Oxidation Numbers: Numbers are very useful. When it comes to redox reactions, there are a few reasonably effective numbers, such as oxidation numbers. In simple words, the oxidation number is the number assigned to the components in a chemical combination. The oxidation number is the total number of electrons that atoms in a molecule can share, lose, or acquire while establishing chemical interactions with atoms of another element.

The oxidation state helps us to ascertain the transfer of electrons. It is defined as the charge that appears to form an ionic bond with other heteroatoms. An atom with higher electronegativity is given a negative oxidation state. In this article, we will focus on the rules and calculation of oxidation numbers. Continue reading this article.

What are Oxidation Numbers?

In a chemical process, how do you identify the oxidised substance, the reduced substance, and the reducing and oxidising agents?

Such identification may be made by knowing which substance is transferring electrons to other substances, according to the concept of redox reaction. However, this description in terms of electron loss and gain applies only to ionic compounds with a complete electron transfer and does not apply to some reactions, such as:

\({{\rm{N}}_2} + {{\rm{O}}_2} \to 2{\rm{NO}}\)

\({{\rm{H}}_2} + {\rm{C}}{{\rm{l}}_2} \to 2{\rm{HCl}}\)

There is no electron transfer from one atom to another during the creation of molecules like \({\rm{NO}}\) and \({\rm{HCl}}\). The preceding definition of a Redox reaction will not serve in this circumstance. Thus chemists find it easier to keep track of the electrons in such processes by using the idea of oxidation number.

“The oxidation number defines the residual charge that an atom has or appears to have in a molecule when all other atoms remove from the molecule as ions”.

The oxidation number is frequently used interchangeably with an Oxidation state. It is because the stock notation of oxidation numbers is the basis of the periodic property, electronegativity.

An atom in a molecule can assign to negative, positive or zero oxidation number by considering its environment. In a few cases, oxidation numbers can even be fractional. Find the oxidation numbers list for rules in the next sections.

Rules for Assignment of Oxidation Numbers to the Elements

The calculation of the oxidation state of all elements in compounds using the oxidation state formula can be made from a knowledge of the following rules.

Oxidation Number Rule 1

The sum of oxidation numbers of all the atoms in a molecule is equal to zero.

Example: In \({\rm{KMn}}{{\rm{O}}_4}\), oxidation number of \({\rm{K}}\) is \(+1\), the oxidation number of \({\rm{Mn}}\) is \(+7\) and oxidation number of oxygen is \(-2\).

Oxidation Number Rule 2

In an elementary form, the oxidation number of an atom is always zero.

Example: The oxidation number of \({\rm{H}},{\rm{O}},{\rm{N}},{\rm{P}},{\rm{S}},{\rm{Se}},{\rm{Cu}},{\rm{Ag}}\) in their element forms is \({{\rm{H}}_2},{{\rm{O}}_2},\;{{\rm{N}}_2},{{\rm{P}}_4},\;{{\rm{S}}_8},{\rm{S}}{{\rm{e}}_8},{\rm{Cu}},{\rm{Ag}}\) respectively, is zero.

Oxidation Number Rule 3

The oxidation number of alkali metals \(({\rm{Li}},{\rm{Na}},{\rm{K}},{\rm{Rb}},{\rm{Cs}})\) in their compounds is always \(+1\).

Example: In \({\rm{ NaCl }}\), the oxidation number of \({\rm{ Na}}\) is \(+1\).

Oxidation Number Rule 4

The oxidation number of alkaline earth metals \(({\rm{Be}},{\rm{Mg}},{\rm{Ca}},{\rm{Sr}},{\rm{Ba}})\) in their compounds is always \(+2\).

Example: The oxidation number of \({\rm{Mg}}\) in \({\rm{MgO}}\) is \(+2\).

Oxidation Number Rule 5

The oxidation number of \({\rm{H}}\) in its compound is always \(+1\) except in metal hydrides.

