Systematic Method for Constructing Lewis Representations

Lahbib Abbas^1*, Lahcen Bih², Khalid Yamni³, Abderrahim Elyahyaouy⁴, Abdelmalik El Attaoui⁵, Zahra Ramzi⁶

¹Materials, Membranes and Nanotechnology, Faculty of Science and Technology, Moulay Ismail University Meknes, Er-Rachidia, Morocco.
²Engineering Sciences and Professions Laboratory, National School of Arts and Crafts, Moulay Ismail University Meknes, Meknes, Morocco.
³Materials, Membranes and Nanotechnology, Chemistry Department, Faculty of Sciences, Moulay Ismail University Meknes, Meknes, Morocco.
⁴Physics Department, Faculty of Sciences, Ibn Tofail University, Kenitra, Morocco.
⁵Laboratory of Molecular Chemistry, Materials and Environment (LCM2E), Multidisciplinary Faculty of Nador, Mohamed 1st University Oujda, Nador, Morocco.
⁶Laboratory of SCiMATOP, Chemistry Department, Faculty of Sciences Semlalia, Cadi Ayyad University, Marrakech, Morocco.

DOI: 10.4236/ojic.2024.141001                

Abstract

The systematic method for constructing Lewis representations is a method for representing chemical bonds between atoms in a molecule. It uses symbols to represent the valence electrons of the atoms involved in the bond. Using a number of rules in a defined order, it is often better suited to complicated cases than the Lewis representation of atoms. This method allows us to determine the formal charge and oxidation number of each atom in the edifice more efficiently than other methods.

Keywords

Systematic Method, Lewis Representation, Chemical Bond, Formal Charge, Oxidation Number

1. Introduction

Polyatomic structures are the result of interactions between the valence electrons of atoms that form chemical bonds [1] - [6] .

In a Lewis diagram, atoms are represented by their chemical symbol, surrounded by one or more dots representing valence electrons. Covalent bonds are represented by a line connecting two atoms, with each atom sharing a pair of electrons. Non-bonding electrons, also known as lone electrons, are represented by pairs of electrons located on a single atom.

The use of Lewis diagrams is useful for understanding the molecular geometry, chemical properties and reactivity of molecules [7] - [12] .

The sequence of the eight rules of the systematic method for constructing Lewis representations of buildings are:

Step 1: Determine the total number of valence electrons (n_E.V) in the molecule, which is equal to the sum of the number of valence electrons n_{E V} in each atom of the molecule minus the building block charge according to the following relationship:

n_E.V(building) = ${\displaystyle {\sum }_i^{\text{numberofatomsinthebuilding}}{\text{n}}_{\text{E}\text{.V}}\left(\text{atom}i\right)}$ – charge of the building (Relationship 1)

Step 2: Determine the number of binding and free doublets in the molecule (n_D):

n_D(building) = 1/2 total number of valence electrons of the building n_{E V} (Relationship 2)

If the number of doublets is not integer but half-integer, we are dealing with a radical (single electron).

Step 3: Determine the number of anti-binding doublets in the molecule (n_D.A.L):

n_D.A.L(building) = ${\displaystyle {\sum }_i^{\text{numberofatomsinthebuilding}}{\text{n}}_{\text{D}\text{.A}\text{.L}}\left(\text{atom}i\right)}$ – charge of the building (Relationship 3)

Step 4: Determine the number of binding doublets in the molecule (n_D.L):

n_D.L(building) = n_D(building) – n_D.A.L(building) (Relationship 4)

Step 5: Determine the geometry according to Gillespie, which tells us about the spatial geometry of the species:

The geometry of a molecule around a given atom will be that for which the repulsions between the doublets are minimal.

Gillespie gave the rules for the orientation of bonds around the central atom of a molecule.

Gillespie postulate:

- All pairs of electrons (binding or non-binding) are at the same distance from the nucleus.

- These pairs are assumed to be located on the surface of a sphere with the nucleus at its center.

- These doublets repel each other, and are furthest apart to minimize electron repulsions.

- The V.S.E.P.R formulation is expressed as follows: AX_nE_n': (Relationship 5).

A: Central atom.

X: atoms or molecules covalently bonded to atom A and n their number.

E: free doublets or single electrons and n' their number on the valence layer of atom A.

Gillespie considers that multiple bonds behave like single bonds, to establish the V.S.E.P.R. formulation, it is necessary to establish the Lewis representation.

Step 6: Position any formal load(s).

The formal charge carried by an atom corresponds to the difference in valence electrons between the isolated atom and the atom when the edifice is formed, all bonds being considered purely covalent.

The sum of all formal charges is equal to the charge of the species.

C.F(atom) = n_E.V(atom in isolated state) – n_E.V(atom linked to the structure) (Relationship 6)

Step 7: Build the structure by positioning the doublets (binding—by single, double or triple bond—or non-binding), respecting the octet rule as far as possible and placing any single electron.

