Chapter 12.16 - Resonance Effect

In the previous section, we saw the details about stability of resonance structures. In this section, we will see resonance effect.

• We know that inorganic molecules like HCl are polar.
    ♦ The Cl atom being electronegative, pulls the electron density in the covalent bond.
    ♦ This gives Cl a partial -ve charge. Also H gets a partial +ve charge.
    ♦ See section 4.10.
• Some organic molecules are also polar. This can be demonstrated using two examples.

Example 1:
In this example, we analyze the resonance structures of aniline. It can be written in 6 steps:
1. The structure I of fig.12.101 below shows aniline. In this structure, all the atoms have octet.

Fig.12.101

• The structure II is obtained by rearranging the electrons present in I
• In I, NㅡC₁ is a single bond. But this same NㅡC₁ becomes a double bond in II.
• This is because of the movement of electrons indicated by the green curved-arrow.
   ♦ Two electrons from the lone pair of N moves to NㅡC₁
   ♦ Those two electrons were singly-owned by N.
   ♦ After the formation of N=C₁, the N co-owns those electrons.
   ♦ So N does not lose any electrons.
• So due to the green curved-arrow,
   ♦ N does not lose any electrons.
   ♦ C₁ gains two electrons.
• Note the + formal charge of N in II. This can be explained in 3 steps:
(i) An independent N atom will have five electrons around it.
(ii) But in II, there are only four electrons around N.
    ♦ Two from the bonds with the H atoms.
    ♦ Two from the double bond.
(iii) That means, N has lost an electron. Thus it gains one +ve formal charge. 
• Due to the green curved-arrow in I, C₁ now possess two new electrons.
• But C₁ already has octet. It does not need any electrons.
• So the magenta curved-arrow comes into play.
   ♦ Two electrons in C₁=C₂ moves to C₂
   ♦ They become a lone pair of C₂
• Thus the two unwanted electrons gained by C₁ through the green curved-arrow, are lost by the action of magenta curved arrow.
• Note the - formal charge of C₂ in II. This can be explained in 3 steps:
(i) An independent C atom will have four electrons around it.
(ii) But in II, there are five electrons around C₂
    ♦ Two from the lone pair.
    ♦ One from C₂ㅡC₁
    ♦ One from C₂ㅡC₃
    ♦ One from C₂ㅡH
(iii) That means, C₂ has gained an electron. Thus it gains one -ve formal charge. 
2. From the above step (1), we know how structure II is formed. So let us analyze the curved-arrows in structure II
• The C₂ in II has two unwanted electrons. So the yellow curved-arrow comes into play.
    ♦ The lone pair of C₂ moves to C₂ㅡC₃.
    ♦ Thus C₂ㅡC₃  becomes C₂=C₃.
    ♦ This double bond is shown in III
• When such a double bond is formed, C₃ will have two unwanted electrons.
    ♦ So the cyan curved-arrow comes into play.
    ♦ Two electrons in C₃=C₄ moves to C₄
    ♦ They become a lone pair of C₄
• Thus the two unwanted electrons gained by C₃ through the yellow curved-arrow, are lost by the action of cyan curved arrow.
• Note the - formal charge of C₄ in III. This can be explained in 3 steps:
(i) An independent C atom will have four electrons around it.
(ii) But in II, there are five electrons around C₄
    ♦ Two from the lone pair.
    ♦ One from C₄ㅡC₃
    ♦ One from C₄ㅡC₅
    ♦ One from C₄ㅡH
(iii) That means, C₄ has gained an electron. Thus it gains one -ve formal charge.
3. From the above step (2), we know how structure III is formed. We can write similar steps to analyze the curved-arrows in structure III.
• Based on those steps, we would see how structure IV is formed.
• The reader is advised to write those steps in his/her own notebook.
4. In effect, we see that:
• Structure I is neutral.
• Structures II, III and IV are not neutral. They are charged.
    ♦ This is due to the movement of electrons as indicated by the curved arrows.
    ♦ Positions with -ve formal charge have high electron densities.
    ♦ Positions with +ve formal charge have low electron densities.
5. So one contributing structure is neutral while the remaining three contributing structures are charged.
• So the hybrid structure will be charged. That means, in a sample of aniline, the molecules will exist in a charged state. This is just like charged molecules in a sample of HCl.
6. In this example, the electrons are transferred away from the substituent group (ㅡNH₂ group).
• If the transfer of electrons is away from an atom or substituent group attached to the conjugated system, it is known as positive resonance effect (+R effect)
• A conjugated system is that system in which single bonds and double bonds occur in an alternating arrangement.

