Chapter 7, Problem 7.48P

(a)

Interpretation Introduction

Interpretation:

The Lewis formula of ${\text{HClO}}_{\text{3}}$ 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 ${\text{HClO}}_{\text{3}}$ is,

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

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

Number of valence electrons in chlorine=$\left(1\right)\left(7\right)\text{=7}$ electrons

The total number of valence electrons is twenty-six.

One chlorine atom forms one bond with hydrogen and oxygen forms three bonds with that mean eight electrons are used to form that bonds and remaining eighteen electrons are used to satisfy the octet rule of oxygen atoms.

The Lewis formula is,

The formal charge for each atom is calculated as,

Formal charge on chlorine=$\text{7-2-}\frac{6}{\text{2}}\text{=+2}$

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 chlorine.  The chlorine 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 ${\text{HNO}}_{\text{2}}$ has to be written.

Concept Introduction:

Refer to part (a).

Explanation of Solution

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

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

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

Number of valence electrons in nitrogen=$\left(1\right)\left(5\right)\text{=5}$ electrons

The total number of valence electrons is eighteen.

One nitrogen atom forms one bond with hydrogen and two bonds with oxygen that means six electrons are used to form that bonds and the remaining twelve 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 nitrogen=$\text{5-2-}\frac{4}{\text{2}}\text{=+1}$

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 one pi bond between oxygen and nitrogen.  The final Lewis structure is,

(c)

Interpretation Introduction

Interpretation:

The Lewis formula of ${\text{HIO}}_{\text{4}}$ has to be written.

Concept Introduction:

Refer to part (a).

Explanation of Solution

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

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

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

Number of valence electrons in iodine=$\left(1\right)\left(7\right)\text{=7}$ electrons

The total number of valence electrons is thirty-two.

One iodine atom forms one bond with hydrogen and four bonds with oxygen that means ten electrons are used to form that bonds and the remaining twenty-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 iodine=$\text{7-0-}\frac{\text{8}}{\text{2}}\text{=+3}$

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 three pi bonds between oxygen and iodine.  The iodine atom can expand the octet rule since it belongs to period five.  The final Lewis structure is,

(d)

Interpretation Introduction

Interpretation:

The Lewis formula of ${\text{HBrO}}_{\text{2}}$ has to be written.

Concept Introduction:

Refer to part (a).

Explanation of Solution

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

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

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

Number of valence electrons in bromine=$\left(1\right)\left(7\right)\text{=7}$ electrons

The total number of valence electrons is twenty.

One bromine atom forms one bond with hydrogen and two bonds with oxygen that means six electrons are used to form that bonds and the remaining twelve 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 bromine=$\text{7-4-}\frac{4}{\text{2}}\text{=+1}$

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 one pi bonds between oxygen and bromine.  The bromine atom can expand the octet rule since it belongs to period four.  The final Lewis structure is,