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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