Chapter 8, Problem 8H.8E

(a)

Interpretation Introduction

Interpretation:

The Lewis structure of oleum has to be drawn.

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.

Answer to Problem 8H.8E

The Lewis structure of oleum is,

Explanation of Solution

Oleum is otherwise known as pyrosulphuric acid and its formula is ${\text{H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{7}}\text{.}$  The total number of valence electrons is calculated as,

The number of valence electrons in hydrogen=$\left(\text{2}\right)\left(\text{1}\right)\text{=2}$ electrons

The number of valence electrons in oxygen=$\left(\text{7}\right)\left(\text{6}\right)\text{=42}$ electrons

The number of valence electrons in sulphur=$\left(\text{2}\right)\left(\text{6}\right)\text{=12}$ electrons

The total number of valence electrons in oleum is fifty-six.

The Lewis structure of oleum is,

(b)

Interpretation Introduction

Interpretation:

The formal charge on sulphur and oxygen in oleum has to be determined.

Concept Introduction:

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.

Answer to Problem 8H.8E

The formal charge in sulphur is zero.

The formal charge in oxygen is zero.

Explanation of Solution

Oleum is otherwise known as pyrosulphuric acid and its formula is ${\text{H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{7}}\text{.}$  The total number of valence electrons is calculated as,

The number of valence electrons in hydrogen=$\left(\text{2}\right)\left(\text{1}\right)\text{=2}$ electrons

The number of valence electrons in oxygen=$\left(\text{7}\right)\left(\text{6}\right)\text{=42}$ electrons

The number of valence electrons in sulphur=$\left(\text{2}\right)\left(\text{6}\right)\text{=12}$ electrons

The total number of valence electrons in oleum is fifty-six.

The Lewis structure of oleum is,

The formal charges on sulphur is,

  $\begin{array}{l}\text{F}\text{.C=6-0-}\frac{\text{12}}{\text{2}}\\ \text{F}\text{.C=6-0-6}\\ \text{F}\text{.C=0}\end{array}$

The formal charge in sulphur is zero.

The formal charges on oxygen $\left(\text{1,4 and 7}\right)$ is,

  $\begin{array}{l}\text{F}\text{.C=6-4-}\frac{\text{4}}{\text{2}}\\ \text{F}\text{.C=6-4-2}\\ \text{F}\text{.C=0}\end{array}$

The formal charge on oxygen $\left(\text{1,4 and 7}\right)$ is zero.

The formal charge on oxygen $\left(\text{2,3,5 and 6}\right)$ is,

  $\begin{array}{l}\text{F}\text{.C=6-4-}\frac{\text{4}}{\text{2}}\\ \text{F}\text{.C=6-4-2}\\ \text{F}\text{.C=0}\end{array}$

The formal charge on oxygen $\left(\text{2,3,5 and 6}\right)$ is zero.

(c)

Interpretation Introduction

Interpretation:

The oxidation number of sulphur in oleum has to be drawn.

Answer to Problem 8H.8E

The oxidation number of sulphur in oleum is $\text{+6}\text{.}$

Explanation of Solution

Oleum is otherwise known as pyrosulphuric acid and its formula is ${\text{H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{7}}\text{.}$  Let x be the oxidation number of sulphur in ${\text{H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{7}}.$  The oxidation number is calculated as,

  $\begin{array}{l}\text{2}\left(\text{+1}\right)\text{+2x+7}\left(\text{-2}\right)\text{=0}\\ \text{2+2x-14=0}\\ \text{2x-12=0}\\ \text{2x=+12}\\ \text{x=+6}\end{array}$

The oxidation number of sulphur in ${\text{H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{7}}$ is $\text{+6}\text{.}$