Chapter 4.3 - Drawing Lewis Dot Structures of Polyatomic Ions

In the previous section, we saw the steps to draw Lewis dot structures of 'polyatomic molecules'. In this section, we will see the steps to draw Lewis dot structures of 'polyatomic ions'

Some examples of polyatomic ions are: NH₄⁺, CO₃^2-
The following 6 steps will help us to make a Lewis dot structure of such ions
Step 1: Finding the number of dots
• In a Lewis dot structure, we see a large number of dots
• So the first step will be to determine the answer to this question:
How many dots will be present in the final structure?
■ The answer is simple:
• All ‘available valence electrons’ will be present in the final structure
    ♦ But anions will have some extra number of electrons. So final number will increase
    ♦ Also, cations have a deficiency of electrons. So final number will decrease
■ But, instead of calculating the 'final number', it is convenient to keep the numbers separate, as items (a) and (b) shown below:
    (a) Total number of ‘available valence electrons’
    (b) Number of electrons to be added/subtracted
• Final number =  (a) ± (b)
■ Let us see an example. We will write it in steps:
(i) Consider the ion CO₃^2-
• C has the electronic configuration 1s²2s²2p²
    ♦ So it has 4 valence electrons
• O has the electronic configuration 1s²2s²2p⁴
    ♦ So it has 6 valence electrons
(ii) So the total number of ‘available valence electrons’ = [4 + (6 × 3)] = 22
(iii) But there are two extra electrons (indicated by the charge of -2)
(iv) So we can write the numbers separately:
    (a) Total number of ‘available valence electrons’ = 22
    (b) Number of electrons to be added = 2
(v) Final number =  [(a) ± (b)] = [(a) + (b)] = [22 + 2] = 24 
Thus, there will be 24 dots in the final structure

■ Let us see another example. We will write it in steps:
(i) Consider the ion NH₄⁺
• N has the electronic configuration 1s²2s²2p³
    ♦ So it has 5 valence electrons
• H has the electronic configuration 1s¹
    ♦ So it has 1 valence electron
(ii) So the total number of ‘available valence electrons’ = [5 + (4 × 1)] = 9
(iii) But there is a deficiency of one electron (indicated by the charge of +1)
(iv) So we can write the numbers separately:
    (a) Total number of ‘available valence electrons’ = 9
    (b) Number of electrons to be subtracted = 1
(v) Final number =  [(a) ± (b)] = [(a) - (b)] = [9 - 1] = 8
 Thus, there will be 8 dots in the final structure
Step 2: The skeletal structure
(We have seen this in the previous section. However, we will write the steps again)
• Draw the skeletal structure of the ion
    ♦ The least electronegative atom will be the central atom
    ♦ The other atoms will be distributed around the central atom
We have seen examples in the previous section
Step 3: Preliminary single bonds
(We have seen this in the previous section. However, we will write the steps again)
• Connect the central atom to all the surrounding atoms with a single bond ('─')
• There will be at least one bond between atoms. So we can confidently put a ('─') at all possible connections
• The single bonds that we put in this step can be called preliminary bond
    ♦ This is because, in later steps, these single bonds may have to be changed to double or triple bonds
Step 4: Preliminary distribution of electrons
• Distribute all the 'available valence electrons' around the atoms
          ✰ Note that, it is the 'available valence electrons' that we distribute in this step
          ✰ We do not consider the extra or deficient electrons in this step  
    ♦ Distribute the electrons in such a way that, the outer atoms get octet (duplet in case of H atoms) first
    ♦ Then give the remaining electrons to the central atom
• The distribution that we do in this step can be called preliminary distribution
    ♦ This is because, in later steps, this distribution may have to be changed
Step 5: Check for duplet and octet
• Check whether all H atoms (if present), have duplet
• Check whether all other atoms have octet
• If the required duplets and octets are not obtained, change the single bonds to double or triple bonds as necessary
■ In this step, we will see an interesting result:
    ♦ In the case of anions:
          ✰ The complete octet/duplet will be obtained only when we put the extra electrons
    ♦ In the case of cations:
          ✰ The complete octet/duplet will be obtained only when we take away the deficient electrons
• By the end of this step, the final structure should emerge  
    ♦ All H atoms (if present), should be having duplet
    ♦ All other atoms should be having octet
Step 6: Check the number of dots
The total number of dots in the final structure must be same as the number obtained in step 1

---

Let us now apply the above rules to some ions:

Example 1: CO₃^2-
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+(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
• C is less electronegative than O
(Electronegativity of C is 2.55 and that of O is 3.44)
    ♦ So C is the central atom
    ♦ The three O atoms will be distributed around the C atom
• The skeletal structure is shown in fig.4.19(a) below:

Fig.4.19

Step 3: Preliminary single bonds
• The four atoms are joined by a '─' as shown in fig.4.19(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' = 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
(i) Rearrangement: Change the preliminary single bond
    ♦ Change the top single bond to double bond as shown in the fig.d
(ii) 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
(iii) All the 22 electrons are used up. Still, complete octet is not achieved
• So, we will need external electrons
(iv) 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
Step 6: Check the number of dots
    ♦ Total number of dots in fig.e = 24
    ♦ Number calculated in step 1 = 24