Step 8: Determine the oxidation number of each atom in the structure.

The oxidation number of an atom corresponds to the difference in valence electrons between the isolated atom and the atom when the structure is formed, all bonds being considered purely ionic.

The sum of the oxidation numbers is equal to the formal charge of the species

d.o(atom) = n_E.V(atom in isolated state) – n_E.V(atom bound to the structure) (Relationship 7)

2. Constructing Lewis Representations of Buildings Using the Systematic Method

In the following, we present the construction of Lewis representations according to the systematic method for some of the following molecules: ${\text{NH}}_4^{+}$ ; ${\text{NH}}_2^{-}$ ; ${\text{NO}}_2^{-}$ ; PCl₅; SO₂; H₂SO₃; ${\text{Cr}}_2{\text{O}}_7^{2-}$ ; Fe₃O₄; Pb₃O₄; H₂O₂.

2.1. The Building $N{H}_4^{+}$

Step 1:

Applying relation 1, we find: n_{E V}( ${\text{NH}}_4^{+}$ ) = n_E.V(N) + 4n_E.V(H) – 1

Z(N) = 7 from where [He] 2s² 2p³ → n_E.V(N) = 5

and Z(H) = 1 from where 1s¹ → n_E.V(H) = 1

n_E.V( ${\text{NH}}_4^{+}$ ) = 5 + 4 × 1 – 1 = 8

Step 2:

Applying relation 2, we find:

n_D( ${\text{NH}}_4^{+}$ ) = 1/2 × n_{E V}( ${\text{NH}}_4^{+}$ ) = 8/2 = 4 binding and non-binding doublets

Step 3:

Applying relation 3, we find: n_D.A.L( ${\text{NH}}_4^{+}$ ) = n_D.N.L(N) + 4n_D.N.L(H) – 1

Z(N) = 7 from where [He] 2s² 2p³ → n_D.N.L(N) = 1

and Z(H) = 1 from where n_D.N.L(H) = 0

n_D.A.L( ${\text{NH}}_4^{+}$ ) = 1 + 4 × 0 – 1 = 0 free doublets

Step 4:

Applying relation 4, we find:

n_D.L( ${\text{NH}}_4^{+}$ ) = n_D( ${\text{NH}}_4^{+}$ ) – n_D.A.L( ${\text{NH}}_4^{+}$ )

so n_D.L( ${\text{NH}}_4^{+}$ ) = 4 – 0 = 4 Bonding doublets

Step 5:

The V.S.E.P.R formulation is expressed as follows: AX_nE_n':

N: central atom

Four hydrogen atoms covalently bonded to the nitrogen atom and n = 4 their number.

E: free doublets and n' = 0, zero their number on the valence layer of the nitrogen atom.

The type of molecule according to Gillespie is NH₄ and the basic geometry tetrahedral, as shown in the diagram opposite.

Step 6:

Applying relation 6, we find:

C.F(N) = n_E.V(N in isolated state) – n_E.V(N linked to the building) = 5 – 4 = + 1

C.F(H) = n_E.V(H in the isolated state) – n_E.V(H bound to the structure) = 1 – 1 = 0 no charge for the four hydrogens

Step 7:

Build the structure using the systematic method:

2.2. The Species $N{H}_2^{-}$

Step 1:

n_E.V( ${\text{NH}}_2^{-}$ ) = n_E.V(N) + 2n_E.V(H) – (–1)

Z(N) = 7 from where n_E.V(N) = 5 and Z(H) = 1 from where n_E.V(H) = 1

n_E.V( ${\text{NH}}_2^{-}$ ) = 5 + 2 × 1 + 1 = 8

Step 2:

n_D( ${\text{NH}}_2^{-}$ ) = 1/2 × n_E.V( ${\text{NH}}_2^{-}$ ) = 8/2 = 4 binding and non-binding doublets.

Step 3:

n_D.A.L( ${\text{NH}}_2^{-}$ ) = n_D.N.L(N) + 2n_D.N.L(H) – (–1)

Z(N) = 7 from where n_D.N.L(N) = 1 and Z(H) = 1 from where n_D.N.L(H) = 0

n_D.A.L( ${\text{NH}}_2^{-}$ ) = 1 + 2 × 0 + 1 = 2 free doublets.

Step 4:

n_D.L( ${\text{NH}}_2^{-}$ ) = n_D( ${\text{NH}}_2^{-}$ ) – n_D.A.L( ${\text{NH}}_2^{-}$ )

so n_D.L( ${\text{NH}}_2^{-}$ ) = 4 – 2 = 2 Bonding doublets.

Step 5:

The V.S.E.P.R formulation is expressed as follows: AX_nE_n':

N: central atom.

H: two atoms covalently bonded to the N atom and n = 2 is the number of covalent bonds that hydrogen can form with nitrogen.

