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How to Find Oxidation Number for JEE

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Last updated date: 29th May 2024
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What is an Oxidation Number?

Different elements form chemical bonds with each other by sharing, losing or gaining electrons. The oxidation number or oxidation state is the number allocated to an element in a compound that represents the number of electrons lost, or gained by that atom in the given compound. It can be positive, negative or zero. A positive oxidation number represents that an atom loses electrons and a negative oxidation number represents that an atom gains electrons. If the oxidation number of elements is zero, it neither gains nor loses electrons. The sum of the oxidation states of all the atoms in a neutral molecule is zero. The oxidation number helps us to keep track of the electrons lost or gained. In this topic, we will learn how to calculate oxidation numbers using a set of different oxidation number rules.


Rules to Assign Oxidation Numbers

  1. In elemental form, every atom has an oxidation number of zero. For example, the oxidation number of each atom in Cu, Cl2, O2 is zero.

  2. The oxidation number and charge are the same for a monatomic ion. For example, the oxidation number of potassium ion (K+) is +1, which is equal to its charge.

  3. The oxidation number of metals in group 1 (Alkali metals) is +1 and of the metals of group 2 (Alkaline earth metals) is +2. Hydrogen is not included in this rule, as it is not a metal.

  4. Hydrogen exhibits two possible oxidation numbers: +1 and -1. When bonded to atoms more electronegative than itself like non-metals, it shows the oxidation number as +1. For example, in HCl, the oxidation number of H is +1.

However, when bonded with metals (less electronegative), H has an oxidation number of -1. For example, in sodium hydride (NaH), the oxidation number of H is -1.

  1. The oxidation number of oxygen in most of its compounds is -2. However, peroxides are exceptions, where the oxidation number of oxygen is -1. Another exception is oxygen difluoride, where oxygen exhibits a +2 oxidation number.

  2. The oxidation number of fluorine is -1 in all its compounds, like hydrogen fluoride (HF), sodium fluoride (NaF), etc.

  3. The rest of the halogens (Cl, Br, I) also have a -1 oxidation number in their compounds, except when they are bonded to oxygen or fluorine. For example, chlorine shows a -1 oxidation number in sodium chloride (NaCl), but +1 in Hypochlorous acid (HOCl).

  4. In a neutral molecule, the sum of the oxidation number of all the elements is equal to 0. For a polyatomic ion, the sum of individual oxidation numbers is equal to the charge of the ion.


Calculation of Oxidation Numbers

Let us calculate the oxidation number for a neutral molecule and an ion, using oxidation number rules.


Sodium Chloride (NaCl): 

  • From the rules, we know that chlorine has an oxidation number of -1 unless it is combined with a more electronegative element. 

  • Since Cl is more electronegative than Na, -1 is assigned to chlorine. We know that the sum of individual oxidation numbers in a neutral molecule is 0.

Na + (-1) = 0

  • Therefore, the oxidation number of Na is +1 in NaCl. We have also learned in the rules that a group 1 metal always shows a +1 oxidation number.


The Oxidation Number of Manganese in MnO4- (Permanganate ion):

  • The overall charge on permanganate ions is -1.

  • The oxidation number of MnO4- (permanganate ion) = Oxidation number of Mn (Manganese) + 4 (Oxidation number of Oxygen).

  • From the rules, we know that the oxidation number of oxygen in its compounds except peroxides and fluorides is -2. As we have to find the oxidation number of manganese, let us consider the oxidation number of Mn = x.

$x + 4 (-2) = -1$

$x – 8 = -1$ or $x = +7$

  • Therefore, the oxidation number of manganese (Mn) in MnO4- is +7.


Conclusion

The oxidation number or oxidation state represents the number of electrons lost, gained or shared by that atom in the given compound. To calculate the oxidation number of an atom in a given molecule or ion, a given set of rules has to be followed. These rules are called ‘oxidation number rules’. The sum of the oxidation states of all the atoms in a neutral molecule is zero. For an ion, it is equal to the charge of an ion. Oxidation numbers are important to study chemical reactions from an electron transfer point of view.

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FAQs on How to Find Oxidation Number for JEE

1. What is the Oxidation Number of Carbon in CO2?

  • CO2 is a neutral molecule. Therefore, the sum of all individual oxidation numbers would be equal to 0.

Oxidation number of carbon + 2 (Oxidation number of oxygen) = 0.
  • From the rules, we know that the oxidation number of oxygen in its compounds except peroxides and fluorides is -2. As we have to find the oxidation number of carbon, let us consider the oxidation number of C = X.

$X + 2(-2) = 0$
$X + (-4) = 0$ or $X = +4$

  • Therefore, the oxidation number of carbon in CO2 is +4.

2. What is the easiest way to find the oxidation number?

The sum of the oxidation number of elements in a neutral molecule is zero and that of an ion is equal to its charge.

  • To calculate the oxidation number of an element in a compound, consider it as ‘x’ and put the values of the known oxidation number of other atoms in a molecule.

  • For example, let us calculate the oxidation number of sulfur in S2O3. We know from the rules that the oxidation number of oxygen is -2, except in its peroxides and fluorides.

$2(x) + 3 (-2) = 0$
$2x = +6$ or $x = +3$

  • Therefore, the oxidation number of S in S2O3 is +6.

3. How do you find the oxidation number of an ion? 

  • The oxidation number of a monatomic ion is equal to its charge. For example, potassium ion (K+) has an oxidation number of +1 and the oxidation number of chloride ion (Cl-) is -1.

  • For polyatomic ions, the sum of the oxidation number of all the individual atoms is equal to the charge on the ion. For example, in sulfate ion (SO42-), we know the oxidation number of oxygen is -2, the oxidation number of sulfur can be calculated by considering it as ‘x’.
    $x + 4 (-2) = -2$
    $x – 8 = -2$ or $x = +6$

  • Therefore, the oxidation number of S in SO42- is +6.