---

Now a question arises:

■ What 'source' would give two electrons so that, all atoms in CO₃^2- have octet?
The answer can be written using an example. We will write it in steps:
(i) Consider the Ca atom
• It can donate two electrons and become Ca²⁺ ion
• The Ca²⁺ ion is stable because, it has octet
(ii) The two electrons donated by Ca can be used to make all the atoms in CO₃^2- octet
• When all the atoms have octet, the 'CO₃^2- ion as a whole' becomes stable
(iii) So we have two stable ions:
• The positive ion: Ca²⁺
• The negative ion: CO₃^2-
(iv) An electrostatic force of attraction comes into effect between the two oppositely charged ions
• So the two ions begin to act together as a single unit
• We will not be able to separate the two ions easily from each other
(v) As a result, we get calcium carbonate (CaCO₃)
• This is similar to the formation of NaCl from Na⁺ and Cl^- ions
    ♦ The only difference is that, in our present case, one of the ions is a 'polyatomic ion'

---

Example 2: NO₂^-
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+(2 × 6)] = 17
• One extra electron is also present
■ We will write the number as two items:
    (a) Total number of ‘available valence electrons’ = 17
    (b) Number of electrons to be added = 1
• Final number =  [(a) ± (b)] = [(a) + (b)] = [17 + 1] = 18 
Step 2: The skeletal structure
• N is less electronegative than O
(Electronegativity of N is 3.04 and that of O is 3.44)
    ♦ So N is the central atom
    ♦ The two O atoms will be distributed around the N atom
• The skeletal structure is shown in fig.4.20(a) below:

Fig.4.20

Step 3: Preliminary single bonds
• The three atoms are joined by a '─' as shown in fig.4.20(b) above
Step 4: Preliminary distribution of electrons
(Remember that, the 'available valence electrons' are distributed in this step)
• First make the two outer O atoms octet
• Then give the remaining electrons to the N 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
• The two 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 = (17-16) = 1
    ♦ This 1 electron is given to the N atom
• There are no more electrons to distribute
Step 5: Check for octet
• The two O atoms have got 8 electrons each
• The N atom has got only 5 electrons
    ♦ So this atom needs 3 more electrons
(i) Rearrangement: Change the preliminary single bond
    ♦ Change the left side single bond to double bond as shown in the fig.d
(ii) Two electrons from the left side O is used for making the new bond
    ♦ Now the left and right side O atoms have octet
    ♦ But the N atom has got only 7 electrons
(iii) All the 17 electrons are used up. Still, complete octet is not achieved
• So, we will need one external electron
(iv) Get one external electron from any suitable source
• Give it to the N atom
• This is shown in fig.e
• Now all atoms have octet
(v) But the one external electron will create a charge of -1
• So we put the structure inside square brackets and put a '-' at the top right corner
Step 6: Check the number of dots
    ♦ Total number of dots in fig.e = 18
    ♦ Number calculated in step 1 = 18

---

Now a question arises:

■ What 'source' would give one electron so that, all atoms in NO₂^- have octet?
The answer can be written using an example. We will write it in steps:
(i) Consider the Na atom
• It can donate one electron and become Na⁺ ion
• The Na⁺ ion is stable because, it has octet
(ii) The electron donated by Na can be used to make all the atoms in NO₂^- octet
• When all the atoms have octet, the 'NO₂^- ion as a whole' becomes stable
(iii) So we have two stable ions:
• The positive ion: Na⁺
• The negative ion: NO₂^-
(iv) An electrostatic force of attraction comes into effect between the two oppositely charged ions
• So the two ions begin to act together as a single unit
• We will not be able to separate the two ions easily from each other
(v) As a result, we get sodium nitrite (NaNO₂)
• This is similar to the formation of NaCl from Na⁺ and Cl^- ions
    ♦ The only difference is that, in our present case, one of the ions is a 'polyatomic ion'

---

• The above discussion will enable us to 'draw Lewis dot structures' of some polyatomic ions
• In the next section, we will see Formal charge

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Chapter 4.2 - Drawing Lewis Dot Structures

In the previous section, we saw how to 'extract information' from any 'given Lewis dot structure'. In this section, we will see the steps to draw Lewis dot structures of 'given molecules'

