Chapter 12.16 - Resonance Effect

In the previous section, we saw the details about stability of resonance structures. In this section, we will see resonance effect.

• We know that inorganic molecules like HCl are polar.
    ♦ The Cl atom being electronegative, pulls the electron density in the covalent bond.
    ♦ This gives Cl a partial -ve charge. Also H gets a partial +ve charge.
    ♦ See section 4.10.
• Some organic molecules are also polar. This can be demonstrated using two examples.

Example 1:
In this example, we analyze the resonance structures of aniline. It can be written in 6 steps:
1. The structure I of fig.12.101 below shows aniline. In this structure, all the atoms have octet.

Fig.12.101

• The structure II is obtained by rearranging the electrons present in I
• In I, NㅡC₁ is a single bond. But this same NㅡC₁ becomes a double bond in II.
• This is because of the movement of electrons indicated by the green curved-arrow.
   ♦ Two electrons from the lone pair of N moves to NㅡC₁
   ♦ Those two electrons were singly-owned by N.
   ♦ After the formation of N=C₁, the N co-owns those electrons.
   ♦ So N does not lose any electrons.
• So due to the green curved-arrow,
   ♦ N does not lose any electrons.
   ♦ C₁ gains two electrons.
• Note the + formal charge of N in II. This can be explained in 3 steps:
(i) An independent N atom will have five electrons around it.
(ii) But in II, there are only four electrons around N.
    ♦ Two from the bonds with the H atoms.
    ♦ Two from the double bond.
(iii) That means, N has lost an electron. Thus it gains one +ve formal charge. 
• Due to the green curved-arrow in I, C₁ now possess two new electrons.
• But C₁ already has octet. It does not need any electrons.
• So the magenta curved-arrow comes into play.
   ♦ Two electrons in C₁=C₂ moves to C₂
   ♦ They become a lone pair of C₂
• Thus the two unwanted electrons gained by C₁ through the green curved-arrow, are lost by the action of magenta curved arrow.
• Note the - formal charge of C₂ in II. This can be explained in 3 steps:
(i) An independent C atom will have four electrons around it.
(ii) But in II, there are five electrons around C₂
    ♦ Two from the lone pair.
    ♦ One from C₂ㅡC₁
    ♦ One from C₂ㅡC₃
    ♦ One from C₂ㅡH
(iii) That means, C₂ has gained an electron. Thus it gains one -ve formal charge. 
2. From the above step (1), we know how structure II is formed. So let us analyze the curved-arrows in structure II
• The C₂ in II has two unwanted electrons. So the yellow curved-arrow comes into play.
    ♦ The lone pair of C₂ moves to C₂ㅡC₃.
    ♦ Thus C₂ㅡC₃  becomes C₂=C₃.
    ♦ This double bond is shown in III
• When such a double bond is formed, C₃ will have two unwanted electrons.
    ♦ So the cyan curved-arrow comes into play.
    ♦ Two electrons in C₃=C₄ moves to C₄
    ♦ They become a lone pair of C₄
• Thus the two unwanted electrons gained by C₃ through the yellow curved-arrow, are lost by the action of cyan curved arrow.
• Note the - formal charge of C₄ in III. This can be explained in 3 steps:
(i) An independent C atom will have four electrons around it.
(ii) But in II, there are five electrons around C₄
    ♦ Two from the lone pair.
    ♦ One from C₄ㅡC₃
    ♦ One from C₄ㅡC₅
    ♦ One from C₄ㅡH
(iii) That means, C₄ has gained an electron. Thus it gains one -ve formal charge.
3. From the above step (2), we know how structure III is formed. We can write similar steps to analyze the curved-arrows in structure III.
• Based on those steps, we would see how structure IV is formed.
• The reader is advised to write those steps in his/her own notebook.
4. In effect, we see that:
• Structure I is neutral.
• Structures II, III and IV are not neutral. They are charged.
    ♦ This is due to the movement of electrons as indicated by the curved arrows.
    ♦ Positions with -ve formal charge have high electron densities.
    ♦ Positions with +ve formal charge have low electron densities.
5. So one contributing structure is neutral while the remaining three contributing structures are charged.
• So the hybrid structure will be charged. That means, in a sample of aniline, the molecules will exist in a charged state. This is just like charged molecules in a sample of HCl.
6. In this example, the electrons are transferred away from the substituent group (ㅡNH₂ group).
• If the transfer of electrons is away from an atom or substituent group attached to the conjugated system, it is known as positive resonance effect (+R effect)
• A conjugated system is that system in which single bonds and double bonds occur in an alternating arrangement.

Example 2:
In this example, we analyze the resonance structures of nitrobenzene. It can be written in 6 steps:
1. The structure I of fig.12.102 below shows nitrobenzene. In this structure, all the atoms have octet.

