The Oxidation Number, in simple terms, can be described as the number that is allocated to elements in a chemical combination. The oxidation number is basically the count of electrons that atoms in a molecule can share, lose or gain while forming chemical bonds with other atoms of a different element.
The oxidation number is also referred to as the oxidation state. However, sometimes these terms can have a different meaning depending on whether we are considering the electronegativity of the atoms or not. The oxidation number term is used frequently in coordination chemistry.
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In general, the oxidation state or number helps us describe the transfer of electrons. However, students have to note that it is different from a formal charge which determines the arrangement of atoms. The oxidation number/state is also used to determine the changes that occur in redox reactions. Besides, it is quite similar to valence electrons.
What Is the Oxidation Number?
The oxidation number of an atom is defined as the charge that an atom appears to have on forming ionic bonds with other heteroatoms. An atom having higher electronegativity (even if it forms a covalent bond) is assigned a negative oxidation state.
The definition assigns an oxidation state to an atom on conditions that the atom –
i) Bonds with heteroatoms.
ii) Always form ionic bonding by either gaining or losing electrons, irrespective of the actual nature of bonding.
Since an atom can have multiple valence electrons and form multiple bonds, all of them will be assumed to be ionic and assigned an oxidation state equal to the number of electrons involved in the bonding. So, oxidation number or state is a hypothetical case of the assumption of atoms forming an ionic bond.
In the complex cation tetroxoplatinum (PtO4)2+, platinum possesses an oxidation state of 10. Ten is the maximum oxidation state exhibited by any atom. It appears to have lost ten electrons to form the ion. But, the ionization energy required for removing an electron from charges positively species increases heavily.
The larger the charge, difficult it is to remove an electron, and so, the higher the ionization energy. So, the removal of ten electrons is highly hypothetical. Similarly, the addition of electrons also becomes difficult with increasing negative charge.
Oxidation states larger than three, whether positive or negative, are practically impossible. In spite of the assumption, it helps in understanding the changes accompanying the atom undergoing a chemical change.
How to Find the Oxidation Number of an Atom?
The oxidation number or state of an atom/ion is the number of electrons an atom/ion that the molecule has either gained or lost compared to the neutral atom. Electropositive metal atoms of groups 1, 2 and 3 lose a specific number of electrons and always have constant positive oxidation numbers.
In molecules, more electronegative atoms gain electrons from less electronegative atoms and have negative oxidation states. The numerical value of the oxidation state is equal to the number of electrons lost or gained.
The oxidation number or oxidation state of an atom or ion in a molecule/ion is assigned by:
i) Summing up the constant oxidation state of other atoms/molecules/ions that are bonded to it
ii) Equating the total oxidation state of a molecule or ion to the total charge of the molecule or ion
Atoms/ Molecules and Ions That Have Constant Oxidation State (Number)
a) The net charge on neutral atoms or molecules is zero. So, their overall oxidation state is zero.
For example, the oxidation state of elemental atoms such as sodium, magnesium, and iron is zero. Similarly, the net oxidation state of neutral molecules such as oxygen, chlorine, water, ammonia, methane, and potassium permanganate is zero.
The oxidation state of atoms in homo-polar molecules is zero. The oxidation number of an atom in an oxygen molecule is zero.
b) The oxidation state of charged ions is equal to the net charge of the ion. So,
- The oxidation number of all alkali metal ions is always = +1
- The oxidation number of all alkaline earth metal ions is always = +2
- The oxidation number of all boron family metal ions is always = +3
- The oxidation number of hydrogen in proton (H+) is +1, and in hydride is -1.
- The oxidation number of oxygen in oxide ion(O2-) is -2, and in peroxide ion(O-O2-) is -1.
