom is calculated as,

Formal charge on xenon=$\text{8-6-}\frac{4}{\text{2}}\text{=0}$

Formal charge on fluorine=$\text{7-6-}\frac{\text{2}}{\text{2}}\text{=0}$

Formal charge on oxygen=$\text{6-4-}\frac{\text{4}}{\text{2}}\text{=0}$

$Xe{O}_2{F}_2:$

The total number of valence electrons in ${\text{XeO}}_{\text{2}}{\text{F}}_{\text{2}}$ is,

Number of valence electrons in xenon=$\left(\text{1}\right)\left(\text{8}\right)\text{=8}$ electrons

Number of valence electrons in fluorine=$\left(2\right)\left(\text{7}\right)\text{=14}$ electrons

Number of valence electrons in oxygen=$\left(\text{2}\right)\left(\text{6}\right)\text{=12}$ electrons

The total number of valence electrons is 34 electrons.

One xenon atom forms two bonds with fluorine and two pi bonds with oxygen that means eight electrons form that bonds and the remaining twenty-six are used to satisfy the octet rule of fluorine atoms.  A pi bond is used because of extra unassigned electrons.

The Lewis formula of ${\text{XeO}}_{\text{2}}{\text{F}}_{\text{2}}$ is,

As Xenon belongs to the fifth period, and it can expand the octet rule.

The formal charge for each atom is calculated as,

Formal charge on xenon=$\text{8-6-}\frac{4}{\text{2}}\text{=0}$

Formal charge on fluorine=$\text{7-6-}\frac{\text{2}}{\text{2}}\text{=0}$

Formal charge on oxygen=$\text{6-4-}\frac{\text{4}}{\text{2}}\text{=0}$