Example: In \({\rm{HCl}}\), the oxidation number of \({\rm{H}}\) is \(+1\).
In \({\rm{NaH}}\) (Sodium hydride), the oxidation number of \({\rm{H}}\) is \(-1\).

Oxidation Number Rule 6

The fluorine oxidation number in all of its compounds is \(-1\).

Example: In \({\rm{NaF}}\), the oxidation number of \({\rm{F}}\) is \(-1\).

Oxidation Number Rule 7

The oxidation number of oxygen in most of its oxides is \(-2\) except in peroxide, superoxides, oxyfluorides, and ozonides.

Example: In \({\rm{N}}{{\rm{a}}_2}{\rm{O}}\), the oxidation number of \({\rm{O}}\) is \(-2\).
In \({\rm{Mg}}0\), the oxidation number of \({\rm{O}}\) is \(-2\).

Exceptional Cases:

1. Peroxides: In peroxides, the oxidation number of oxygen is \(-1\). Examples, \({{\rm{H}}_2}{{\rm{O}}_2},{\rm{N}}{{\rm{a}}_2}{{\rm{O}}_2}\).
   In compounds with fluorine, the oxidation number of oxygen is \(+2\). Examples \({{\rm{F}}_2}{\rm{O}}\) or \({\rm{O}}{{\rm{F}}_2},\) etc.
2. Super oxides: In super oxides, the oxidation number of oxygen is \( – \frac{1}{2}\).
3. Ozonide’s: The oxidation number of each oxygen atom in ozonide’s is \( – \frac{1}{3}\).
   a. Example: In \({\rm{K}}{{\rm{O}}_3}\), let \({\rm{X}}\) be the oxidation number of each oxygen atom.
   b. \({\rm{3}}\left( {\rm{X}} \right){\rm{ = }}\,{\rm{ – 1}}\)
   c. \({\rm{X}} = \, – \frac{1}{3}\)

Oxidation Number Rule 8

In an ionic compound, the sum of the oxidation states of all atoms is equal to its charge.

Example: In \({\rm{SO}}_4^{2 – }\), the oxidation number of Sulphur \(+6\). Oxidation number of oxygen = \(-2\).

Oxidation Number Rule 9

The maximum oxidation number of any element is equal to its group number except in the case of oxygen and fluorine.

Example: The oxidation number of sulphur in \({{\rm{H}}_2}\;{{\rm{S}}_2}{{\rm{O}}_8},\;{{\rm{K}}_2}\;{{\rm{S}}_2}{{\rm{O}}_8},\;{{\rm{S}}_2}{\rm{O}}_8^{ – 2}\) and \({{\rm{H}}_2}{\rm{S}}{{\rm{O}}_5}\) is \(+6\) due to the presence of a peroxide bond.

Oxidation Number Rule 10

In some compounds, all the atoms of the same element may not have the same oxidation number. When we calculate the oxidation number for that element in such components, we get the average value.

Example: In \({\rm{N}}{{\rm{a}}_2}\;{{\rm{S}}_2}{{\rm{O}}_3}\), the oxidation number of one sulfur atom is \(+6\) and that of the other sulfur atoms is \(-2\). So the average Oxidation state number of sulphur in \({\rm{N}}{{\rm{a}}_2}\;{{\rm{S}}_2}{{\rm{O}}_3}\), is \(+2\).

In \({\rm{CaOC}}{{\rm{l}}_2}\) the oxidation number of one chlorine is \(-1\) and another chlorine \(+1\). So the average oxidation number of chlorine is zero.

Structure of \({\rm{CaOC}}{{\rm{l}}_2}\) and \({\rm{N}}{{\rm{a}}_2}\;{{\rm{S}}_2}{{\rm{O}}_3}\)

Oxidation Number Rule 11

Inorganic compounds, carbon can have any oxidation number from \(- 4\) to \(+4\).

For example, in \({\rm{ HCHO}}\), the oxidation number of carbon is zero.