Fig.12.102

• The arrow between N and O indicates that, the bond between N and that O is a coordinate bond. Both electrons in that bond originally belonged to N. However, that coordinate bond has no role in our present discussion.
• The structure II is obtained by rearranging the electrons present in I
• Consider the N=O double bond in I. This same N=O becomes a single bond in II.
• This is because of the movement of electrons indicated by the green curved-arrow.
   ♦ Two electrons in the N=O moves to O.
   ♦ They become a lone pair of O
• Due to this movement of electrons, N has lost two electrons. So the magenta curved-arrow comes into play.
   ♦ Two electrons in C₁=C₂ moves to NㅡC₁
   ♦ Thus C₁=C₂ becomes C₁ㅡC₂ and NㅡC₁ becomes N=C₁
• Note the - formal charge of O in II. This can be explained in 3 steps:
(i) An independent O atom will have six electrons around it.
(ii) But in II, there are seven electrons around that O
    ♦ Six from three lone pairs.
    ♦ One from NㅡO
(iii) That means, O has gained an electron. Thus it gains one -ve formal charge.
• Note the + formal charge of C₂ in II. This can be explained in 3 steps:
(i) An independent C atom will have four electrons around it.
(ii) But in II, there are only three electrons around C₂.
    ♦ One from bond with the H atom.
    ♦ One from C₂ㅡC₁
    ♦ One from C₂ㅡC₃
(iii) That means, C₂ has lost an electron. Thus it gains one +ve formal charge. 
• Thus the two electrons lost by N through the green curved-arrow, are gained by the action of magenta curved arrow.
2. From the above step (1), we know how structure II is formed. So let us analyze the curved-arrow in structure II
• The C₂ in II needs two electrons. So the yellow curved-arrow comes into play.
   ♦ Two electrons in C₃=C₄ moves to C₂ㅡC₃
   ♦ Thus C₃=C₄ becomes C₃ㅡC₄ and C₂ㅡC₃ becomes C₂=C₃.
   ♦ This is shown in III.
• Note the + formal charge of C₄ in III. This can be explained in 3 steps:
(i) An independent C atom will have four electrons around it.
(ii) But in III, there are only three electrons around C₄.
    ♦ One from bond with the H atom.
    ♦ One from C₄ㅡC₃
    ♦ One from C₄ㅡC₅
(iii) That means, C₄ has lost an electron. Thus it gains one +ve formal charge. 
3. From the above step (2), we know how structure III is formed. We can write similar steps to analyze the curved-arrow in structure III.
• Based on those steps, we would see how structure IV is formed.
• The reader is advised to write those steps in his/her own notebook.
4. In effect, we see that:
• Structure I is neutral.
• Structures II, III and IV are not neutral. They are charged.
    ♦ This is due to the movement of electrons as indicated by the curved arrows.
    ♦ Positions with -ve formal charge have high electron densities.
    ♦ Positions with +ve formal charge have low electron densities.
5. So one contributing structure is neutral while the remaining three contributing structures are charged.
• So the hybrid structure will be charged. That means, in a sample of nitrobenzene, the molecules will exist in a charged state. This is just like charged molecules in a sample of HCl.
6. In this example, the electrons are transferred towards the substituent group (ㅡNO₂ group).
• If the transfer of electrons is towards an atom or substituent group attached to the conjugated system, it is known as negative resonance effect (-R effect).

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◼ Note that:
• In example 1 (+R effect), the C atoms in the chain acquired greater electron densities. This is because of the movement of electrons away from the substituent group.
• In example 2 (-R effect), the C atoms in the chain lost electron densities. This is because of the movement of electrons towards the substituent group.

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• The atoms or substituent groups which cause +R effect are:
ㅡ halogen, ㅡOH, ㅡOR, ㅡOCOR, ㅡNH₂, ㅡNHR, ㅡNR₂, ㅡNHCOR
• The atoms or substituent groups which cause -R effect are:
ㅡCOOH, ㅡCHO, >C=O, ㅡCN, ㅡNO₂

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• As mentioned before, the resonance effect (+R effect and -R effect) occur in conjugated systems. In conjugated systems, single bonds and double bonds in the chain are present in an alternating arrangement.
• Benzene and substituted benzene compounds are examples of a closed chain conjugated system.
• 1,3-butadiene is an example of an open chain conjugated system. It’s condensed formula is shown below:
CH₃=CH₂ㅡCH₂=CH₃

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◼ Why is it that, the conjugated systems are subjected to +R effect and -R effects ?
• The answer can be written in 3 steps:
1. We know that, in a double bond, there is a 𝜎 bond and a π bond.
• In a 𝜎 bond, there is linear overlap of orbitals. The electrons are locked in position.
• But in a π bond, there is lateral overlap of orbitals. So the electrons are free to move.
2. So the electrons of the π bond move towards an adjacent single bond. That single bond will then become a double bond. We indicated such movements using curved-arrows in the above examples.
• The electrons of the π bond can also move towards an adjacent atom. Those atoms will then become regions of high electron densities. We indicated such movements also using curved-arrows in the above examples.
3. Regions of ‘electron richness’ and ‘electron deficiency’, will make the molecule polar.

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In the next section we will see electromeric effect.

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Chapter 12.15 - Relative Stability of Resonance Structures

In the previous section, we saw the basics about resonance in organic molecules. In this section, we will see how to compare various resonance structures of a molecule.

Rule 1:
    ♦ The resonance structure which has more number of covalent bonds
    ♦ is more stable than
    ♦ The resonance structure which has lesser number of covalent bonds.
• For calculating the number of covalent bonds, a double bond is considered as two bonds. A triple bond is considered as three bonds.
• An example is shown in fig.12.92 below.