E: two free doublets on the valence layer of the N atom.

So the Gillespie type of molecule is NH₂E₂ and the basic geometry tetrahedral, as shown in the diagram opposite.

Step 6:

C.F(N) = n_E.V(N in isolated state) – n_E.V(N linked to the building) = 5 – 6 = –1

C.F(H) = n_E.V(H in the isolated state) – n_E.V(H bound to the structure) = 1 – 1 = 0 no charge for the two hydrogens.

Step 7:

Build the structure using the systematic method:

2.3. The Species $N{O}_2^{-}$

Step 1:

n_E.V( ${\text{NO}}_2^{-}$ ) = n_E.V(N) + 2n_E.V(O) – (–1)

Z(N) = 7 from where n_E.V(N) = 5 and Z(O) = 8 from where [He] 2s² 2p⁴ → n_E.V (O) = 6

n_E.V( ${\text{NO}}_2^{-}$ ) = 5 + 2 × 6 + 1 = 18

Step 2:

n_D( ${\text{NO}}_2^{-}$ ) = 1/2 × n_E.V( ${\text{NO}}_2^{-}$ ) = 18/2 = 9 binding and non-binding doublets.

Step 3:

n_D.A.L( ${\text{NO}}_2^{-}$ ) = n_D.N.L(N) + 2n_D.N.L(O) – (–1)

Z(N) = 7 from where n_D.N.L(N) = 1 and Z(O) = 8 from where n_D.N.L(O) = 2

n_D.A.L( ${\text{NO}}_2^{-}$ ) = 1 + 2 × 2 + 1 = 6 free doublets.

Step 4:

n_D.L( ${\text{NO}}_2^{-}$ ) = n_D( ${\text{NO}}_2^{-}$ ) – n_D.A.L( ${\text{NO}}_2^{-}$ )

so n_D.L( ${\text{NO}}_2^{-}$ ) = 9 – 6 = 3 bonding doublets.

Step 5:

The V.S.E.P.R formulation is expressed as follows: AX_nE_n':

N: central atom.

O: two atoms covalently bonded to the N atom and n = 2 is the number of covalent bonds oxygen can form with nitrogen.

E: one free doublet on the valence layer of the N atom.

So the Gillespie type of molecule is NO₂E₁ and the basic geometry triangular planar as shown in the diagram opposite.

Step 6:

C.F(N) = n_E.V(N in isolated state) – n_E.V(N linked to the building) = 5 – 5 = O no formal charge on N.

C.F(O₁) = n_E.V(O₁ in the isolated state) – n_E.V(O₁ bound to the structure) = 6 – 6 = 0 no formal charge on O₁.

C.F(O₂) = n_E.V(O₂ in isolated state) – n_E.V(O₂ bound to the structure) = 6 – 7 = –1.

Step 7:

Build the structure using the systematic method:

2.4. The Species PCl₅

Step 1:

n_{E V}(PCl₅) = n_E.V(P) + 5n_E.V(Cl) – 0

Z(P) = 15 from where [Ne] 2s² 2p³ → n_E.V(P) = 5

and Z(Cl) = 17 from where [Ne] 2s² 2p⁵ → n_E.V(Cl) = 7

n_{E V}(PCl₅) = 5 + 5 × 7 – 0 = 40

Step 2:

n_D(PCl₅) = 1/2 × n_{E V}(PCl₅) = 40/2 = 20 binding and non-binding doublets.

Step 3:

n_D.A.L(PCl₅) = n_D.N.L(P) + 5n_D.N.L(Cl) – 0

Z(P) = 15 from where n_D.N.L(P) = 1 and Z(Cl) = 17 from where n_D.N.L(Cl) = 3

n_D.A.L(PCl₅) = 1 + 5 × 3 – 0 = 16 free doublets.

Step 4:

n_D.L(PCl₅) = n_D(PCl₅) – n_D.A.L(PCl₅)

So n_D.L(PCl₅) = 20 − 16 = 4 Binding doublets.

Step 5:

The V.S.E.P.R formulation is expressed as follows: AX_nE_n':

P: central atom.

Cl: four atoms covalently bonded to the P atom and n = 4 is the number of covalent bonds that chlorine can form with phosphorus.

E: one free doublet on the valence layer of atom P.

So the Gillespie type of molecule is PCl₄E and the basic geometry trigonal bipyramid, as shown in the diagram opposite.

Step 6:

C.F(P) = n_E.V(P in isolated state) – n_E.V(P linked to the building) = 5 − 5 = 0 no formal charge on P.

C.F(Cl) = n_E.V(Cl in isolated state) – n_E.V(Cl bound to the structure) = 7 − 7 = 0 no formal charge on the five Cl atoms.

