1. Height, weight and oxidation state
Like how everyone can be described by their height and weight, every element has an oxidation state. It describes how oxidised an element is in a substance.
If we interpret oxidation as the loss of electrons, the oxidation state indirectly tells us how deprived an element is, of electrons of course! The more positive the oxidation state, the more electrons the element has lost.
2. Starting from scratch: free elements have an oxidation state of zero
Hold up! What are free elements? Before you chiong to collect them as freebies, free elements are simply pure elements: just one type of atoms. They can be metals, like sodium and iron. They can also be non-metals that exist as simple molecules or giant molecules.
We shall define the oxidation state of free elements as zero. They are seen as the default state, before atoms have gained or lost any electron.
3. Same old, same old: some elements form typical, fixed oxidation state
When elements react to form compounds, their oxidation state changes.
| Element | Oxidation State in Compounds |
|---|---|
| Hydrogen | +1 (except in metal hydride) |
| Group I metals | +1 |
| Group II metals | +2 |
| Oxygen | -2 (except in peroxide) |
| Fluorine | -1 |
Some elements mainly form one type of oxidation state in compounds. We say that they have fixed oxidation state. For example, hydrogen has a fixed oxidation state of +1 in all covalent compounds.
Likewise, fluorine is pretty boring. In all fluorine-containing compounds, fluorine has a fixed oxidation state of -1.
4. You change your state, like a girl changes clothes: variable oxidation state
On the other hand, many elements have variable oxidation state. In other words, their oxidation state depends on what compound they are found in. These elements include transition metals, carbon, nitrogen, and non-metals in Period 3 and below.
For example, copper has an oxidation state of +1 in red copper(I) oxide. Copper can also take up the oxidation state of +2 in compounds like blue copper(II) sulfate.
5. Into the woods: calculating unknown oxidation state from the known
If the oxidation state of these elements are so variable, how then can we find their exact oxidation state?
The trick is to know that the combined oxidation state of all elements in a compound is zero. For ions, the combined oxidation state is equal to the charge of the ion.
We can work from the above rule to find the unknown oxidation state. Let’s attempt this by finding the oxidation state of manganese in potassium manganate, KMnO4.
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STEP 1: Consider the combined oxidation state
As KMnO4 is a compound, the combined oxidation state = 0
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STEP 2: State the oxidation state of elements with fixed oxidation state
K: +1, as it is a Group I metal that has 1 valence electron to lose
O: -2 -
STEP 3: Form an equation by adding up all the oxidation states
Let x be the unknown oxidation state of manganese in KMnO4
Combined OS = OS of K + OS of Mn + 4(OS of O)
0 = +1 + x + 4(-2) -
STEP 4: Solve for x and find the unknown oxidation state
0 = x – 7
x = +7Therefore, the oxidation of manganese in KMnO4 is +7.
6. Weird oxidation states?
The oxidation state of iron in magnetite, Fe3O4, may seem weird at first blush. It is not a whole number. Calculate it yourself with the method above!
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STEP 1: Consider the combined oxidation state
As Fe3O4 is a compound, the combined oxidation state = 0
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STEP 2: State the oxidation state of elements with fixed oxidation state
O: -2
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STEP 3: Form an equation by adding up all the oxidation states
Let x be the unknown oxidation state of iron in Fe3O4
Combined OS = 3(OS of Fe) + 4(OS of O)
0 = 3x + 4(-2) -
STEP 4: Solve for x and find the unknown oxidation state
0 = 3x – 8
x = 2.67Therefore, the oxidation of iron in Fe3O4 is 2.67.
The oxidation state of iron in magnetite is fractional because it is an average value. As magnetite contains both iron(II) and iron(III) ions, the oxidation state is a weighted average of the respective oxidation states of +2 and +3.
