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 - Chemical Bonding and Molecular Structure

In the previous section 3.11, we completed the chapter on classification of elements and periodic trends. In this chapter, we will see chemical bonding and molecular structure

1. Taking two samples:
■ We know that, ‘any thing which occupies space’ is called matter
• So consider two samples (Sample A and Sample B) of ‘two different things that occupy space’
    ♦ Suppose that, ‘Sample A’ is a sample of any one of the noble gases
    ♦ Suppose that, ‘Sample B’ does not contain any one of the noble gases
2. Then we can write two points:
(i) Sample A will contain independent atoms
(ii) Sample B will not contain even a single independent atom
■ How can we be so sure about Sample B?
• Answer can be written in just one sentence:
No element (except noble gases) can exist as independent atoms
3. So in what form do the ‘elements other than noble gases’ exist?
• The answer can be written in 2 steps:
(i) ‘Elements other than noble gases’ exist as independent molecules
(ii) Each molecule will contain two or more atoms
• In some cases, ‘those atoms in a molecule’ will be of the same type
    ♦ For example, in the molecule O₂, all are O atoms
• In some cases, ‘those atoms in a molecule’ will be of different types
    ♦ For example, in the molecule H₂O, there are H and O atoms
4. Then the next question arises:
■ How do ‘those atoms in a molecule’ stick together?
• The answer can be written in steps:
(i) There exists a ‘force of attraction’ between individual atoms in a molecule
(ii) Due to the presence of this 'attractive force', the atoms cannot separate away from each other
(iii) This 'attractive force between atoms' is called chemical bond
■ We can write the definition in a single sentence:
The attractive force which holds various constituents (atoms, ions, etc.) together in different chemical species is called a chemical bond 
5. So it is clear that, individual atoms of various elements combine together to form a molecule
■ But this information leads to several more questions:
• Why do atoms combine?
• Why are only certain combinations possible? 
• Why do some atoms combine while certain others do not?
• Why do molecules possess definite shapes?
In this chapter we will try to find the answers to these questions

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• In the year 1916 Kossel and Lewis succeeded in presenting a satisfactory 'model of molecule'

• This model gave a basic explanation of 'how various atoms stick together in a molecule'
• G. N. Lewis was an American scientist
• Walther Kossel was a German scientist
• Though the model is known as Kossel-Lewis model, the two scientists had worked independently
• Let us write the salient features of this model:

Feature 1. Parts of an atom
• Consider an atom
• It will consist of two parts:
(i) An inner ‘kernel’
    ♦ The 'kernel' is positively charged
    ♦ But it consists of the nucleus as well as the inner electrons
(ii) The 'outer shell'
    ♦ The 'outer shell' is the shell which contains the outermost electrons (valence electrons)  
Let us see some examples:
Example 1:
• The electronic configuration of Na is 1s²2s²2p⁶3s¹
    ♦ This is same as [Ne]3s¹
• The 'kernel' of Na will consist of two items:
    ♦ The nucleus of Na
    ♦ The electrons of [Ne]
• The 'outer shell' of Na will be the 'shell which contains the last single electron'
• The 'kernel' and the 'outer shell' together constitute the atom
Example 2:
• The electronic configuration of Cl is 1s²2s²2p⁶3s²3p⁵
    ♦ This is same as [Ne]3s²3p⁵
• The 'kernel' of Cl will consist of two items:
    ♦ The nucleus of Cl
    ♦ The electrons of [Ne]
• The 'outer shell' of Cl will be the 'shell which contains the last 7 electrons'
• The 'kernel' and the 'outer shell' together constitute the atom

Feature 2: The number of electrons in the 'outer shell'
• There can be a maximum of 8 electrons in the 'outer shell'

Feature 3: Shape of the 'outer shell'
• The 'outer shell' is in the shape of a cube
• This cube surrounds the 'kernel'
• This is shown in fig.4.1 below:

Fig.4.1

Feature 4: Positions of the electrons in the 'outer shell'
• The electrons are situated at the corners of the cube
• Any cube will have 8 corners
• So the 'outer shell' can accommodate a maximum of 8 electrons
Examples:
    ♦ The one and only outer electron of Na will be situated at one of the total 8 corners
          ✰ This is shown in fig.4.2(a) below
          ✰ Note that, 7 corners of the cube are vacant
    ♦ The 7 outer electrons of Cl will be situated at 7 of the total 8 corners
          ✰ This is shown in fig.4.2(b) below
          ✰ Note that, only 1 corner of the cube is vacant

