Chapter 8.2 - Oxidation Number

In the previous section we saw the electrochemical series. In this section, we will see oxidation number.

Oxidation number can be explained using some examples. It can be written in 17 steps. We will also see the rules on the way:
1. Consider the reaction between Na and Cl to form NaCl:
$\mathbf\small{\rm{2Na +Cl_2\rightarrow 2NaCl}}$
2. We know that Na donates one electron and attains a positive charge.
• So we write ‘+1’ above Na in the final product NaCl. This is shown in the product side of the equation below:
$\mathbf\small{\rm{2\overset{0}{Na} +\overset{0}{Cl_2}\rightarrow 2\overset{+1}{Na}\overset{-1}{Cl}}}$
• This ‘+1’ is the oxidation number of Na in NaCl.
3. We know that Cl accepts one electron and attains a negative charge.
• So we write ‘-1’ above Cl in the final product NaCl. This is also shown in the product side of equation above.
• This ‘-1’ is the oxidation number of Cl in NaCl
4. We know that, in the Cl₂ molecule, the shared pair of electrons are equally attracted by the two atoms. (see fig.4.9 in section 4.1)
• As a result, none of the Cl atoms will have any extra charge. So we write ‘0’ above Cl in Cl₂ on the reactant side of the equation in (2). This ‘0’ is the oxidation number of Cl in Cl₂.
5. Similarly, the Na on the reactant side is neutral. It does not have any extra charge. So we write '0' above Na on the reactant side of the equation in (2). This ‘0’ is the oxidation number of Na.
6. Based on steps (4) and (5) above, we can write the first rule related to the calculation of oxidation number.
Rule I:
In elements, in the free or the uncombined state, each atom bears an oxidation number of zero. For example, each atom in H₂, O₂, Cl₂, O₃, P₄, S₈, Na, Mg, Al has the oxidation number zero.
7. Based on steps (2) and (3), we can write the second rule.
Rule II:
This rule can be written in four steps:
(i) For ions composed of only one type of atom, the oxidation number is equal to the charge on the ion. Thus we can write some examples:
   ♦ Na⁺ ion has an oxidation number of +1.
   ♦ Mg²⁺ ion has an oxidation number of +2.
   ♦ Fe³⁺ ion has an oxidation number of +3.
   ♦ Cl^- ion has an oxidation number of -1.
   ♦ O^2- ion has an oxidation number of -2.
so on.
(ii) In their compounds, all alkali metals (group 1 elements) have oxidation number of +1.
(iii) In their compounds, all alkaline earth metals (group 2 elements) have an oxidation number of +2.
(iv) Aluminum is regarded to have an oxidation number of +3 in all its compounds.
8. NaCl that we considered above, is an ionic compound. Next let us consider a covalent compound. It can be written in 4 steps:
(i) Consider the Lewis dot structure of H₂O (see fig.4.10 in section 4.1)
    ♦ We have one H atom, on either sides of the O atom.
(ii) The shared pair of electrons on the left side belongs to both left H atom and O atom. But the O atom, being more electronegative than H, pulls the pair. One electron already belonged to the O. So due to the pull, the O attains one partial negative charge.
(iii) In the same way, on the right side also, O attains one partial negative charge. So in total, O has two partial negative charges.
(iv) For the purpose of calculating oxidation number, we assume that, the partial charge is a 'full charge'. That is., the extra electron which causes the partial negative charge is completely shifted towards the O atom. Thus O atom has a charge of -2. So it’s oxidation number is -2.
9. Based on the above step (8), we can write the third rule:
Rule III:
For the purpose of calculating oxidation number, it is assumed that, the electron pair in the covalent bond belongs entirely to the more electronegative element.
10. Step (8) gives us the impression that, oxidation number of oxygen atom will be always -2. This is true in general. But there are three exceptions (a), (b) and (c).
(a) The first exception can be written in 4 steps:
(i) Consider the peroxide H₂O₂. It is a covalent compound.
• It’s structural formula is: H-O-O-H
(ii) We see that, all atoms have octet. Consider the left side O atom.
   ♦ It can pull the electron pair from the H atom.
   ♦ It cannot pull the electron pair from the O atom.
   ♦ So the oxidation number of that O atom will be -1.
(iii) Using the same steps as in (ii), we can find that:
