How To Calculate Oxidation Number ?

Before the calculation of oxidation number of an atom in a given compound, it is important to know about oxidation numbers first. The oxidation number of an atom is a number that indicates the extent to which an atom in a reaction has suffered oxidation and reduction.
Oxidation number of an atom may be zero, positive, negative or fraction.

Calculation of Oxidation Numbers

Oxidation number can be calculated in a given molecule or compound by the following rules-
1. Oxidation number of an atom in free state is equal to zero.
Examples- Oxidation number of Na, K, Ca, H₂, S etc is zero.
2. Charge on an ion is equal to its oxidation number.
Examples-
oxidation number of Na⁺ is +1
oxidation number of Ca⁺² is +2
oxidation number of NH₄⁺ is +1
oxidation number of SO⁻³ is −3.
3. Oxidation number of alkali metals (group 1 elements) is +1 in their compounds.
Examples-
oxidation number of Na in NaH is +1
4. Oxidation number of alkaline earth metals (group 2 elements) is +2 in their compounds.
Examples-
oxidation number of Ca in CaCl₂ is +2
4. Oxidation number of hydrogen atom is +1 in its compounds but −1 when bonded with an element with less electronegativity than it.
Example-
oxidation number of H in NaH is −1
5. Oxidation number of Oxygen is −2 in most of its compounds. However, in the case of peroxides, the oxidation number is −1, In F₂O is +2, in F₂O₂ is +1, in KO₂ is −1/2 and in ozonide (O₃⁻¹) is −1/3
6. Oxidation number of halogens (group 17 elements) is −1. But other halogens except 'F' shows positive oxidation number too.
Examples-
oxidation number of Br in BrCl₃ is +3
oxidation number of I in IF₇ is +7
7. Sum of the oxidation numbers of all the atoms in a given compound is zero and in a radical it is equal to be its charge.

NOTE: In any compound oxidation number of an atom can never be more than number of electrons in its outermost orbit or valence cell electrons or valence electrons.

Examples-

Oxidation number of Chromium in CrO₅

According to the rule, the oxidation number of 'C'r in CrO₅ is +10 which is not possible as Cr has only six electrons in the outermost orbit. So, its oxidation number will assign by their structure. CrO₅ has a butterfly structure in which four oxygen atoms are in peroxide linkage and one oxygen atom is in normal covalent bond with Cr atom.

Therefore, its oxidation umber is +6
Let the oxidation number of 'Cr' is 'x'
so, x + (−4) + (−2) = 0
or, x = +6

Oxidation number of Chlorine in CaOCl₂

According to the rule, the oxidation number of Cl -
Let the oxidation number of 'Cl' is 'x'
so, +2 + (−2) + (x) = 0
or, x = 0
Oxidation number of Cl is zero is wrong.
Oxidation number of Cl can be determine by their bonding or structure-

From the structure the oxidation number of Cl is +1 and −1.

Oxidation number of Carbon in C₃O₂

The structure of carbon suboxide C₃O₂ is O = C = C = C = O.
Each of the two terminal oxygen atoms has an oxidation state of −2 and the carbons to which they are attached have an oxidation state of +2.
The central carbon atom in C₃O₂ has zero oxidation state and the two terminal carbon atoms are present in +2 oxidation state each, whereas the third one is present in zero oxidation state.
Therefore, the average oxidation state of carbon is 4/3.

Oxidation number of Boron in NaBH₄

In this case oxidation number of 'H' is negative becasue, both 'Na' and 'B' are less electronegative atom than 'H'.
Let the oxidation number of 'B' is 'x'
+1 + x + (−1 x 4) = 0
or, x = +3

Oxidation number of Carbon in CH₂Cl₂

a. +1
b. +4
c. −2
d. 0

Oxidation number of Oxygen in CaO₂

a. −1/2
b. −1
c. −1
d. 0