In chemistry, compounds are discussed very often and extensively. The reason is obvious, because these compounds can be used to explain how and why substances exist and how atoms can react with each other. Many scientists have considered various theories such as the octet rule, orbital models or bond models such as ionic bonds and covalent bonds.
Why we need oxidation numbers
One of these theories enables you to recognize and describe redox reactions or their partial reactions, reduction and oxidation. The oxidation numbers (also called oxidation states or states) are a model that helps you to find out whether the atom in question has more or fewer valence electrons available in a bond with other atoms than it would have in the neutral state. The result of this consideration is then denoted by hypothetical ionic charges, which represent the oxidation numbers!
With «hypothetical charge» here simply means a charge, as you already know it with ions. So the single negatively charged chloride ion (Cl-) or the doubly positively charged magnesium ion (Mg2+) each have a charge.
It is important, however, that these are really only hypothetical, i.e. theoretical charges. In order to distinguish these from the actual ionic charges, the hypothetical charges are written in the form of oxidation numbers, starting with the sign. For example +3 or -4. The charge of an ion, on the other hand, is written starting with the number and ending with the sign. For example with a sulfide ion (S2-) or the aluminum ion (Al3+).It’s best to remember this difference quickly, because it’s really important!
This may sound more complicated than it actually is, but by the end of this explanation you will see that it is not that difficult and at the same time it will explain an important aspect of chemistry to you.
Application of oxidation numbers
Imagine that all the polyatomic molecules in a reaction equation existed as ions rather than as covalent bonds. You then assign the aforementioned hypothetical charges to these “ions”. Let’s look at an example where you consider the following reaction equation:
Mg + 2HCl → MgCl + H2
The task now is to find out whether the reaction is a redox reaction and if so, what is reduced and what is oxidized. The first step is to imagine the theoretical ions. Here is the whole thing written down:
Mg + 2 H+ + 2 Cl- → Mg2+ + 2 Cl- + H2
In this example, the compounds are also present as real ions because the reaction actually takes place in solution. You can often find out the charges of the actual ions directly when listing the oxidation numbers. As a second step, you assign the theoretical ions their hypothetical charge. There are some very helpful rules that you need to follow.
Oxidation Numbers – Rules
In fact, there are numerous rules that apply to oxidation numbers. There are also some exceptions to each. In the following you will get to know the most important rules and their exceptions step by step, so that the implementation does not become a problem for you.
1. Atoms in their elemental form have the oxidation number ±0.
You certainly remember magnesium in the previous equation. Magnesium is found here in its elemental form, meaning it stands alone. Other examples are:
Cu0, Zn0, Al0, Na0, Ca0, …
2. Monatomic ions are given an oxidation number according to the charge on the ion.
Back to the example from before. There you will also find H+, among other things. In this case, hydrogen has a single positive charge and therefore has an oxidation number of +1.
It will also happen that oxidation numbers are written in Roman numerals. This is no longer the norm, and the oxidation numbers are being replaced by Arabic numerals.
Other examples of monatomic ions with a corresponding oxidation number are:
K++1, S2++2, Cl–1, Ca2++2, Al3++3, …
3. Nonmetals mostly have negative oxidation numbers, metals have positive oxidation numbers.
The following table shows you the most important elements with their corresponding oxidation number. It is important that you do not forget that the rules only apply to the respective elements in combination.
Elementoxidation numberoxygenoxygen with fluorineoxygen in hydrogen peroxide (H2O2)–2+2–1hydrogen with nonmetalshydrogen with metals+1–1halogen–1alkali metals+1alkaline earth metals+2nonmetalshighest oxidation number corresponding to the main group
When two rules interfere with each other, you always apply the rule that applies to the element with the higher electronegativity first. This attracts the electrons more strongly and must therefore be considered first. The second element will be adjusted accordingly.
4. The sum of the oxidation numbers in a compound equals the charge on the compound
As soon as complex molecules appear or the ions are polyatomic, oxidation numbers do not appear feasible at first glance. With the help of the rules presented to you, you will quickly become a pro here too. It is important that you always remember that the sum of the oxidation numbers is equal to the charge of the compound. If a molecule is neutral, the sum is 0.
The first example you’ll look at is Ammonia (NH4+). As you can see, the ion is simply positively charged. So the sum must be +1. The oxidation number for nitrogen (N) is not known, so it is variable. For that, you know the oxidation number for hydrogen with a nonmetal: +1. A quadruple positive charge results. However, since the result should only be a single positive, it requires a -3, which the nitrogen then receives.
