How to Calculate Oxidation Number - Explanation, Examples, FAQs

The concept of Oxidation Number and oxidation state was not developed by any scientist at a particular time it was created in the broader study of redox reactions and the chemical bondings. Oxidation numbers are a very essential component of redox reaction and also in balancing the chemical reactions, in studying the behavior of the element in various components.

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Oxidation Number And Oxidation State

Oxidation
It is a process that involves the loss of electrons by the atoms or ions.

Reduction

It is a process that involves the gain of electrons by the atoms or ions.

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Any reaction, in which the electrons are exchanged between atoms or ions, represents a simultaneous process of oxidation and reduction and is called a Redox Reaction.

In a Redox Reaction, the species that loses electrons (i.e., gets oxidized) is known as the reducing agent or reductant, (since it causes the reduction of other species), and the species which accepts electrons from reductant (i.e., gets reduced) is known as oxidizing agent or oxidant (as it causes oxidation of other species).

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Oxidation State (O.S.):

It refers to the hypothetical charge on atoms in a compound if all the bonds were assumed to be 100% ionic.

Oxidation state, many times, is also referred to as Oxidation Number.
This means the oxidation number of an element in a compound is equal to the oxidation state of that element multiplied by the total atoms of that element in a particular compound.
(i) In ionic compounds, it is simply the charge on the corresponding cation and anion which is expressed as the oxidation state of that particular element. For example, the oxidation state of potassium and chlorine in potassium chloride (KCl) is simply +1 and –1 respectively as KCl is treated as K⁺Cl^–.

Refer to the following examples where oxidation states are written above the atoms:

NOTE: (a) In MgCl₂ and AlCl₃, -1 is the oxidation state of Cl.
(b) In each of the cases, the sum of the oxidation number of all atoms of all kinds is equal to zero since the compound is neutral.

(ii) In Covalent Compounds, it is not easy to assign an atom's oxidation state. To simplify the concept, we are going to define a set of rules that would enable us to assign an oxidation state to every element in any compound.

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Rules for Assigning Oxidation State (O.S.) and Oxidation Number (O.N.):

• Any element in a free state is assigned an oxidation state of zero. For example, the O.S. of H, P, S, and O in H₂, P₄, S₈, and O₂ respectively is zero.
• The oxidation state of any cation or anion (of form A⁺ or B^-) is equal to the magnitude of its charge. For example: O.S of Ca in Ca²⁺ = +2 and O.S of Al in Al³⁺ = +3.

• The algebraic sum of the oxidation number of all atoms in a neutral compound is equal to 0. The algebraic sum of the oxidation numbers of all atoms in an ion (like PO₄^3-) is equal to the charge on the ion.

• The oxidation state of Alkali Metals (Group IA) is +1 in all of their compounds and that of Alkaline Earth elements (Group IIA) is +2 in all of their compounds.

• Hydrogen in almost all of its compounds is assigned an oxidation state of +1. The exception occurs when hydrogen forms compounds with strong metals like KH, NaH, MgH₂, CaH₂, etc. In all of these, the oxidation state of hydrogen is -1.

• Oxygen in almost all of its compounds is assigned an oxidation state of -2. But in certain compounds like Peroxides(H₂O₂), the oxidation state of oxygen is -1. Another exception is OF₂, where O.S. is +2. O₂F₂, where O.S. is +1, and KO₂ in which O.S. is -1/2.

• Fluorine is the most electronegative element and is assigned an O.S. of -1, in all its compounds. For other halogens, O.S. is generally -1 except when they are bonded to a more electronegative halogen or oxygen. O.S. of iodine in IF₇ is +7, O.S. of chlorine in KClO₃ is +5.

• Generally, an element with greater electronegativity is assigned -1 by the hypothetical breaking of one covalent bond.

Recommended video on (Oxidation Number )

Some Solved Examples

Example.1

1. Which of the following substances is oxidized in the following reaction:

$\mathrm{Cl}_2+2 \mathrm{Br}^{-} \rightarrow 2 \mathrm{Cl}^{-}+\mathrm{Br}^{-}$

Cl2+2Br−→2Cl−+Br−

1)Cl₂

2) (correct)Br

3)Cl^-Cl−

4)Br₂

Solution

As we have learned,

Loss of electron is Oxidation

Since Br−Br^- loses electrons, it gets oxidized.

