The number or oxidation state indicates the electrons that an atom gains or loses to unite with other atoms and form chemical compounds. Let's apply this concept to the formula unit NaCl. Sodium loses an electron, which chlorine gains, becoming Na + and Cl - . The oxidation state of sodium is +1 and that of chlorine is -1.
In MgCl 2 magnesium loses two electrons to Mg 2+ , these electrons are captured by two chlorine atoms that are transformed into Cl - . Magnesium therefore has oxidation state +2.
To assign the oxidation number we must consider the following rules. In case of contradiction, the rule that goes before in the list prevails.
- The oxidation state of a neutral element or molecule is 0. Cl 2 (EO = 0); H(EO = 0); No 2 (EO = 0)
- The sum of the oxidation states of the atoms that make up a neutral molecule is zero. NaCl (+1-1=0); MgCl 2 (+2-1-1=0)
- Ions have an oxidation number equal to their charge. SO 4 2- (EO = -2); NO 3 - (EO = -1)
- The metals of group 1 (alkali) have an oxidation number of +1 and those of group 2 (alkaline earth) of +2.
- The fluorine atom presents oxidation state -1 in its compounds.
- The oxidation state of hydrogen is +1, except when combined with metals it becomes -1.
- The oxidation state of oxygen in its compounds is -2
- When combined with metals, the oxygen group has EO = -2, the nitrogen group EO = -3, and the halogen group EO = -1
Example 1. Indicate the oxidation state of sulfur in each of the following compounds SO 3 2- , HSO 4 - , S 2 O 8 2- , S 4 O 6 2-
- In SO 3 2- the sum of the oxidation states must be equal to -2 (charge of the anion). The oxygen in its compounds has EO -2 (rule 7), as we have three oxygens in the molecule -6. Therefore, sulfur has an EO of 4.
- In HSO 4 - the sum of the oxidation states is -1. Oxygen has EO -2, all four oxygens -8. Hydrogen has EO +1 (rule 6). Therefore, the EO of sulfur is +6.
- In S 2 O 8 2- the sum of the EOs is -2. The eight oxygens have EO -16, so the EO of the sulfurs will add up to +14. Sulfur has an EO of +7
- In S 4 O 6 2- the sum of the EOs is -2. The EO of the six oxygens is -12, which implies that the sum of the EO of the sulfurs is +10. Each sulfur has an EO of 10/4.
Example 2. Indicate the oxidation state of the following elements: a) Cl in perchloric acid; b) N in nitric acid; c) P in pyrophosphorous acid; d) I in diiodine oxide; e) Mn in H 2 MnO 4 (manganic acid); f) Cu in copper sulfate; g) If in metasilicic acid; h) N in nitrogen monoxide; i) Mn in MnO 3 .
a) In perchloric acid, HClO4, the sum of the oxidation states of the elements that compose it is 0 (rule 1). Oxygen in its compounds presents EO -2 (rule 7), while hydrogen has EO +1 (rule 6). The sum of EO of the four oxygens is -8, implying an EO for Cl of -7.
b) In HNO 3 nitric acid, the sum of EO is 0 (rule 1). Oxygen has EO -2 and the sum of EO for all three oxygens is -6. Hydrogen has EO +1, therefore the oxidation state of nitrogen is +5.