Oxidation Number Rule 12

From IA to IV A group, the common oxidation number of an element is equal to its group number. From V A group to VIII A the common oxidation number of any element is given by the formula, (Group number \(– 8\)).

Example:
1. The oxidation number of I A group elements \({\rm{ = + 1}}\).
2. The oxidation number of II A group elements \({\rm{ = + 2}}\).
3. The oxidation number of III A group elements \({\rm{ = + 3}}\).
4. The oxidation number of IV A group elements \({\rm{ = + 4}}\).
4. The oxidation number of V A group elements \({\rm{ = – 3}}\).
5. The oxidation number of VI A group elements \({\rm{ = – 2}}\).
6. The oxidation number of VII A group elements \({\rm{ = – 1}}\).
7. The oxidation number of VIII A group elements \({\rm{ = 0}}\).

Oxidation Number Rule 13

In all carbides, nitrides, phosphides, sulphides, the oxidation number of \({\rm{C}},{\rm{N}},{\rm{P}}\) and \({\rm{S}}\) are \({\rm{4, – 3, – 3}}\) and \( – 2\), respectively.

Example: In \({\rm{M}}{{\rm{g}}_3}\;{{\rm{N}}_2}\) the oxidation number of Nitrogen is \(-3\).

Oxidation Number Rule 14

The oxidation number of metals in all Metal carbonyls is zero.

Example: In \({\rm{Ni}}{({\rm{CO}})_4}\), the oxidation number of \({\rm{Ni}}\) is zero.

Note: In \({\rm{NC}}{{\rm{l}}_3}\), nitrogen oxidation number is \(-3\), and \({\rm{ Cl }}\) is \(+1\) because nitrogen is smaller in size when compared to chlorine.

Calculation of Oxidation Numbers

The calculation of oxidation numbers is explained below:

Example of Oxidation Number Calculation of Sulphur in \({{\rm{S}}_8}\)

Solution: \({{\rm{S}}_8}\) is a polyatomic molecule, but it is in the elementary state. Therefore, the oxidation number of sulphur in this molecule is zero.

Example of Oxidation Number Calculation of Sulphur in \({{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\).

Solution: let the oxidation number of sulphur in \({{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\) be equal to \({\rm{ X}}\).
Oxidation number of hydrogen \({\rm{ = + 1}}\)
Oxidation number of oxygen is \({\rm{ = – 2}}\)
\(2( + 1) + (X) + 4 \times ( – 2) = 0\)
\({\rm{2 + X – 8 = 0}}\)
\({\rm{X = 8 – 2 = + 6}}\).
Oxidation number of sulphur in \({{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\) is \(+6\).

Calculate the Oxidation Number of Chromium in \({\rm{C}}{{\rm{r}}_2}{\rm{O}}_7^{2 – }\) ion.

Solution: Oxidation number of oxygen is \({\rm{ = – 2}}\)
Oxidation number of chromium \({\rm{ = X}}\)
\({\rm{2 X + 7( – 2) = – 2}}\)
\({\rm{2 X – 14 = – 2}}\)
\({\rm{2 X = – 2 + 14 = 12}}\)
Therefore \({\rm{X = + 6}}\).
The oxidation number of chromium in \({\rm{C}}{{\rm{r}}_2}{\rm{O}}_7^{2 – }\) is \(+6\).

Example of Oxidation Number Calculation of Manganese in \({\rm{MnO}}_4^{2 – }\) ion.