Example 2:
In this example, we analyze the resonance structures of nitrobenzene. It can be written in 6 steps:
1. The structure I of fig.12.102 below shows nitrobenzene. In this structure, all the atoms have octet.

Fig.12.102

• The arrow between N and O indicates that, the bond between N and that O is a coordinate bond. Both electrons in that bond originally belonged to N. However, that coordinate bond has no role in our present discussion.
• The structure II is obtained by rearranging the electrons present in I
• Consider the N=O double bond in I. This same N=O becomes a single bond in II.
• This is because of the movement of electrons indicated by the green curved-arrow.
   ♦ Two electrons in the N=O moves to O.
   ♦ They become a lone pair of O
• Due to this movement of electrons, N has lost two electrons. So the magenta curved-arrow comes into play.
   ♦ Two electrons in C₁=C₂ moves to NㅡC₁
   ♦ Thus C₁=C₂ becomes C₁ㅡC₂ and NㅡC₁ becomes N=C₁
• Note the - formal charge of O in II. This can be explained in 3 steps:
(i) An independent O atom will have six electrons around it.
(ii) But in II, there are seven electrons around that O
    ♦ Six from three lone pairs.
    ♦ One from NㅡO
(iii) That means, O has gained an electron. Thus it gains one -ve formal charge.
• Note the + formal charge of C₂ in II. This can be explained in 3 steps:
(i) An independent C atom will have four electrons around it.
(ii) But in II, there are only three electrons around C₂.
    ♦ One from bond with the H atom.
    ♦ One from C₂ㅡC₁
    ♦ One from C₂ㅡC₃
(iii) That means, C₂ has lost an electron. Thus it gains one +ve formal charge. 
• Thus the two electrons lost by N through the green curved-arrow, are gained by the action of magenta curved arrow.
2. From the above step (1), we know how structure II is formed. So let us analyze the curved-arrow in structure II
• The C₂ in II needs two electrons. So the yellow curved-arrow comes into play.
   ♦ Two electrons in C₃=C₄ moves to C₂ㅡC₃
   ♦ Thus C₃=C₄ becomes C₃ㅡC₄ and C₂ㅡC₃ becomes C₂=C₃.
   ♦ This is shown in III.
• Note the + formal charge of C₄ in III. This can be explained in 3 steps:
(i) An independent C atom will have four electrons around it.
(ii) But in III, there are only three electrons around C₄.
    ♦ One from bond with the H atom.
    ♦ One from C₄ㅡC₃
    ♦ One from C₄ㅡC₅
(iii) That means, C₄ has lost an electron. Thus it gains one +ve formal charge. 
3. From the above step (2), we know how structure III is formed. We can write similar steps to analyze the curved-arrow in structure III.
• Based on those steps, we would see how structure IV is formed.
• The reader is advised to write those steps in his/her own notebook.
4. In effect, we see that:
• Structure I is neutral.
• Structures II, III and IV are not neutral. They are charged.
    ♦ This is due to the movement of electrons as indicated by the curved arrows.
    ♦ Positions with -ve formal charge have high electron densities.
    ♦ Positions with +ve formal charge have low electron densities.
5. So one contributing structure is neutral while the remaining three contributing structures are charged.
• So the hybrid structure will be charged. That means, in a sample of nitrobenzene, the molecules will exist in a charged state. This is just like charged molecules in a sample of HCl.
6. In this example, the electrons are transferred towards the substituent group (ㅡNO₂ group).
• If the transfer of electrons is towards an atom or substituent group attached to the conjugated system, it is known as negative resonance effect (-R effect).

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◼ Note that:
• In example 1 (+R effect), the C atoms in the chain acquired greater electron densities. This is because of the movement of electrons away from the substituent group.
• In example 2 (-R effect), the C atoms in the chain lost electron densities. This is because of the movement of electrons towards the substituent group.

