How To Calculate Oxidation Number

How To Calculate Oxidation Number

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.
Oxidation number of Cl in CrO5
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-
Oxidation number of Cl in CaOCl2
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