Chapter 4.5 - Limitations of Octet Rule

In the previous section, we saw the steps to find the formal charge. In this section, we will see Limitations of the octet rule

■ Recall that, the octet rule states:
Atoms can combine either by transfer of electrons from one atom to another or by sharing of valence electrons in order to have an octet in their valence shells
• We have learnt the details of the rule. But there are three points to note:
(i) Those details that we learnt can be used effectively for understanding the structures of organic compounds
(ii) Unfortunately, the successful application of octet rule is possible only in the case of the ‘elements of the 2nd period’
• That is., whenever two elements of the 2nd period combine with each other, we can draw the ‘Lewis dot structure’ based on the octet rule
(iii) From the 3rd period onwards, the ‘electrons in the d orbitals’ also may take part in the chemical reactions. In such cases, we will not be able to apply the octet rule

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■ So we can write:

There are exceptions to the octet rule
■ What are those exceptions?
Answer:
There are three types of exceptions to the octet rule. They are:
(i) The incomplete octet of the central atom
(ii) Odd-electron molecules
(iii) The expanded octet
We will now see each of them in detail:

The incomplete octet of the central atom

This exception happens when the central atom has less than 4 valence electrons. It can be explained using some examples:
Example 1:
1. Fig.4.23 below shows the formation of LiCl (Lithium chloride)
    ♦ On the left side of the arrow, we see the Lewis symbols of Li and Cl
          ✰ Li has only one valence electron
    ♦ On the right side of the arrow, we see the Lewis dot structure of LiCl

Fig.4.23

• We notice that:
    ♦ Cl has attained octet
    ♦ But Li has not attained octet
          ✰ It needs 6 more electrons for octet
• Even with the 'Li without octet', the LiCl is a stable molecule
• The octet rule says that, atoms 'enter into bonds' to attain stable configuration (octet)
    ♦ But here, the Li enters into a bonding with Cl, even when it does not attain octet
■ So the octet rule fails to explain the formation of this molecule

Example 2:
1. Fig.4.24 below shows the formation of BeH₂ (Beryllium dihydride)
    ♦ On the left side of the arrow, we see the Lewis symbols of H and Be
          ✰ Be has only two valence electrons
    ♦ On the right side of the arrow, we see the Lewis dot structure of BeH₂

Fig.2.24

• We notice that:
    ♦ Both H atoms have attained octet
    ♦ But Be has not attained octet
          ✰ It needs 4 more electrons for octet
• Even with the 'Be without octet', the BeH₂ is a stable molecule
• The octet rule says that, atoms 'enter into bonds' to attain stable configuration (octet)
    ♦ But here, the Be enters into a bonding with H, even when it does not attain octet
■ So the octet rule fails to explain the formation of this molecule

Example 3:
1. Fig.4.25 below shows the formation of BCl₃ (Boron trichloride)
    ♦ On the left side of the arrow, we see the Lewis symbols of Cl and B
          ✰ B has only three valence electrons
    ♦ On the right side of the arrow, we see the Lewis dot structure of BCl₃

Fig.4.25

• We notice that:
    ♦ All Cl atoms have attained octet
    ♦ But B has not attained octet
          ✰ It needs 2 more electrons for octet
• Even with the 'B without octet', the BCl₃ is a stable molecule
• The octet rule says that, atoms 'enter into bonds' to attain stable configuration (octet)
    ♦ But here, the B enters into a bonding with Cl, even when it does not attain octet
■ So the octet rule fails to explain the formation of this molecule

Odd-electron molecules

• In some molecules, the 'sum of valence electrons of the atoms' will be an odd number. In such molecules, the octet rule will not be satisfied
• NO (Nitric oxide) is an example. It is shown in fig.4.26 below:

Fig.4.26

    ♦ On the left side of the arrow, we see the Lewis symbols of N and O
    ♦ On the right side of the arrow, we see the Lewis dot structure of NO

• We notice that:
    ♦ The O atom has attained octet
    ♦ But N has not attained octet
          ✰ It needs 1 more electron for octet
• Even with the 'N without octet', the NO is a stable molecule
• The octet rule says that, atoms 'enter into bonds' to attain stable configuration (octet)
    ♦ But here, the B enters into a bonding with Cl, even when it does not attain octet
■ So the octet rule fails to explain the formation of this molecule
Note that, the sum of valence electrons, which is [(5+6) = 11], is an odd number