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• The atoms or substituent groups which cause +R effect are:
ㅡ halogen, ㅡOH, ㅡOR, ㅡOCOR, ㅡNH₂, ㅡNHR, ㅡNR₂, ㅡNHCOR
• The atoms or substituent groups which cause -R effect are:
ㅡCOOH, ㅡCHO, >C=O, ㅡCN, ㅡNO₂

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• As mentioned before, the resonance effect (+R effect and -R effect) occur in conjugated systems. In conjugated systems, single bonds and double bonds in the chain are present in an alternating arrangement.
• Benzene and substituted benzene compounds are examples of a closed chain conjugated system.
• 1,3-butadiene is an example of an open chain conjugated system. It’s condensed formula is shown below:
CH₃=CH₂ㅡCH₂=CH₃

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◼ Why is it that, the conjugated systems are subjected to +R effect and -R effects ?
• The answer can be written in 3 steps:
1. We know that, in a double bond, there is a 𝜎 bond and a π bond.
• In a 𝜎 bond, there is linear overlap of orbitals. The electrons are locked in position.
• But in a π bond, there is lateral overlap of orbitals. So the electrons are free to move.
2. So the electrons of the π bond move towards an adjacent single bond. That single bond will then become a double bond. We indicated such movements using curved-arrows in the above examples.
• The electrons of the π bond can also move towards an adjacent atom. Those atoms will then become regions of high electron densities. We indicated such movements also using curved-arrows in the above examples.
3. Regions of ‘electron richness’ and ‘electron deficiency’, will make the molecule polar.

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In the next section we will see electromeric effect.

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Resonance structures of Sulfur trioxide molecule and Nitrate ion

We were learning about resonance structures in section 4.9, We saw the examples of ozone molecule, carbon dioxide molecule and carbonate ion. In this section, we will see two more examples

Resonance structures of Sulfur trioxide

1. Let us first draw the Lewis dot structure of SO₃ (Sulfur trioxide)
(We have seen the steps in an earlier section 4.2)
Step 1: Finding the number of dots
• Number of valence electrons of C = 6
• Number of valence electrons of O = 6
• So total number of valence electrons = [6+(3 × 6)] = 24
Step 2: The skeletal structure
• The skeletal structure is shown in fig.4.56(a) below:

Fig.4.56

Step 3: Preliminary single bonds
• The four atoms are joined by '─' as shown in fig.4.55(b) above
Step 4: Preliminary distribution of electrons
• First make the three outer O atoms octet
• This is shown in fig.c
• In the fig.c, we see that:
    ♦ The valence electrons of the O atoms are shown in green color
    ♦ The valence electrons of the C atom are shown in red color
• All the O atoms have 8 electrons (including the one red dot in the single bonds)
    ♦ So the three O atoms use up (3 × 8) = 24 electrons
    ♦ The number of remaining electrons = (24-24) = 0
• There are no more electrons to distribute
Step 5: Check for octet
• All the O atoms have got 8 electrons each
• The S atom has got only 6 electrons
    ♦ So this atom needs 2 more electrons
■ Rearrangement: Change the preliminary single bond
    ♦ Take a lone pair from the top O atom
    ♦ Using those electrons, change the top single bond to double bond as shown in the fig.d
• Now all atoms have octet
• The structure in fig.4.56(d) is stable
2. Another possible rearrangement
• We already know how to obtain the structure in fig.4.56(d) above
    ♦ We obtained it by working from fig.4.56(c)
• That same fig.4.56(c) is shown again in fig.4.57(c) below:

Fig.4.57

• Earlier, we took a pair from the top O atom
    ♦ This time, we take a pair from the left O atom and make a double bond
    ♦ This is shown in fig.4.57(d')
• The structure in fig.4.57(d') is stable
3. Yet another possible rearrangement
• We already know how to obtain the structure in fig.4.56(d) above
    ♦ We obtained it by working from fig.4.56(c)
• That same fig.4.56(c) is shown again in fig.4.58(c) below:

Fig.4.58

• Earlier, we took a pair from the top O atom
    ♦ This time, we take a pair from the right O atom and make a double bond
    ♦ This is shown in fig.4.58(d'')
• The structure in fig.4.58(d'') is stable
4. So we have three possible structures of SO₃
    ♦ The structure in fig.4.56(d)
    ♦ The structure in fig.4.57(d')
    ♦ The structure in fig.4.58(d'')
• They are shown together in fig.4.59 below:

Fig.4.59

5. Now the next question arises:
■ In reality, which is the correct form in which SO₃ exists? Fig.4.59 (d), (d') or (d'')?
• Let us try to find the answer:
(i) A S-O single bond will have a certain length
(ii) A S=O double bond will have a different length
(iii) With this information, we examine the bond lengths in an actual SO₃ molecule
• Surprisingly, the actual values are different from (i) and (ii)
• In fact there are no 'values'. There is only one value
• The distance between S and O atoms in all the three pairs are the same
6. The structures in figs (d) (d') and (d'') are called canonical structures of SO₃
• They are also called resonance structures of SO₃
7. Resonance structures are indicated by giving double headed arrows between them

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Resonance structures of nitrate ion

1. Let us first draw the Lewis dot structure of NO₃^2- (nitrate ion)
(We have seen the steps in an earlier section 4.3)
Step 1: Finding the number of dots
• Number of valence electrons of N = 5
• Number of valence electrons of O = 6
• So total number of valence electrons = [5+(3 × 6)] = 23
• One extra electron is also present
■ We will write the number as two items:
    (a) Total number of ‘available valence electrons’ = 23
    (b) Number of electrons to be added = 1
• Final number =  [(a) ± (b)] = [(a) + (b)] = [23 + 1] = 24 
Step 2: The skeletal structure
• The skeletal structure is shown in fig.4.60(a) below:

Fig.4.60

Step 3: Preliminary single bonds
• The four atoms are joined by '─' as shown in fig.4.60(b) above
Step 4: Preliminary distribution of electrons
(Remember that, the 'available valence electrons' are distributed in this step)
• First make the three outer O atoms octet
    ♦ For that (3×8) = 24 electrons will be required
    ♦ But the number of 'available valence electrons' = 23
• So first, we will make the left and right O atoms octet 
• Then give the remaining electrons to the top O atom
• This is shown in fig.c
• In the fig.c, we see that:
    ♦ The valence electrons of the O atoms are shown in green color
    ♦ The valence electrons of the N atom are shown in red color
• Left and right side O atoms have 8 electrons (including the one red dot in the single bonds)
    ♦ So the two O atoms use up (2 × 8) = 16 electrons
    ♦ The number of remaining electrons = (23-16) = 7
    ♦ These 7 electrons are given to the top O atom
• There are no more electrons to distribute
Step 5: Check for octet
• The left and right side O atoms have got 8 electrons each
• The top O atom has got only 7 electrons
    ♦ So this atom needs 1 more electron
• The N atom has got only 6 electrons
    ♦ So this atom also needs 2 more electrons
■ Rearrangement: Change the preliminary single bond
    ♦ Take a lone pair from the top O atom
    ♦ Using those electrons, change the top single bond to double bond as shown in the fig.d
    ♦ Now the left and right side O atoms have octet
    ♦ The N atom also has octet
    ♦ But the top O atom has got only 7 electrons
• All the 23 electrons are used up. Still, complete octet is not achieved
• So, we will need an external electron
• Get one external electron from any suitable source
• Give it to the top O atom
• This is shown in fig.e
• Now all atoms have octet
(v) But the external electron will create a charge of -1
• So we put the structure inside square brackets and put a -1 at the top right corner
• The structure in fig.4.60(e) is stable
2. Another possible rearrangement
• We already know how to obtain the structure in fig.4.60(e) above
    ♦ We obtained it by working from fig.4.60(b)
• That same fig.4.60(b) is shown again in fig.4.61(b) below:

Fig.4.61

• Earlier, we made the left and right O atoms octet
    ♦ This time, we make the top and right O atoms octet. This is shown in fig.4.61(c')
• Next we take two electrons from the left O atom and make a double bond
    ♦ This is shown in fig.4.61(d')
• Finally, we add the extra electron to the left side O atom to attain octet
    ♦ This is shown in fig.4.61(e')
• The structure in fig.4.61(e') is stable
3. Yet another possible rearrangement
• We already know how to obtain the structure in fig.4.60(e) above
    ♦ We obtained it by working from fig.4.60(b)
• That same fig.4.60(b) is shown again in fig.4.62(b) below:

Fig.4.62

• Earlier, we made the left and right O atoms octet
    ♦ This time, we make the top and left O atoms octet. This is shown in fig.4.62(c'')
• Next we take two electrons from the right O atom and make a double bond
    ♦ This is shown in fig.4.62(d'')
• Finally, we add the extra electron to the right side O atom to attain octet
    ♦ This is shown in fig.4.62(e'')
• The structure in fig.4.62(e'') is stable
4. So we have three possible structures of NO₃^-
    ♦ The structure in fig.4.60(e)
    ♦ The structure in fig.4.61(e')
    ♦ The structure in fig.4.62(e'')
• They are shown together in fig.4.63 below:

Fig.4.63

5. Now the next question arises:
■ In reality, which is the correct form in which NO₃^- exists? Fig.4.63 (e), (e') or (e'')?
• Let us try to find the answer:
(i) A N-O single bond will have a certain length
(ii) A N=O double bond will have a different length
(iii) With this information, we examine the bond lengths in an actual NO₃^- ion
• Surprisingly, the actual values are different from (i) and (ii)
• In fact there are no 'values'. There is only one value
• The distance between N and O atoms in all the three pairs are the same
6. The structures in figs (e) (e') and (e'') are called canonical structures of NO₃^-
• They are also called resonance structures of NO₃^-
7. Resonance structures are indicated by giving double headed arrows between them

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Chapter 4.9 - Resonance Structures

In the previous section, we saw the basics about bond order. In this section, we will see resonance structures. We will explain it using some examples

Example 1: Resonance structures of ozone
1. Let us first draw the Lewis dot structure of O₃ (ozone) molecule
(We have seen the steps in an earlier section 4.2)
Step 1: Finding the number of dots
• Number of valence electrons of O = 6
• So total number of valence electrons = (3 × 6) = 18
Step 2: The skeletal structure
• The skeletal structure is shown in fig.4.44(a) below:

Fig.4.44

Step 3: Preliminary single bonds
• The three atoms are joined by '─' as shown in fig.4.44(b) above
Step 4: Preliminary distribution of electrons
• First make the two outer O atoms octet
• Then give the remaining electrons to the central O atom
• This is shown in fig.c
• In the fig.c, we see that:
    ♦ The valence electrons of the outer O atoms are shown in green color
    ♦ The valence electrons of the central O atom are shown in red color
• Both the outer O atoms have 8 electrons (including the one red dot in the single bonds)
    ♦ So the number of electrons used up for making those O atoms octet = 16
    ♦ So the number of remaining electrons = (18-16) = 2
    ♦ These 2 electrons are given to the central O atom 
• There are no more electrons to distribute
Step 5: Check for octet
• Both the outer O atoms have got 8 electrons each. They have attained octet
• The central O atom has got only 6 electrons. It has not attained octet
• Thus, the preliminary distribution needs to be changed
■ Rearrangement:
• In fig.c, take a lone pair from the left side O atom
• Using those two electrons, convert the left side single bond to a double bond
• This is shown in fig.d
• In fig.d, all the atoms have octet. So it is a stable O₃ molecule
2. Another possible rearrangement
• We already know how to obtain the structure in fig.4.44(d) above
    ♦ We obtained it based on fig.4.44(c)
• That same fig.4.44(c) is shown again in fig.4.45(c) below:

Fig.4.45

• In fig.4.45(c), take a lone pair from the right side O atom
• Using those two electrons, convert the right side single bond to a double bond
• This is shown in fig.4.45(d')
• In fig.d', all the atoms have octet. So it is also a stable O₃ molecule
3. So we have two possible structures of O₃
    ♦ One is the structure in fig.4.44(d)
    ♦ The other is the structure in fig.4.45(d')
• They are shown together in figs.4.46 below:

Fig.4.46

4. Now the next question arises:
■ In reality, which is the correct form in which O₃ exists? Fig.4.46 (d) or (d')?
• The answer can be written in 5 steps:
(i) An O-O single bond will have a length of 148 pm.
    ♦ This is shown in figs.4.47(a) and (b) below.
(ii) An O=O double bond will have a length of 121 pm
    ♦ This is shown in figs.4.47(a) and (b) below.
(iii) With this information, we examine an actual O₃ molecule.
    ♦ We would expect the distance between one pair of O atoms to be 148 pm
    ♦ We would expect the distance between the other pair of O atoms to be 121 pm
(iv) But surprisingly, the actual values are different from both 148 and 121
• In fact there are no 'values'. There is only one value. It is 128 pm
• The distance between atoms in both pairs is 128 pm.
• This is shown in fig.4.47(c) below:

Fig.4.47

• (a) and (b) are the two possible structures that we saw in fig.4.46
• (c) is different from both (a) and (b)
(v) Note the bonds in fig.c. They are neither single bonds nor double bonds
• This is indicated by the dashed lines
■ Fig.c represents the structure of O₃ more accurately
5. The structures in (a) and (b) are called canonical structures
    ♦ They are also called resonance structures
(For some molecules, there will be more than two resonance structures)
6. Resonance structures are indicated by giving double headed arrows between them
• The structure in (c) is called the hybrid of the resonance structures
    ♦ It is also called the resonance hybrid