Fig.12.92

• This can be explained in 4 steps:
(i) I and II are resonance structures of the same molecule.
• In I, the number of bonds is four. But in II, the number of bonds is three.
• So I will be more stable than II
(ii) Consequently, the hybrid structure will have a greater resemblance to I than II.
• We can say:
Structure I will contribute more towards the hybrid structure.
(iii) Let us check the formal charges. We have seen how to calculate formal charges in section 4.4.
We know that, any independent O atom will have six valence electrons.
• In I, we see that, the bottom O atom has gained an electron.
    ♦ That means, O has gained a -ve charge.
    ♦ Thus the formal charge of that O is ‘-’.
• In II, we see that, both top O and bottom O has gained an electron each.
    ♦ That means, those O atoms have gained a -ve charge each.
    ♦ Thus the formal charge of each of those O atoms is ‘-’.
(iv) We know that, any independent C atom will have four valence electrons.
• In II, we see that, the C atom has lost an electron.
    ♦ That means, C has gained a +ve charge.
    ♦ Thus the formal charge of that C is ‘+’.

Rule 2:
    ♦ The resonance structure in which all atoms have octet
    ♦ is more stable than
    ♦ The resonance structure in which one or more atoms have incomplete octet.
• An example is shown in fig.12.93 below.

Fig.12.93

This can be explained in 3 steps:
(i) I and II are resonance structures of the same molecule.
• In I, all atoms have octet.
• In II, the C atom has only six electrons around it.
• So I will be more stable than II
(ii) Consequently, the hybrid structure will have a greater resemblance to I than II.
• We can say:
Structure I will contribute more towards the hybrid structure.
(iii) We know that, any independent C atom will have four valence electrons.
• In II, we see that, the C has lost one of it’s electrons.
    ♦ That means, C has gained a +ve charge.
    ♦ Thus the formal charge of that C is ‘+’.

Rule 3:
    ♦ The resonance structure which has the least 'number of formal charges'
    ♦ is more stable than
    ♦ The resonance structure which has greater 'number of formal charges'.
• An example is shown in fig.12.94 below:

Fig.12.94

This can be explained in 4 steps:
(i) I and II are resonance structures of the same molecule.
• In II, there are two formal charges, while in I, there are none.
• So I will be more stable than II
(ii) Consequently, the hybrid structure will have a greater resemblance to I than II.
• We can say:
Structure I will contribute more towards the hybrid structure.
(iii) We know that, any independent C atom will have four valence electrons.
• In II, we see that, the C atom of methyl branch has lost one of it’s electrons.
    ♦ That means, C has gained a +ve charge.
    ♦ Thus the formal charge of that C is ‘+’.
(iv) We know that, any independent O atom will have six valence electrons.
• In II, we see that, the O atom has gained one electron.
    ♦ That means, O has gained a -ve charge.
    ♦ Thus the formal charge of that O is ‘-’.

Rule 4:
    ♦ The resonance structure which has the -ve charge on more electronegative atom
    ♦ is more stable than
    ♦ The resonance structure which has the -ve charge on less electronegative atom.
• An example is shown in fig.12.95 below:

Fig.12.95

This can be explained in 4 steps:
(i) I and II are resonance structures of the same molecule.
• In II, the -ve charge is at the O atom. This is acceptable.
• In I, the -ve charge is at a C atom. This is also acceptable.
• But the -ve charge tends to be at a more electronegative atom. When O and C are compared, O is more electronegative.
• So II will be more stable than I
(ii) Consequently, the hybrid structure will have a greater resemblance to II than I.
• We can say:
Structure II will contribute more towards the hybrid structure.
(iii) We know that, any independent C atom will have four valence electrons.
• In I, we see that, the C atom of CH₂ group has gained an electron.
    ♦ That means, C has gained a -ve charge.
    ♦ Thus the formal charge of that C is ‘+’.
(iv) We know that, any independent O atom will have six valence electrons.
• In II, we see that, the O atom has gained one electron.
    ♦ That means, O has gained a -ve charge.
    ♦ Thus the formal charge of that O is ‘-’.

Rule 5:
This is the reverse of rule 3.
    ♦ The resonance structure which has the +ve charge on more electropositive atom
    ♦ is more stable than
    ♦ The resonance structure which has the +ve charge on less electropositive atom.

Rule 6:
    ♦ The resonance structure in which 'distance between charges' is lesser
    ♦ is more stable than
    ♦ The resonance structure in which 'distance between charges' is greater.
• An example is shown in fig.12.96 below:

Fig.12.96

 

• This can be explained in 3 steps:
(i) I and II are resonance structures of the same molecule.
• In I, the charges are close together (on adjacent atoms).
• In II, the distance between the charges is larger.
• So I will be more stable than II
(ii) Consequently, the hybrid structure will have a greater resemblance to I than II.
• We can say:
Structure I will contribute more towards the hybrid structure.
(iii) We know that, any independent C atom will have four valence electrons.
• In I, we see that, the first C atom from left has lost an electron.
    ♦ That means, C has gained a +ve charge.
    ♦ Thus the formal charge of that C is ‘+’.
• In I, we see that, the second C atom from left has gained an electron.
    ♦ That means, C has gained a -ve charge.
    ♦ Thus the formal charge of that C is ‘-’.
• In II, we see that, the first C atom from left has lost an electron.
    ♦ That means, C has gained a +ve charge.
    ♦ Thus the formal charge of that C is ‘+’.
• In II, we see that, the last C atom from left has gained an electron.
    ♦ That means, C has gained a -ve charge.
    ♦ Thus the formal charge of that C is ‘-’.

Rule 7:
Resonance structures which are equivalent, will have the same stability.
• An example is shown in fig.12.97 below:

Fig.12.97

• This can be explained in 3 steps:
(i) I and II are resonance structures of the same molecule.
If we rotate I about an axis passing through the CㅡH bond, we will get II.
(ii) We can apply any of the six rules written above. We will see that both are equivalent.
(iii) We know that, any independent O atom will have six valence electrons.
• In I, we see that, the bottom O atom has gained an electron.
    ♦ That means, O has gained a -ve charge.
    ♦ Thus the formal charge of that O is ‘-’.
• In II, we see that, the top O atom has gained an electron.
    ♦ That means, O has gained a -ve charge.
    ♦ Thus the formal charge of that O is ‘-’.

