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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