N-3H+14+
The same goes for that sulfate ion (SO42-). As a tip for determination: Oxygen has the oxidation number -2. Now all you have to do is calculate and you get the following oxidation numbers:
S+6O4-242-
Tin is a special example here because, like some other elements, it can also have different oxidation numbers in different compounds. You don’t have to memorize this, however, because how the tin is made is always stated in the name, or you can easily find out with the rules.
Tin(II) Oxide (ZnO): You can again assign a -2 to the oxygen and know that the tin must have a +2 since the entire molecule is neutral.
Zn+2O-2
Tin(IV) oxide (ZnO2): Since we have two oxygen atoms that together have the oxidation number -4, the zinc must get a +4 in this case.
Zn+4O2-2
Tin(II) chloride (ZnCl2): You know from the third rule that halogens tend to get negative oxidation numbers. And since zinc gets a +2 in this case, each chloride must get a -1.
Zn+2Cl2-1
The table below gives you a few more examples. 3 molecules are considered and they each consist of two elements. The first element in the table is always what you determine by one of the rules. The second element was found arithmetically by making sure that the sum of the oxidation numbers equaled the charge on the molecule (Rule 4).
For hydrogen sulfide (H2S) it can be seen directly that the hydrogen must have a value of +1. Since there are two hydrogens, and the molecule is neutral overall, sulfur in the compound must have a value of -2. The other two examples were calculated using the same principle.
connection
1st item
2nd itemoverall equationH2SH = +1 S = -2 1 * 2 + (-2) * 1 = 0 Cr2O7-2O = -2 Cr = +6(-2) * 7 + 6 * 2 = -2SCl2
Cl = -1
S = +2-1 * 2 + 2 * 1 = 0
Application of the rules in reactions
But back to the example above and how you can determine the oxidation numbers there.
Mg + 2 H+ + 2 Cl- → Mg2+ + 2 Cl- + H2
Magnesium (Mg) and hydrogen (H2) are in elemental form in the equation, which is why they each have a 0 charge. (1st rule) The marked elements are no longer present as molecules but as hypothetical ions. You can easily find the valences of the elements in the periodic table of the elements based on the main groups, which allows you to find out the charges on the ions and use the second rule to determine their oxidation numbers. You can see an overview of the numbers in Figure 1.
Figure 1: Equation specifying the oxidation numbers
To solidify this, here is another example of the distribution of oxidation numbers. This time you just imagine the ions and don’t write them down to get a little faster.
CuO + Fe → FeO + Cu
Iron (Fe) and copper (Cu) are elementary, which is why their oxidation number is 0 in each case. Copper oxide (CuO) has an oxygen that is rated -2 according to the 2nd rule. Since the entire molecule has no charge, the oxidation numbers must add up to 0 (4th rule). Therefore, the Cu has the oxidation number +2. Iron oxide (FeO) can be calculated in the same way. Oxygen has an oxidation number of -2, so the iron must be given an oxidation number of +2 (Figure 2).
Figure 2: Example equation with oxidation numbers and detection of reduction and oxidation
In this reaction, the reduction and the oxidation have already been labeled directly as an example. However, this will be looked at in more detail in the next section.
There are some elements that by their very nature can have different oxidation numbers depending on the compound in which they are present. Hydrogen and oxygen have already been shown here as examples. So don’t be surprised if you sometimes get a different oxidation number for the same element in a different compound. Iron, for example, has the oxidation number +2 in the compound FeSO4 and the oxidation number +3 in Fe2O3. Just follow the rules and it will work.
Organic Compounds and the Lewis Formula
The previous examples only referred to inorganic compounds and were also kept quite simple. However, there are situations in which more complex molecules such as organic compounds are considered or you are simply given a Lewis formula.
The Lewis formula shows us the chemical structure of atoms and molecules. The atoms themselves and their valence electrons are also shown. For clarity, these valence electrons are represented as pairs of electrons whenever possible. This goes either through two points or as a continuous line (see Figure 3).
Figure 3: Representation of the bonding electrons in the Lewis formula
However, keep in mind that pairs of electrons between two atoms can also be used by both. Which atom in the bond can now use these electrons depends on the electronegativity of the atom and is shown in Figure 4. The higher the electronegativity is, the more likely it is that the electron is in the vicinity of the more electronegative atom. This creates a dipole moment and the atom becomes more negative than the atom that gave up the electron.
Figure 4: Representation of the…