Solved example 12.18
Write the resonance structures of CH₂=CHㅡCHO. Indicate the relative stability of the contributing structures.
Solution:
1. The three resonance structures are shown in fig.12.98 below:

Fig.12.98

  
2. First we apply rule 1:
(a) The number of bonds in I is 9
(b) The number of bonds in II is 8
(c) The number of bonds in III is 8
(d) Based on the number of bonds:
    ♦ I is more stable than II
    ♦ I is more stable than III
• So I is the most stable among the three.
3. Based on the number of bonds, II and III has the same stability.
• So we have to apply the other rules to find which one among II and III is more stable.
4. Applying rule 2, we get:
(a) In II, the first C (from left) has an incomplete octet. All other atoms have octet.
(b) In III, the O has an incomplete octet. All other atoms have octet.
(c) So applying rule 2, both II and III have the same stability.
• We have to apply the other rules.
5. Applying rule 3, we get:
(a) The number of formal charges in II is 2  
(b) The number of formal charges in III is also 2
(c) So applying rule 3, both II and III have the same stability.
• We have to apply the other rules.
6. Applying rule 4, we get:
(a) In II, the -ve charge is on the more electronegative atom, which is O
(b) In III, the -ve charge is on the lesser electronegative atom, which is C
(c) So applying rule 4, II is more stable than III
7. So the order of stability in the decreasing order is: I > II > III
• I makes the greatest contribution towards the hybrid structure.
• III makes the least contribution towards the hybrid structure.
• The hybrid structure will have more resemblance to I

Solved example 12.19
Explain why the following two structures I and II cannot be the major contributors to the real structure of CH₃COOCH₃

Fig.12.99

Solution:
1. In this problem, we are not asked to compare the two given structures. We are asked why both are not major contributors.
2. Consider I. The middle C atom do not have octet. So this structure cannot be a major contributor.
3. Consider II. A +ve charge is present on the lower O atom. This will make the structure very unstable because, O is highly electronegative. So this structure also cannot be a major contributor.

Solved example 12.20
Indicate the relative stability of the following resonance structures:

Fig.12.100

Solution:
1. First we apply rule 1:
(a) The number of bonds in I is 6
(b) The number of bonds in II is 5
(c) The number of bonds in III is 6
(d) Based on the number of bonds:
    ♦ I and III are more stable than II
• So the least stable structure is II.
2. Based on the number of bonds, I and III has the same stability.
• So we have to apply the other rules to find which one among I and III is more stable.
3. Applying rule 2, we get:
(a) In I, all atoms have octet.
(b) In III also, all atoms have octet.
(c) So applying rule 2, both I and III have the same stability.
• We have to apply the other rules.
4. Applying rule 3, we get:
(a) The number of formal charges in I is 2  
(b) The number of formal charges in III is also 2
(c) So applying rule 3, both II and III have the same stability.
• We have to apply the other rules.
5. Applying rule 4, we get:
(a) In I, the -ve charge is on the more electronegative atom, which is N
(b) In III, the -ve charge is on the lesser electronegative atom, which is C
(c) So applying rule 4, I is more stable than III
6. So the order of stability in the decreasing order is: I > III > II
• I makes the greatest contribution towards the hybrid structure.
• II makes the least contribution towards the hybrid structure.
• The hybrid structure will have more resemblance to I.

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In the next section, we will see resonance effect.

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Chapter 8.7 - The Paradox of Fractional Oxidation Number

In the previous section, we completed a discussion on different types of redox reactions. In this section, we will see the Paradox of fractional oxidation number. This can be explained using some examples.

Example 1:
This can be written in 8 steps:
1. Consider C₃O₂
• We know that, the oxidation number of O will be -2. We want to find the oxidation number of C.
2. For that, we write: $\mathbf\small{\rm{\overset{x}{C}_3\,\overset{-2}{O}_2}}$
• Thus we get: 3x - 2 × 2 = 0 ⇒ 3x - 4 = 0 ⇒ x = +⁴⁄₃
3. This indicates that, each C has lost ⁴⁄₃ electrons.
• But electrons are whole objects. They cannot be cut into fractions.
◼ Then why do we get '⁴⁄₃' ?
The following steps from (4) to (8) will give the answer:
4. The structure of C₃O₂ is shown in fig.8.4(a) below:

Fig.8.4

• There are four bonds in total.
• In fig.b,
   ♦ the two C＝C bonds are marked in red color. They are named as R₁ and R₂.
   ♦ the two C＝O bonds are marked in green color. They are named as G₁ and G₂.
5. Consider G₁. It is a double bond. So there will be four electrons
• Originally,
   ♦ two electrons belong to C
   ♦ two electrons belong to O
• But O being more electronegative, will pull the electrons belonging to C also. So C loses two electrons
• Consider R₁. This bond is between two C atoms. The pulls will be equal.
   ♦ Since the pulls are equal, the carbon will not lose any electrons at R₁
• So the net effect is:
The carbon on the right side of G₁, loses two electrons and gets an oxidation number of +2
6. By symmetry, the same steps in (5) can be written about G₂ and R₂. We will get:
• The carbon on the left side of G₂, loses two electrons and gets an oxidation number of +2
7. Consider the middle C atom. It is surrounded by C atoms on either sides. So the pulls will be equal. It neither loses nor gains electrons. Thus it’s oxidation number will be 0.
8. Thus, in the structural formula of C₃O₂, we see that, different C atoms have different oxidation numbers.
• But in the molecular formula C₃O₂, we write C only once. So we write the average oxidation number. The average can be calculated in 3 steps:
(i) Total oxidation number of C atoms = 2 + 0 + 2 = 4
(ii) Number of C atoms = 3
(iii) So average oxidation number = ⁴⁄₃