Example 2: Resonance structures of carbon dioxide
1. Let us first draw the Lewis dot structure of CO₂ (carbon dioxide) molecule
(We have seen the steps in an earlier section 4.2)
Step 1: Finding the number of dots
• Number of valence electrons of C = 4
• Number of valence electrons of O = 6
• So total number of valence electrons = [4+(2 × 6)] = 16
Step 2: The skeletal structure
• The skeletal structure is shown in fig.4.48(a) below:

Fig.4.48

Step 3: Preliminary single bonds
• The three atoms are joined by '─' as shown in fig.4.48(b) above
Step 4: Preliminary distribution of electrons
• First make the two outer O atoms octet
• Then give the remaining electrons to the central C atom
• This is shown in fig.c
• In the fig.c, we see that:
    ♦ The valence electrons of the outer O atoms are shown in green color
    ♦ The valence electrons of the central O atom are shown in red color
• Both the outer O atoms have 8 electrons (including the one red dot in the single bonds)
    ♦ So the number of electrons used up for making those O atoms octet = 16
    ♦ So the number of remaining electrons = (16-16) = 0
• There are no more electrons to distribute
Step 5: Check for octet
• Both the outer O atoms have got 8 electrons each. They have attained octet
• The central C atom has got only 4 electrons. It has not attained octet
• Thus, the preliminary distribution needs to be changed
■ Rearrangement:
• In fig.c, take a lone pair from the left side O atom
    ♦ Using those two electrons, convert the left side single bond to a double bond
• Again, in fig.c, take a lone pair from the right side O atom
    ♦ Using those two electrons, convert the right side single bond to a double bond
• This is shown in fig.d
• In fig.d, all the atoms have octet. So it is a stable CO₂ molecule
2. Another possible rearrangement
• We already know how to obtain the structure in fig.4.48(d) above
    ♦ We obtained it based on fig.4.48(c)
• That same fig.4.48(c) is shown again in fig.4.49(c) below:

Fig.4.49

• In fig.4.49(c), take two lone pairs from the left side O atom
• Using those four electrons, convert the left side single bond to a triple bond
• This is shown in fig.4.49(d')
• In fig.4.49(d'), all the atoms have octet. So it is also a stable CO₂ molecule
3. Yet another possible rearrangement
• The above rearrangement in fig.4.49(d') was based on fig.4.48(c)
• That same fig.4.48(c) is shown again in fig.4.50(c) below:

Fig.4.50

• In fig.4.50(c), take two lone pairs from the right side O atom
• Using those four electrons, convert the right side single bond to a triple bond
• This is shown in fig.4.50(d'')
• In fig.4.50(d''), all the atoms have octet. So it is also a stable CO₂ molecule
4. So we have three possible structures of CO₂
    ♦ The structure in fig.4.48(d)
    ♦ The structure in fig.4.49(d')
    ♦ The structure in fig.4.50(d'')
• They are shown together in fig.4.51 below:

Fig.4.51

5. Now the next question arises:
■ In reality, which is the correct form in which CO₂ exists? Fig.4.51 (d), (d') or (d'')?
• The answer can be written in 4 steps:
(i) A C-O single bond will have a length of 134 pm
(ii) A C=O double bond will have a length of 121 pm
(iii) A C≡O triple bond will have a length of 110 pm
(iv) With this information, we examine the bond lengths in an actual CO₂ molecule
• Surprisingly, the actual values are different from 134, 121 and 110
• In fact there are no 'values'. There is only one value. It is 115 pm
• The distance between C and O atoms in both pairs is 115 pm
6. The structures in figs (d) (d') and (d'') are called canonical structures of CO₂
• They are also called resonance structures of CO₂
7. Resonance structures are indicated by giving double headed arrows between them