Calculation of Oxidation Number of an Atom in a Molecule/Ion
Oxidation number of potassium permanganate (KMnO4) = Sum of oxidation number of (K + Mn + 4O) = 0
Oxidation number of permanganate ion (MnO4)– = Sum of oxidation number of ( Mn + 4O)= -1
Calculation of Oxidation Number of Atoms Occurring Only Once in a Molecule
Examples 1: Oxidation state of chlorine in KCl
KCl is neutral, and so net charge = 0
Oxidation state of KCl = Oxidation state of potassium + oxidation state of chlorine = 0.
Oxidation state of potassium = +1
Oxidation states → +1 + x = 0: x = -1
Atoms in the species → K Cl
Oxidation state of chlorine in KCl = -1
Example 2: Oxidation number of manganese in permanganate ion MnO4–
The charge on the permanganate ion is -1
Oxidation state of permanganate ion = Oxidation state of manganese + 4 oxidation state of oxygen = -1.
Oxidation state of oxygen = -2
Oxidation states → x + (4*-2) = -1: x = +7
Atoms in the species → Mn 4O
Oxidation state of manganese = +7
Example 3: Oxidation number of a metal ion in a complex.
i) Ni(CO) 4.
The total charge of the complex is zero. CO is a neutral molecule.
Oxidation states → x + (4*0) = 0: x = 0
Atoms in the species → Ni 4 CO
Nickel is also in a zero oxidation state.
ii) [CoCl2(NH3)4]Cl.
The complex can be written in the ionic forms as [CoCl2(NH3)4]+Cl–.
Metal is in a cationic complex with a unitary positive charge. Ammonia is a neutral ligand, and chlorine has a unit negative charge.
Oxidation number of [CoCl2(NH3)4]+ = Oxidation number of (Co + 2Cl + 4×0) = +1.
Oxidation states → x + (2*-1) + 4*0 = +1: x = +3
Atoms in the species → Co 2Cl 4 NH3
Oxidation number of cobalt in the complex = +3
Calculation of Oxidation Number of Atoms Occurring More Than Once in a Molecule and Having Identical Bonding
Atoms occurring more than once in a molecule may be bonded in an identical way or not. If they are identically bonded, then there is no difference between them, and all the atoms will have the same oxidation numbers. The oxidation state of such an atom in a molecule can be calculated by the normal method.
The average oxidation number will be the same as calculated individually and as a whole number.
Example 1: The number of atoms of chlorine is two in the molecules Cl2O, Cl2O5 and Cl2O7. But, the environment of both atoms of chlorine is the same, as shown by their structures. The oxidation number of the atoms calculated either individually or from the whole molecule is the same.
i) Cl2O:
Cl2O is neutral, and so net charge = 0.
Net oxidation state of Cl2O = 2 x Oxidation state of chlorine + 1x Oxidation state of oxygen = 0.
Oxidation state of oxygen = -2.
Oxidation states → 2 x + (-2) = 0: x = +1
Atoms in the species → 2Cl O
Oxidation state of chlorine in Cl2O= 2/2 = +1
ii) Cl2O5:
Cl2O5 is neutral, and so net charge = 0
Oxidation state of Cl2O5 = 2 x Oxidation state of chlorine + 5 x oxidation state of oxygen = 0.
Therefore,
Oxidation state of oxygen = -2.
Oxidation states → 2x + (5*-2) = 0: x = +5
Atoms in the species → 2Cl 5O
Oxidation state of chlorine in Cl2O5 = 10/2 = +5
iii) Cl2O7:
Cl2O7 is neutral, and so net charge = 0
Oxidation state of Cl2O7 = 2 x Oxidation state of chlorine + 7 x oxidation state of oxygen = 0.
⸪ Oxidation state of oxygen = -2.
Oxidation states → 2x + (7*-2) = 0: x = +7
Atoms in the species → 2Cl 7O
Oxidation state of chlorine in Cl2O = 14/2 = +7
Note: Except the atoms/molecules/ions mentioned, as having a constant oxidation state, the oxidation state of other atoms/molecules and ions will vary depending on the molecule they are present.
