Step 1:

Applying relation 1, we find: n_{E V}( 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( NH 4 + ) = 5 + 4 × 1 – 1 = 8

Step 2:

Applying relation 2, we find:

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

Step 3:

Applying relation 3, we find: n_D.A.L( 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( NH 4 + ) = 1 + 4 × 0 – 1 = 0 free doublets

Step 4:

Applying relation 4, we find:

n_D.L( NH 4 + ) = n_D( NH 4 + ) – n_D.A.L( NH 4 + )

so n_D.L( 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:

Step 1:

n_E.V( 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( NH 2 − ) = 5 + 2 × 1 + 1 = 8

Step 2:

n_D( NH 2 − ) = 1/2 × n_E.V( NH 2 − ) = 8/2 = 4 binding and non-binding doublets.

Step 3:

n_D.A.L( 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( NH 2 − ) = 1 + 2 × 0 + 1 = 2 free doublets.

Step 4:

n_D.L( NH 2 − ) = n_D( NH 2 − ) – n_D.A.L( NH 2 − )

so n_D.L( 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:

Step 1:

n_E.V( 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( NO 2 − ) = 5 + 2 × 6 + 1 = 18

Step 2:

n_D( NO 2 − ) = 1/2 × n_E.V( NO 2 − ) = 18/2 = 9 binding and non-binding doublets.

Step 3:

n_D.A.L( 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( NO 2 − ) = 1 + 2 × 2 + 1 = 6 free doublets.

Step 4:

n_D.L( NO 2 − ) = n_D( NO 2 − ) – n_D.A.L( NO 2 − )

so n_D.L( 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:

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:

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:

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:

Step 1:

n_E.V( Cr 2 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( Cr 2 O 7 2 − ) = 2 × 6 + 7 × 6 − (−2) = 56

Step 2:

n_D( Cr 2 O 7 2 − ) = 1/2 × n_E.V( Cr 2 O 7 2 − ) = 56/2 = 28 binding and non-binding doublets.

Step 3:

n_D.A.L( Cr 2 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}( Cr 2 O 7 2 − ) = 2 × 1 + 7 × 2 – (−2) = 18 free doublets.

Step 4:

n_D.L( Cr 2 O 7 2 − ) = n_D( Cr 2 O 7 2 − ) – n_D.A.L( Cr 2 O 7 2 − )

so n_D.L( Cr 2 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( Cr 2 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( Cr 2 O 7 2 − ) = −2

Step 7:

Build the structure using the systematic method:

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.

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:

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:

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( NH 4 + ) = d.o(N) + 4d.o(H) = −3 + 4 × 1 = +1

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 ( NH 2 − ) = d.o (N) + d.o (H₁) + d.o (H₂) = −3 + 2 × 1 = −1

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( NO 2 − ) = d.o(N) + d.o(O₁) + d.o(O₂) = 3 + 2 × (−2) = −1

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

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

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

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( Cr 2 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( Cr 2 O 7 2 − ) = −2

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

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

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] .