Chapter 7, Problem 7.49P

(a)

Interpretation Introduction

Interpretation:

The Lewis formula of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ 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

The total number of valence electrons in ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is,

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

Number of valence electrons in oxygen=$\left(\text{4}\right)\left(\text{6}\right)\text{=24}$ electrons

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

The total number of valence electrons is thirty-two.

One sulphur atom forms four bonds oxygen and two oxygen forms a bond with hydrogen atoms that means twelve electrons are used to form that bonds and remaining twenty electrons are used to satisfy the octet rule of atoms.

The Lewis formula is,

The formal charge for each atom is calculated as,

Formal charge on sulphur=$\text{6-0-}\frac{\text{8}}{\text{2}}\text{=+2}$

Formula charge on oxygen (1)=$\text{6-4-}\frac{4}{\text{2}}\text{=0}$

Formula charge on oxygen (2)=$\text{6-6-}\frac{\text{2}}{\text{2}}\text{=-1}$

Formula charge on hydrogen=$\text{1-0-}\frac{\text{2}}{\text{2}}\text{=0}$

The Lewis formula is,

A large charge separation can be seen in the structure, this can be avoided by adding two pi bonds between oxygen and sulphur.  The sulphur atom can expand the octet rule since it belongs to period three.  The final Lewis structure is,

(b)

Interpretation Introduction

Interpretation:

The Lewis formula of ${\text{H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ has to be written.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The total number of valence electrons in ${\text{H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ is,

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

Number of valence electrons in oxygen=$\left(3\right)\left(\text{6}\right)\text{=18}$ electrons

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

The total number of valence electrons is thirty-two.

One sulphur atom forms one bond with other sulphur atom and three bonds oxygen and two oxygen forms a bond with hydrogen atoms that means twelve electrons are used to form that bonds and remaining twenty electrons are used to satisfy the octet rule of atoms.

The Lewis formula is,

The formal charge for each atom is calculated as,

Formal charge on sulphur (3)=$\text{6-6-}\frac{\text{2}}{\text{2}}\text{=-1}$

Formal charge on sulphur (4)=$\text{6-0-}\frac{\text{8}}{\text{2}}\text{=+2}$

Formula charge on oxygen (1)=$\text{6-4-}\frac{4}{\text{2}}\text{=0}$

Formula charge on oxygen (2)=$\text{6-6-}\frac{\text{2}}{\text{2}}\text{=-1}$

Formula charge on hydrogen=$\text{1-0-}\frac{\text{2}}{\text{2}}\text{=0}$

The Lewis formula is,

A large charge separation can be seen in the structure, this can be avoided by adding two pi bonds between oxygen and sulphur.  The sulphur atom can expand the octet rule since it belongs to period three.  The final Lewis structure is,

(c)

Interpretation Introduction

Interpretation:

The Lewis formula of ${\text{H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{7}}$ has to be written.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The total number of valence electrons in ${\text{H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{7}}$ is,

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

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

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

The total number of valence electrons is fifty-six.

One $\text{S-O-S}$ bond is seen.  One oxygen atom forms two bonds with two sulphur atoms.  Each of two sulphur forms another three bonds with oxygen atoms and two oxygen atoms form a bond with hydrogen atom that means twenty electrons form that bonds and the remaining thirty-six are used to satisfy the octet rule of atoms.

The Lewis formula is,

The formal charge for each atom is calculated as,

Formal charge on sulphur=$\text{6-0-}\frac{\text{8}}{\text{2}}\text{=+2}$

Formula charge on oxygen (1)=$\text{6-4-}\frac{4}{\text{2}}\text{=0}$

Formula charge on oxygen (2)=$\text{6-6-}\frac{\text{2}}{\text{2}}\text{=-1}$

Formula charge on hydrogen=$\text{1-0-}\frac{\text{2}}{\text{2}}\text{=0}$

The Lewis formula is,

A large charge separation can be seen in the structure, this can be avoided by adding four pi bonds between oxygen and sulphur.  The sulphur atom can expand the octet rule since it belongs to period three.  The final Lewis structure is,

(d)

Interpretation Introduction

Interpretation:

The Lewis formula of ${\text{H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_6$ has to be written.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The total number of valence electrons in ${\text{H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_6$ is,

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

Number of valence electrons in oxygen=$\left(6\right)\left(\text{6}\right)\text{=36}$ electrons

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

The total number of valence electrons is fifty.

One S-S bonds is formed.  One sulphur atom forms one bond with other sulphur.  Each of two sulphur atom forms another three bonds with oxygen atoms and two oxygen atoms form a bond with hydrogen atom that means eighteen electrons form that bonds and the remaining thirty-two are used to satisfy the octet rule of atoms.

The Lewis formula is,

The formal charge for each atom is calculated as,

Formal charge on sulphur=$\text{6-0-}\frac{\text{8}}{\text{2}}\text{=+2}$

Formula charge on oxygen (1)=$\text{6-4-}\frac{4}{\text{2}}\text{=0}$

Formula charge on oxygen (2)=$\text{6-6-}\frac{\text{2}}{\text{2}}\text{=-1}$

Formula charge on hydrogen=$\text{1-0-}\frac{\text{2}}{\text{2}}\text{=0}$

The Lewis formula is,

A large charge separation can be seen in the structure, this can be avoided by adding four pi bonds between oxygen and sulphur.  The sulphur atom can expand the octet rule since it belongs to period three.  The final Lewis structure is,

(e)

Interpretation Introduction

Interpretation:

The Lewis formula of ${\text{H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}$ has to be written.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The total number of valence electrons in ${\text{H}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}$ is,

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

Number of valence electrons in oxygen=$\left(8\right)\left(\text{6}\right)\text{=48}$ electrons

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

The total number of valence electrons is sixty-two.

One O-O bond is formed from one oxygen atoms that forms one bond with other oxygen atom.  Each of the oxygen atom forms one bond with one sulphur tom.  Each of two sulphur forms another three bonds with oxygen atoms and two oxygen atoms form two bonds with hydrogen atom that means twenty-two electrons form that bonds and the remaining forty are used to satisfy the octet rule of atoms.

The Lewis formula is,

The formal charge for each atom is calculated as,

Formal charge on sulphur=$\text{6-0-}\frac{\text{8}}{\text{2}}\text{=+2}$

Formula charge on oxygen (1)=$\text{6-4-}\frac{4}{\text{2}}\text{=0}$

Formula charge on oxygen (2)=$\text{6-6-}\frac{\text{2}}{\text{2}}\text{=-1}$

Formula charge on hydrogen=$\text{1-0-}\frac{\text{2}}{\text{2}}\text{=0}$

The Lewis formula is,

A large charge separation can be seen in the structure, this can be avoided by adding four pi bonds between oxygen and sulphur.  The sulphur atom can expand the octet rule since it belongs to period three.  The final Lewis structure is,