Example 2:
This can be written in 8 steps:
1. Consider Br₃O₈
• We know that, the oxidation number of O will be -2. We want to find the oxidation number of Br.
2. For that, we write: $\mathbf\small{\rm{\overset{x}{Br}_3\,\overset{-2}{O}_8}}$
• Thus we get: 3x - 2 × 8 = 0 ⇒ 3x - 16 = 0 ⇒ x = +¹⁶⁄₃
3. This indicates that, each C has lost ¹⁶⁄₃ electrons.
• But electrons are whole objects. They cannot be cut into fractions.
◼ Then why do we get '¹⁶⁄₃' ?
The following steps from (4) to (8) will give the answer:
4. The structure of Br₃O₈ is shown in fig.8.5(a) below:

Fig.8.5

• There are ten bonds in total.
• In fig.b,
   ♦ the two Br一Br bonds are marked in red color. They are named as R₁ and R₂.
   ♦ the eight Br＝O bonds are marked in green color. They are named G₁ to G₈.
5. Consider G₁. It is a double bond. So there will be four electrons
• Originally,
   ♦ two electrons belong to Br.
   ♦ two electrons belong to O.
• But O being more electronegative, will pull the electrons belonging to Br also. So Br loses two electrons.
• This happens in G₂ and G₃ also.
   ♦ So there is a total loss of 6 electrons.
• Consider R₁. This bond is between two Br atoms. The pulls will be equal.
   ♦ Since the pulls are equal, the Br will not lose any electrons at R₁
• So the net effect is:
The left most Br, loses six electrons and gets an oxidation number of +6.
6. By symmetry, the same steps in (5) can be written about the right most Br atom. We will get:
• The right most Br, loses six electrons and gets an oxidation number of +6.
7. Consider the middle Br atom. It is surrounded by Br atoms on either sides. So the pulls will be equal towards the sides. It neither loses nor gains electrons towards the sides.
• But the O atom at top will pull two electrons.
• The O atom at bottom will also pull two electrons.
• Thus it’s oxidation number will be +4.
8. Thus, in the structural formula of Br₃O₈, we see that, different Br atoms have different oxidation numbers.
• But in the molecular formula Br₃O₈, we write Br only once. So we write the average oxidation number. The average can be calculated in 3 steps:
(i) Total oxidation number of Br atoms = 6 + 4 + 6 = 16
(ii) Number of Br atoms = 3
(iii) So average oxidation number = ¹⁶⁄₃

Example 3:
This can be written in 8 steps:
1. Consider Na₂S₄O₆
• It is an ionic compound. In aqueous solution, it dissociates into Na⁺ and S₄O₆^2- ions
• The S₄O₆^2- ions are stable and exist independently. So we will analyze it.
• We know that, the oxidation number of O will be -2. We want to find the oxidation number of S.
2. For that, we write: $\mathbf\small{\rm{\overset{x}{S}_4\,\overset{-2}{O}_6}}$
• Thus we get: 4x - 2 × 6 = -2 ⇒ 4x - 12 = -2 ⇒ x = +¹⁰⁄₄ = +2.5
3. This indicates that, each S has lost 2.5 electrons.
• But electrons are whole objects. They cannot be cut into fractions.
◼ Then why do we get '2.5' ?
The following steps from (4) to (8) will give the answer:
4. The structure of S₄O₆^2- is shown in fig.8.6(a) below:

Fig.8.6

• There are nine bonds in total.
• In fig.b,
   ♦ the three S一S bonds are marked in red color. They are named R₁ to R₃.
   ♦ the two S一O bonds are marked in yellow color. They are named as Y₁ and Y₂.
   ♦ the four S＝O bonds are marked in green color. They are named G₁ to G₄.
5. Consider G₁. It is a double bond. So there will be four electrons
• Originally,
   ♦ two electrons belong to S
   ♦ two electrons belong to O
• But O being more electronegative, will pull the electrons belonging to S also. So S loses two electrons.
   ♦ This happens in G₂ also.
• So there is now a loss of 4 electrons.
• Consider Y₁. It is a single bond. So there will be two electrons
• Originally,
   ♦ one electron belong to S
   ♦ one electron belong to O
• But O being more electronegative, will pull the electron belonging to S also. So S loses one electron.
• So there is now a loss of 5 electrons.
• Consider R₁. This bond is between two S atoms. The pulls will be equal.
   ♦ Since the pulls are equal, the S will not lose any electrons at R₁
• So the net effect is:
The left most S, loses five electrons and gets an oxidation number of +5.
6. By symmetry, the same steps in (5) can be written about the right most S atom. We will get:
The right most S, loses five electrons and gets an oxidation number of +5.
7. Consider the two middle S atoms. They are surrounded by S atoms on either sides. So the pulls will be equal towards the sides. They neither lose nor gain electrons towards the sides.
   ♦ Thus their oxidation numbers will be 0.
8. Thus, in the structural formula of S₄O₆^2-, we see that, different S atoms have different oxidation numbers.
• But in the molecular formula, S₄O₆^2-, we write S only once. So we write the average oxidation number. The average can be calculated in 3 steps:
(i) Total oxidation number of S atoms = 5 + 0 + 0 + 5 = 10
(ii) Number of S atoms = 4
(iii) So average oxidation number = ¹⁰⁄₄ = 2.5

