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Oxidation Number | Definition, How To Find, Examples

Last Updated : 13 Feb, 2024
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Oxidation number is defined as the total number of electrons that an atom either gains or loses to form a chemical bond with another atom. 

Let’s learn about oxidation number in detail, including its rules and steps to calculate it with the help of examples.

Oxidation Number Definition

Oxidation Number is defined as the charge, whether positive or negative gained by an atom of an element when they form bonds with atoms of other elements either by sharing or transferring electrons.

  • The charge that appears as an oxidation number is the number of electrons lost, gained or shared by the atom during the formation of a bond with atoms of another element in a molecule.
  • It can be positive, negative, zero or fraction depending upon the situation.
  • The oxidation number of an atom is often referred to as its oxidation state.
  • It is generally assigned to atoms in the reaction in which they undergo oxidation and reduction. Such a reaction in which atoms either get oxidised or get reduced is called a redox reaction. The word redox is an acronym of two words reduction and oxidation. 

Note – We always assume the reaction to be ionic for the calculation of oxidation number. Since every molecule possesses partial ionic and partially covalent character we assume ionic character for ease of calculation. In a molecule, it is assumed that more electronegative element gain electrons in whole numbers irrespective of sharing and less electronegative elements lose electrons in whole numbers thus making the bond ionic hypothetically. We can understand this by the following example in the case of Carbon monoxide(CO).

How to Find Oxidation Number

The oxidation number of an atom is the number of electrons lost or gained to form bonds with heteroatoms in a particular molecule. The molecule can be neutral (its overall charge is zero) or may be charged (it has some overall positive or negative charge).

To find the oxidation number of an atom we need to follow the following steps:

Step 1: Assume the oxidation number of the atom to be X which you need to calculate.

Step 2:Mention the oxidation state of other bonded atoms and multiply it with the number of such atoms present in one molecule

Step 3: Write the oxidation number of all the atoms in the molecule in a linear sum format and equate it to the overall charge of the molecule.

Step 4: Solve for X.

We can learn it with the help of examples below.

Oxidation Number of Sulphur in H2SO4

Oxidation Number of Sulphur in H2SO4, Sulphuric Acid

Oxidation Number of Sulphur in H2SO4

Solution:

Step 1: Assume the oxidation number of sulphur to be x

Step 2: The oxidation number for Hydrogen is +1 and for O is -2.

Step 3: Since the overall charge on the molecule is 0, therefore 2(+1) + X + 4(-2) = 0

Step 4: 2 + X – 8 = 0 ⇒ X – 6 = 0 ⇒ X = +6

Hence, the oxidation number of Sulphur in H2SO4 is +6

Oxidation Number of Chromium in Cr2O72-

Oxidation Number of Chromium in Cr2O72-

Oxidation Number of Chromium in Cr2O72-

Solution:

Step 1: Assume the oxidation number for Chromium be X

Step 2: The oxidation number for oxygen is -2

Step 3: Since the molecule has an overall charge of -2 the equation can be written as 2X + 7(-2)=-2

Step 4: 2X – 14 = -2 ⇒ 2X = +12 ⇒ X = +6

Hence the oxidation number of chromium in Cr2O72- is +6.

It should be noted that in one molecule there can be different oxidation numbers for the same atom. This can be understood from the following example

Multiple Oxidation Numbers of an Atom in a Molecule

Now will learn how to calculate multiple oxidation numbers of an atom in a molecule discussed below:

Oxidation Number of Nitrogen in Ammonium Nitrate i.e. NH4NO3

Solution:

Step 1: In this case, we will split the molecule into two ions Ammonium ion (NH4+) and Nitrate ion (NO3)

Step 2: We will find the oxidation number of Nitrogen in each ion. Assume the oxidation number for Nitrogen to be X in each case

Step 3: For the Ammonium ion, since the molecule has an overall +1 charge, hence equation can be written as X + 4(+1) = +1

Step 4: Solving the equation X + 4 = 1 ⇒ X = -3

Step 5: In the case of nitrate ion, the oxidation number for oxygen is -2 and the molecule has an overall charge of -1.

Step 6: Hence equation, in this case, will be X + 3(-2) = -1

Step 7: Solving the equation we get X – 6 = -1 ⇒ X = +5

Hence the oxidation number of Nitrogen in ammonium nitrate is -3 and +5

In the above examples, we saw that some atoms have fixed oxidation numbers. To understand why some atoms have fixed atomic numbers we need to look at the following mentioned rules of oxidation number.