We know that, a Lewis dot structure give us the following 3 information:
1. How atoms are bonded together in a molecule
2. How many pairs of electrons are shared between atoms
3. How is  the octet attained by each atom in the molecule
• But what if we want to make a ‘new’ Lewis dot structure?
• Let us see the steps:
■ We will encounter two types:
(i) Polyatomic molecules
Examples: H₂, CO₂, H₂O, NF₃
(ii) Polyatomic ions
Examples: NH₄⁺, CO₃^2-
• First we will see polyatomic molecules
The following 6 steps will help us to make a Lewis dot structure of such molecules
Step 1: Finding the number of dots
• In a Lewis dot structure, we see a large number of dots
• So the first step will be to determine the answer to this question:
How many dots will be present in the final structure?
• The answer is simple:
All ‘available valence electrons’ will be present in the final structure
■ Let us see an example. We will write it in steps:
(i) Consider the molecule CH₄
• C has the electronic configuration 1s²2s²2p²
    ♦ So it has 4 valence electrons
• H has the electronic configuration 1s¹
    ♦ So it has 1 valence electron
(ii) So the total number of ‘available valence electrons’ = [4 + (4 × 1)] = 8
Thus, there will be 8 dots in the final structure
Step 2: The skeletal structure
• Draw the skeletal structure of the molecule
    ♦ The least electronegative atom will be the central atom
    ♦ The other atoms will be distributed around the central atom
An example:
• Consider the NF₃ molecule
• N is less electronegative than F
(Electronegativity of N is 3.0 and that of F is 4.0)
    ♦ So N is the central atom
    ♦ The three F atoms will be distributed around the N atom
Step 3: Preliminary single bonds
• Connect the central atom to all the surrounding atoms with a single bond ('─')
• There will be at least one bond between atoms. So we can confidently put a ('─') at all possible connections
• The single bonds that we put in this step can be called preliminary bond
    ♦ This is because, in later steps, these single bonds may have to be changed to double or triple bonds
Step 4: Preliminary distribution of electrons 
• Distribute all the 'available valence electrons' around the atoms
    ♦ Distribute in such a way that, the outer atoms get octet (duplet in case of H atoms) first
    ♦ Then give the remaining electrons to the central atom
• The distribution that we do in this step can be called preliminary distribution
    ♦ This is because, in later steps, this distribution may have to be changed
Step 5: Check for duplet and octet
• Check whether all H atoms (if present), have duplet
• Check whether all other atoms have octet
• If the required duplets and octets are not obtained, change the single bonds to double or triple bonds as necessary
• By the end of this step, the final structure should emerge  
    ♦ All H atoms (if present), should be having duplet
    ♦ All other atoms should be having octet
Step 6: Check the number of dots
The total number of dots in the final structure must be same as the number obtained in step 1

---

Let us now apply the above rules to some molecules:

Example 1: H₂
Step 1: Finding the number of dots
• Number of valence electrons of H = 1
• So total number of valence electrons = (2 × 1) = 2
Step 2: The skeletal structure
• The skeletal structure is shown in fig.4.13(a) below:

Fig.4.13

Step 3: Preliminary single bonds
• The two H atoms are joined by a '─' as shown in fig.4.13(b) above
Step 4: Preliminary distribution of electrons
• Both are H atoms. So we need not look for the 'central atom' or 'outer atoms'
• Let us take the left side H atom
    ♦ We will give it duplet first
    ♦ And give the remaining electrons to the right side H atom
• This is shown in fig.4.13(c)  
• In the fig.c, we see that:
    ♦ The valence electron of the 1st H atom are shown in green color
    ♦ The valence electron of the 2nd H atom are shown in red color
• The left side H atom has 2 electrons (including the one red dot in the single bond)
    ♦ The number of remaining electrons = (2-2) = 0
• There are no more electrons to distribute
Step 5: Check for duplet
• The 1st H atom has attained duplet
• The 2nd H atom has attained duplet
• So the '─' need not be changed to double or triple bonds
• Also, the preliminary distribution need not be changed
Step 6: Check the number of dots
    ♦ Total number of dots in fig.c = 2
    ♦ Number calculated in step 1 = 2

Example 2: NF₃
Step 1: Finding the number of dots
• Number of valence electrons of N = 5
• Number of valence electrons of F = 7
• So total number of valence electrons = [5+(3 × 7)] = 26
Step 2: The skeletal structure
• N is less electronegative than F
(Electronegativity of N is 3.0 and that of F is 4.0)
    ♦ So N is the central atom
    ♦ The three F atoms will be distributed around the N atom
• The skeletal structure is shown in fig.4.14(a) below:

Fig.4.14

Step 3: Preliminary single bonds
• The four atoms are joined by a '─' as shown in fig.4.14(b) above
Step 4: Preliminary distribution of electrons
• First make the three outer F atoms octet
• Then give the remaining electrons to the central N atom
• This is shown in fig.c
• In the fig.c, we see that:
    ♦ The valence electrons of the F atoms are shown in green color
    ♦ The valence electrons of the N atom are shown in red color
• All the F atoms have 8 electrons (including the one red dot in the single bonds)
    ♦ So the F atoms use up (3 × 8) = 24 electrons
    ♦ The number of remaining electrons = (26-24) = 2
    ♦ These 2 electrons are given to the N atom 
• There are no more electrons to distribute
Step 5: Check for octet
• Each of the F atoms have got 8 electrons. They have attained octet
• The N atom has got 8 electrons. It has attained octet
• Thus, the preliminary distribution need not be changed
Step 6: Check the number of dots
    ♦ Total number of dots in fig.c = 26
    ♦ Number calculated in step 1 = 26

Example 3: H₂O
• We have already seen the finished structure of H₂O in the previous section. Now we will see how that finished structure is obtained in steps:
Step 1: Finding the number of dots
• Number of valence electrons of H = 1
• Number of valence electrons of O = 6
• So total number of valence electrons = [(2 × 1)+6] = 8
Step 2: The skeletal structure
• H is less electronegative than O
(Electronegativity of H is 2.2 and that of O is 3.4)
    ♦ So H must be the central atom
    ♦ But there are two H atoms but only one O atom
    ♦ So we will make O the central atom
• The two H atoms will be distributed around the O atom
• The skeletal structure is shown in fig.4.15(a) below:

Fig.4.15

Step 3: Preliminary single bonds
• The three atoms are joined by a '─' as shown in fig.4.15(b) above
Step 4: Preliminary distribution of electrons
• First make the two outer H atoms duplet
• 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 H atoms are shown in green color
    ♦ The valence electrons of the O atom are shown in red color
• All the H atoms have 2 electrons (including the one red dot in the single bonds)
    ♦ So the number of electrons used up for making the H atoms duplet = 4
    ♦ So the number of remaining electrons = (8-4) = 4
    ♦ These 4 electrons are given to the O atom 
• There are no more electrons to distribute
Step 5: Check for octet
• Each of the H atoms have got 2 electrons. They have attained duplet
• The O atom has got 8 electrons. It has attained octet
• Thus, the preliminary distribution need not be changed
Step 6: Check the number of dots
    ♦ Total number of dots in fig.c = 8
    ♦ Number calculated in step 1 = 8

Example 4: O₂
Step 1: Finding the number of dots
• Number of valence electrons of O = 6
• So total number of valence electrons = (2 × 6) = 12
Step 2: The skeletal structure
• The skeletal structure is shown in fig.4.16(a) below:

Fig.4.16

Step 3: Preliminary single bonds
• The two O atoms are joined by a '─' as shown in fig.4.16(b) above
Step 4: Preliminary distribution of electrons
• Both are O atoms. So we need not look for the 'central atom' or 'outer atoms'
• Let us take the left side O atom
    ♦ We will give it octet first
    ♦ And give the remaining electrons to the right side O atom
• This is shown in fig.4.16(c)  
• In the fig.c, we see that:
    ♦ The valence electrons of the 1st O atom are shown in green color
    ♦ The valence electrons of the 2nd O atom are shown in red color
• The left side O atom has 8 electrons (including the one red dot in the single bond)
    ♦ The number of remaining electrons = (12-8) = 4
    ♦ These 4 electrons are given to the right side O atom 
• There are no more electrons to distribute
Step 5: Check for octet
• The left side O atom has got 8 electrons
• The right side O atom has got only 6 electrons
    ♦ So this atom needs 2 more electrons
(i) Rearrangement: Change the preliminary single bond
    ♦ Change the single bond to double bond as shown in the fig.d
• Let us take the left side O atom
    ♦ We will give it octet first
    ♦ And give the remaining electrons to the right side O atom
• This is shown in fig.4.16(e)  
• In the fig.e, we see that:
The left side O atom has 8 electrons (including the two red dots in the double bond)
    ♦ The number of remaining electrons = (12-8) = 4
    ♦ These 4 electrons are given to the right side O atom 
• There are no more electrons to distribute
(ii) Check for octet:
    ♦ The left side O atom has 8 electrons
    ♦ The right side O atom has 8 electrons
Step 6: Check the number of dots
    ♦ Total number of dots in fig.e = 12
    ♦ Number calculated in step 1 = 12

Example 5: N₂
Step 1: Finding the number of dots
• Number of valence electrons of N = 5
• So total number of valence electrons = (2 × 5) = 10
Step 2: The skeletal structure
• The skeletal structure is shown in fig.4.17(a) below:

Fig.4.17

Step 3: Preliminary single bonds
• The two N atoms are joined by a '─' as shown in fig.4.17(b) above
Step 4: Preliminary distribution of electrons
• Both are N atoms. So we need not look for the 'central atom' or 'outer atoms'
• Let us take the left side N atom
    ♦ We will give it octet first
    ♦ And give the remaining electrons to the right side N atom
• This is shown in fig.4.17(c)  
• In the fig.c, we see that:
    ♦ The valence electrons of the 1st N atom are shown in green color
    ♦ The valence electrons of the 2nd N atom are shown in red color
• The left side N atom has 8 electrons (including the one red dot in the single bond)
    ♦ The number of remaining electrons = (10-8) = 2
    ♦ These 2 electrons are given to the right side N atom 
• There are no more electrons to distribute
Step 5: Check for octet
• The left side N atom has got 8 electrons
• The right side N atom has got only 4 electrons
    ♦ So this atom needs 4 more electrons
(i) Rearrangement: Change the preliminary single bond
    ♦ Change the single bond to double bond as shown in the fig.d
• Let us take the left side N atom
    ♦ We will give it octet first
    ♦ And give the remaining electrons to the right side N atom
• This is shown in fig.4.17(e)  
• In the fig.e, we see that:
The left side N atom has 8 electrons (including the two red dots in the double bond)
    ♦ The number of remaining electrons = (10-8) = 2
    ♦ These 2 electrons are given to the right side N atom 
• There are no more electrons to distribute
(ii) Check for octet:
• The left side N atom has got 8 electrons
• The right side N atom has got only 6 electrons
    ♦ So this atom needs 2 more electrons
(i) Rearrangement: Change the double bond
    ♦ Change the double bond to triple bond as shown in the fig.f
• Let us take the left side N atom
    ♦ We will give it octet first
    ♦ And give the remaining electrons to the right side N atom
• This is shown in fig.4.17(g)  
• In the fig.g, we see that:
The left side N atom has 8 electrons (including the three red dots in the triple bond)
    ♦ The number of remaining electrons = (10-8) = 2
    ♦ These 2 electrons are given to the right side N atom 
• There are no more electrons to distribute
(ii) Check for octet:
    ♦ The left side N atom has 8 electrons
    ♦ The right side N atom has 8 electrons
Step 6: Check the number of dots
    ♦ Total number of dots in fig.g = 10
    ♦ Number calculated in step 1 = 10