Step 7:

Build the structure using the systematic method:

2.5. The Species SO₂

Step 1:

n_E.V(SO₂) = n_E.V(S) + 2n_E.V(O) − 0

Z(S) = 16 hence [Ne] 2s² 2p⁴ → n_E.V(S) = 6 and Z(O) = 8 hence n_E.V(O) = 6

n_E.V(SO₂) = 6 + 2 × 6 − 0 = 18

Step 2:

n_D(SO₂) = 1/2 × n_E.V(SO₂) = 18/2 = 9 binding and non-binding doublets.

Step 3:

n_D.A.L(SO₂) = n_D.N.L(S) + 2n_D.N.L(O) − 0

Z(S) = 16 hence n_D.N.L(S) = 2 and Z(O) = 8 hence n_D.N.L(O) = 2

n_D.A.L(SO₂) = 2 + 2 × 2 − 0 = 6 free doublets.

Step 4:

n_D.L(SO₂) = n_D(SO₂) – n_D.A.L(SO₂)

so n_D.L(SO₂) = 9 − 6 = 3 Binding doublets.

Step 5:

The V.S.E.P.R formulation is expressed as follows: AX_nE_n':

S: central atom.

O: two atoms covalently bonded to the S atom and n = 2 is the number of covalent bonds oxygen can form with sulfur.

E: one free doublet on the valence layer of atom S.

So the Gillespie type of molecule is SO₂E and the basic geometry triangular planar, as shown in the diagram opposite.

Step 6:

C.F(S) = n_E.V(S in isolated state) − n_E.V(S linked to the building) = 6 − 5 = +1

C.F(O₁) = n_E.V(O₁ in isolated state) – n_E.V(O₁ bound to the structure) = 6 − 6 = 0 no formal charge on O₁

C.F(O₂) = n_E.V(O₂ in isolated state) – n_E.V(O₂ bound to the structure) = 6 − 7 = −1

C.F(SO₂) = C.F(S) + C.F(O₁) + C.F(O₂) = 1 + 0 − 1 = 0

Step 7:

Build the structure using the systematic method:

2.6. The Species H₂SO₃

Step 1:

n_E.V(H₂SO₃) = n_E.V(S) + 3n_E.V(O) + 2n_E.V(H) − 0

Z(S) = 16 hence n_E.V(S) = 6; Z(H) = 1 hence n_E.V(H) = 1

and Z(O) = 8 hence n_E.V(O) = 6

n_E.V(H₂SO₃) = 6 + 3 × 6 + 2 × 1 − 0 = 26

Step 2:

n_D(H₂SO₃) = 1/2 × n_E.V(H₂SO₃) = 26/2 = 13 binding and non-binding doublets.

Step 3:

n_D.A.L(H₂SO₃) = n_D.N.L(S) + 3n_D.N.L(O) + 2n_D.N.L(H) − 0

Z(S) = 16 hence n_D.N.L(S) = 2; Z(H) = 1 hence n_D.N.L(H) = 0

and Z(O) = 8 hence n_D.N.L(O) = 2

n_D.A.L(H₂SO₃) = 2 + 3 × 2 + 2 × 0 − 0 = 8 free doublets.

Step 4:

n_D.L(H₂SO₃) = n_D(H₂SO₃) – n_D.A.L(H₂SO₃)

so n_D.L(H₂SO₃) = 13 − 8 = 5 Bonding Doublets.

Step 5:

The V.S.E.P.R formulation is expressed as follows: AX_nE_n'’:

S: central atom.

O: three oxygen atoms covalently bonded to the S atom and n = 3 is the number of covalent bonds that oxygen can form with sulfur.

E: two free doublets on the valence layer of atom S.

So the Gillespie type of molecule is H₂SO₃E₂ and the energetically most favorable trigonal bipyramid base geometry in T, as shown in the diagram opposite.

Step 6:

C.F(S) = n_E.V(S in isolated state) – n_E.V(S linked to the building) = 6 − 6 = 0

C.F(O₁) = n_E.V (O₁ in isolated state) – n_E.V(O₁ bound to the structure) = 6 − 6 = 0 no formal charge on O₁

C.F(O₂) = n_E.V(O₂ in isolated state) – n_E.V(O₂ bound to the structure) = 6 − 6 = 0 no formal charge on O₂

C.F(O₃) = n_E.V(O₃ in isolated state) – n_E.V(O₃ bound to the structure) = 6 − 6 = 0 no formal charge on O₃

C.F(H) = n_E.V(H in isolated state) – n_E.V(H linked to the building) = 1 − 1 = 0 no formal charge on twice H

C.F(H₂SO₃) = C.F(S) + C.F(O₁) + C.F(O₂) + C.F(O₃) + 2 C.F(H) = 0 + 3 × 0 + 2 × 0 = 0

Step 7:

Build the structure using the systematic method:

2.7. The Species $C{r}_2{O}_7^{2-}$

Step 1:

n_E.V( ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ ) = 2n_E.V(Cr) + 7n_E.V(O) – (−2)

Z(Cr) = 24 where Cr’s valence layer is: [Ar] 3d⁵ 4s¹ → n_E.V(Cr) = 6;

Z(O) = 8 from which the valence layer of O is: [He] 2s² 2p⁴ n_E.V(O) = 6;

n_E.V( ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ ) = 2 × 6 + 7 × 6 − (−2) = 56

Step 2:

n_D( ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ ) = 1/2 × n_E.V( ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ ) = 56/2 = 28 binding and non-binding doublets.

