Chapter 7.13 - Lewis Acids and Bases

In the previous section, we saw conjugate acids and bases. In this section, we will see a few more solved examples. Later in this section, we will see Lewis acids and bases

Solved example 7.48
The species: H₂O, HCO₃^-, HSO₄^- and NH₃ can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and conjugate base.
Solution:
From the solved examples 7.46 and 7.47 of the previous section, we obtained an easy method to find conjugate acid and conjugate base. We wrote it in two steps:
1. Given an acid. It's conjugate base can be written by removing H⁺ from that acid
2. Given a base. It's conjugate acid can be written by adding H⁺ to that base 

• The present problem can be solved using this easy method
Part (i): H₂O
1. When H₂O is an acid, the corresponding conjugate base can be written by removing H⁺ from H₂O
• We get: OH^-
2. When H₂O is a base, the corresponding conjugate acid can be written by adding H⁺ to H₂O
• We get: H₃O⁺

Part (ii): HCO₃^-
1. When HCO₃^- is an acid, the corresponding conjugate base can be written by removing H⁺ from HCO₃^-
• We get: CO₃^2-
2. When HCO₃^- is a base, the corresponding conjugate acid can be written by adding H⁺ to HCO₃^-
• We get: H₂CO₃

Part (iii): HSO₄^-
1. When HSO₄^- is an acid, the corresponding conjugate base can be written by removing H⁺ from HSO₄^-
• We get: SO₄^2-
2. When HSO₄^- is a base, the corresponding conjugate acid can be written by adding H⁺ to HSO₄^-
• We get: H₂SO₄

Part (iv): NH₃
1. When NH₃ is an acid, the corresponding conjugate base can be written by removing H⁺ from NH₃
• We get: NH₂^-
2. When NH₃ is a base, the corresponding conjugate acid can be written by adding H⁺ to NH₃
• We get: NH₄⁺

Solved example 7.49
Find the conjugate acid/base for the following species:
HNO₂, CN^-, HClO₄, F^-, OH^-, CO₃^2-, and S^2-
Solution:
Part (i): HNO₂
1. This species contain an H atom. So it can donate a H⁺
• Thus it is an acid. The conjugate base can be written by removing H⁺
• We get: NO₂^-

Part (ii): CN^-
1. This species does not contain any H atom. So it can only accept a H⁺
• Thus it is a base. The conjugate acid can be written by adding H⁺
• We get: HCN

Part (iii): HClO₄
1. This species contain an H atom. So it can donate a H⁺
• Thus it is an acid. The conjugate base can be written by removing H⁺
• We get: ClO₄^-

Part (iv): F^-
1. This species does not contain any H atom. So it can only accept a H⁺
• Thus it is a base. The conjugate acid can be written by adding H⁺
• We get: HF

Part (v): OH^-
1. This species contain an H atom. But it already has one excess electron. So it can not donate any more proton. It can only accept proton
• Thus it is a base. The conjugate acid can be written by adding H⁺
• We get: H₂O

Part (vi): CO₃^2-
1. This species does not contain any H atom. So it can only accept a H⁺
• Thus it is a base. The conjugate acid can be written by adding H⁺
• We get: HCO₃^-

Part (vii): S^2-
1. This species does not contain any H atom. So it can only accept a H⁺
• Thus it is a base. The conjugate acid can be written by adding H⁺
• We get: HS

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Lewis Acids and bases

• In the case of Brönsted-Lowry acids and bases, we saw that:
    ♦ Acids donate a proton
    ♦ Bases accept a proton
• But donating a proton is equivalent to accepting two electrons
    ♦ This can be explained using an example. It can be written in 7 steps:
1. Fig.7.10 below shows the reaction between BF₃ and NH₃