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◼ Based on the above three examples, we can write a conclusion. It can be written in 4 steps:
1. Whenever, we see a fractional oxidation number, we must remember that, it is only an average value.
2. Atoms cannot lose or gain fractional electrons.
3. The actual loss or gain can be obtained only from the structural formula. The structural formula will reveal that, the atom is present in more than one oxidation states.
4. The fraction is obtained when we take the average of those oxidation states.

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• Just like the three examples, we can explain the fractional oxidation numbers in some other compounds like Fe₃O₄, Mn₃O₄, Pb₃O₄ etc.,

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◼ Fractional oxidation states can occur in some ions also.
• In O₂⁺, even though there are no other atoms, the oxidation number of O is +¹⁄₂. This can be explained in steps:
(i) The oxidation number of O in O₂ is 0
(ii) But O₂⁺ has lost one electron. This loss is indicated by +1.
(iii) The loss is shared by two O atoms. So each O will have an oxidation number of +¹⁄₂.
• In O₂^-, even though there are no other atoms, the oxidation number of O is -¹⁄₂. This can be explained in steps:
(i) The oxidation number of O in O₂ is 0
(ii) But O₂^- has gained one electron. This gain is indicated by -1.
(iii) The gain is shared by two O atoms. So each O will have an oxidation number of -¹⁄₂.

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In the next section, we will see balancing of redox reactions.


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Chapter 3.9 - Periodic Trends in Electronegativity

In the previous section 3.8, we completed a discussion on periodic trends in electron gain enthalpy. In this section, we will see the periodic trends in electronegativity

1. In our previous classes, we have learned about covalent bonds (Details here)
• In this type of bonds, two atoms share ‘two electrons’ (a pair of electrons)
• Both the atoms have claim over both the electrons
2. There are two possibilities:
(i) Both the atoms are of the same type (eg: O-O, Cl-Cl)
    ♦ In this case, the ‘shared pair of electrons’ is equally attracted by the two atoms
(ii) The two atoms are of different types
    ♦ In this case, the ‘shared pair of electrons’ may get attracted towards one of the atoms
3. Some atoms have greater ability to attract the ‘shared pair of electrons’
■ A qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself is called electronegativity
4. Properties like ionization enthalpy, electron gain enthalpy etc., can be measured. But electronegativity cannot be measured
• That is why we use the word ‘qualitative’ 
5. Let us see an example of ‘qualitative measure’
We will write it in steps:
(i) We want a color which is close to red
(ii) We examine different available colors
    ♦ Some colors are very reddish
    ♦ Some colors are moderately red
    ♦ Some colors have only a reddish tint
(iii) In this situation, we assume a convenient value of say ‘10’ for ‘perfect red’
    ♦ ‘Very reddish’ colors will be given values like 8 or 9
    ♦ ‘Moderate reddish’ colors will be given values like 4 or 5
    ♦ ‘Colors with only reddish tints’ will be given values like 2 or 3
(iv) In this way, we make a ‘qualitative measure’ of the color
6. Another example would be the ‘quality of work’ done by a mechanic when he repairs a car
    ♦ The owner of the car may give a ‘rating of 10’ if the work is ‘excellent’
    ♦ Rating may be ‘5’ or ‘6’ if the work ‘satisfactory’
    ♦ Rating may be ‘1’ or ‘2’ if the work is ‘poor’
7. In this way, electronegativity can be measured only in a qualitative way
■ Scientists have developed different scales for measuring electronegativity
• Some of them are:
    ♦ Pauling scale
    ♦ Mulliken-Jaffe scale
    ♦ Allred-Rochow scale
8. Pauling scale is the most widely used scale
• It was developed in 1922 by the American scientist Linus Pauling
9. In the Pauling scale, fluorine is considered to have the ‘greatest ability to attract electrons’
■ Pauling assigned an arbitrary value of ‘4’ to fluorine
• ‘Arbitrary’ means: 
Based on random choice, rather than any reason or system
(The dictionary meaning can be seen here)
10. Approximate values for the electronegativity of a few elements are given in the tables 3.6 and 3.7 below:

Table 3.6

Table 3.7

• Note that, fluorine (which has the ‘greatest ability to attract electrons’) is given a value of ‘4’. So all other elements will be having values less than ‘4’
11. Electronegativity of a given element is not constant
• It depends on the other elements to which the ‘given element’ is bound

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• Now we know the basics about electronegativity