Example 3: Resonance structures of carbonate ion

1. Let us first draw the Lewis dot structure of CO₃^2- (carbonate ion)
(We have seen the structure in an earlier section 4.3. But there we did not explore the various possible arrangements)
Step 1: Finding the number of dots
• Number of valence electrons of C = 4
• Number of valence electrons of O = 6
• So total number of valence electrons = [4+(3 × 6)] = 22
• Two extra electrons are also present
■ We will write the number as two items:
    (a) Total number of ‘available valence electrons’ = 22
    (b) Number of electrons to be added = 2
• Final number =  [(a) ± (b)] = [(a) + (b)] = [22 + 2] = 24 
Step 2: The skeletal structure
• The skeletal structure is shown in fig.4.52(a) below:

Fig.4.52

Step 3: Preliminary single bonds
• The four atoms are joined by '─' as shown in fig.4.52(b) above
Step 4: Preliminary distribution of electrons
(Remember that, only the 'available valence electrons' are distributed in this step)
• First make the three outer O atoms octet
    ♦ For that (3×8) = 24 electrons will be required
    ♦ But the number of 'available valence electrons' = 22
• So first, we will make the left and right O atoms octet 
• Then give the remaining electrons to the top O atom
• This is shown in fig.c
• In the fig.c, we see that:
    ♦ The valence electrons of the O atoms are shown in green color
    ♦ The valence electrons of the C atom are shown in red color
• Left and right side O atoms have 8 electrons (including the one red dot in the single bonds)
    ♦ So the two O atoms use up (2 × 8) = 16 electrons
    ♦ The number of remaining electrons = (22-16) = 6
    ♦ These 6 electrons are given to the top O atom
• There are no more electrons to distribute
Step 5: Check for octet
• The left and right side O atoms have got 8 electrons each
• The top O atom has got only 6 electrons
    ♦ So this atom needs 2 more electrons
• The C atom has got only 6 electrons
    ♦ So this atom also needs 2 more electrons
■ Rearrangement: Change the preliminary single bond
    ♦ Change the top single bond to double bond as shown in the fig.d
• Two electrons from the top O is used for making the new bond
    ♦ Now the left and right side O atoms have octet
    ♦ The C atom also has octet
    ♦ But the top O atom has got only 6 electrons
• All the 22 electrons are used up. Still, complete octet is not achieved
• So, we will need external electrons
• Get two external electrons from any suitable source
• Give them to the top O atom
• This is shown in fig.e
• Now all atoms have octet
(v) But the two external electrons will create a charge of -2
• So we put the structure inside square brackets and put a -2 at the top right corner
• The structure in fig.4.52(e) is stable
2. Another possible rearrangement
• We already know how to obtain the structure in fig.4.52(e) above
    ♦ We obtained it by working from fig.4.52(b)
• That same fig.4.52(b) is shown again in fig.4.53(b) below:

Fig.4.53

• Earlier, we made the left and right O atoms octet
    ♦ This time, we make the top and right O atoms octet. This is shown in fig.4.53(c')
• Next we take two electrons from the left O atom and make a double bond
    ♦ This is shown in fig.4.53(d')
• Finally, we add the extra two electrons to the left side O atom to attain octet
    ♦ This is shown in fig.4.53(e')
• The structure in fig.4.53(e') is stable
3. Yet another possible rearrangement
• We already know how to obtain the structure in fig.4.52(e) above
    ♦ We obtained it by working from fig.4.52(b)
• That same fig.4.52(b) is shown again in fig.4.54(b) below:

Fig.4.54

• Earlier, we made the left and right O atoms octet
    ♦ This time, we make the top and left O atoms octet. This is shown in fig.4.54(c'')
• Next we take two electrons from the right O atom and make a double bond
    ♦ This is shown in fig.4.54(d'')
• Finally, we add the extra two electrons to the right side O atom to attain octet
    ♦ This is shown in fig.4.54(e'')
• The structure in fig.4.54(e'') is stable
4. So we have three possible structures of CO₃^2-
    ♦ The structure in fig.4.52(e)
    ♦ The structure in fig.4.53(e')
    ♦ The structure in fig.4.54(e'')
• They are shown together in fig.4.55 below:

Fig.4.55

5. Now the next question arises:
■ In reality, which is the correct form in which CO₃^2- exists? Fig.4.55 (e), (e') or (e'')?
• The answer can be written in 3 steps:
(i) A C-O single bond will have a length of 134 pm
(ii) A C=O double bond will have a length of 121 pm
(iii) With this information, we examine the bond lengths in an actual CO₃^2- molecule
• Surprisingly, the actual values are different from 134 and 121
• In fact there are no 'values'. There is only one value. It is 128 pm
• The distance between C and O atoms in all the three pairs is 128 pm
6. The structures in figs (e) (e') and (e'') are called canonical structures of CO₃^2-
• They are also called resonance structures of CO₃^2-
7. Resonance structures are indicated by giving double headed arrows between them