Step 3:

n_D.A.L( ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ ) = 2n_D.N.L(Cr) + 7n_D.N.L(O) – (−2)

Z(Cr) = 24 hence n_D.N.L(Cr) = 1 and Z(O) = 8 hence n_D.N.L(O) = 2

n_{D A L}( ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ ) = 2 × 1 + 7 × 2 – (−2) = 18 free doublets.

Step 4:

n_D.L( ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ ) = n_D( ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ ) – n_D.A.L( ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ )

so n_D.L( ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ ) = 28 − 18 = 10 Bonding Doublets.

Step 5:

The V.S.E.P.R formulation is expressed as follows: AX_nE_n':

Cr: central atom.

O: four oxygen atoms covalently bonded to the Cr atom and n = 4 is the number of covalent bonds that oxygen can form with chromium.

E: one free doublet on the valence layer of the Cr atom.

The Gillespie type of molecule is therefore CrO₄E and the basic geometry trigonal bipyramid, as shown in the diagram opposite.

Step 6:

C.F(Cr) = n_E.V(Cr in isolated state) − n_E.V(Cr bound to the structure) = 6 − 5 = +1

C.F(O₁) = n_E.V(O₁ in isolated state) – n_E.V(O₁ bound to the structure) = 6 − 6 = 0 no formal charge on O₁

C.F(O₂) = n_E.V(O₂ in isolated state) − n_E.V(O₂ bound to the structure) = 6 − 6 = 0 no formal charge on O₂

C.F(O₃) = n_E.V(O₃ in isolated state) – n_E.V(O₃ bound to the structure) = 6 − 7 = −1

C.F(O₄) = n_E.V(O₄ in isolated state) – n_E.V(O₄ bound to the structure) = 6 − 7 = −1

C.F( ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ ) = 2C.F (Cr) + C.F(O₁) + 2C.F(O₂) + 2C.F(O₃) + 2C.F(O₄) = 2 × 1 + 1 × 0 + 2 × 0 +2 × (−1) + 2 × (−1)

C.F( ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ ) = −2

Step 7:

Build the structure using the systematic method:

2.8. The Species Fe₃O₄

Step 1:

n_{E V}(Fe₃O₄) = 3n_E.V(Fe) + 4n_E.V(O) – 0

Z(Fe) = 26 from which the valence layer of Fe is: [Ar] 3d⁶ 4s² → n_E.V(Cr) = 8;

Z(O) = 8 from which the valence layer of O is: [He] 2s² 2p⁴ → n_E.V(O) = 6;

n_E.V(Fe₃O₄) = 3 × 8 + 4 × 6 − 0 = 48

Step 2:

n_D(Fe₃O₄) = 1/2 ×n_E.V(Fe₃O₄) = 48/2 = 24 binding and non-binding doublets.

Step 3:

n_D.A.L(Fe₃O₄) = 3n_D.N.L(Fe) + 4n_D.N.L(O) – 0

Z(Fe) = 26 from where n_D.N.L(Fe) = 2 and Z(O) = 8 from where n_D.N.L(O) = 2

n_D.A.L(Fe₃O₄) = 3 ×2 + 4 × 2 – 0 = 14 free doublets.

Step 4:

n_D.L(Fe₃O₄) = n_D(Fe₃O₄) – n_D.A.L(Fe₃O₄)

Therefore n_D.L(Fe₃O₄) = 24 − 14 = 10 Bonding doublets.

Step 5:

The V.S.E.P.R formulation is expressed as follows: AX_nE_n':

Fe₁: central atom

Two oxygen atoms and one Fe₂ atom covalently bonded to the Fe₁ atom and n = 3 is their number.

E: two free doublets on the valence layer of the Fe₁ atom.

The Gillespie type of molecule is therefore Fe₁O₂Fe₂E₂ and the basic geometry trigonal bipyramid.

Fe₂: central atom

Two iron atoms and one oxygen atom covalently bonded to the central Fe₂ atom and n = 3 their number.

E: two free doublets on the valence shell of the central Fe atom.

The Gillespie type of molecule is therefore Fe₂Fe₁Fe₃OE₂ and the basic geometry trigonal bipyramid, as shown in the diagram opposite.