• We can start the discussion on it’s periodic trends
First we will see the trend along periods. We will write it in steps:
1. Imagine that, we are moving from left to right along a period
• We will see that, the electronegativity increases
2. For example, consider the 2nd period
• In this period,
    ♦ Li will have the smallest electronegativity
    ♦ F will have the largest electronegativity 
3. Recall that, along the periods, ‘atomic radii’ also has a similar (but opposite) trend:
■ When we move from left to right, the radii decreases
• So naturally, we are inclined to make a comparison between two items:
    ♦ ‘Ability to attract electrons (electronegativity)’ increases towards the right
    ♦ ‘Atomic radii’ decreases towards the right
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The atomic radii decreases because, the outermost electrons are attracted more and more tightly towards the nucleus
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted more and more tightly towards the nucleus
4. Again recall that, along the periods, the ‘ionization enthalpy’ also has a similar trend:
■ When we move from left to right, the ‘ionization enthalpy’ increases
• That is., it is more and more difficult to remove an electron
• So naturally, we are inclined to make a comparison between two items:
    ♦ ‘Ability to attract electrons (electronegativity)’ increases towards the right
    ♦ ‘Ionization enthalpy’ increases towards the right
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The ionization enthalpy increases because, the outermost electrons are attracted more and more tightly towards the nucleus
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted more and more tightly towards the nucleus
5. Again recall that, along the periods, the ‘electron gain enthalpy’ also has a similar trend:
■ When we move from left to right, the ‘electron gain enthalpy’ becomes more and more negative
• That is., elements are more and more happier to gain an electron
• So naturally, we are inclined to make a comparison between two items:
    ♦ ‘Ability to attract electrons (electronegativity)’ increases towards the right
    ♦ ‘Electron gain enthalpy’ becomes more and more negative towards the right
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The electron gain enthalpy becomes more negative because, the elements are happier to add a new electron
    ♦ That is., the newly added electron will be held more tightly towards the nucleus
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted more and more tightly towards the nucleus
6. So we have seen four items along a period:
(i) Atomic radii   (ii) Ionization energy   (iii) Electron gain enthalpy   (iv) Electronegativity
• We saw that, all the four are closely related
• Since we compared their trends along periods (horizontal rows), we will use four horizontal arrows to show their increasing and decreasing trends
• This is shown in fig.3.14 below:
    ♦ This image is obtained from Wikimedia commons
    ♦ The link is given below:
    ♦ Link to image

Fig.3.14

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Next we will see the trend along groups. We will write it in steps:

1. Imagine that, we are moving from top to bottom along a period
• We will see that, the electronegativity decreases
2. For example, consider the 17th group
• In this period,
    ♦ F will have the largest electronegativity
    ♦ At will have the smallest electronegativity
3. Recall that, along the groups, ‘atomic radii’ also has a similar (but opposite) trend:
■ When we move from top to bottom, the radii increases
• So naturally, we are inclined to make a comparison between two items:
    ♦ ‘Ability to attract electrons (electronegativity)’ decreases towards the bottom
    ♦ ‘Atomic radii’ increases towards the bottom
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The atomic radii increases because, the outermost electrons are attracted less and less tightly towards the nucleus
    ♦ With the passage of each element, one more main-shell is added
    ♦ The repulsion from the inner electron increases
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted less and less tightly towards the nucleus
4. Again recall that, along the groups, the ‘ionization enthalpy’ also has a similar trend:
■ When we move from top to bottom, the ‘ionization enthalpy’ decreases
• That is., it is more and more easy to remove an electron
• So naturally, we are inclined to make a comparison between two items:
    ♦ ‘Ability to attract electrons (electronegativity)’ decreases towards the bottom
    ♦ ‘Ionization enthalpy’ decreases towards the bottom
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The ionization enthalpy decreases because, the outermost electrons are attracted less and less tightly towards the nucleus
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted less and less tightly towards the nucleus
5. Again recall that, along the periods, the ‘electron gain enthalpy’ also has a similar trend:
■ When we move from top to bottom, the ‘electron gain enthalpy’ becomes less and less negative
• That is., elements are less and less happier to gain an electron
• So naturally, we are inclined to make a comparison between two items:
    ♦ ‘Ability to attract electrons (electronegativity)’ decreases towards the bottom
    ♦ ‘Electron gain enthalpy’ becomes less and less negative towards the bottom
Are these two, related to each other?
• We can easily see the relation. It can be explained in 2 steps:
(i) The electron gain enthalpy becomes less negative because, the elements are less happier to add a new electron
    ♦ That is., the newly added electron will be held less tightly towards the nucleus
(ii) In the same way, the ‘shared pair of electrons’ will also be attracted less and less tightly towards the nucleus
6. So we have seen four items along a group:
(i) Atomic radii   (ii) Ionization energy   (iii) Electron gain enthalpy   (iv) Electronegativity
• We saw that, all the four are closely related
• Since we compared their trends along groups (vertical columns), we will use four vertical arrows to show their increasing and decreasing trends
• This is shown in fig.3.15 below:
    ♦ This image is obtained from Wikimedia commons
    ♦ The link is given below:
    ♦ Link to image

Fig.3.15

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• Next we will see the following relation:

Relation between electronegativities and 'non-metallic character of elements'
1. Non-metallic elements have a strong tendency to gain electrons
2. We have seen that ‘electronegativity’ indicates the tendency to gain electrons
• So elements having ‘greater electronegativity’ will be ‘more non-metallic' in nature
3. Now we will bring ‘periodic trends in electronegativity’ into this discussion:
(i) Consider the ‘horizontal arrow of electronegativity’ in fig.3.14 above
• We see that, it points towards the right
    ♦ That means, electronegativity increases towards the right
    ♦ That means, elements towards the right will be more non-metallic
(ii) Consider the ‘vertical arrow of electronegativity’ in fig.3.15 above
• We see that, it points upwards
    ♦ That means, electronegativity increases towards the top
    ♦ That means, elements towards the top will be more non-metallic
4. So we have two arrows
• They are shown in figs.3.16 (a) and (b) below
    ♦ Fig.a indicates that, the elements towards the right will be more non-metallic
    ♦ Fig.b indicates that, the elements towards the top will be more non-metallic

