What are they?
In chemistry, the description of different oxidizing-reducing processes is impossible without oxidation states – special conditional numbers which are used to determine the charge of an atom of any element.
If you imagine an oxidation state (not to be confused with valence! In some cases, they do not coincide) as a note in an exercise book, we will simply see numbers with the figure zero (0 in a simple substance), plus (+) or minus (-) above the substance that interests us. As it happens, they play an enormous role in chemistry, and the ability to determine the OS (oxidation state) is essential in the study of this subject, without which further actions are pointless.
We use the OS for the description of the chemical properties of a substance (or separate element), the correct writing of its international name (understandable for any country and nation regardless of the language used) and formula, and also for classification by properties.
There can be three types of state: highest (to determine this we must know the group that the element is located in), intermediary, and lowest (we must subtract the number of the group in which the element is located from the number 8; naturally, we take the number 8 because there are a total of 8 groups in the Periodic Table). We will look at how to determine the oxidation state and to balance it correctly in more detail below.
How to determine the oxidation state: constant OS
Firstly, the OS may be variable or constant. Determining a constant oxidation state is not very difficult, so it’s best to start the lesson with this: all we need is to be able to use the periodic table. There are a number of certain rules:
Zero state. We mentioned this above – only simple substances have it: S, O2, Al, K and so on.
If molecules are neutral (in other words, they do not have an electric charge), the sum of their oxidation state is equal to zero. However, in the case of ions the sum should be equal to the charge of the ion itself.
Primarily metals are located in groups I, II and III of the Periodic Table. Elements of these groups have a positive charge, the value of which corresponds to the number of the group (+1, +2, or +3). An exception is several metals mainly from side groups – for example, iron (Fe) – its OS can be +2 or +3.
The OS of hydrogen (H) is usually +1 (in interaction with non-metals: HCl, H₂S), but in some cases it is -1 (in the formation of hydrides in compounds with metals: KH, MgH₂).
The OS of oxygen (O) is +2. Compounds with this element form oxides (MgO, Na₂O, H₂O – water). However, there are also cases when oxygen has an oxidation state of -1 (in the formation of peroxides), or acts as a reducer (in compounds with fluorine F, because the oxidizing properties of oxygen are weaker). On the basis of this information, we can balance the oxidation states in many complex substances, describe oxidizing-reducing reactions and so on, but more about this later.
Some elements are distinguished by the fact that they do not have a single oxidation state and change depending on the formula they are in. According to the rules, the sum of all oxidation states should also be equal to zero, but to find this, certain calculations must be made. In written form, this looks like a simple algebraic equation, but over time we “get the hang of it”, and easily learn to perform the entire operation in our heads. It will not be so easy to describe this in words, so let’s move straight to practice:
HNO₃ – determine the oxidation state of nitrogen (N) in this formula.
In chemistry we also read the names of elements, and approach the placing of oxidation states from the end as well. So, we know that the OS of oxygen is 02. We must multiply the oxidation state by the coefficient on the right (if there is one): -2•3=-6. We then move to hydrogen (H): its CO in the equation will be +1. So for the sum of the OS to be zero, we must add 5. We make sure: +1+5-6=-0.
Additional exercises can be found at the end, but first we must determine which elements have a variable oxidation state. In principle, all elements, not counting the first three groups, change their states. The most striking example is the halogens (elements of group VII, not counting fluorine F), group IV and noble gases. Below you will see a list of several metals and non-metals with a variable state:
- H (+1, -1);
- B (-3, +3);
- C (-4, -2, +2, +4);
- N (-3, -2, -1, +1, +2, +3, +4, +5);
- O (-2, -1);
- Si (-4, -2, +2, +4);
- P (-3, +1, +3, +5);
- S (-2, +2, +4, +6);
- Cl (-1, +1, +3, +5, +7).
This is just a small selection of elements. To learn to determine the OS, you will need study and practice, but this does not mean that you need to learn all the constant and variable OS by heart: just remember that the first and last numbers are encountered much more often. An important role is often played by the coefficient, and what substance is under discussion – for example, in sulfides sulfur (S) takes a negative oxidation state, in oxides - oxygen (O), and in chlorides – chlorine (Cl). Accordingly, in these salts a different element takes a positive state (and is said to be oxidized in this situation).
Solving problems to determine oxidation states
Now we have come to the most important part – practice. Try to solve the following problems yourself, and then look at the solution and check your answers:
K₂Cr₂O₇ –find the oxidation state of chromium The OS of oxygen is -2, for potassium it is +1, while we indicate chromium by the unknown variable x for the meantime. Accordingly, we draw up the equation: +1•2+2•x-2•7=0. After solving it, we receive the answer 6. We check this, and everything matches up, so the problem has been solved.
H₂SO₄ – find the OS of sulfur According to the same concept we draw up an equation: +1•2+x-2•4=0. Then: 2+x-8=0.x=8-2; x=6. Click here to do entertaining experiments for better understanding how to determine oxidation states.
To learn to determine the oxidation state independently, you must not only be able to draw up equations, but also study the properties of elements of different groups thoroughly, remember algebra lessons, drawing up and solving equations with an unknown variable. Don’t forget that rules have exceptions, and you shouldn’t forget them: this concerns elements with a variable OS. You must also learn to balance coefficients to solve many problems and equations (and know why you are doing so).