Step 6:

C.F(Fe₁) = n_E.V(Fe1 in isolated state) − n_E.V(Fe₁ bound to the structure) = 8 − 8 = 0

C.F(Fe₂) = n_E.V(Fe2 in isolated state) − n_E.V(Fe₂ bound to the structure) = 8 − 8 = 0

C.F(Fe₃) = n_E.V(Fe3 in isolated state) − n_E.V(Fe₃ bound to the structure) = 8 − 8 = 0

C.F(O) = n_E.V(O in isolated state) − n_E.V(O bound to the structure) = 6 − 6 = 0 no formal charge on the three O.

C.F(Fe₃O₄) = 3C.F(Fe) + 4C.F(O) = 3 × 0 + 4 × 0

C.F(Fe₃O₄) = 0

Step 7:

Build the structure using the systematic method: same layout as described for geometry.

2.9. The Species Pb₃O₄

Step 1:

n_E.V(Pb₃O₄) = 3 n_E.V(Pb) + 4n_E.V(O) – 0

Z(Pb) = 82 from which the valence layer of Pb is [Xe] 4f¹⁴ 5d¹⁰ 6s²6p² → n_E.V(Pb) = 4;

Z(O) = 8 from which the valence layer of O is: [He] 2s² 2p⁴ → n_E.V(O) = 6;

n_E.V(Pb₃O₄) = 3 × 4 + 4 × 6 − 0 = 36

Step 2:

n_D(Pb₃O₄) = 1/2 × n_E.V(Pb₃O₄) = 36/2 = 18 binding and non-binding doublets.

Step 3:

n_D.A.L(Pb₃O₄) = 3n_D.N.L(Pb) + 4n_D.N.L(O) – 0

Z(Pb) = 82 from where n_D.N.L(Pb) = 1 and Z(O) = 8 from where n_D.N.L(O) = 2

n_D.A.L(Pb₃O₄) = 3 × 1 + 4 × 2 – 0 = 11 free doublets.

Step 4:

n_D.L(Pb₃O₄) = n_D(Pb₃O₄) – n_D.A.L(Pb₃O₄)

so n_D.L(Pb₃O₄) = 18 − 11 = 7 Bonding Doublets.

Step 5:

The V.S.E.P.R formulation is expressed as follows: AX_nE_n':

Pb₁: central atom

two oxygen atoms and one Pb₂ atom covalently bonded to the

central Pb₁ atom and n = 3 in number.

E: zero free doublet on the valence layer of the central atom Pb₁.

So the Gillespie type of molecule is Pb₁O₂Pb₂ and the basic geometry triangular planar.

Pb₂: central atom

two lead atoms and one oxygen atom covalently bonded to the central Pb₂ atom and n = 3 their number.

E: zero free doublets on the valence layer of the central atom Pb₂.

So the Gillespie type of molecule is Pb₂Pb₁Pb₃O and the basic geometry triangular planar, as shown in the diagram opposite.

Step 6:

C.F(Pb₁) = n_E.V(Pb₁ in isolated state) − n_E.V(Pb₁ linked to the building) = 4 − 4 = 0

C.F(Pb₂) = n_E.V(Pb₂ in isolated state) − n_E.V(Pb₂ linked to the building) = 4 − 4 = 0

C.F(Pb₃) = n_E.V(Pb₃ in isolated state) − n_E.V(Pb₃ linked to the building) = 4 − 4 = 0

C.F(O) = n_E.V(O in isolated state) − n_E.V(O bound to the structure) = 6 − 6 = 0 no formal charge on the three O

C.F(Pb₃O₄) = 3C.F(Pb) + 4 C.F(O) = 3 × 0 + 4 × 0

C.F(Pb₃O₄) = 0

Step 7:

Build the structure using the systematic method:

2.10. The Species H₂O₂

Step 1:

n_E.V(H₂O₂) = 2n_E.V(O) + 2n_E.V(H) − 0

Z(O) = 8 from where n_E.V(O) = 6 and Z(H) = 1 from where n_E.V(H) = 1

n_E.V(H₂O₂) = 2 × 6 + 2 × 1 − 0 = 14

Step 2:

n_D(H₂O₂) = 1/2 ×n_E.V(H₂O₂) = 14/2 = 7 binding and non-binding doublets.

Step 3:

n_D.A.L(H₂O₂) = 2n_D.N.L(O) + 2n_D.N.L(H) − 0

Z(O) = 8 from where n_D.N.L(O) = 2 and Z(H) = 1 from where n_D.N.L(H) = 0

n_D.A.L(H₂O₂) = 2 × 2 + 2 × 0 – 0 = 4 free doublets.

Step 4:

n_D.L(H₂O₂) = n_D(H₂O₂) – n_D.A.L(H₂O₂)

Therefore n_D.L(H₂O₂) = 7 − 4 = 3 Bonding Doublets.

