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Oxidation State – Rules for Calculating Oxidation State, Maximum and Minimum Oxidation States, Practice Problems and FAQ

We use the term oxidation state to denote the number of electrons lost or gained by a chemical compound. But have you ever wondered why we use the term ‘oxidation state’ and not ‘reduction state’?

To answer this interesting question we’ll have to rewind the clock to the late 18th century. Antoine Lavoisier, a French chemist and biologist, coined the term ‘oxidation’ to describe the reaction of oxygen with other elements. In his footsteps, Wendell Mitchell Latimer, an American chemist, introduced the concept of oxidation state in its current form in the year 1938.

On this concept page, we will get to know more about oxidation states!

• Oxidation State
• Rules for Calculating Oxidation State
• Maximum and Minimum Oxidation States
• Average Oxidation State
• Practice Problems
• Frequently Asked Questions – FAQ

Oxidation State

The oxidation state of an atom in a chemical compound sheds light on how many electrons were lost from it or gained by it. The oxidation state of an atom is the hypothetical charge that it would possess if all of its bonds with other atoms were fully ionic in character.

In a redox reaction, oxidation state represents the number of electrons an atom can gain or lose. Oxidation state is used to identify which species is oxidised and which is reduced in a redox reaction. If the oxidation state of an atom increases, it is oxidised. On the contrary, when the oxidation state of an atom decreases, it is reduced. Oxidation state can be denoted by integers. An atom or an ion’s oxidation state is not the real charge on the atom.

Rules for Calculating Oxidation State

We must comprehend and adhere to specific guidelines in order to determine the oxidation state of an atom in a compound.

1. Each atom in an element holds up a zero oxidation state in either its free or uncombined state. The oxidation state of each atom in H2,Cl2,Na,Mg,Al,Ca,O2,O3,P4,S8 is zero.
1. Ions with a single atom in them have an oxidation state equal to their actual charge.

For example, the oxidation state of iron in Fe2+ is +2.

1. The oxidation state is always (+1) when alkali metals (Li,Na,K,Rb,Cs) form a compound.
1. The oxidation state is always (+2) when alkaline earth metals (Be,Mg,Ca,Sr,Ba) form a compound.
1. Halide ions in their compounds, fluorine and other halogens have an oxidation state of (-1). Iodine, chlorine, and bromine all have positive oxidation states (+1,+3,+5,+7) when they are mixed with oxygen.
1. The oxidation state of oxygen is typically -2 in compounds. There are three exceptions.
1. Every oxygen atom in peroxides has an oxidation state of (-1).

For instance, in Na2O2, the oxidation state of oxygen is

2(+1)+2(x)=0

x=-1

1. Every oxygen atom in superoxides has an oxidation state of (-1/2).

For instance, in KO2, the oxidation state of oxygen is

+1+2(x)=0

2x=-1

x=-1/2

1. The oxidation state of oxygen when there is an oxygen-fluorine bond is positive.

For instance, the oxidation state of oxygen in oxygen difluoride O2F2

2(x)+(-2)=0

x=+1

1. Hydrogen has an oxidation state of (-1) when it is connected to metals and (+1) when it is connected to non-metals.

For example, the oxidation state of H in CaH2

+2+2(x)=0

x=-1

The oxidation state of H in H2O

2(x)+(-2)=0

x=+1

The algebraic sum must be equal to zero when the oxidation states of the constituent atoms of a compound are combined together.

For example, the oxidation state of Cr in K2Cr2O7

2(+1)+2x+7(-2)=0

2x-12=0; x=+6

When the oxidation states of the atoms that make up an ion are added together, the resulting algebraic total must equal the charge of the ion in the case of polyatomic ions.

For example, the oxidation state of Mn in MnO4-

x+4(-2)=-1

x-8=-1; x=+7

Maximum and Minimum Oxidation States

Oxidation is the process in which the oxidation state of a species increases, and reduction is the process in which the oxidation state of a species decreases. The lowest oxidation state observed is -4 for carbon in CH4 (methane) and the highest oxidation state observed is +9 for iridium in [IrO4]+ (tetraoxo iridium cation).

Average Oxidation State

Let us understand the concept of average oxidation state with the help of an example. In the structure of H2S4O6, there are four sulphur atoms present in the compound. So, the average oxidation state of sulphur can be calculated as

The oxidation state of the First sulphur S1x-2-2-1=0,

x=+5

The oxidation state of the second sulphur S2x=0 because the second sulphur is surrounded by the first and third sulphur and where there is bond with the same atom, the contribution from them taken to be zero. The oxidation state of the third sulphur S3x=0 because the third sulphur is surrounded by the second and fourth sulphur and where there is a bond with the same atom, the contribution from them taken be zero.