Oxidation Number Rules

These are the rules to assign constant oxidation numbers to atoms of an element:

Rule 1: The total of all the atoms in a molecule’s oxidation numbers equals zero.

e.g. In KMnO4, the oxidation number of K is +1, the oxidation number of Mn is +7 and the oxidation number of oxygen is −2.

Rule 2: An atom’s oxidation number is always 0 in its most basic form.

e.g. The oxidation number of H, O, N, P, S, Se, Cu, Ag in their element forms is H2, O2, N2, P4, S8, Se8, Cu, Ag respectively, is zero.

Rule 3: Alkali metals (Li, Na, K, Rb, Cs) have an oxidation number of +1 in their compounds.

e.g. In NaCl, the oxidation number of Na is +1.

Rule 4: Alkaline earth metals (Be, Mg, Ca, Sr, Ba) usually have a +2 oxidation number in their compounds.

e.g. The oxidation number of Mg in MgO is +2.

Rule 5: Except in metal hydrides, the oxidation number of H in its compound is always +1.

e.g. In HCl, the oxidation number of H is +1 and in NaH (Sodium hydride), the oxidation number of H is −1.

Rule 6: The fluorine oxidation number is 1 in all its compounds.

e.g. In NaF, the oxidation number of F is −1.

Rule 7: The charge of an ionic compound is equal to the sum of all oxidation states of all the atoms in the compound. 

e.g. In SO42- the oxidation number of Sulphur is +6. The oxidation number of oxygen is −2. Hence the sum is +6+4(-2) = 6-8 = -2.

Rule 8: Except for oxygen and fluorine, the maximum oxidation number of any element is equal to its group number.

e.g. The oxidation number of sulphur in H2S2O8, K2S2O8, S2O–28 and H2SO5 is +6 due to the presence of a peroxide bond.

Rule 9: It’s possible that not all atoms of the same element have the same oxidation number in certain compounds. We get the average result when we compute the oxidation number for that element in such components.

e.g. One sulphur atom in Na2S2O3 has an oxidation number of +6, whereas the other sulphur atoms have an oxidation number of -2. As a result, the average sulphur oxidation state number in Na2S2O3 is +2.

Oxidation Number in Na2S2O3, Sodium thiosulfate

Rule 10: Carbon in organic molecules can have any oxidation number between -4 and +4.

e.g. in HCHO, the oxidation number of carbon is zero.

Rule 11: The common oxidation number of an element is equivalent to its group number from IA to IV A. The formula (Group number –8) gives the common oxidation number of any element from V A to VIII-A.

e.g. 

  • Oxidation number of I A group elements = +1.
  • Oxidation number of II A group elements = +2.
  • Oxidation number of III A group elements = +3.
  • Oxidation number of IV A group elements = +4.
  • Oxidation number of V A group elements = –3.
  • Oxidation number of VI A group elements = –2.
  • Oxidation number of VII A group elements = –1.
  • Oxidation number of VIII A group elements = 0.

Rule 12: C, N, P, and S have oxidation numbers of 4,–3,–3 and –2 in all carbides, nitrides, phosphides, and sulphides, respectively.

e.g. In Mg3N2, the oxidation number of Nitrogen is −3.

Rule 13: In all Metal carbonyls, the oxidation number of metals is zero. 

e.g. In Ni(CO)4, the oxidation number of Ni is zero.

Rule 14: Except for peroxide, superoxides, oxyfluorides, and ozonides, oxygen has an oxidation number of -2 in most of its oxides.

e.g. In Na2O, the oxidation number of O is −2. 

Exceptional Cases for Oxidation Number of Oxygen

There are some exceptional cases for the oxidation number of Oxygen, which are

  • Peroxides: The oxidation number of oxygen in peroxides is 1.  Examples, are H2O2, and Na2O2.
  • Oxidation number of oxygen in fluorine compounds is +2. Examples are F2O or OF2, etc.
  • Superoxides: The oxidation number of oxygen in superoxides is –1/2.
  • Ozonides: Each oxygen atom in an ozonide has an oxidation number of –1/3.

Oxidation Number Examples

As of now, we know the basic rules to calculate the Oxidation Number, hence we will find the Oxidation Number of some common compounds which we come across frequently in chemistry.