Fig.3.16

5. Combining the two information, we get:
■ Elements towards the top-right will be more non-metallic
• So we can draw an arrow towards the top-right corner of the periodic table
6. We can also write it in this way:
• A horizontal force acts towards the right as indicated by fig.3.16(a)
• A vertical force acts towards the top as indicated by fig.3.16(b)
■ The resultant of the two forces will obviously be a slanting force acting towards the top-right. This is shown in fig.3.16(c)
7. Now we know the periodic trend in non-metallic character
• We can show it in the periodic table. This is shown as the brown arrow in fig.3.17 below:

Fig.3.17

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• Next we will see the following relation:

Relation between electronegativities and 'metallic character of elements'
1. Metallic elements have a strong tendency to lose electrons
2. We have seen that ‘electronegativity’ indicates the tendency to gain electrons
• So elements having ‘lesser electronegativity’ will be ‘more metallic' in nature
3. Now we will bring ‘periodic trends in electronegativity’ into this discussion:
(i) Consider the ‘horizontal arrow of electronegativity’ in fig.3.14 above
• We see that, it points towards the right
    ♦ That means, electronegativity increases towards the right
    ♦ That means, electronegativity decreases towards the left
    ♦ That means, elements towards the left will be more metallic
(ii) Consider the ‘vertical arrow of electronegativity’ in fig.3.15 above
• We see that, it points upwards
    ♦ That means, electronegativity increases towards the top
    ♦ That means, electronegativity decreases towards the bottom
    ♦ That means, elements towards the bottom will be more metallic
4. So we have two arrows
• They are shown in figs.3.18 (a) and (b) below
    ♦ Fig.a indicates that, the elements towards the left will be more metallic
    ♦ Fig.b indicates that, the elements towards the bottom will be more metallic

Fig.3.18

5. Combining the two information, we get:
■ Elements towards the bottom-left will be more metallic
• So we can draw an arrow towards the bottom-left corner of the periodic table
6. We can also write it in this way:
• A horizontal force acts towards the left as indicated by fig.3.18(a)
• A vertical force acts towards the bottom as indicated by fig.3.18(b)
■ The resultant of the two forces will obviously be a slanting force acting towards the bottom-left. This is shown in fig.3.18(c)
7. Now we know the periodic trend in metallic character
We can show it in the periodic table. This is shown  as the magenta arrow in fig.3.17 above

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Now we will see some solved examples

Solved example 3.19
Considering the elements B, Al, Mg, and K, the correct order of their metallic character is :
(a) B > Al > Mg > K     (b) Al > Mg > B > K
(c) Mg > Al > K > B     (d) K > Mg > Al > B
Solution:
1. All the given four options are in the decreasing order
• That means:
    ♦ The most metallic element is written first
    ♦ The next most metallic element is written second
    ♦ so on . . .
2. We know that, the metallic character increases towards the bottom-left
• Out of the four given elements, the 'left-most' and 'bottom-most' is K
• So, out of the four given elements, K is the most metallic. It should be written first
3. Out of the four given elements, the 'right-most' and 'top-most' will be the least metallic
• We can see that, it is B
• So, out of the four given elements, B is the least metallic. It should be written last
4. Now, two elements remain: Al and Mg
• They belong to the same period
    ♦ So there is no question of 'being on top or bottom' of the other
• The element on the left will be more metallic
• So we get: Mg is more metallic. It should be written before Al
5. So the fourth option gives the correct order: K > Mg > Al > B

Solved example 3.20 
Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is :
(a) B > C > Si > N > F     (b) Si > C > B > N > F
(c) F > N > C > B > Si     (d) F > N > C > Si > B
Solution:
1. All the given four options are in the decreasing order
• That means:
    ♦ The most non-metallic element is written first
    ♦ The next most non-metallic element is written second
    ♦ so on . . .
2. We know that, the non-metallic character increases towards the top-right
• Out of the five given elements, the 'right-most' and 'top-most' is F
• So, out of the five given elements, F is the most non-metallic. It should be written first
3. Out of the five given elements, the 'left-most' and 'bottom-most' will be the least non-metallic
    ♦ We see that, the left-most is B
    ♦ But the 'bottom-most' is Si
• So we have to pick the correct one from B and Si
• Which one is least non-metallic?
• To find the answer, we compare the electronegativities
    ♦ The electronegativity of B is 2.0
    ♦ The electronegativity of Si is 1.8
• B has greater electronegativity
    ♦ So B is more non-metallic than Si
    ♦ So the element with least non-metallic character is Si
    ♦ It should be written last
■ Note: To compare B and Si, we cannot seek the help of C. This is because, C is more non-metallic than both B and Si 
4. Three elements remain: N, C and Si
• Out of them, the 'right-most' and 'top-most' is N
• So N should be written next to F
5. Two elements remain: C and Si
• They both belong to the same group
    ♦ So there is no question of being on left or right of the other
• The element on the top will be more non-metallic
• So C is more non-metallic than Si
• C must be written before Si
6. Thus the third option gives the correct order: F > N > C > Si > B

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• We have completed a discussion on the basics of seven physical properties:
(i) Atomic radius
(ii) Ionic radius
(iii) Ionization energy
(iv) Electron gain enthalpy
(v) Electronegativity
(vi) Metallic character
(vii) Non-metallic character
• We also saw the basics about their periodic trends
• In the next section, we will see some chemical properties

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