Solution: Oxidation number of oxygen is \({\rm{ = – 2}}\)
Oxidation number of manganese \( = {\rm{X}}\)
\({\rm{X + 4( – 2) = – 2}}\)
\({\rm{X – 8 = – 2}}\).
\({\rm{X = – 2 + 8 = + 6}}\)
The oxidation number of manganese is \(+6\).
In the above worked out examples, one can observe that the oxidation number of chromium in \({\rm{C}}{{\rm{r}}_2}{\rm{O}}_7^{2 – }\) is \(+6\) while it is \(+3\) in \({\rm{C}}{{\rm{r}}_2}{{\rm{O}}_3}\). Thus, certain elements show different oxidation numbers in different compounds.
Nitrogen may be given as an example

Example of Oxidation Number Calculation of Nitrogen in Ammonium Nitrite

Ammonium Nitrite is an ionic compound containing \({\rm{NH}}_4^ + \) and \({\rm{NO}}_2^ – \) ions.
The oxidation number of nitrogen in \({\rm{NH}}_4^ + \) ion \({\rm{ = – 3}}\)
The oxidation number of nitrogen in \({\rm{NO}}_2^ – = + 3\).
Thus, one atom of Nitrogen in Ammonium Nitrite is in \({\rm{a}}\,{\rm{ – }}\,{\rm{3}}\) – oxidation state, while the other nitrogen atom is in \({\rm{a + 3}}\) oxidation state.

Summary

From this article, we learned the importance of oxidation numbers in chemistry, rules for oxidation numbers, oxidation numbers in different elements and their calculation.

FAQs on Oxidation Number

Below are the frequently asked questions about oxidation number

Q.1: How to find oxidation numbers?
Ans: We are calculating the oxidation number of sulphur in \({{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\).
Let the oxidation number of sulphur in \({{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\) be equal to \({\rm{X}}\).
Oxidation number of hydrogen \({\rm{ = + 1}}\)
Oxidation number of oxygen is \({\rm{ = – 2}}\)
\(2( + 1) + ({\rm{X}}) + 4 \times ( – 2) = 0\)
\({\rm{2 + X – 8 = 0}}\)
\({\rm{X = 8 – 2 = + 6}}\).

Q.2: What are the rules for oxidation numbers?
Ans: 1. The sum of oxidation numbers of all the atoms in a molecule is equal to zero.
2. The oxidation number of an atom in its elementary form is always zero.
3. The oxidation number of alkali metals \(({\rm{Li}},{\rm{Na}},{\rm{K}},{\rm{Rb}},{\rm{Cs}})\) in their compounds is always \(+1\).
4. The oxidation number of alkaline earth metals \(({\rm{Be}},{\rm{Mg}},{\rm{Ca}},{\rm{Sr}},{\rm{Ba}})\) in their compounds is always \(+2\).
5. The oxidation number of \({\rm{H}}\) in its compound is always \(+1\) except in metal hydrides.
6. The oxidation number of fluorine in all of its compounds is \(-1\).
7. The oxidation number of oxygen in most of its oxides is \(-2\) except in peroxide, superoxides, oxyfluorides, and ozonides.
8. Inorganic compounds, carbon can have any oxidation number from \(- 4\) to \(+4\).

Q.3: Write the oxidation number of \({\rm{Hg}}\) in amalgam.
Ans: The oxidation number of mercury in amalgam is zero. The oxidation number of each constituent in an alloy or an amalgam is zero.

Q.4: Write the oxidation number of oxygen in \({\rm{ (a) }}{{\bf{O}}_3},({\bf{b}}){\bf{MgO}},({\bf{c}}){\bf{K}}{{\bf{O}}_2}\).
Ans: The oxidation number of \({\rm{O}}\) in \({{\rm{O}}_3}{\rm{ }} = {\rm{ }}0\)
The oxidation number of \({\rm{O}}\) in \({\rm{MgO}} = – 2\)
The oxidation number of \({\rm{O}}\) in \({\rm{K}}{{\rm{O}}_2} = – \frac{1}{2}\)

Q.5: What is the common oxidation number for inert gases?
Ans: The common oxidation number for inert gases is zero.

Q.6: What is the oxidation number method?
Ans: In the oxidation number method, you determine the oxidation numbers of all atoms. Then you multiply the atoms that have changed by small whole numbers.

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