Hence, the answer is the option (2).

Example.2

2. Select the redox change in the following reaction

$\mathrm{Hg} \rightarrow \mathrm{Hg}^{2+}+2 \mathrm{e}^{-}$

Hg→Hg2++2e−

1) (correct)Oxidation

2)reduction

3)both

4)neither

Solution

Since the mercury atom loses electrons, it gets oxidized.

Hence, the answer is the option (1).

Example.3

3. Oxidants are substances which

1)are reduced themselves

2)show a decrease in oxidation number

3)show electron motion

4) (correct)All of these

Solution

Oxidation -

It is defined as the addition of oxygen / electronegative element to a substance or the removal of hydrogen / electropositive element from a substance.

Oxidants do all the above options.

Hence, the answer is the option (4).

Example.4

4. In which compound does Vanadium have an oxidation number of +4?

1)NH₄VO₂

2)$\mathrm{K}_4\left[\mathrm{~V}(\mathrm{CN})_6\right]$K4[ V(CN)6]

3)$\mathrm{VSO}_4$VSO4

4) (correct)$\mathrm{VOSO}_4$VOSO4

Solution

The structure of VOSO4 $\mathrm{VOSO}_4$ is

O=*1OS(=O)(=O)O1

VOSO4 $\mathrm{VOSO}_4$dissociates as

$\mathrm{VOSO}_4 \rightleftharpoons \mathrm{VO}^{2+}+\mathrm{SO}_4$

VOSO4⇌VO2++SO42−

Let the oxidation state of Vanadium be x, charge balance on the cation VO2+ $\mathrm{VO}^{2+}$gives

$x-2=+2$

$x=4$

x−2=+2

x=4

Thus in VOSO₄ VOSO4, Vanadium has an oxidation number of +4

Hence, option number (4) is correct.

Example.5

5. Which of the following is reduced in the following reaction?

$\mathrm{Cl}_2+2 \mathrm{Br}^{-} \rightarrow 2 \mathrm{Cl}^{-}+\mathrm{Br}_2$

Cl2+2Br−→2Cl−+Br2

1) (correct)Cl₂Cl2

2)Br₂Br2

3)Cl^-Br−

4)Br^-Cl−

Solution

As we have learned,

The addition of an electron to an element is known as Reduction.

Since Cl2Cl₂ gains electrons, it is reduced.

Hence, the answer is an option (1).

Example.6

6. When KMnO₄ acts as an oxidizing agent and ultimately forms $\left[\mathrm{MnO}_4\right]^{2-}, \mathrm{MnO}_2, \mathrm{Mn}_2 \mathrm{O}_3, \mathrm{Mn}^{2+}$[MnO4]2−,MnO2,Mn2O3,Mn2+ then the number of electrons transferred in each case respectively is

1)4, 3, 1, 5

2)1, 5, 3, 7

3) (correct)1, 3, 4, 5

4)3, 5, 7, 1

Solution

Let us take the constituents one by one and state their oxidation states.

$\mathrm{KMnO}_4, \mathrm{O} . \mathrm{S} .=+7$

$\left[\mathrm{MnO}_4\right]^{2-}, \mathrm{O} . \mathrm{S} .=+6$

$\mathrm{MnO}_2, \mathrm{O} . \mathrm{S} .=+4$

$\mathrm{Mn}_2 \mathrm{O}_3, \mathrm{O} . \mathrm{S} .=+3$

$\mathrm{Mn}^{2+}, \mathrm{O} . \mathrm{S} .=+2$

KMnO4,O.S.=+7[MnO4]2−,O.S.=+6MnO2,O.S.=+4Mn2O3,O.S.=+3Mn2+,O.S.=+2

The number of electrons transferred in each case respectively is 1,3,4,5

Hence, the answer is the option (3).

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Summary

In The redox reaction, it is used to balance the reaction. oxidation states provide the more appropriate pathways for tracing back the electron transfer on the redox (reduction-oxidation) reactions. If the gain in electron Oxidation occurs and there is the loss in the electron then reduction occurs after the reaction. This property helps to predict the productivity of the reaction in the different substances.

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