Oxidation Numbers and Oxidation States Chemistry Tutorial

• An oxidation number is used to indicate the oxidation state of an atom in a compound.
• An oxidation number is defined as the charge an atom would carry if the molecule or polyatomic ion were completely ionic.

  When calculating the oxidation number of an element in a compound, treat all the elements present as if they are present as ions, EVEN if they are clearly part of a covalent molecule.

• The oxidation state of an atom can be represented as an oxidation number in either of two ways:

  	IUPAC recommended (1) oxidation number	Alternative oxidation number
  General form:	Roman numeral	Arabic number
  Positive oxidation state examples:	I	+1
  	II	+2
  	III	+3
  Oxidation state of zero:	0	0
  Negative oxidation state examples:	-I	-1
  	-II	-2
  	-III	-3

  NOTE: the sign (+ or -) is always located to the left of the number in an oxidation number.

• There are different methods for assigning oxidation numbers to atoms in compounds.⁽²⁾
  The table below lists 4 rules that can be used to assign oxidation numbers to atoms in compounds.
  

  Rules for Assigning Oxidation Numbers
  Rule 1:	The oxidation number of the atoms making up an element in the free state, including its standard state(3), is 0.
  Rule 2:	The oxidation number for the ion of an element is equal to the charge on the ion (its charge number).
  Rule 3:	When found in compounds, the oxidation state of some elements are always the same (with a few exceptions) so their oxidation numbers will always be the same. These elements and their oxidation numbers are listed below: Element (in compound) oxidation number Exceptions to this Rule IUPAC alternative Group 1 element I +1   Group 2 element II +2   hydrogen I +1 metal hydrides (eg NaH), oxidation number of hydrogen is -I or -1 fluorine -I -1   oxygen -II -2 (a) peroxides, O22- [IUPAC name: dioxide(2-)], eg H2O2 oxidation number of oxygen is -I or -1 (b) superoxides, O2- [IUPAC name: dioxide(1-)], eg KO2 oxidation number of oxygen(4) is -½ (c) oxygen fluorides: ⚛ in OF2 the oxidation number of oxygen is II or +2 ⚛ in O2F2 the oxidation number of oxygen is I or +1
  Rule 4:	The oxidation numbers of other elements in compounds can be calculated using the overall charge on the compound: (a) For a compound or molecule with no overall net charge(5), the sum of the oxidation numbers for each atom of each element in the molecule equals 0. (b) For a charged molecule (a polyatomic ion), the sum of the oxidation numbers for each atom of each element in the ion equals the charge on the ion (the ion's charge number).

• Unless stated otherwise, assume that all the atoms of the same element in a molecular formula have the same oxidation number.⁽⁶⁾
• An oxidation number, or oxidation state, is assigned to help us determine whether or not an element in a reaction has been oxidised or reduced.

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What is an Oxidation Number?

Before we consider what an oxidation number is, it is worth recalling what an ion is, and what a charge number is.

Consider an atom of oxygen, O, which has 6 valence electrons.
It has a strong tendency to pull 2 electrons towards itself in order to complete its octet of electrons (it is said to be electronegative). If the oxygen atom gains 2 electrons it forms an oxide ion, O^2-, with a charge of 2- (charge number is 2-).
This happens when our oxygen atom ecounters an atom that is much, much less electronegative (that is, an atom that is more electropositive), such as an atom of magnesium, Mg.
If we react oxygen gas and magnesium metal together in an oxidation reaction then the oxygen atom pulls off 2 of the magnesium atom's electrons forming the magnesium ion, Mg²⁺, and the oxide ion, O^2-.
Together they make up the compound magnesium oxide in which the ratio of Mg²⁺ to O^2- is 1:1 so the compound can be represented as Mg²⁺O^2- but is usually given as a "molecular" formula, MgO.
Note that the compound, Mg²⁺O^2- or MgO, has no overall net charge because the number of positive charges is equal to the number of negative charges in the compound. The charge number of the compound is 0.

But what happens if our oxygen atom encounters another atom that is not capable of pulling the electrons off the oxygen atom?
An atom like nitrogen for example.

Imagine that we could react nitrogen gas and oxygen gas together in an oxidation reaction to produce new compounds which are oxides of nitrogen.
These new compounds are all covalent, no ions are present.
These new compounds are all electrically neutral, there is no net overall charge, the charge number of each compound is 0.
Consider these possible compounds:

• NO (nitrogen monooxide)
• NO₂ (nitrogen dioxide)

Nitrogen monooxide (NO) is the least oxidised form of nitrogen, it contains the lowest number of oxygen atoms. We can say that the nitrogen atom in NO is in a lower state of oxidation, a lower oxidation state.
Nitrogen dioxide (NO₂) is the most oxidised form of nitrogen, it contains the greatest number of oxygen atoms. We can say that the nitrogen atom in NO₂ is in the highest state of oxidation, the highest oxidation state.