Fig.7.10

• On the left side of the ‘+’ sign, we have the Lewis structure of BF₃
    ♦ We see that, the B atom needs two more electrons to complete octet
    ♦ (Recall the method to draw Lewis structure given in section 4.2)
• On the right side of the ‘+’ sign, we have the Lewis structure of NH₃
    ♦ We see that, the N atom has two lone electrons
2. The BF₃ accepts the two lone electrons of NH₃
    ♦ This is indicated by the curved arrow
• Those two electrons are used to form a new bond between B and N
    ♦ This is shown in the final product on the right side of the straight arrow
        ✰ Note the 'bond' between B and N
        ✰ Both the dots on that bond are white in color
        ✰ Those white dots originally belonged to the N atom
• In the final product, all atoms have octet
3. We know how to calculate formal charges (section 4.4)
• In the final product, we see that:
    ♦ B atom has a formal charge of -1
    ♦ N atom has a formal charge of +1
• So the final product as a whole, is neutral
4. We can say:
• The B atom donated a proton and thus acquired a formal charge of -1
    ♦ So according to Brönsted-Lowry theory, BF₃ is an acid
• The N atom accepted that proton and thus acquired a formal charge of +1
    ♦ So according to Brönsted-Lowry theory, NH₃ is a base
5. We know that the American scientist G N Lewis formulated the method of using Lewis structures for describing bonds
◼ He was the first scientist to notice that:
Instead of using the concept of 'proton transfer', we can use the concept of ‘electron pair transfer’ to describe acid-base reactions
6. According to Lewis theory:
    ♦ Acids accept an electron pair
    ♦ Bases donate an electron pair
7. In fig.7.9 above, we see that:
• BF₃ accepts an electron pair. So it is the acid
• NH₃ donates an electron pair. So it is the base

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Let us see another example. It can be written in 7 steps:
1. Fig.7.11 below shows the reaction between H₂O and NH₃

Fig.7.11

• On the left side of the ‘+’ sign, we have the Lewis structure of H₂O
    ♦ We see that, the O atom has two pairs of lone electrons
• On the right side of the ‘+’ sign, we have the Lewis structure of NH₃
    ♦ We see that, the N atom has a single pair of lone electrons
2. So none of the reactants needs electrons. Then how does the reaction take place?
The steps below will give the answer
3. In fig.7.12 below, ionization of H₂O molecule is shown using Lewis structures

Fig.7.12

• We see that:
    ♦ One of the H atoms, donates it's electron and leaves the molecule
        ✰ When the electron is donated, it becomes H⁺
    ♦ The OH portion accepts the electron and becomes OH^-
• Consider the products:
    ♦ On the left side of the ‘+’ sign, we have the Lewis structure of H⁺
    ♦ On the right side of the ‘+’ sign, we have the Lewis structure of OH^-
• Also note the formal charge of O atom in OH^-
    ♦ This formal charge is the overall negative charge of the OH^-
4. So now we can replace the H₂O in fig.7.11
• Instead of H₂O, we can put the combination of H⁺ and OH^-
    ♦ This is shown in fig.7.13 below
    ♦ The combination is shown inside the red ellipse

Fig.7.13

• We see that:
    ♦ The lone electron pair of N atom is donated to make a bond with H⁺ ion
        ✰ (Both dots in the new bond between N and H are white dots)
        ✰ So NH₃ is the base
    ♦ The H⁺ accepts the electron pair
        ✰ So H⁺ is the acid
5. The N atom on the product side, has a formal charge of +1
    ♦ This is the overall +1 charge for NH₄⁺
6. Recall that, the Brönsted-Lowry theory also explains the reaction between NH₃ and H₂O
• In that theory,
    ♦ H₂O donates the H⁺ and hence, H₂O is the acid
    ♦ NH₃ accepts the H⁺ and hence, NH₃ is the base
◼ We can say:
• The O atom donated a proton and thus acquired a formal charge of -1
    ♦ So according to Brönsted-Lowry theory, H₂O is an acid
• The N atom accepted that proton and thus acquired a formal charge of +1
    ♦ So according to Brönsted-Lowry theory, NH₃ is a base
7. In our present Lewis theory also,
    ♦ H₂O [which produces the H⁺(which accepts the electron pair)] is the acid
    ♦ NH₃ [which donates the electron pair] is the base

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Let us see one more example. It can be written in 7 steps:
1. Fig.7.14 below shows the reaction between H₂O and HCl

Fig.7.14

• On the left side of the ‘+’ sign, we have the Lewis structure of H₂O
    ♦ We see that, the O atom has two pairs of lone electrons
• On the right side of the ‘+’ sign, we have the Lewis structure of HCl
    ♦ We see that, the Cl atom has three pairs of lone electrons
2. So none of the reactants need electrons. Then how does the reaction take place?
The steps below will give the answer
3. In fig.7.15 below, ionization of HCl molecule is shown using Lewis structures

