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# Oxidation Number – Definition, Rules, Calculation, Examples

• Last Updated : 28 Nov, 2021

Redox is a chemical process that involves changing the oxidation states of atoms. The actual or formal transfer of electrons between chemical species is defined by redox reactions, which usually include one species (the reducing agent) suffering oxidation (losing electrons) while another species (the oxidising agent) experiences reduction (gains electrons). The chemical species that loses an electron is said to have been oxidised, whereas the chemical species that gains an electron is said to have been reduced. To put it another way:

The loss of electrons or a rise in the oxidation state of an atom, an ion, or specific atoms in a molecule is referred to as oxidation.

The gain of electrons or a drop in the oxidation state of an atom, an ion, or specific atoms in a molecule is referred to as reduction (a reduction in the oxidation state).

### Oxidation Numbers

We can identify the oxidised material, the reduced substance, and the reducing and oxidising agents in a chemical reaction. According to the notion of a redox reaction, such identification can be done by determining which material is transferring electrons to other compounds. This explanation in terms of electron loss and gain, on the other hand, only applies to ionic compounds with full electron transfer and excludes other reactions, such as:

N2+O2→2NO

H2+Cl2→2HCl

“When all other atoms in a molecule leave as ions, the oxidation number determines the leftover charge that an atom has or seems to have.”

The terms oxidation number and oxidation state are commonly interchanged. It’s because the periodic quality of electronegativity is based on the stock notation of oxidation numbers. By evaluating its surroundings, an atom in a molecule can assign a negative, positive, or zero oxidation number. Oxidation values can even be fractional in some circumstances.

Rules for assigning oxidation numbers to the elements:

Knowledge of the following criteria can be used to calculate the oxidation number of elements in compounds.

Rule 1: The total of all the atoms in a molecule’s oxidation numbers equals zero.

e.g. In KMnO4, the oxidation number of K is +1, the oxidation number of Mn is +7 and the oxidation number of oxygen is −2.

Rule 2: An atom’s oxidation number is always 0 in its most basic form.

e.g. The oxidation number of H ,O ,N ,P ,S ,Se ,Cu, Ag in their element forms is H2,O2,N2,P4,S8,Se8,Cu,Ag respectively, is zero.

Rule 3: Alkali metals (Li, Na, K, Rb, Cs) have an oxidation number of +1 in their compounds.

e.g. In NaCl, the oxidation number of Na is +1.

Rule 4: Alkaline earth metals (Be, Mg, Ca, Sr, Ba) usually have a +2 oxidation number in their compounds.

e.g. The oxidation number of Mg in MgO is +2.

Rule 5: Except in metal hydrides, the oxidation number of H in its compound is always +1.

e.g. In HCl, the oxidation number of H is +1 and in NaH (Sodium hydride), the oxidation number of H is −1.

Rule 6: The fluorine oxidation number is 1 in all its compounds.

e.g. In NaF, the oxidation number of F is −1.

Rule 7: Except for peroxide, superoxides, oxyfluorides, and ozonides, oxygen has an oxidation number of -2 in most of its oxides.

e.g. In Na2O, the oxidation number of O is −2.

Exceptional Cases:

• Peroxides: The oxidation number of oxygen in peroxides is 1.  Examples, H2O2 , Na2O2 .

The oxidation number of oxygen in fluorine compounds is +2. Examples F2O or OF2 , etc.

• Super oxides: The oxidation number of oxygen in super oxides is –1/2.
• Ozonide’s: Each oxygen atom in an ozonide has an oxidation number of –1/3.

Rule 8: The charge of an ionic compound is equal to the sum of all atoms oxidation states.

e.g. In SO2–4, the oxidation number of Sulphur is +6. The oxidation number of oxygen is −2.

Rule 9: Except for oxygen and fluorine, the maximum oxidation number of any element is equal to its group number.

e.g. The oxidation number of sulphur in H2S2O8, K2S2O8, S2O–28 and H2SO5 is +6 due to the presence of a peroxide bond.