Step 5:

The V.S.E.P.R formulation is expressed as follows: AX_nE_n':

O: central atom.

An oxygen atom and a hydrogen atom covalently bonded to the O atom and n = 2 is the number of covalent bonds that oxygen 1 can form with oxygen 2 and hydrogen.

E: free doublets and 2 their number on the valence layer of the O₁ atom.

So the Gillespie type of molecule is O₁H₁E₂ and the basic geometry is tetrahedral angled, as shown in the diagram opposite.

Step 6:

C.F(O) = n_E.V(O in isolated state) – n_E.V(O bound to the structure) = 6 − 6 = 0 no charge for both oxygens

C.F(H) = n_E.V(H in the isolated state) – n_E.V(H bound to the structure) = 1 − 1 = 0 no charge for the two hydrogens

C.F(H₂O₂) = 2C.F(O) + 2C.F(H) = 2 × 0 + 2 × 0 = 0

Step 7:

Build the structure using the systematic method:

3. Oxidation Number of Atoms in Buildings Using the Systematic Method

We determine the degree of oxidation of the various atoms in the structure using the systematic method for some of the following molecules: ${\text{NH}}_4^{+}$ ; ${\text{NH}}_2^{-}$ ; ${\text{NO}}_2^{-}$ ; PCl₅; SO₂; H₂SO₃; ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ ; Fe₃O₄; Pb₃O₄; H₂O₂.

3.1. The Species $N{H}_4^{+}$

Step 8:

We have χ(N) > χ(H), so the nitrogen atom will attract towards itself the electrons involved with hydrogen.

d.o(N) = n_E.V(N atom in isolated state) – n_E.V(N atom bound to the structure) = 5 − 8 = −3

d.o(H) = n_E.V(H atom in isolated state) – n_E.V(H atom bound to the structure) = 1 − 0 = 1

C.F( ${\text{NH}}_4^{+}$ ) = d.o(N) + 4d.o(H) = −3 + 4 × 1 = +1

3.2. The Species $N{H}_2^{-}$

Step 8:

We have χ(N) > χ(H), so the nitrogen atom will attract towards itself the electrons involved with hydrogen.

d.o (N) = n_E.V (N atom in isolated state) – n_E.V (N atom bound to the structure) = 5 − 8 = −3

d.o (H) = n_E.V (H atom in isolated state) – n_E.V (H atom bound to the structure) = 1 − 0 = 1

C.F ( ${\text{NH}}_2^{-}$ ) = d.o (N) + d.o (H₁) + d.o (H₂) = −3 + 2 × 1 = −1

3.3. The Species $N{O}_2^{-}$

We have χ(O) > χ(N), so the oxygen atom will attract to itself the electrons involved with nitrogen.

d.o(N) = n_E.V(N atom in isolated state) – n_E.V(N atom bound to the structure) = 5 − 2 = 3

d.o(O₁) = n_E.V(O atom1 in isolated state) – n_E.V(O atom1 bound to the edifice) = 6 − 8 = −2

d.o(O₂) = n_E.V(O atom2 in isolated state) – n_E.V(O atom2 bound to the edifice) = 6 - 8 = - 2

C.F( ${\text{NO}}_2^{-}$ ) = d.o(N) + d.o(O₁) + d.o(O₂) = 3 + 2 × (−2) = −1

3.4. The Species PCl₅

We have χ(Cl) > χ(P), so the chlorine atom will attract to itself the electrons involved with the phosphorus.

d.o(Cl) = n_E.V(Cl atom in isolated state) – n_E.V(Cl atom bound to the structure) = 7 − 8 = −1

d.o(P) = n_E.V(P atom in isolated state) − n_E.V(P atom bound to the structure) = 5 − 0 = +5

C.F(PCl₅) = d.o(S) + 5d.o(Cl) = 5 + 5 × (−1) = 0

3.5. The Species SO₂

We have χ(O) > χ(S), so the oxygen atom will attract to itself the electrons involved with sulfur.

d.o(S) = n_E.V(S atom in isolated state) – n_E.V(S atom bound to the structure) = 6 − 2 = +4

d.o(O₁) = n_E.V(O atom1 in isolated state) – n_E.V(O atom1 bound to the edifice) = 6 − 8 = −2

d.o(O₂) = n_E.V(O atom2 in isolated state) – n_E.V(O atom2 bound to the edifice) = 6 − 8 = −2

C.F(SO₂) = d.o(S) + d.o(O₁) + d.o(O₂) = 4 + 2 × (−2) = 0

3.6. The Species H₂SO₃

We have χ(O) > χ(S) and χ(O) > χ(S), so the oxygen atom will attract the electrons involved with sulfur and hydrogen.