We will find the Oxidation Number of the following compounds:

  • Potassium Permanganate (KMnO4)
  • Sodium Carbonate (Na2CO3)
  • Ammonium Nitrite (NH4NO2)
  • Cyanide Ion (CN)
  • Ammonia (NH3)

Oxidation Number of Manganese in KMnO4

Solution:

 Let the oxidation number of Manganese in KMnO4 be equal to X

The oxidation number of Potassium = +1

The oxidation number of oxygen is =  –2

⇒1 + (X) + 4×(–2)=0

⇒1 + X – 8 = 0

⇒X -7 = 0

⇒X = +7

The oxidation number of Manganese in KMnO4 is +7.

Oxidation Number of Carbon in Na2CO3

Solution:

Oxidation number of oxygen is = –2

Oxidation number of Sodium = +1

Let the oxidation number of carbon be X

⇒2(+1) + X + 3(–2) = 0

⇒2 + X – 6 = 0

⇒X – 4 = 0

⇒X = +4

Therefore X = +4

Oxidation Number of Carbon in Na2CO3 is +4 

Oxidation Number of Nitrogen in Ammonium Nitrite

Solution:

Ammonium Nitrite is an ionic compound containing NH4+ and NO2 ions.

Oxidation number of nitrogen in NH4+

Let oxidation number of N atom is x

⇒ x + 4 = +1

⇒ x = 1 – 4 = -3

Oxidation number of nitrogen in NO2

Let oxidation number of Nitrogen atom be Y

Y + 2(-2) = -1

⇒ Y – 4 = -1

⇒ Y = -1 + 4 = +3

Thus, one atom of Nitrogen in Ammonium Nitrite is in the -3 oxidation state, while the other nitrogen atom is in the +3 oxidation state.

Oxidation Number of Carbon in CN

Solution:

Let Oxidation Number of C atom in CN be x, Oxidation Number for N atom is -3 and the overall charge on the molecule is -1. 

⇒ x – 3 = -1

⇒ x = -1 + 3 = +2

Hence, Carbon atom has oxidation number of +2 in CN

Oxidation Number of N in NH3

Solution:

Let Oxidation Number of N in NH3 be x and we know that Oxidation Number of H is +1.

⇒ x + 3(+1) = 0

⇒ x + 3 = 0

⇒ x = -3

Hence, the Oxidation Number of N atom in NH3 is -3.

Difference between Oxidation and Reduction

Oxidation: The loss of electrons or a rise in the positive oxidation state or decrease in the negative oxidation state of an atom, an ion, or specific atoms in a molecule is referred to as oxidation.

Reduction: The gain of electrons or a drop in the positive oxidation state or increase in the negative oxidation state of an atom, an ion, or specific atoms in a molecule is referred to as reduction . It is referred to as reduction in the oxidation state.

Related:

FAQs on Oxidation Number

What is Oxidation Number?

The charge that appears on an atom of an element when it bonds with atoms of other elements in a molecule is called Oxidation Number.

What is the Common Oxidation Number of Inert Gases?

Inert gases have an oxidation number of zero.

What are the Rules for Calculating Oxidation Numbers?

The rules for oxidation numbers are:

  1. The total of all the atoms in a molecule’s oxidation numbers equals zero.
  2. An atom’s oxidation number in its most basic form is always zero.
  3. In their compounds, the oxidation number of alkali metals (Li, Na ,K ,Rb ,Cs) is always +1.
  4. In their compounds, the oxidation number of alkaline earth metals (Be ,Mg ,Ca ,Sr ,Ba) is always +2.
  5. Except in metal hydrides, the oxidation number of H in its compound is always +1.
  6. Fluorine has an oxidation number of 1 in all of its compounds.
  7. Except for peroxide, super oxides, oxyfluoride, and ozonides, the oxidation number of oxygen in most of its oxides is -2.
  8. Carbon in organic molecules can have any oxidation number between -4 and +4.

What is the Oxidation Number of Mercury(Hg) in Amalgam?

Mercury amalgam has a zero oxidation number. Each element in an alloy or amalgam has an oxidation number of zero.

What is the Oxidation Number of Oxygen in O3 and in MgO?

  • Oxidation number of O in O3 = 0
  • Oxidation number of O in MgO = –2

What are the Oxidation Numbers of Oxide and Hydride ions?

The oxidation number for oxide ion is -2 and that of hydride ion is -1.

What is the Oxidation Number of Sulphur in S8?

S8 is a polyatomic molecule, it is in its most basic form. As a result, sulphur’s oxidation number in this molecule is zero.



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