Introduction to the Oxidation State
In chemistry, oxidation-reduction reactions are those in which one atom loses an electron and the other one accepts it. These reactions are also called redox reactions and keep on occurring all-time in our surroundings.
To understand these reactions, we have to study their balanced chemical equations since a balanced chemical equation can fully and precisely describe the entire event in a single glimpse.
So in this article, we will discuss two stepwise methods to determine the oxidation states of all the elements present in a chemical compound but first, let's recall the definition of oxidation state/number.
"Oxidation state or oxidation number of an element is the net charge it carries either being while a part of a compound or in a free state"
Related: Learn about the general chemical equations of endothermic and exhotermic reactions.
Stepwise Method to Determine the Oxidation State of Elements in a Chemical Compound
The process of assigning the oxidation number to an element just by seeing the formula could be quite complicated or extremely simple. It all depends upon how complex the composition of the compound is.
Even in some cases, a single element shows more than one oxidation state. Luckily we have some easy-to-understand and well-defined oxidation state rules which can help us in assigning the oxidation state more easily. What are these rules and how to find oxidation state by using these rules effectively, let's find out!
Related: Learn about metal displacement reaction in aqueous medium.
Determining the Oxidation State of Elements Using General Rules
Below are the steps to learn how to determine oxidation number of elemented using general rules. These steps of oxidation state rules are:
Step 1: Determine the nature of the substance (free state or combined)
Always remember that whenever an element is present in free or combined elemental form, its oxidation state is always zero.
This rule applies to those atoms who possess a lone pair as well as those who are in diatomic and polyatomic elemental form.For example, the oxidation state of both Cl2 and Al2 is zero since both are present in combined elemental form.
Similarly, all the other bigger substances, even the octahedron sulfur S8, that is present in combined elemental form would always possess the oxidation state equivalent to zero.
Related: Learn about different Gas laws and the impact in real life.
Step 2: Determine the nature of the substance (inert or ionic)
As we know, an ion is formed whenever an atom gains or loses an electron, so any substance which is present in the ionic form must possess either some negative or positive charges.
For example, in the synthesis reaction of Na and Cl, the oxidation states changes such as
Since sodium loses one electron to form an ionic bond with chlorine, it develops a positive charge, and its oxidation state changes from 0 to +1. Similarly, when chlorine accepts that electron from the sodium, it develops a negative charge, and its oxidation state changes from 0 to -1.
Step 3: Determine oxidation states of metallic ions
Don't forget that metallic ions can show two or more oxidation states. For example, Al when found in the form of ions, might possess oxidation numbers +1, +2, or +3.
The question which arises here is how to determine when the metallic ion of Al would show +1, +2, and +3? It depends upon the charge of other atoms present in the compound.
For example, Al in the compound AlCl3 possesses a +3 oxidation state. Since one Cl- carries one negative charge, and there are three Cl atoms. The Al must possess +3 to keep the compound neutral.
Step 4: Look for the exceptional cases
Some things you should always need to remember are
Oxygen atoms always have oxidation number -2 unless it is
- Present in elemental form (here its oxidation state will be zero as per rule no. 1.
- Part of the peroxide compound since in H2O2, the oxidation state of oxygen will be always -1.
- Part of superoxides where the oxidation state of oxygen is always -½.
- Bound to F, since fluorine is more electronegative and changes the oxidation state of oxygen to +2 or +1.
The oxidation state of hydrogen is always +1 unless it is
- A part of hydrides since here the oxidation state of hydrogen will be -1.
- The fluorine always possesses the oxidation state -1 and it remains the same regardless of the nature of the compound it is a part of.
Determining the oxidation state of elements without general rules
This method is usually used to understand how to determine oxidation number/state of an unknown substance in a chemical compound. Suppose we have a compound Na2SO4 and we have to find the oxidation state of S. For this purpose, we have to follow these simple steps
Step 1: Write down the oxidation states of known substances
According to rule no. 2, the oxidation state of sodium is +1 and that of oxygen is -2 as per rule no. 4. Since S in Na2SO4 is not present in free or combined elemental form, its oxidation state would be a non-zero value.
Step 2: Multiply the numerical coefficient of each atom with its oxidation state.
Na2SO4 = (2x - 1) + S + (4x - 2)
Na2SO4 = 2 + S + (-8)
Step 3: Add the known values together
2 + S + (-8)
S + (-6)
S + (-6) = 0
S - 6 = 0
S = +6
There are many useful blogs in this website. You can find the most recent blogs in the blog section. Also you can use balancing chemical equations calculator for the purpose of balancing an chemical reaction.
Thus, the oxidation state of S in Na2SO4 is +6.
Both of these methods of determining oxidation states of elements are equally effective and interlinked with each other. If a person doesn't know the general rules of assigning the oxidation state, he would also not be able to find the oxidation state of an unknown element using the algebraic method.