Oxidation number of right side O atom is also -1.
(iv) So in peroxides (eg., H₂O₂, Na₂O₂), the oxidation number of O atoms will be -1.
(b) The second exception can be written in 4 steps:
(i) Consider the superoxide KO₂
• It is an ionic compound. The ions are K⁺ and O₂^-
(ii) K can donate only one electron. This electron is possessed by the O^2- ion. So the electron belongs to two O atoms.
(iii) That means, the one extra negative charge belongs to two O atoms. Thus the oxidation number of each O atom will be -¹⁄₂
(iv) So in superoxides (eg., KO₂, RbO₂), the oxidation number of O atoms will be -¹⁄₂
(c) The third exception can be written in 6 steps:
(i) Consider oxygen difluoride OF₂. It is a covalent compound.
• It’s structural formula is: F-O-F
(ii) We see that, all atoms have octet. Consider the O atom.
   ♦ It cannot pull the electron pair from the left F atom.
   ♦ It cannot pull the electron pair from the right F atom either.
   ♦ This is because, O is less electronegative than F
• In fact, the F atoms will pull the pairs towards either sides.
   ♦ Thus the O atom loses two electrons.
• So in oxygen fluoride, the oxidation number of O atom will be +2.
(iii) Consider dioxygen difluoride F₂O₂. It is a covalent compound.
• It’s structural formula is: F-O-O-F
(iv) We see that, all atoms have octet. Consider the left side O atom.
   ♦ It will lose one electron to the F atom.
   ♦ It will not lose any electron to the O atom.
• So the oxidation number of that O atom will be +1.
(v) Using the same steps as in (iv), we can find that:
Oxidation number of right side O atom is also +1.
(vi) So in dioxygen difluoride, the oxidation number of O atoms will be +1.
11. Based on the above step (10), we can write the fourth rule. It is related to oxygen:
Rule IV:
• Oxygen generally has an oxidation number of -2
• But there are three exceptions:
(i) In peroxides (eg., H₂O₂, Na₂O₂), the oxidation number of O atoms will be -1.
(ii) In superoxides (eg., KO₂, RbO₂), the oxidation number of O atoms will be -¹⁄₂
(iii) When oxygen is bonded to fluorine, the oxidation number becomes positive
   ♦ In oxygen difluoride OF₂, the oxidation number of O atom will be +2.
   ♦ In dioxygen difluoride O₂F₂, the oxidation number of O atom will be +1.
12. Consider the element hydrogen
• It forms covalent bond with non metals. The electronegativity of those non metals is greater than that of hydrogen (eg., HCl). So H will lose the electron and will thus have an oxidation number of +1.
• But when hydrogen combines with metals, the electronegativity of H will be higher. So in compounds like LiH, NaH, CaH₂, which have H and another metal only (binary compounds having two elements), H will have an oxidation number of -1.
13. Based on the above step (8), we can write the fourth rule. It is related to hydrogen:
Rule V:
• Hydrogen generally has an oxidation number of +1
• But there is an exception:
When H is bonded to metals in binary compounds, the oxidation number will be -1.
14. The sixth rule is related to halogens. It can be written in 3 steps:
Rule VI:
(i) F will always have an oxidation number of -1. Because it is highly electronegative.
(ii) In general, Cl, Br and I also will have an oxidation number of -1.
(iii) But when Cl, Br and I combines with O, the oxidation numbers will be positive.
16. So now we have the basic knowledge about the oxidation numbers of some common elements. Using those known oxidation numbers, we can calculate unknown oxidation numbers. This can be explained using an example. It can be written in 5 steps:
(i) In FeCl₃, we know that each Cl atom will have an oxidation number of -1.
(ii) We want to find the oxidation number of Fe. Let us assume that Fe has an oxidation number of x
(iii) Then we can represent the compound as: $\mathbf\small{\rm{\overset{x}{Fe}\,\overset{-1}{Cl}_3}}$
(iv) The algebraic sum of the oxidation number of all the atoms in a compound must be zero.
• In our present case,
   ♦ there is one Fe atom. It has an oxidation number of x
   ♦ there are three Cl atoms. Each has an oxidation number of -1
• So the algebraic sum is (x + 3 × -1) = (x - 3)
(v) This sum must be zero. So we can write: x - 3 = 0
• Thus we get: x = 3
• That is., the oxidation number of Fe in FeCl₃ is +3 