Example 6: CO
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+6) = 10
Step 2: The skeletal structure
• The skeletal structure is shown in fig.4.18(a) below:

Fig.4.18

Step 3: Preliminary single bonds
• The two atoms are joined by a '─' as shown in fig.4.18(b) above
Step 4: Preliminary distribution of electrons
• There are only two atoms. So we need not look for the 'central atom' or 'outer atoms'
• Let us take the C atom
    ♦ We will give it octet first
    ♦ And give the remaining electrons to the O atom
• This is shown in fig.4.18(c)  
• In the fig.c, we see that:
    ♦ The valence electrons of the C atom are shown in green color
    ♦ The valence electrons of the O atom are shown in red color
• The C atom has 8 electrons (including the one red dot in the single bond)
    ♦ The number of remaining electrons = (10-8) = 2
    ♦ These 2 electrons are given to the O atom 
• There are no more electrons to distribute
Step 5: Check for octet
• The C atom has got 8 electrons
• The O atom has got only 4 electrons
    ♦ So this atom needs 4 more electrons
(i) Rearrangement: Change the preliminary single bond
    ♦ Change the single bond to double bond as shown in the fig.d
• Let us take the C atom
    ♦ We will give it octet first
    ♦ And give the remaining electrons to the O atom
• This is shown in fig.4.18(e)  
• In the fig.e, we see that:
The C atom has 8 electrons (including the two red dots in the double bond)
    ♦ The number of remaining electrons = (10-8) = 2
    ♦ These 2 electrons are given to the O atom 
• There are no more electrons to distribute
(ii) Check for octet:
• The C atom has got 8 electrons
• The O atom has got only 6 electrons
    ♦ So this atom needs 2 more electrons
(i) Rearrangement: Change the double bond
    ♦ Change the double bond to triple bond as shown in the fig.f
• Let us take the C atom
    ♦ We will give it octet first
    ♦ And give the remaining electrons to the O atom
• This is shown in fig.4.18(g)  
• In the fig.g, we see that:
The C atom has 8 electrons (including the three red dots in the triple bond)
    ♦ The number of remaining electrons = (10-8) = 2
    ♦ These 2 electrons are given to the O atom 
• There are no more electrons to distribute
(ii) Check for octet:
    ♦ The C atom has 8 electrons
    ♦ The O atom has 8 electrons
Step 6: Check the number of dots
    ♦ Total number of dots in fig.g = 10
    ♦ Number calculated in step 1 = 10

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Some more examples are given below:

1. CCl₄ Carbon tetra chloride

2. SiCl₄ Silicon tetra chloride

3. H₂S Hydrogen sulfide

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• The above discussion will enable us to 'draw Lewis dot structures' of some polyatomic molecules
• In the next section, we will see the steps to draw Lewis dot structures of given polyatomic ions 

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Chapter 4.1 - Covalent Bonds and Lewis Dot Structures

In the previous section, we saw Lewis symbols. In this section, we will see Octet rule. We will also see details about covalent bonding and Lewis dot structures

The octet rule can be written in 5 steps:
1. Atoms combine with each other. This combination can be achieved by two methods:
(i) Ionic bond
• In this method, there is 'transfer of one or more valence electrons' from one atom to the other
    ♦ One of the atom loses one or more of it’s valence electrons
          ✰ The other atom accepts those electrons
(ii) Covalent bond
In this method, there is 'sharing of one or more pairs of electrons' between atoms
    ♦ All the ‘member electrons’ of all the ‘shared pairs’ will belong to both the atoms 
          ✰ If one pair is shared, the two electrons in that pair will belong to both the atoms
          ✰ If two pairs are shared, the four electrons in those two pairs will belong to both the atoms
          ✰ So on . . .
2. When the combination of the atoms is completed, ‘each atom in the combination’ will have 8 electrons in their outermost shell
3. When the 8 electrons are attained, we say this: The atom has attained octet
    ♦ An atom which has attained octet is stable
    ♦ In other words, the atom which has octet, has a stable electronic configuration
4. Every atom tries to attain stability
• For attaining stability, the atoms need to attain octet
• For attaining octet, the atoms need to combine with other atoms
■ So we can write:
The atoms combine with other atoms in order to attain octet and stability. This is known as octet rule
5. Kossel and Lewis put forward this theory in 1916. It is known as: the electronic theory of chemical bonding

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Covalent bond

• Now we will see some details about the covalent bond
• The theory about covalent bond was developed in 1919 by the American scientist Irving Langmuir
• The works of Langmuir was based upon the earlier theory put forward by Kossel and Lewis
• The theory of covalent bonding can be easily understood if we take the Cl₂ molecule as an example. We will write it in steps:

1. Consider a Cl atom
• The electronic configuration of Cl is [Ne]3s²3p⁵
• It is clear that, Cl requires one more electron to attain octet
2. Consider the situation where there is a second Cl atom nearby
• If this 2nd Cl atom can donate an electron, the 1st Cl can attain octet
• But then, the 2nd Cl atom will become 'more electron deficient'. So it will not donate it's electron
3. In such a situation, the solution is this:
• The 2nd Cl atom allows the 1st Cl atom to share an electron
• That is., the 1st Cl atom is allowed to use one electron of the 2nd Cl atom
4. But the 2nd Cl atom is still in need of one electron
• So the 2nd Cl atom is allowed to use one electron of the 1st Cl atom
5. In short, we can write:
    ♦ The 1st Cl atom has claim on one electron of the 2nd Cl atom
    ♦ The 2nd Cl atom has claim on one electron of the 1st Cl atom
• That is., one pair of electrons is shared by two Cl atoms
    ♦ Remember that, a 'pair' means 'two'
    ♦ So 'two electrons' are shared
    ♦ This can be demonstrated using Lewis symbols. It is shown in fig.4.9(a) below:

Fig.4.9

• All electrons of the first Cl atom are shown in green color
• All electrons of the second Cl atom are shown in red color
6. In the above fig.4.9(a), on the right side of the arrow, we have the 'product'
• This 'product' gives a clear idea about the structure of the 'resulting atom'
• The dots represent electrons. Such structures are referred to as: Lewis dot structures
■ In the Lewis dot structure, we notice the following 5 points:
(i) The first Cl atom is a green region
(ii) The second Cl atom is a red region
(iii) The two regions overlap at a small central portion
    ♦ We can call this overlapping portion as: the 'shared region'
          ✰ 'shared region' is the 'region common to two atoms'
(iv) The 'electrons in the shared region' are the 'electrons which are shared'
(v) The 'electrons outside the shared region' do not take part in any sharing
• There is a special name for these 'outside electrons': lone pairs
• This is because, 'number of these electrons' will be always even. So they can be grouped into 'pairs'
• For example:
    ♦ If there are 2 'outside electrons', we have one lone pair
    ♦ If there are 4 'outside electrons', we have two lone pairs
    ♦ If there are 6 'outside electrons', we have three lone pairs
    ♦ so on . . .
7. Note that, both the Cl atoms have equal claims on both the electrons in the 'shared pair'
• So the two Cl atoms cannot separate away from each other
8. Once the sharing has taken place, we say this:
■ The two Cl atoms are connected together by a single covalent bond
9. Using appropriate symbols:
• When 'one pair of electrons' is shared, it results in a single covalent bond or single bond
    ♦ We put a '─' between the two atoms 
• When 'two pairs of electrons' is shared, it results in a double covalent bond or double bond
    ♦ We put a '＝' between the two atoms
• When 'three pairs of electrons' is shared, it results in a triple covalent bond or triple bond
    ♦ We put a '☰' between the two atoms
8. So the structure of Cl₂ molecule can be represented as Cl─Cl
• While using such 'shortened form' to represent the structure, we must add some additional information
• This can be explained in 2 steps:
(i) We have electrons inside the 'shared region'
    ♦ All information about these 'shared electrons' can be conveyed using '─' OR '＝' OR '☰'
(i) We have electrons outside the 'shared region'. They are the 'lone pairs'
    ♦ We must covey the information about these 'lone pairs' also
    ♦ For that, we must put appropriate number of dots around the symbols of the atom
    ♦ This is shown in fig.4.9(b) 
• In the fig.b, we have 6 green dots
    ♦ They represent the '3 lone pairs' in the green region in fig.a
• In the fig.b, we have 6 red dots
    ♦ They represent the '3 lone pairs' in the red region in fig.a

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• So we have completed a discussion on the basics of covalent bonds. Next we will see some more details about Lewis dot structures

• We can draw Lewis dot structures of a large number of compounds
• We will encounter two types:
Type 1: Same atoms are present in the compound
• Examples:
    ♦ All atoms in O₂ are O
    ♦ All atoms in F₂ are F
Type 2: Different atoms are present in the compound
Examples:
    ♦ H and O atoms are present in H₂O
    ♦ C and Cl atoms are present in CCl₄
■ While studying Lewis dot structures, the following 4 points must be kept in mind:
1. Number of bonds:
• When we see a single bond (‘─’), it indicates ‘sharing of two electrons (a pair)’
    ♦ Conversely, when we see ‘two electrons (a pair) in the shared region’, it indicates a ‘─’
          ✰ 'shared region' is the 'region common to two atoms'
• When we see a double bond (‘＝’), it indicates ‘sharing of four electrons (two pairs)’
    ♦ Conversely, when we see ‘four electrons (two pairs) in the shared region’, it indicates a ‘＝’
• When we see a triple bond (‘☰’), it indicates ‘sharing of six electrons (three pairs)’
    ♦ Conversely, when we see ‘six electrons (three pairs) in the shared region’, it indicates a ‘☰’
2. Contribution from each atom
• If we see a ‘─’ between two atoms, it is clear that, each of those two atoms have contributed exactly one electron to make that pair
• If we see a ‘＝’ between two atoms, it is clear that, each of those two atoms have contributed exactly two electrons to make those two pairs
• If we see a ‘☰’ between two atoms, it is clear that, each of those two atoms have contributed exactly three electrons to make those three pairs
3. Number of electrons possessed by each atom
• When the Lewis dot structure of a molecule is completed, we must do a check. This check can be done in 3 steps:
(i) Take an atom in the Lewis dot structure
(ii) Count the number of dots around that atom
(iii) This number must be 8
Do this check for each atom in the molecule
4. The shortened form:
• All information about the 'shared electrons' can be conveyed using '─' OR '＝' OR '☰'
• All information about the 'lone pairs' should be conveyed using dots around the symbols