Fig.4.2

Feature 5: Arrangement in the case of noble gases
• In the case of noble gases, all the 8 corners will be occupied
• When all the 8 corners are occupied, 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

Feature 6: Octet of atoms other than noble gases
• Atoms other than noble gases try to attain octet
• They attain octet by 'entering into chemical bonds' with other atoms

Feature 7: The two methods for 'entering into chemical bonds'
Method 1:
• One or more electrons will be transferred from one atom to the other atom
    ♦ After the transfer, 'the atom which loses electrons' will be having 8 electrons at the 8 corners
    ♦ After the transfer, 'the atom which gains electrons' also will be having 8 electrons at the 8 corners
An example:
• Na loses it's one and only electron and becomes Na⁺
    ♦ As a result, [Ne]3s¹ becomes [Ne]
    ♦ [Ne] has 8 electrons in the outermost shell
• Cl gains the 'electron lost by Na' and becomes Cl^-
    ♦ As a result, [Ne]3s²3p⁵ becomes [Ne]3s²3p⁶
    ♦ [Ne]3s²3p⁶ has 8 electrons in the outermost shell
• Na⁺ is positively charged and Cl^- is negatively charged
    ♦ As a result, an electrostatic force of attraction comes into effect between the two ions
    ♦ So we will not be able to separate the two ions from each other
    ♦ This is shown in fig.4.3 below:

Fig.4.3

Method 2:
• A 'pair of electrons' (two electrons) is shared between two atoms
• When two electrons are shared in this way, both the atoms will be having 8 electrons at the respective 8 corners
An example:
• Fig.4.4 below shows two independent Cl atoms

Fig.4.4

• The 7 outermost electrons of the first Cl atom are shown in red color
• The 7 outermost electrons of the second Cl atom are shown in green color
• Make a note of the electron marked as 'A' in the first Cl atom
    ♦ It has an adjacent vacant corner
• Make a note of the electron marked as 'B' in the first Cl atom
    ♦ It also has an adjacent vacant corner
• Now consider fig.4.5 below:

Fig.4.5

• The two Cl atoms are now combined to form a Cl molecule
    ♦ The electron 'A' occupies the corner which was vacant in the second Cl atom
    ♦ The electron 'B' occupies the corner which was vacant in the first Cl atom
■ From the view point of the first Cl atom, all it's 8 corners are now occupied
    ♦ Thus this Cl atom has attained octet
■ From the view point of the second Cl atom, all it's 8 corners are now occupied
    ♦ Thus this Cl atom also has attained octet
• Electrons 'A' and 'B' constitute the 'shared pair'
    ♦ Both the Cl atoms have equal claim on both 'A' and 'B'
    ♦ So the two Cl atoms cannot move away from each other   
■ 'Both the electrons in the pair' belongs to both the atoms

Feature 8: Using symbols
• It is not easy to draw 3D models of the cube for every atoms
• So Lewis developed a simplified method
• Only the electrons in the outer shell will take part in chemical reactions
• The electrons in the inner shells are well protected. In most cases, they do not take part in chemical reactions
■ Lewis noticed that, we need to show the outermost electrons only
    ♦ There electrons are shown as dots
    ♦ The dots are marked around the ‘symbol of atoms’
■ This notation is called Lewis symbol
• The fig.4.6 below shows the Lewis symbols of elements of the 2nd period

Fig.4.6

Feature 9: Significance of Lewis symbols
• The ‘number of dots’ in the Lewis symbol can be used to calculate the common valence or group valence
• When the number of dots is less than or equal to 4:
    ♦ Common valence = Number of dots
• When the number of dots is greater than 4:
    ♦ Common valence = 8 – number of dots

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• The above given are the nine main features of the Lewis-Kossel model