The expanded octet

This can be explained in 6 steps:
(i) We know that, from the 3rd period onwards, the elements can possess the 3d orbitals also
(ii) The general electronic configuration of elements are in the pattern: 1s²2s²2p⁶3s²3p⁶3d¹⁰4s . . .
(iii) Note the terms with coefficient '2'. They are: 2s² and 2p⁶
    ♦ From those two sub-shells, we get 8 electrons
(iv) But from the 3rd period onwards, d orbitals can also be present
    ♦ Some of the electrons in those 3d orbitals may take part in chemical bondings
(v) In such cases, we will see more than eight electrons around the central atom
■ This is called the expanded octet
(vi) 'oct'et is related to eight
• If there are more than eight electrons, it is obvious that, octet rule is not applicable
Let us see some examples:
Example 1:
1. Fig.4.27 below shows the formation of PF₅ (Phosphorus pentafluoride)
    ♦ On the left side of the arrow, we see the Lewis symbols of F and P
          ✰ F has seven valence electrons. It needs one more electron
          ✰ P has five valence electrons. It needs three more electrons
    ♦ On the right side of the arrow, we see the Lewis dot structure of PF₅

Fig.4.27

• We notice that:
    ♦ All F atoms have attained octet
    ♦ The P atom needed only 3 more electrons. But it has acquired 5 more
          ✰ If it had attained 3 more, the total would have been (5+3) = 8
          ✰ But since it had attained 5 more, the total now is (5+5) = 10
• Compare this situation with N atom (belonging to the same group as P)
    ♦ N atom also needs 3 more electrons
    ♦ It can never acquire '5 more' because, there are no d orbitals available
■ So P atom has an expanded octet

Example 2:
1. Fig.4.28 below shows the formation of SF₆ (Sulfur hexafluoride)
    ♦ On the left side of the arrow, we see the Lewis symbols of F and S
          ✰ F has seven valence electrons. It needs one more electron
          ✰ S has six valence electrons. It needs three more electrons
    ♦ On the right side of the arrow, we see the Lewis dot structure of SF₆

Fig.4.28

• We notice that:
    ♦ All F atoms have attained octet
    ♦ The S atom needed only 2 more electrons. But it acquired 6 more
          ✰ If it had attained 2 more, the total would have been (6+2) = 8
          ✰ But since it had attained 6 more, the total now is (6+6) = 12
• Compare this situation with O atom (belonging to the same group as S)
    ♦ O atom also needs 2 more electrons
    ♦ It can never acquire '6 more' because, there are no d orbitals available
■ So S atom has an expanded octet

Example 3:
1. Fig.4.29 below shows the formation of H₂SO₄ (Sulfuric acid)
    ♦ On the left side of the arrow, we see the Lewis symbols of H, O and S
          ✰ H has one valence electron. It needs one more electron
          ✰ O has six valence electrons. It needs two more electrons
          ✰ S has six valence electrons. It needs two more electrons
    ♦ On the right side of the arrow, we see the Lewis dot structure of H₂SO₄

Fig.4.29

• We notice that:
    ♦ All H atoms have attained duplet
    ♦ All O atoms have attained octet
    ♦ The S atom needed only 2 more electrons. But it acquired 6 more
          ✰ If it had attained 2 more, the total would have been (6+2) = 8
          ✰ But since it had attained 6 more, the total now is (6+6) = 12
• Compare this situation with O atom (belonging to the same group as S)
    ♦ O atom also needs 2 more electrons
    ♦ It can never acquire '6 more' because, there are no d orbitals available
■ So S atom has an expanded octet

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However, S is a special case. It does obey octet rule in many compounds. An example is given below:

1. Fig.4.30 below shows the formation of SCl₂ (Sulfur dichloride)
    ♦ On the left side of the arrow, we see the Lewis symbols of Cl and S
          ✰ Cl has seven valence electrons. It needs one more electron
          ✰ S has six valence electrons. It needs two more electrons
    ♦ On the right side of the arrow, we see the Lewis dot structure of SCl₂

Fig.2.29

• We notice that:
    ♦ All Cl atoms have attained octet
    ♦ The S atom needed only 2 more electrons. Indeed, it has acquired only 2
■ So S atom obeys the octet rule in this case

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■ Note: The 'expanded octet' needs more detailed study. We want to know 'how the d orbitals get into bonds' with other atoms. We will see those details in later sections

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Other drawbacks of the octet rule
1. The octet rule is based on the chemical inertness of noble gases
• That is., octet rule says that:
Every element tries to attain the stable electronic configuration of the nearest noble gas
• So the octet rule assumes that, the noble gases does not take part in chemical reactions because, they already have octet
• But in reality, some noble gases do react with elements like oxygen and fluorine
    ♦ Examples are: XeF₂, KrF₂, XeOF₂ etc.,
• The octet rule is unable to give an explanation for such reactions
2. The molecules that we see around us have different shapes
    ♦ Some have planar shapes
    ♦ Some have pyramidal shapes
    ♦ So on . . .
• The octet rule is unable to give reasons for such different shapes
3. We know that elements react with each other to attain stability
• That means, molecules are more stable than atoms
• The octet rule is not able to give reasons for that ‘greater stability’
• This is mainly because, the octet rule is silent about ‘energy changes’ involved in reactions

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In the next section, we will see ionic bonds

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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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