Let us see another example. It can be written in 5 steps:
(i) In (CO₃)^2-, we know that each O atom will have an oxidation number of -2.
(ii) We want to find the oxidation number of C. Let us assume that C has an oxidation number of x
(iii) Then we can represent the compound as: $\mathbf\small{\rm{\left (\overset{x}{C}\,\overset{-2}{O}_3 \right )^{-2}}}$
(iv) In polyatomic ion, the algebraic sum of all the oxidation numbers of atoms of
the ion must equal the charge on the ion.
• In our present case,
   ♦ there is one C atom. It has an oxidation number of x
   ♦ there are three O atoms. Each has an oxidation number of -2
• So the algebraic sum is (x + 3 × -2) = (x - 6)
(v) This sum must be equal to the charge of the ion, which is -2.
So we can write: x - 6 = -2
• Thus we get: x = 4
• That is., the oxidation number of C in (CO₃)^2- is +4.
17. Based on the above step (16), we can write the seventh rule. It is related to the actual mathematical calculation of oxidation number. It can be written in two steps:
Rule VII:
(i) The algebraic sum of the oxidation number of all the atoms in a compound must be zero.
(ii) In polyatomic ion, the algebraic sum of all the oxidation numbers of atoms of
the ion must equal the charge on the ion.

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A simplified set of rules can be written as follows:
1. In elements, in the free or the uncombined state, each atom bears an oxidation number of zero.
2. For ions composed of only one type of atom, the oxidation number is equal to the charge on the ion.
   ♦ For group 1 elements, oxidation number is +1
   ♦ For group 2 elements, oxidation number is +2
3. Common oxidation number of O is -2
   ♦ But in H₂O₂, the oxidation number of O is -1
   ♦ Also, when O combines with F, the oxidation number changes.
4. The oxidation number of F is always -1
5. Common oxidation number of H is +1
   ♦ But when H combines with metals, the oxidation number becomes -1
   ♦ Also when H combines with boron, the oxidation number becomes -1
6. Group 17 elements (F, Cl, Br, I) have a common oxidation number of -1
   ♦ But when Cl, Br and I combines with O or F, the oxidation number changes.
7. For a neutral molecule, the sum of all oxidation numbers will be zero.
   ♦ For a polyatomic ion, the sum of all oxidation numbers will be equal to the charge of the ion.

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• By the application of above rules, we can find out the unknown oxidation number of any element in a molecule or in an ion.
• It is clear that:
   ♦ Metallic elements have positive oxidation number.
        ✰ This is because, they tend to donate electrons.
   ♦ Nonmetallic elements have negative or positive oxidation number.
        ✰ This is because, they tend to accept electrons.
        ✰ They also tend to donate electrons to more electronegative elements.
   ♦ Transition elements usually display several positive oxidation states.
        ✰ This is because, the electrons in the penultimate subshell also some times take part in the reactions (details here). It depends on the circumstances.

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• Next we will see how the highest oxidation number is related to groups and periods in the periodic table. It can be written in steps:
1. We know that, when the transition elements are avoided, the remaining elements are called representative elements.
• The s and p blocks together constitute the representative elements.
• In the s-block, there are two groups: group 1 and group 2.
    ♦ For the elements in group 1, the highest oxidation number will be +1.
    ♦ For the elements in group 2, the highest oxidation number will be +2.
2. In the p-block, there are six groups: group 13 to group 18.
    ♦ For the elements in group 13, the highest oxidation number will be +3.
        ✰ Note that, this ‘+3’ is equal to (group number 13 - 10)
    ♦ For the elements in group 14, the highest oxidation number will be +4.
        ✰ Note that, this ‘+4’ is equal to (group number 14 - 10)
    ♦ For the elements in group 15, the highest oxidation number will be +5.
        ✰ Note that, this ‘+5’ is equal to (group number 15 - 10)
    ♦ For the elements in group 16, the highest oxidation number will be +6.
        ✰ Note that, this ‘+6’ is equal to (group number 16 - 10)
    ♦ For the elements in group 17, the highest oxidation number will be +7.
        ✰ Note that, this ‘+7’ is equal to (group number 17 - 10)
    ♦ For the elements in group 18, we need not consider the oxidation number because, they take part in reactions only rarely.
3. So, as we move from left to right along a period, the highest oxidation number increases. This can be confirmed if we examine the elements of period 3.
• The first element of this period is Na. Consider the compound NaCl.
    ♦ We know that, oxidation number of Na in NaCl is +1.
• The next element of this period is Mg. Consider the compound MgCl₂.
    ♦ We know that, oxidation number of Mg in MgCl₂ is +2.
• The Next element of this period is Al. Consider the compound AlF₃.
    ♦ We can write this as: $\mathbf\small{\rm{\overset{x}{Al}\,\overset{-1}{F}_3}}$
    ♦ So we get: (x + 3 × -1) = 0  ⇒ x = 3
    ♦ Thus, oxidation number of Al in AlF₃ is +3.
• The next element of this period is Si. Consider the compound SiCl₄.
    ♦ We can write this as: $\mathbf\small{\rm{\overset{x}{Si}\,\overset{-1}{Cl}_4}}$
    ♦ So we get: (x + 4 × -1) = 0  ⇒ x = 4
    ♦ Thus, oxidation number of Si in SiCl₄ is +4.
• The next element of this period is P. Consider the compound P₄O₁₀.
    ♦ We can write this as: $\mathbf\small{\rm{\overset{x}{P}\,\overset{-2}{O}_{10}}}$
    ♦ So we get: (4x + 10 × -2) = 0  ⇒ x = 5
    ♦ Thus, oxidation number of P in P₄O₁₀ is +5.
• The next element of this period is S. Consider the compound SF₆.
    ♦ We can write this as: $\mathbf\small{\rm{\overset{x}{S}\,\overset{-1}{F}_6}}$
    ♦ So we get: (x + 6 × -1) = 0  ⇒ x = 6
    ♦ Thus, oxidation number of S in SF₆ is +6.
• The next element of this period is Cl. Consider the compound HClO₄.
    ♦ We can write this as: $\mathbf\small{\rm{\overset{+1}{H}\,\overset{x}{Cl}\,\overset{-2}{O}_4}}$
    ♦ So we get: (+1 + x + 4 × -2) = 0  ⇒ x = 7
    ♦ Thus, oxidation number of Cl in HClO₄ is +7.