In the given examples, the oxidation state of chlorine is not constant, but variable (+1, +5 and +7)
Example 2: Oxidation state of chromium in dichromate anion.
Dichromate ion is Cr2O72-.
The charge on the ion is -2.
Oxidation state of dichromate ion = 2 x Oxidation state of chromium + 7 x oxidation state of oxygen = -2.
Oxidation state of oxygen = -2.
Oxidation states → 2x + (7*-2) = -2: x = +6
Atoms in the species → 2Cr 7O
Oxidation state of chromium= 12 / 2 = 6
Calculation of Oxidation Number of Atoms Occurring More Than Once in a Molecule and Having a Difference in Bonding
Atoms having different bond structures will have different oxidation states. Hence, their oxidation state has to be individually determined from their molecular structure. The average oxidation state can be calculated by assuming them to be equal. In such a case, the average oxidation could be fractional rather than a whole integer.
Example 1: Cl2O4
i) The average oxidation state of chlorine
Cl2O4 is neutral, and so net charge = 0
Oxidation state of Cl2O4 = 2 x Oxidation state of chlorine + 4 x oxidation state of oxygen = 0. ⸪
Oxidation state of oxygen = -2.
Oxidation states → 2x + (4*-2) = 0: x = +4
Atoms in the species → 2Cl 4O
Oxidation state of chlorine in Cl2O5 = 8/2 = +4
ii) Structure of Cl2O4
Individual oxidation state of oxygen ‘a’ is +7
Individual oxidation state of oxygen ‘b’ is +1
None of the oxygen has a +4 oxidation state.
Example 2: Cyclopentadienyl anion C5H5–
i) Average oxidation number of carbon
Five carbon atoms share the five electrons from five hydrogen atoms and an additional electron of the negative charge by resonance. So, six electrons are shared by five-carbon.
The average oxidation state of each carbon = 6/5 = fraction
ii) Without resonance, four carbon has a -1 oxidation state, and one carbon has a -2 oxidation state.
Oxidation Number of Atoms in a Diatomic Molecule
A diatomic molecule can be either homo or heteronuclear.
i) Homonuclear diatomic molecule
The oxidation number concept is applicable only to heteroatoms forming a molecule. Hence, in a homonuclear diatomic molecule, the oxidation number of the atoms is zero. Also, the oxidation number of hydrogen or oxygen, nitrogen, and chlorine in respective molecules is zero.
ii) Heteronuclear diatomic molecule
In hetero diatomic molecules, all bonds formed between the atoms are considered as ionic.
More electronegative atoms are assumed to take away the bonding electrons from the less electronegative atom. So, the electronegative atom will have a negative oxidation state, and the magnitude is equal to the number of electrons taken by it.
The less electronegative atom is supposed to have lost its electron to the more electronegative atom. So, the less electronegative atom will have a positive oxidation state equal to the number of electrons lost by it.
Example 1: HCl
Chlorine is more electronegative than hydrogen. So, chlorine is assumed to take away the electron from hydrogen. Chlorine, which receives one electron, has an oxidation number of -1, while hydrogen losing one electron, has an oxidation state of +1.
Example 2: H2O
Oxygen is more electronegative than hydrogen. So, the oxygen atom receives one electron each from the two-hydrogen atom and will have an oxidation number of -2. Both hydrogens losing one electron each will have an oxidation number of +1 each.
Fractional Oxidation States
The oxidation state is the number of electrons assumed to have either been lost or taken by heteroatoms during their bonding. Since the numbers of electrons are whole numbers, the oxidation number of individual atoms also has to be a whole integer.
But, there are molecules that contain an atom more than once and each bonded differently. Such atoms shall have different oxidation state at different positions and hence has to be, calculated individually, taking into consideration the atoms it bonds.