The oxidation state of the fourth sulphur S4x-2-2-1=0,

x=+5

Average oxidation state of S = 5+0+0+54 =2.5

Practice Problems

1. Pick the correct statement out of the following statements.

a. Iron has the same oxidation state in both FeCl3 and FeCl2
b. Fluorine has the same oxidation state in both OF2 and O2F2
c. Manganese has the same oxidation state in both KMnO4 and K2MnO4
d. Phosphorus has the same oxidation state in both H3PO4 and H3PO2.

1. Fe in FeCl3: x+3(-1)=0

x=+3

Fe in FeCl2: x+2(-1)=0

x=+2

Thus, the oxidation state of iron in FeCl3 and FeCl2 is not the same. Therefore, the statement given in option A is incorrect.

1. Fluorine is the most electronegative element in the periodic table. So, it always has -1 oxidation state in all its compounds.

F in OF2: +2+2x=0

x=-1

F in O2F2: 2(+1)+2x=0

x=-1

Thus, the oxidation state of fluorine in OF2 and O2F2 is the same. Therefore, the statement given in option B is correct.

1. Mn in KMnO4: (+1)+x+4(-2)=0

x=+7

Mn in K2MnO4: 2(+1)+x+4(-2)=0

x=+6

Thus, the oxidation state of manganese in KMnO4 and K2MnO4 is not the same. Therefore, the statement given in option C is incorrect.

1. P in H3PO4: 3(+1)+x+4(-2)=0

x=+5

P in H3PO2: 3(+1)+x+2(-2)=0

x=+1

Thus, the oxidation state of phosphorus in H3PO4 and H3PO2 is not the same. Therefore, the statement given in option D is incorrect.

Thus, the statements given in options A, C and D are correct, but the statement given in option B is correct.

So, option B is the correct answer.

2. Calculate the oxidation state of chlorine in ClO4-,ClO3-,ClO2-,ClO-.

a. +1, +3, +5, +7
b. +7, +5, +3, +1
c. -1, -1, -1, -1
d. +8, +6, +4, +2

Solution:

The oxidation state of Cl in ClO4-

x+4(-2)=-1

x=+7

The oxidation state of Cl in ClO3-

x+3(-2)=-1

x=+5

The oxidation state of Cl in ClO2-

x+2(-2)=-1

x=+3

The oxidation state of Cl in ClO-

x+1(-2)=-1

x=+1

Hence, the oxidation state of chlorine in ClO4-,ClO3-,ClO2-,ClO- is +7, +5, +3, +1 respectively.

So, option B is the correct answer.

3. Calculate the oxidation state of Rb in RbO2, Rb2O2, Rb2O.

a. +4, +2, +1
b. +2, +2, +2
c. +2, +4, +1
d. +1, +1, +1

Solution: Rubidium is an alkali metal and hence, has the tendency to donate an electron. Thus, it exists in +1 oxidation state.

In Rb2O2, oxygen exists in a peroxide form as O22-. The oxidation state of oxygen in peroxide is -1.

Thus, the oxidation state of rubidium in Rb2O2 is

2x+2(-1)=0

x= +1

In RbO2, oxygen exists in a superoxide form as O2-. The oxidation state of oxygen in superoxide is -12.

Thus, the oxidation state of rubidium in RbO2 is

x+2(-1/2)=0

x= +1

In Rb2O, oxygen exist in an oxide form as O2-. The oxidation state of oxygen in oxide is -2.

Thus, the oxidation state of rubidium in Rb2O is

2x+(-2)=0

x= +1

So, option D is the correct answer.

4. Calculate the average oxidation state of C in CH2FCOOH.

a. +4
b. +3
c. +2
d. +1

Solution: In the given structure of CH2FCOOH, there are two carbon atoms present in the compound. So, the average oxidation state of carbon can be calculated as

The oxidation state of the first carbon C1x-1+1+1=0,

x=-1

The oxidation state of the second carbon C2x-2-1=0,

x=+3

The average oxidation state of C = +3-12 = +1

So, option D is the correct answer.

1. Although oxygen is a very electronegative element, it possesses a positive oxidation state in the O-F bond. Why?
Answer: This is the only case in which oxygen has a positive oxidation state because fluorine is more electronegative than oxygen. Hence, oxygen has to lose electrons and possess a positive oxidation state.

2. What does oxidation mean?
Answer: Oxidation is the process of addition of oxygen or an electronegative element or group or the removal of hydrogen or an electropositive element or group. Oxidation also refers to the loss of electrons from the element in consideration or the increase in the oxidation state of the element in consideration.

Early on, the term "oxidation" was used to describe the reaction of an element with oxygen. The definition was later modified to include all processes in which electrons are lost, regardless of whether oxygen was present after it was discovered that the element loses electrons when it is oxidised.

3. Does fluorine exhibit a positive oxidation state?
Answer: Fluorine is the most electronegative element in the periodic table. Therefore, fluorine cannot have a positive oxidation state and always has a negative oxidation state.

4. Does oxygen have a role in oxidation always?