SS2 Chemistry Lesson Note on Rules for Determining Oxidation Number

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OXIDATION NUMBERS OF CENTRAL ELEMENTS IN SOME COMPOUNDS

Oxidation number is the electrical change assigned to an atom in accordance with some prescribed set of rules

RULES FOR DETERMINING OXIDATION NUMBER

The following sets of rules are used to determine the oxidation state or number of substances.

1. The oxidation number of an element in an un-combined state is zero, for example the oxidation number of Hydrogen, Oxygen and Sodium atom in a free state or un-combined with another element is zero.
2. In most compounds containing hydrogen, the oxidation number of hydrogen is +1 except in hybrids where it is -1
3. Electrons shared between two unlike atoms are counted with the more electro negative atom. For example, in water molecule the electron are regarded as being with the more electronegative oxygen. Thus in H₂O, each hydrogen atom is in +1 oxidation state, and the oxygen atom is in -2 oxidation state.
4. In most compounds containing oxygen, the oxidation number of each oxygen atom is -2 except in peroxides where it is -1. E.g. hydrogen peroxides, (H₂O₂), sodium peroxide (Na₂O₂), barium peroxide (BaO₂)
5. The oxidation number of each halogen is -1, except when bonded with fluorine which is the most electro negative. For instance, in IF₇, each fluorine atom is in oxidation state of -1 and iodine is in oxidation state of +7
6. The sum of all the oxidation numbers of elements in a compound is zero and with this simple relationship, the oxidation number of each element in a compound can be calculated.
7. In simple ions, i.e. ions containing one atom, the oxidation number is equal to the change on the ion. For example, the ion Al³⁺ has on the oxidation number of +3, the ion Cu²⁺ has the oxidation number is -2
8. What about complex ion? In a complex ion (i.e. ion consisting of more than one element) the oxidation is the algebraic sum of all the oxidation numbers of all the elements in the ion. This will be the sign on the ion and of the same size. For example, in tetraoxosulphate (vi) ion (SO₄2^–) the overall charge is -2 which its oxidation number is. It is obtained as follows:

+ 4  =

+4 x (-2)                                    =       -2

For other ions such as OH^–, NO₃^‑, NO^2-₃, SO^2-₃, PO^3-₄, NH₄⁺ their oxidation numbers are -1, -1, -2, -3 and +1 respectively.

Calculation of oxidation number

With the rules for determining oxidation numbers in our memory, it is possible to calculate the oxidation number of any given element in an ion or a compound.

Worked examples

1. Calculate the oxidation number of chromium in

Solution

This can be solved by simple linear equation by making the unknown subject of formula. We have 4 oxygen and its oxidation is -2

Let the unknown (oxidation number of Cr be x)

But   = -2 (because the sign on the ion is -2)

x + (-24) = -2

x – 8 = -2

x = -2 + 8

x= +6

Thus, the oxidation number of Cr in   is +6

Note: oxidation number is never written as a neutral number, i.e. it is either written as a neutral number, i.e .It is either written as a positive or negative number.

2. Calculate the oxidation number of sulphur in H₂SO₄

Solution

The oxidation number (O.N) of hydrogen is +1 and the two hydrogen atoms will give+2. The four oxygen atoms will give (-2 x4) = -8, since -2 is the oxidation number (O, N) of an atom of oxygen

2×(+1) + x + (4 × -2) =0

2 + x + -8 =0

=0 +8 -2

=+8-2

=+6

CONNECTION OF OXIDATION NUMBERS WITH IUPAC NAMES

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