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Now we will see some examples:

Example 1:
• Fig.4.10(a) below shows the Lewis dot structure of H₂O

Fig.4.10

• The 1 valence electron of H is shown in green color
• The 6 valence electrons of O are shown in red color
• We can write the following 4 points:
1. Type of bond
• Consider the ‘shared region’ between the first H and the O
    ♦ There are two dots in this region
    ♦ So we put a ‘─’ between the first H and O
    ♦ This is a single bond
• Consider the ‘shared region’ between the second H and the O
    ♦ There are two dots in this region
    ♦ So we put a ‘─’ between the second H and O
    ♦ This is a single bond
■ So the shortened form is H─O─H
This is shown below the Lewis dot structure in fig.4.10(a)
2. Contribution from each atom
• Consider the ‘─’ between the first H and the O
    ♦ It is clear that:
          ✰ One electron in the '─' belongs to the H
          ✰ The other electron in the '─' belongs to the O
• Consider the ‘─’ between the second H and the O
    ♦ It is clear that:
          ✰ One electron in the '─' belongs to the H
          ✰ The other electron in the '─' belongs to the O
3. Checking the number of electrons in the Lewis dot structure:
• The first H has 2 dots around it
• The O has 8 dots around it
• The second H has 2 dots around it

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• The H atom needs only 2 electrons to fill it's 1s shell

• When the H attains 2 electrons in the 1s shell, we cannot call it an 'octet'
    ♦ The word 'oct' is related to '8'. For example, an octagon has 8 sides
■ When H attains the required 2 electrons, we say this:
The H has attained duplet

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4. The shortened form must show the 'lone pairs' also

• The two H atoms do not have any 'lone pairs'
• The two 'lone pairs' of O are indicated by four red dots in the shortened form  

Example 2:
• Fig.4.10(b) above shows the Lewis dot structure of CCl₄
• The 7 valence electrons of Cl are shown in green color
• The 4 valence electrons of C are shown in red color
• We can write the following 4 points:
1. Type of bond
• Consider the ‘shared region’ between the left Cl and the C
    ♦ There are two dots in this region
    ♦ So we put a ‘─’ between the left Cl and C
    ♦ This is a single bond
• Consider the ‘shared region’ between the right Cl and the C
    ♦ There are two dots in this region
    ♦ So we put a ‘─’ between the right Cl and the C
    ♦ This is a single bond
• Consider the ‘shared region’ between the top Cl and the C
    ♦ There are two dots in this region
    ♦ So we put a ‘─’ between the top Cl and the C
    ♦ This is a single bond
• Consider the ‘shared region’ between the bottom Cl and the C
    ♦ There are two dots in this region
    ♦ So we put a ‘─’ between the bottom Cl and the C
    ♦ This is a single bond
■ The shortened form is shown in the fig.4.10(c)
2. Contribution from each atom
• Consider the ‘─’ between the left Cl and the C
    ♦ It is clear that:
          ✰ One electron in the '─' belongs to the Cl
          ✰ The other electron in the '─' belongs to the C
• Consider the ‘─’ between the right Cl and the C
    ♦ It is clear that:
          ✰ One electron in the '─' belongs to the Cl
          ✰ The other electron in the '─' belongs to the C
• Consider the ‘─’ between the top Cl and the C
    ♦ It is clear that:
          ✰ One electron in the '─' belongs to the Cl
          ✰ The other electron in the '─' belongs to the C
• Consider the ‘─’ between the bottom Cl and the C
    ♦ It is clear that:
          ✰ One electron in the '─' belongs to the Cl
          ✰ The other electron in the '─' belongs to the C
3. Checking the number of electrons in the Lewis dot structure:
• The left Cl has 8 dots around it
• The right Cl has 8 dots around it
• The top Cl has 8 dots around it
• The bottom Cl has 8 dots around it
• The C has 8 dots around it
4. The shortened form must show the 'lone pairs' also
• The C atom do not have any lone pairs
• The 'three lone pairs' of each Cl are indicated by six green dots around each Cl in the shortened form