d.o(S) = n_E.V(S atom in isolated state) – n_E.V(S atom bound to the structure) = 6 − 2 = +4

d.o(O₁) = n_E.V(O atom1 in isolated state) – n_E.V(O atom1 bound to the edifice) = 6 − 8 = −2

d.o(O₂) = n_E.V(O atom2 in isolated state) – n_E.V(O atom2 bound to the edifice) = 6 − 8 = −2

d.o(O₃) = n_E.V (O atom3 in isolated state) – n_E.V(O atom3 bound to the edifice) = 6 − 8 = −2

d.o(H) = n_E.V (H atom in isolated state) – n_E.V(H atom bound to the structure) = 1 − 0 = +1

C.F(H₂SO₃) = d.o(S) + d.o(O₁) + d.o(O₂) + d.o(O₃) + 2d.o(H) = 4 + 3 × (−2) + 2 × 1 = 0

3.7. The Species $C{r}_2{O}_7^{2-}$

We have χ(O) > χ(Cr), so the oxygen atom will draw the electrons involved with chromium towards itself.

d.o(Cr) = n_E.V(Cr atom in the isolated state) – n_E.V(Cr atom bound to the structure) = 6 − 0 = + 6

d.o(O₁) = n_E.V(O atom1 in isolated state) - n_E.V(O atom1 bound to the edifice) = 6 − 8 = −2

d.o(O₂) = n_E.V(O atom2 in isolated state) – n_E.V(O atom2 bound to the edifice) = 6 − 8 = −2

d.o(O₃) = n_E.V(O atom3 in isolated state) – n_E.V(O atom3 bound to the edifice) = 6 − 8 = −2

d.o(O₄) = n_E.V(O atom4 in isolated state) – n_E.V(O atom4 bound to the edifice) = 6 − 8 = −2

C.F( ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ ) = 2d.o(Cr) + d.o(O₁) + 2d.o(O₂) + 2d.o(O₃) + 2d.o(O₄) = 2 × 6 + 1 × (−2) + 2 × (−2) + 2 × (−2)

C.F( ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ ) = −2

3.8. The Species Fe₃O₄

We have χ(O) > χ(Fe), so the oxygen atom will attract towards itself the electrons involved with the iron.

d.o(Fe₁) = n_E.V(Fe atom1 in isolated state) – n_E.V(Fe atom1 bound to the structure) = 8 − 5 = +3

d.o(Fe₂) = n_E.V(Fe atom2 in isolated state) – n_E.V(Fe atom2 bound to the structure) = 8 − 6 = +2

d.o(Fe₃) = n_E.V(Fe atom3 in isolated state) – n_E.V(Fe atom3 bound to the structure) = 8 − 5 = +3

d.o(O) = n_E.V(O atom in isolated state) – n_E.V(O atom bound to the structure) = 6 − 8 = −2

C.F(Fe₃O₄) = d.o(Fe₁) + d.o(Fe₂) + d.o(Fe₃) + 4d.o(O) = 3 + 2 + 3 + 4 × (−2) C.F(Fe₃O₄) = 0

3.9. The Species Pb₃O₄

We have χ(O) > χ(Pb), so the oxygen atom will attract towards itself the electrons involved with lead.

d.o(Pb₁) = n_E.V(Pb atom1 in isolated state) – n_E.V(Pb atom1 bound to the structure) = 4 − 1 = +3

d.o(Pb₂) = n_E.V(Pb atom2 in isolated state) – n_E.V(Pb atom2 bound to the structure) = 4 − 2 = +2

d.o(Pb₃) = n_E.V(Pb atom3 in isolated state) – n_E.V(Pb atom3 bound to the structure) = 4 − 1 = +3

d.o(O) = n_E.V(O atom in isolated state) – n_E.V(O atom bound to the structure) = 6 − 8 = −2

C.F(Pb₃O₄) = d.o(Pb₁) + d.o(Pb₂) + d.o(Pb₃) + 4d.o(O) = 3 + 2 + 3 + 4 × (−2) C.F(Pb₃O₄) = 0

3.10. The Species H₂O₂

We have χ(O) > χ(H), so the oxygen atom will attract towards itself the electrons brought into play with hydrogen.

d.o(O₁) = d.o(O₂) = n_E.V(O atom in isolated state) - n_E.V(O atom bound to the structure) = 6 − 7 = −1

d.o(H₁) = d.o(H₂) = n_E.V(H atom in isolated state) – n_E.V(H atom bound to the structure) = 1 − 0 = +1

C.F(H₂O₂) = 2d.o(O) + 2d.o(H) = 2 × (−1) + 2 × 1 = 0

The importance of this systematic method for the Lewis representation of molecules lies in determining the oxidation state of each atom in the structure [7] - [12] .

4. Conclusion

The systematic method for constructing Lewis representations of molecules is an excellent method for simultaneously determining the Gillespie geometry, formal charge and oxidation state of molecules, and is better than the other methods Lewis representations of atoms as well as the linear combination method of atomic orbitals [1] - [6] .

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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