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• We will practice using 2 more examples: SO₃ and NO₃^-
    ♦ The steps can be seen here

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■ Now we can write the definition of resonance structures. It can be written in 3 steps:

(i) Sometimes, a single Lewis structure cannot describe a molecule accurately
• We may have to show two or more structure
(ii) Those structures will have similar energies
• Also positions of atoms will be similar in those structures
• But 'lone pairs' and 'bonds' will be different
• Those structures are called canonical structures or resonance structures
(iii) None of the resonance structures can be used to represent the actual structure
• The actual structure is more accurately described by a structure called hybrid of the resonance structures
• This structure is also called the resonance hybrid

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■ We must always remember two important points related to resonance:

1. Resonance stabilizes the molecule as the energy of the resonance hybrid is less than the energy of any single resonance structure
• This can be explained in 5 steps:
(i) Consider a resonance hybrid and it’s various resonance structures 
(ii) Each of the resonance structures will have it’s own ‘quantity of energy’
(iii) The resonance hybrid will also have it’s own ‘quantity of energy’
(iv) The energy in (iii) will be less than any of the energies in (ii)
(v) So the 'phenomenon of resonance' helps the molecule to attain greater stability 
2. Resonance averages the bond characteristics as a whole
• This can be explained in 6 steps:
(i) We have seen some of the bond characteristics:
Bond length, Bond angle, Bond enthalpy, Bond order
(ii) Take any one of them, say bond length
(iii) Each of the resonance structures will have 'it’s own bond lengths' for it’s various bonds
(iv) The resonance hybrid will also have 'it’s own bond lengths' for it’s various bonds
(v) The values in (iv) will be the averages of the corresponding values in (iii)
(vi) Here, we have considered bond length. The same can be written about other bond characteristics also

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Many misconceptions are associated with resonance. We will analyze four misconceptions and get to know the facts
1. The existence of resonance structures in a sample
• This can be analysed in 3 steps:
(i) Let us take the example of ozone
    ♦ We saw that there are 2 resonance structures for ozone
(ii) Take any sample of ozone
• The fact is that, we will never find any of those two resonance structures in any ozone sample
(iii) The resonance structures exist only in drawings. The true structure is the hybrid of those resonance structures
2. The existence of resonance structures based on time
• This can be analysed in 3 steps:
(i) Let us take the example of ozone
    ♦ We saw that there are 2 resonance structures for ozone
(ii) Take any sample of ozone
• There is a popular belief:
    ♦ At some instances of time, the sample will contain the O₃ molecules in one canonical form
    ♦ At some other instances of time, the sample will contain the O₃ molecules in the other canonical form
(iii) This is totally wrong
• At any instant that we take, there will be only one form, which is the hybrid
3. Equilibrium between various canonical forms
• This can be analysed in 3 steps:
(i) Let us take the example of ozone
    ♦ We saw that there are 2 resonance structures for ozone
(ii) Take any sample of ozone
• There was a popular belief:
    ♦ The sample contains both the canonical forms in equal quantities
    ♦ If one canonical form is in excess quantity, it will be gradually converted into the other form
          ✰ This conversion will continue until both forms are in equal quantities
(iii) This is totally wrong
• The canonical forms do not even exist. So there is no question of attaining an equilibrium between them
4. Representation of the molecule
• This can be analysed in steps:
(i) Let us take the example of ozone
    ♦ We saw that there are 2 resonance structures for ozone
(ii) One may think that he/she can represent the O₃ by drawing 'any one' of it's Lewis structures
(iii) But the fact is this:
• If we draw just any one, we will be conveying the 'wrong information' that, one bond in O₃ is a single bond and the other is a double bond
(iii) It is impossible to represent such molecules by a single Lewis dot structure
• So it is compulsory to draw all the canonical structures and show double headed arrows between them

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Now we will see a solved example

Solved example 4.5
H₃PO₃ can be represented by structures (a) and (b) shown in fig.4.64 below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H₃PO₃ ?

Fig.4.64

If not, give reasons for the same.
Solution:
In the two given structures, the positions of atoms are not the same. So they are not the canonical forms of the resonance hybrid representing H₃PO₃

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In the next section, we will see polarity of bonds

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