Calculation of the oxidation state of the atom using the normal method assumes all the same atoms as equal and will give only an average of the different oxidation states of the same atom in the molecule. This average oxidation state is mostly a fraction instead of a whole number.
So, the fractional oxidation state is always an average oxidation number of the same atoms in a molecule and does not reflect the true state of the oxidation state of atoms.
Example 1: Superoxide -KO2
Potassium ion has an oxidation number of +1. Potassium superoxide molecule being neutral, the oxidation state of two oxygen atoms together is -1.
So, the average oxidation number of oxygen in superoxide is -1/2.
The structure of superoxide ion is given below:
As per the structure, one oxygen atom has a zero oxidation state. The second oxygen atom is negatively charged and has a -1 oxidation state. So, the true oxidation state of oxygen atoms is not minus half each but 0 and -1.
Example 2: Fe3O4
Considering the oxidation state of oxygen as -2, the average oxidation state of iron atoms will be +8/3.
But the molecule is a mixture of two compounds of FeO and Fe2O3.
In FeO and Fe2O3, iron is in +2 and +3 oxidation states. So, in Fe3O4, one iron has +2, and the other iron has +3 oxidation states.
Average oxidation state is
\(\begin{array}{l}= +\frac{2+3+3}{3} = +\frac{8}{3}\end{array} \)
Example 3: Tetra-thionate ion
Tetrathionate ion has four sulphur atoms bonded to oxygen as in the structure. Out of the four sulphur atoms, the two-terminal sulphur atoms are connected to three oxygen heteroatoms and one homo sulphur atom. Each terminal sulphur atom forms five bonds with oxygen heteroatoms, so the oxidation state will be +5. The bridging sulphur atoms being homo-nuclear, have zero oxidation state.
The total oxidation of the entire four sulphur atoms is ten.
So, the average oxidation state of sulphur = 10/4 = 2.5
Oxidation and Reduction, Redox Reactions
Atoms and molecules react to form products. The reactions are classified into many types based on the nature of change in the reactants to form products. Whatever may be the reaction types, reactant and product atoms/ions in the reaction may either have the same or a different number of valence electrons.
Reactions, where the number of valence electrons in the reactant atom/ion is different from the product side, are called reduction-oxidation or simply redox reactions. Atom/ion might have either lost or gained electrons during the reaction, and accordingly, atom/ion is said to be either oxidized or reduced.
Atoms/ions in the reactions are represented by their atomic symbol with a superscript. The superscript represents the difference in the number of electrons of the atom/ion compared to the neutral atom. The superscript also has a positive sign if the electron is lost and a negative sign if the electron is gained compared to the neutral atom.
The superscript, along with the sign, is called the ‘oxidation state’ of the atom. The atom may have different oxidation states depending upon the number of electrons either gained or lost. Neutral atoms have zero oxidation state.
| Reaction | Reaction Type | The Oxidation State of the Atom in the Reactant Side | The Oxidation State of the Atom in the Product Side |
| M0 → M+ + e– | Oxidation reaction | 0 | +1 |
| M+ + e– → M0 | Reduction | +1 | 0 |
| M2+ → M3+ + e– | Oxidation reaction | +2 | +3 |
| M3+ + e– → M2+ | Reduction | +3 | +2 |
In redox reactions:
In redox reactions, atoms or ions either lose or gain electrons and have different oxidation states before and after the reaction.
- The oxidation number can be positive or zero, or negative
- The oxidation number has to be an integer, as the number of electrons can only be an integer.
- The oxidation number cannot be fractional
- The oxidation number is the same as the oxidation state.
Frequently Asked Questions on Oxidation Number
Q1
What is the oxidation number of alkali and alkaline earth metals?
The oxidation number of alkali and alkaline earth metals are +1 and +2, respectively.
Q2
What is the oxidation number of oxide ions?
The oxidation number of oxide ions is -2.
Q3
What is the oxidation number of hydrogen in HCl?
The oxidation number of hydrogen in HCl is +1.