Fig.7.15

• We see that:
    ♦ The H atoms, donates it's electron and leaves the molecule
        ✰ When the electron is donated, it becomes H⁺
    ♦ The Cl accepts the electron and becomes Cl^-
• Consider the products:
    ♦ On the left side of the ‘+’ sign, we have the Lewis structure of H⁺
    ♦ On the right side of the ‘+’ sign, we have the Lewis structure of Cl^-
4. So now we can replace the HCl in fig.7.14
• Instead of HCl, we can put the combination of H⁺ and Cl^-
    ♦ This is shown in fig.7.16 below
    ♦ The combination is shown inside the red ellipse

Fig.7.16

• We see that:
    ♦ One lone electron pair of H₂O molecule is donated to make a bond with H⁺ ion of the HCl
        ✰ (Both dots in the new bond between O and H are white dots)
        ✰ So H₂O is the base
    ♦ The H⁺ accepts the electron pair
        ✰ So H⁺ is the acid
5. The O atom on the product side, has a formal charge of +1
    ♦ This is the overall +1 charge of H₃O⁺
6. Recall that, the Brönsted-Lowry theory also explains the reaction between HCl and H₂O
• In that theory,
    ♦ HCl donates the H⁺ and hence, HCl is the acid
    ♦ H₂O accepts the H⁺ and hence, H₂O is the base
◼ We can say:
• The Cl atom donated a proton and thus acquired a formal charge of -1
    ♦ So according to Brönsted-Lowry theory, HCl is an acid
• The O atom accepted that proton and thus acquired a formal charge of +1
    ♦ So according to Brönsted-Lowry theory, H₂O is a base
7. In our present Lewis theory also,
    ♦ HCl [which produces the H⁺(which accepts the electron pair)] is the acid
    ♦ H₂O [which donates the electron pair] is the base

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• It is interesting to note that:
    ♦ When reacting with NH₃, the H₂O acts as a Lewis acid
    ♦ When reacting with HCl, the H₂O acts as a Lewis base
• We saw the same situation in Brönsted-Lowry theory also:
    ♦ When reacting with NH₃, the H₂O acts as a Lewis acid
    ♦ When reacting with HCl, the H₂O acts as a Lewis base
◼  We will see the reason in later sections

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◼ Based on the above discussion, we can write:
• Electron deficient species like AlCl³, Co³⁺, Mg²⁺, etc. can act as Lewis acids
• Species like H₂O, NH₃, OH etc. which can donate a pair of electrons, can act as Lewis bases

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Now we will see some solved examples
Solved example 7.50
Classify the following species into Lewis acids and Lewis bases and show how
these act as such:
(a) HO^- (b)F^- (c) H⁺ (d) BCl₃
Solution:
Part (a):
1. Fig.7.17(a) below shows the Lewis structure of HO^-

Fig.7.17

• We see that, O has octet and H has duplet. So the species is stable
2. But O can donate one of it's lone pairs to form a bond with another species
• Even after such a donation, O will have it's octet
3. Since HO^- can donate a pair of electrons in this way, it is a Lewis base

Part (b):
1. Fig.7.17(b) above shows the Lewis structure of F^-
• We see that, F has octet
   ♦ The red dot indicates the extra electron acquired from some other atom
2. But F can donate one of it's lone pairs to form a bond with another species
• Even after such a donation, F will have it's octet
3. Since F^- can donate a pair of electrons in this way, it is a Lewis base

Part (c):
1. We know that, H⁺ does not have any electrons around it
2. But it can accept a pair of electrons from another species to form a bond
• After bond formation, H will have duplet
3. For example, H⁺ can form a bond with OH^- by accepting a pair of electrons from O
4. Since H⁺ can accept a pair of electrons in this way, it is a Lewis acid

Part (d):
1. Fig.7.17(c) above shows the Lewis structure of BCl₃
• It is similar to the structure of BF₃ that we saw earlier in fig.7.10
• Remember that, F and Cl belongs to the same group of the periodic table
2. We see that, B needs two more electrons to get octet
• So it can accept a pair of electrons
• Recall that, in fig. 7.10, B accepts a pair from NH₃
3. Since BCl₃ can accept a pair of electrons in this way, it is a Lewis acid

Solved example 7.51
Which of the followings are Lewis acids? H₂O, BF₃, H⁺, and NH₄⁺
Solution:
1. H₂O can act both as Lewis acid and Lewis base
(see the discussion below fig.7.16)
2. BF₃ is a Lewis acid
(see fig.7.10)
3. H⁺ is a Lewis acid
(see the previous solved example Part c)
4. NH₄⁺ is electron deficient. It can accept an electron pair. So it is a Lewis acid
(see the backward reaction in fig.7.13)

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In the next section, we will see a ionization of acids and bases


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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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