• In addition to the above, Kossel gave a few more information. They are known as Kossel's Postulates
(A postulate is 'something which is assumed to be true'. So that, it can be used as a basis for reasoning or discussion. The dictionary meaning can be seen here) 
• They can be written in 6 steps
1. We know that:
• Alkali metals (group 1) are highly electropositive
• Halogens (group 17) are highly electronegative
2. Also we know that, in the periodic table,
• Alkali metals are at the left end
• Halogens are near the right end
3. Alkali metals being electropositive, readily lose their outermost single electron
• When that electron is lost, the atom becomes a +ve ion
    ♦ The +ve ion thus formed will be having the electronic configuration of a noble gas
    ♦ ‘Electronic configuration of a noble gas’ is a very stable configuration
    ♦ That means, the +ve ion formed from the ‘alkali metal atom’ will be very stable
4. Halogens being electronegative, readily accepts one more electron
• When that electron is gained, the atom becomes an -ve ion
    ♦ The -ve ion thus formed will be having the electronic configuration of a noble gas
    ♦ ‘Electronic configuration of a noble gas’ is a very stable configuration
    ♦ That means, the -ve ion formed from the ‘halogen atom’ will be very stable
5. So we have a 'stable +ve ion' and a 'stable -ve ion'
• An electrostatic force of attraction comes into play between the two oppositely charged ions
• Due to this electrostatic force of attraction, we will not be able to separate the two ions
    ♦ The two ions will always stick together
    ♦ Thus a molecule is formed
• We will see two examples:
Example 1: Formation of sodium chloride (NaCl)
• Na is an alkali metal
• It loses one electron as shown below:
$\mathbf\small{\rm{Na\longrightarrow Na^{+}+e^{-}}}$
    ♦ This is same as [Ne]3s¹ becoming [Ne]
• Cl is a halogen
• It gains one electron as shown below:
$\mathbf\small{\rm{Cl+e^{-}\longrightarrow Cl^{-}}}$
    ♦ This is same as [Ne]3s²3p⁵ becoming [Ne]3s²3p⁶ or [Ar]
■ The Na+ and Cl- thus formed will stick together (due to electrostatic force of attraction) as shown below:
$\mathbf\small{\rm{Na^{+}+Cl^{-}\longrightarrow NaCl\,\,\;OR\;\,\,Na^{+}Cl^{-}}}$
• The fig.4.7 below shows the above result using Lewis symbols:

Fig.4.7

Example 2: Formation of calcium fluoride (CaF₂)
• Ca is an alkaline earth metal (group 2)
• It loses two electrons as shown below:
$\mathbf\small{\rm{Ca\longrightarrow Ca^{2+}+2e^{-}}}$
    ♦ This is same as [Ar]4s² becoming [Ar]
• F is a halogen
• It gains one electron as shown below:
$\mathbf\small{\rm{F+e^{-}\longrightarrow F^{-}}}$
    ♦ This is same as [He]2s²2p⁵ becoming [He]2s²2p⁶ or [Ne]
■ The Ca+ and 2F- thus formed will stick together (due to electrostatic force of attraction) as shown below:
$\mathbf\small{\rm{Ca^{2+}+2F^{-}\longrightarrow CaF_2\,\,\;OR\;\,\,Ca^{2+}(F^{-})_2}}$
• The fig.4.8 below shows the above result using Lewis symbols:

Fig.4.8

■ Note: 
    ♦ One Ca atom loses two electrons
    ♦ But one F atom can accept only one electron
    ♦ So 'two F atoms' will be required to accept 'the two electrons' lost by 'the one Ca'
6. In the above two examples, we see a 'chemical bonding' between two ions
• This chemical bonding helps in the formation of a molecule
• This chemical bonding is possible because of the electrostatic force of attraction between +ve and -ve ions 
■ So Kossel called it: electrovalent bond
■ We can write the definition in one sentence:
The bond formed, as a result of the electrostatic attraction between the positive and negative ions was termed (by Kossel) as electrovalent bond
■ Kossel gave the definition for electrovalence also:
The charge possessed by an ion, when that ion is part of an electrovalent bond is called electrovalence
Some examples:
    ♦ Electrovalence of Na is +1
    ♦ Electrovalence of Ca is +2 
    ♦ Electrovalence of Cl is -1
    ♦ Electrovalence of F is -1
■ Kossel's Postulates provided a strong foundation for further studies about 'structure of ionic compounds'. However, Kossel and other scientists of that time knew that, the 'structures of a large number of compounds' cannot be explained using these postulates. We will see them in later sections

Now we will see a solved example

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Solved example 4.1

Write Lewis dot symbols for the atoms of the following elements:
Mg. Na, B, O, N, Br
Solution:
The required Lewis dot symbols are shown in fig. below:

Sample explanation:
• Consider Br. It has the electronic configuration: 1s²2s²2p²3s²3p⁶3d¹⁰4s²4p⁵ OR [Ar]3d¹⁰4s²4p⁵
• So the outermost main-shell has 7 electrons. That means, there are 7 valence electrons
• Thus there will be 7 dots in the Lewis dot symbol of Br

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• In the next section, we will see the Octet rule. We will also see covalent bonds and Lewis dot structures

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