Rule 10: It’s possible that not all atoms of the same element have the same oxidation number in certain compounds. We get the average result when we compute the oxidation number for that element in such components.

e.g. One sulphur atom in Na2S2O3 has an oxidation number of +6, whereas the other sulphur atoms have an oxidation number of -2. As a result, the average sulphur oxidation state number in Na2S2O3 is +2.

Rule 11: Carbon in organic molecules can have any oxidation number between -4 and +4.

e.g. in HCHO, the oxidation number of carbon is zero.

Rule 12: The common oxidation number of an element is equivalent to its group number from IA to IV A. The formula (Group number –8) gives the common oxidation number of any element from V A to VIII-A.

e.g.

• The oxidation number of I A group elements =+1.
• The oxidation number of II A group elements =+2.
• The oxidation number of III A group elements =+3.
• The oxidation number of IV A group elements =+4.
• The oxidation number of V A group elements =–3.
• The oxidation number of VI A group elements =–2.
• The oxidation number of VII A group elements =–1.
• The oxidation number of VIII A group elements =0.

Rule 13: C, N, P, and S have oxidation numbers of 4,–3,–3 and –2 in all carbides, nitrides, phosphides, and sulphides, respectively.

e.g. In Mg3N2, the oxidation number of Nitrogen is −3.

Rule 14: In all Metal carbonyls, the oxidation number of metals is zero.

e.g. In Ni(CO)4, the oxidation number of Ni is zero.

### Calculation of Oxidation Numbers

• Oxidation Number of Sulphur in H2SO4

Let the oxidation number of sulphur in H2SO4 be equal to X

The oxidation number of hydrogen =+1

The oxidation number of oxygen is =–2

2(+1)+(X)+4×(–2)=0

2+X–8=0

X = 8–2=+6.

The oxidation number of sulphur in H2SO4 is +6.

• Oxidation Number of Chromium in Cr2O2–7 ion.

The oxidation number of oxygen is =–2

The oxidation number of chromium =X

2X+7(–2)=–2

2X–14=–2

2X=–2+14=12

Therefore X=+6.

The oxidation number of chromium in Cr2O2–7 is +6.

• Oxidation Number of Nitrogen in Ammonium Nitrite

Ammonium Nitrite is an ionic compound containing NH+4 and NO–2 ions.

The oxidation number of nitrogen in NH+4 ion =–3

The oxidation number of nitrogen in NO2=+3.

Thus, one atom of Nitrogen in Ammonium Nitrite is in a–3 – oxidation state, while the other nitrogen atom is in a+3 oxidation state.

### Sample Questions

Question 1: What is the common oxidation number for inert gases?

Inert gases have an oxidation number of zero.

Question 2: What are the rules for oxidation numbers?

The rules for oxidation numbers are:

1. The total of all the atoms in a molecule’s oxidation numbers equals zero.
2. An atom’s oxidation number in its most basic form is always zero.
3. In their compounds, the oxidation number of alkali metals (Li, Na ,K ,Rb ,Cs) is always +1.
4. In their compounds, the oxidation number of alkaline earth metals (Be ,Mg ,Ca ,Sr ,Ba) is always +2.
5. Except in metal hydrides, the oxidation number of H in its compound is always +1.
6. Fluorine has an oxidation number of 1 in all of its compounds.
7. Except for peroxide, super oxides, oxyfluoride, and ozonides, the oxidation number of oxygen in most of its oxides is -2.
8. Carbon in organic molecules can have any oxidation number between -4 and +4.

Question 3: What is the oxidation number of Hg in amalgam.

Mercury amalgam has a zero oxidation number. Each element in an alloy or amalgam has an oxidation number of zero.

Question 4: What is the oxidation number of oxygen in O3 and in MgO?

The oxidation number of O in O3=0

The oxidation number of O in MgO=–2

Question 5: What is the Oxidation Number of Chromium in Cr2O2–7 ion?

Oxidation number of oxygen is =–2

Oxidation number of chromium =X

2X+7(–2)=–2

2X–14=–2

2X=–2+14=12

Therefore X=+6.

The oxidation number of chromium in Cr2O2–7 is +6.

Question 6: What is the Oxidation Number of Sulphur in S8?