Chapter 7, Problem 7.62P

Interpretation Introduction

Interpretation:

The Lewis structure, formal charges and resonance structures of nitrogen oxide molecules 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.

Resonance:  Each of the individual Lewis formulas is said to be a resonance form and the use of multiple Lewis formula is called resonance.  The two dashed lines taken together represent a pair of bonding electrons that spread over the two bonds.  Such a superimposed formula is called resonance hybrid because it is a hybrid of the various resonance forms.

Explanation of Solution

$N_{2}O:$

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

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

Number of valence electrons in nitrogen=$\left(\text{2}\right)\left(\text{5}\right)\text{=10}$ electrons

The total number of valence electrons is sixteen.

One nitrogen atom forms one bond with other nitrogen atom and one bond is formed with oxygen atom that means four electrons are used to form that bond and the remaining twelve electrons are used to satisfy the octet rule of atoms.

The Lewis formula is,

There is an incomplete octet on nitrogen atoms that can be avoided by adding two pi bonds between both nitrogen atoms.  The Lewis formula becomes,

The formal charge for each atom is calculated as,

Formal charge on nitrogen (1)=$\text{5-2-}\frac{6}{\text{2}}\text{=0}$

Formal charge on nitrogen (2)=$\text{5-0-}\frac{8}{\text{2}}\text{=+1}$

Formal charge on oxygen=$\text{6-6-}\frac{\text{2}}{\text{2}}\text{=-1}$

The Lewis formula is,

The equivalent resonance structure can be written as,

$NO:$

The total number of valence electrons in $\text{NO}$ is,

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

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

The total number of valence electrons is eleven.

One nitrogen atom forms one bond with oxygen atom that means two electrons are used to form that bond and the remaining nine electrons are used to satisfy the octet rule of atoms.

The Lewis formula is,

There is an incomplete octet on nitrogen and oxygen atoms that can be avoided by adding one-pi bond between both nitrogen and oxygen atoms.  The Lewis formula becomes,

The formal charge for each atom is calculated as,

Formal charge on nitrogen=$\text{5-3-}\frac{4}{\text{2}}\text{=0}$

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

The resonance structures of NO are,

$N_2{O}_3:$

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

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

Number of valence electrons in nitrogen=$\left(2\right)\left(\text{5}\right)\text{=10}$ electrons

The total number of valence electrons is twenty-eight.

Two nitrogen atoms forms bond with each other.  One nitrogen atom forms a bond with one oxygen atom and other with two oxygen atoms that means eight electrons from that bonds and the remaining twenty are used to satisfy the octet rule of atoms.

The Lewis formula is,

There is an incomplete octet on nitrogen and oxygen atoms that can be avoided by adding one-pi bond between both nitrogen and oxygen atoms.  The Lewis formula becomes,

The formal charge for each atom is calculated as,

Formal charge on nitrogen (3)=$\text{5-0-}\frac{8}{\text{2}}\text{=+1}$

Formal charge on nitrogen (4)=$\text{5-2-}\frac{6}{\text{2}}\text{=0}$

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

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

The Lewis formula is,

The resonance structures are,

$N{O}_2:$

The total number of valence electrons in ${\text{NO}}_{\text{2}}$ is,

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

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

The total number of valence electrons is seventeen.

One nitrogen atom forms bonds with two electrons that means four oxygen electrons helps in forming bonds and the remaining thirteen are used to satisfy the octet rule of atoms.

The Lewis formula is,

There is an incomplete octet on nitrogen and oxygen atoms that can be avoided by adding one-pi bond between nitrogen and one oxygen atoms.  The Lewis formula becomes,

The formal charge for each atom is calculated as,

Formal charge on nitrogen=$\text{5-2-}\frac{6}{\text{2}}\text{=0}$

Formal charge on oxygen (1)=$\text{6-5-}\frac{2}{\text{2}}\text{=0}$

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

The Lewis formula is,

The resonance structures are,

$N_2{O}_4:$

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

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

Number of valence electrons in nitrogen=$\left(2\right)\left(\text{5}\right)\text{=10}$ electrons

The total number of valence electrons is thirty-four.

Two nitrogen atoms forms bonds with each other and both of them form a bond with two oxygen atoms that means ten electrons helps in forming bonds and the remaining twenty-four are used to satisfy the octet rule of atoms.

The Lewis formula is,

There is an incomplete octet on nitrogen and two oxygen atoms that can be avoided by adding one-pi bond between nitrogen and oxygen atoms.  The Lewis formula becomes,

The formal charge for each atom is calculated as,

Formal charge on nitrogen=$\text{5-0-}\frac{8}{\text{2}}\text{=+1}$

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

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

The Lewis formula is,

The resonance structures are,

$N_2{O}_5:$

The total number of valence electrons in ${\text{N}}_{\text{2}}{\text{O}}_5$ is,

Number of valence electrons in oxygen=$\left(5\right)\left(\text{6}\right)\text{=30}$ electrons

Number of valence electrons in nitrogen=$\left(2\right)\left(\text{5}\right)\text{=10}$ electrons

The total number of valence electrons is forty.

Two nitrogen atoms are bonded by one oxygen atom and both of the nitrogen atoms have bonds with two oxygen atoms that means twelve electrons are used to form bonds and remaining twenty-eight are used to satisfy the octet rule of atoms.

The Lewis formula is,

There is an incomplete octet on nitrogen and two oxygen atoms that can be avoided by adding one-pi bond between both nitrogen and oxygen atoms.  The Lewis formula becomes,

The formal charge for each atom is calculated as,

Formal charge on nitrogen=$\text{5-0-}\frac{8}{\text{2}}\text{=+1}$

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

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

Formal charge on oxygen (3)=$\text{6-4-}\frac{4}{\text{2}}\text{=0}$

The Lewis formula is,

The resonance structures are,