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• The term oxidation state can be used interchangeably with oxidation number.
• For example:
   ♦ The oxidation number of C in CO₂ is +4.
   ♦ The oxidation number of O in CO₂ is -2.
• We can write this as:
   ♦ The oxidation state of C in CO₂ is +4.
   ♦ The oxidation state of O in CO₂ is -2.
• That means:
   ♦ The oxidation number of an element in a compound,
   ♦ denotes the
   ♦ oxidation state of that element in that compound.

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Solved example 8.2
Assign oxidation number to the underlined elements in each of the following species:
(a) NaH₂PO₄   (b) NaHSO₄   (c) H₄P₂O₇   (d) K₂MnO₄
(e) CaO₂   (f) NaBH₄   (g) H₂S₂O₇   (h) KAl(SO₄)₂.12H₂O
Solution:
Part (a): NaH₂PO₄
We can write this as: $\mathbf\small{\rm{\overset{+1}{Na}\,\overset{+1}{H}_2\,\overset{x}{P}\,\overset{-2}{O}_4}}$
    ♦ So we get: [+1 + (2 × +1) + x + (4 × -2)] = 0  ⇒ x = 5
    ♦ Thus, oxidation number of P in NaH₂PO₄ is +5.
Part (b): NaHSO₄
We can write this as: $\mathbf\small{\rm{\overset{+1}{Na}\,\overset{+1}{H}\,\overset{x}{S}\,\overset{-2}{O}_4}}$
    ♦ So we get: [+1 + +1 + x + (4 × -2)] = 0  ⇒ x = 6
    ♦ Thus, oxidation number of S in NaHSO₄ is +6.
Part (c): H₄P₂O₇
We can write this as: $\mathbf\small{\rm{\overset{+1}{H}_4\,\overset{x}{P}_2\,\overset{-2}{O}_7}}$
    ♦ So we get: [(4 × +1) + 2x + (7 × -2)] = 0  ⇒ x = 5
    ♦ Thus, oxidation number of P in H₄P₂O₇ is +5.
Part (d): K₂MnO₄
We can write this as: $\mathbf\small{\rm{\overset{+1}{K}_2\,\overset{x}{Mn}\,\overset{-2}{O}_4}}$
    ♦ So we get: [(2 × +1) + x + (4 × -2)] = 0  ⇒ x = 6
    ♦ Thus, oxidation number of Mn in K₂MnO₄ is +6.
Part (e): CaO₂
We can write this as: $\mathbf\small{\rm{\overset{+2}{Ca}\,\overset{x}{O}_2}}$
    ♦ So we get: [+2 + 2x] = 0  ⇒ x = -1
    ♦ Thus, oxidation number of O in CaO₂ is -1.
Part (f): NaBH₄
We can write this as: $\mathbf\small{\rm{\overset{+1}{Na}\,\overset{x}{B}\,\overset{-1}{H}_4}}$
(Here oxidation number of H is -1 because, it is a hydride)
    ♦ So we get: [+1 + x + (4 × -1)] = 0  ⇒ x = 3
    ♦ Thus, oxidation number of B in NaBH₄ is +3.
Part (g): H₂S₂O7
We can write this as: $\mathbf\small{\rm{\overset{+1}{H}_2\,\overset{x}{S}_2\,\overset{-2}{O}_7}}$
    ♦ So we get: [(2 × +1) + 2x + (7 × -2)] = 0  ⇒ x = 6
    ♦ Thus, oxidation number of S in H₂S₂O7 is +6.
Part (h): KAl(SO₄)₂.12H₂O
We can write this as: $\mathbf\small{\rm{\overset{+1}{K}\,\overset{+3}{Al}\,\overset{x}{S}_2\,\overset{-2}{O}_8\,\overset{+1}{H}_{24}\,\overset{-2}{O}_{12}}}$
    ♦ So we get: [+1 + +3 + 2x + (8 × -2) + (24 × +1) + (12 × -2)] = 0  ⇒ x = 6
    ♦ Thus, oxidation number of S in KAl(SO₄)₂.12H₂O is +6.

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• We have seen the basic details about oxidation number. In the next section, we will see Stock notation. We will also see how oxidation number can be used to define redox reactions.


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