The oxidation number of an atom changes during oxidation or reduction. So, to identify the loss of electrons (oxidation) and gain of electrons (reduction), we must be aware of oxidation number or oxidation state.
What is oxidation number?
Oxidation number: It is the number that appears on an element in a chemical combination or molecule. It shows the count of electrons that are lost, gained or shared by an atom when it reacts with other atoms to form molecules or other chemical species. So, the numerical value of the oxidation state is equal to the number of electrons lost or gained by the atoms. It is also known as the oxidation state.
We can say that it is a charge that an atom gains after forming an ionic bond with other atoms. This charge is positive if electrons are lost and negative if electrons are gained, and is equal to the number of electrons an atom has lost or gained as compared to the neutral atom.
Let us understand it with the following example;
Let us find the oxidation number of magnesium (Mg) in a molecule of magnesium oxide (MgO), which contains one atom of magnesium and one atom of oxygen. The formation of magnesium oxide is shown below;
2Mg + O2 → 2MgO
The atomic number of oxygen is 8. Its electronic configuration is 2, 6, so, It needs 2 electrons to complete its octet, so it will tend to gain two electrons, so its valency is 2. After gaining two electrons, it becomes O2- as one electron has a -1 charge. It takes two electrons from Magnesium which gains 2+ charge on losing two electrons.
So, whenever, oxygen forms an oxide with an element, the oxygen in that oxide will have -2 charge. So, the oxidation number of -2 tells gain of electrons by oxygen and loss of two electrons by magnesium, so there is the reduction of oxygen and oxidation of magnesium.
Magnesium oxide is electrically neutral. So, magnesium must have + 2 charge as oxygen has - 2 charge. So, the oxidation number of magnesium is + 2 as it has lost two electrons. So, the oxidation number changes in a chemical reaction. If oxidation number increases there is oxidation (loss of electrons) and if it decreases, it shows reduction (gain of electrons).
So, in a neutral molecule, if you know the charge on one atom, another atom in this molecule will have an equal and opposite charge as shown below;
Mg + O2 → Mg+2O-2
In this reaction, Mg changes to Mg2+. A positively charged ion that shows it has lost electrons. So, Mg is oxidised in this reaction and its oxidation number of Mg is +2. In this reaction, neutral O2 changes to O -2. A negatively charged ion that shows it has gained electrons. So, reduction of oxygen takes place here.
Besides this, oxidation number can be positive or negative as described below;
Atoms or ions with a constant oxidation state
i) Some atoms and ions have a constant oxidation state. For example, the net charge on neutral atoms or molecules is zero. So, their oxidation state or number is zero (0). For example, the oxidation state of neutral atoms like sodium, iron and magnesium is zero. Similarly, the oxidation number of neutral molecules like water, ammonia, chlorine, oxygen is zero.
ii) The atoms in homo-polar molecules (made of the same atoms) also have a zero oxidation sate. For example, the oxidation number of oxygen atoms in an oxygen molecule (O2) is zero.
iii) The oxidation state of charged ions is equal to the net charge on these ions. For example;
Calculation of Oxidation Number of an Atom in a molecule or ion
Calculation of oxidation no. of permanganate ion (MnO4) in potassium permanganate (KMnO4);
Oxidation no. of potassium permanganate (KMnO4) = Sum of oxidation numbers of different atoms present in it (K+Mn+4O) = 0, as it is a neutral molecule.
Let the oxidation no. of permanganate ion = X
In this molecule, potassium ion (K+) has +1 oxidation state, so we have;
Oxidation no. of K + oxidation no. of MnO4 = 0
1 + X = 0
X = -1
Let us see how to calculate of oxidation number of an atom that occurs once in a molecule;
i) Find the oxidation number of chlorine in KCl?
KCl is a neutral molecule, so, the net charge on this molecule is 0 as the oxidation state of potassium + oxidation state of chlorine = 0
Let the oxidation state of Cl = x
Oxidation state of K = +1
So, we have +1 + x = 0
X = -1
So, Chlorine's oxidation state in KCl is -1.
ii) Find oxidation no. of Manganese in permanganate ion MnO4-
The charge of this ion is -1.
Oxidation no. of Oxygen = -2
Let the oxidation state of Mn = x
Now, we have x+ (4x-2) = -1
X - 8 = -1
X = -1 + 8
= + 7
iii) Let us find the oxidation no. of nickel (Ni) in Ni (CO) 4
i) Ni (CO) 4
The net charge on this molecule is zero.
CO is also a neutral molecule, so it also has 0 charge.
Let the charge on nickel is x.
So, we have;
X + (4x0) = 0
X = 0
So, the oxidation state of nickel in this molecule is 0.
iv) Oxidation state of Cr2 in the formula K2Cr2O7?
Let the oxidation state of Cr2 is Y.
Now, let us write the oxidation state of different atoms in this molecule.
K2=+2; Cr2=2xY; O7= (-2x7)
As it is neutral molecule so net charge on it is 0.
So, we have the following equation;
2 + 2Y - 14 = 0
2Y = 14 - 2
Y = +6
So, in this molecule, the oxidation state or number of chromium = 6
Characteristics of oxidation number