Example 3:
• Fig.4.11(a) below shows the Lewis dot structure of CO₂

Fig.4.11

• The 6 valence electrons of O are shown in green color
• The 4 valence electrons of C are shown in red color
• We can write the following 4 points:
1. Type of bond
• Consider the ‘shared region’ between the first O and the C
    ♦ There are 4 dots in this region. '4' indicates '2 pairs'
    ♦ So we put a ‘＝’ between the first O and the C 
    ♦ This is a double bond
• Consider the ‘shared region’ between the second O and the C
    ♦ There are 4 dots in this region. '4' indicates '2 pairs'
    ♦ So we put a ‘＝’ between the second O and the C 
    ♦ This is a double bond
■ So the shortened form is O＝C＝O
This is shown below the Lewis dot structure in fig.4.11(a)
2. Contribution from each atom
• Consider the ‘＝’ between the first O and the C 
    ♦ It is clear that:
          ✰ Two electrons in the '＝' belongs to the O
          ✰ The remaining two electrons in the '＝' belongs to the C
• Consider the ‘＝’ between the second O and the C 
    ♦ It is clear that:
          ✰ Two electrons in the '＝' belongs to the O
          ✰ The remaining two electrons in the '＝' belongs to the C
3. Checking the number of electrons in the Lewis dot structure:
• The first O has 8 dots around it
• The C has 8 dots around it
• The second O has 8 dots around it
4. The shortened form must show the 'lone pairs' also
• The C atom do not have any lone pairs
• The 'two lone pairs' of each O are indicated by four green dots around each O in the shortened form


Example 4:
• Fig.4.11(b) above shows the Lewis dot structure of C₂H₄
• The 1 valence electron of H is shown in green color
• The 4 valence electrons of C are shown in red color
• We can write the following 4 points:
1. Type of bond
• Consider the ‘shared region’ between the top-left H and the first C
    ♦ There are two dots in this region
    ♦ So we put a ‘─’ between this H and C
    ♦ This is a single bond
• Consider the ‘shared region’ between the bottom-left H and the first C
    ♦ There are two dots in this region
    ♦ So we put a ‘─’ between this H and C
    ♦ This is a single bond
• Consider the ‘shared region’ between the top-right H and the second C
    ♦ There are two dots in this region
    ♦ So we put a ‘─’ between this H and C
    ♦ This is a single bond
• Consider the ‘shared region’ between the bottom-right H and the second C
    ♦ There are two dots in this region
    ♦ So we put a ‘─’ between this H and C
    ♦ This is a single bond
• Consider the ‘shared region’ between the two C atoms
    ♦ There are 4 dots in this region. '4' indicates '2 pairs'
    ♦ So we put a ‘＝’ between the two C atoms 
    ♦ This is a double bond
■ The shortened form is shown in fig.4.11(c) 
2. Contribution from each atom
• Consider the ‘─’ between the top-left H and the first C
    ♦ It is clear that:
          ✰ One electron in the '─' belongs to the H
          ✰ The other electron in the '─' belongs to the C
• Consider the ‘─’ between the bottom-left H and the first C
    ♦ It is clear that:
          ✰ One electron in the '─' belongs to the H
          ✰ The other electron in the '─' belongs to the C
• Consider the ‘─’ between the top-right H and the second C
    ♦ It is clear that:
          ✰ One electron in the '─' belongs to the H
          ✰ The other electron in the '─' belongs to the C
• Consider the ‘─’ between the bottom-right H and the second C
    ♦ It is clear that:
          ✰ One electron in the '─' belongs to the H
          ✰ The other electron in the '─' belongs to the C
• Consider the ‘＝’ between the two C atoms 
    ♦ It is clear that:
          ✰ Two electrons in the '＝' belongs to the first C
          ✰ The remaining two electrons in the '＝' belongs to the second C
3. Checking the number of electrons in the Lewis dot structure:
• Each of the four H atoms have 2 dots around them
• Each of the C atoms have 8 dots around them
(Remember that, H needs to attain duplet only)
4. The shortened form must show the 'lone pairs' also
• The C atoms do not have any lone pairs
• The H atoms also do not have any lone pairs

Example 5:
• Fig.4.12(a) below shows the Lewis dot structure of N₂

Fig.4.12

• The 5 valence electrons of the first N are shown in red color
• The 5 valence electrons of the second N are shown in green color
• We can write the following 4 points:
1. Type of bond
• Consider the ‘shared region’ between the two N atoms
    ♦ There are 6 dots in this region. '6' indicates '3 pairs'
    ♦ So we put a ‘☰’ between the two N s
    ♦ This is a triple bond
■ So the shortened form is N☰N
This is shown the Lewis dot structure in fig.4.12(a)
2. Contribution from each atom
• Consider the ‘☰’ between the two N atoms
    ♦ It is clear that:
          ✰ Three electrons in the '☰' belongs to the first N
          ✰ The remaining three electrons in the '☰' belongs to the second N
3. Checking the number of electrons in the Lewis dot structure:
• The first N has 8 dots around it
• The second N has 8 dots around it
4. The shortened form must show the 'lone pairs' also
• The 'one lone pair' of each N are indicated by two dots around each N in the shortened form

Example 6:
• Fig.4.12(b) above shows the Lewis dot structure of C₂H₂
• The 1 valence electron of H is shown in green color
• The 4 valence electrons of C are shown in red color
• The reader may write all the 4 points in his/her own note books as an exercise

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• The above discussion will enable us to 'extract information' from any 'given Lewis dot structure'. In the next section, we will see the steps to draw Lewis dot structures of given molecules

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