But wait, what about N₂O (dinitrogen oxide)? Is the nitrogen atom in N₂O less oxidised than NO or more oxidised?

In order to answer that question we will use the concept of an oxidation number to indicate the oxidation state (state of oxidation) of each atom of nitrogen in each molecule above.

An oxidation number is kind of "account keeping", helping us to understand how electrons (negative charges) are distributed within a molecule.

We start by saying, "Let's pretend that each of these covalent compounds with no net overall charge is actually ionic, that is, in each compound we will pretend that the oxide ion, O^2-, is present. Then we can work out what the charge on each of these pretend 'nitrogen ions' would be".

In this scenario, NO is an electrically neutral molecule, it has zero overall charge, charge number = 0
An oxide ion, O^2- has a charge number of 2-
So what "charge" must there be on a "nitrogen ion" to balance the charge on the oxide ion?

Let's call the unknown charge x.

For the molecule to have a net charge of 0 the sum of the charges on all the "anions" and "cations" must be 0

0 = charge on oxide ion + charge on "nitrogen ion"

Substitute 2- for the charge on the oxide ion:

0 = 2- + x

Add 2+ to both sides of the equation:

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

2+ = x

If the nitrogen atom carried a charge then the charge would be 2+.
But this is NOT an ionic compound, so the nitrogen atom DOES NOT carry a charge.
Instead we refer to the nitrogen atom as being in an oxidised state (NOT a charged state), and we refer to the "apparent charge" on the atom as its oxidation number.
Now clearly we can't use 2+ to indicate the oxidation number because this would indicate the charge number on an ion, so, we use either:

• a Roman numeral (no sign required for a positive oxidation state, minus sign, -, required before the Roman numeral for a negative oxidation number)
• an Arabic number (+ or - sign located before the number)

Some Roman numerals and their corresponding Arabic numbers are given in the table below:

We need to convert the "charge" on the nitrogen atom calculated above (2+) to an oxidation number:

• Roman numeral II (no sign is required for a positive oxidation number)
• Arabic number: + sign precedes the number, that is, +2

The oxidation number of nitrogen in NO is II (or, alternatively, +2)

Since the compound NO is NOT ionic, we should never have referred to the oxygen atom in the molecule as an "oxide ion" with a charge number of 2-
But we CAN refer to each oxygen atom in the compound as having an oxidation number of -II (or, alternatively, -2).

Let's calculate the oxidation number of nitrogen in the most oxidised form, NO₂.
NO₂ is a covalent molecule with no overall net charge, charge number = 0
Let each oxygen atom have an oxidation number of -II (we will use -2 in the calculations)
Let x = oxidation number of the nitrogen atom

Write an expression for calculating the oxidation number of the nitrogen atom:

charge number of molecule = total of the oxidation numbers for oxygen + total of the oxidation numbers for nitrogen

Since there are 2 oxygen atoms and 1 nitrogen atom in the molecular formula:

charge number of molecule = 2 × oxygen's oxidation number + nitrogen's oxidation number

Substitue the oxidation numbers into the equation:

0 = (2 × -2) + x

0 = -4 + x

Add +4 to both sides of the equation:

+4 + 0 = +4 +-4 + x

+4 = x

The oxidation number of nitrogen in NO₂ is IV (or, alternatively, +4)

Note that NO is a less oxidised form of nitrogen than NO₂, because :

• it contains less oxygen per atom of nitrogen
  (so nitrogen's valence electrons experience less pull from the oxygen atoms)
• nitrogen has a lower oxidation number than in NO₂

Now, what about N₂O?

The oxidation number of nitrogen in N₂O is I (or, alternatively, +1), which is less than the oxidation number of nitrogen in NO (which is II, or, +2), so N₂O is a less oxidised form of nitrogen than NO.

You will see two different methods used to indicate the oxidation state of an atom in a compound as set out below:

• IUPAC recommended oxidation numbers use Roman numerals.

  (a) If the oxidation state is positive, it is indicated by an oxidation number using the Roman numeral ONLY (no + sign), for example, I, II

  (b) If the oxidation state is negative, it is indicated by an oxidation number using a minus sign (-) immediately before the Roman numeral, for example, -I, -II

  (c) If the oxidation state is zero, it is indicated by an oxidation number using the arabic number 0

• Alternative method for indicating the oxidation state of an atom uses signs and arabic numbers for the oxidation number (it is easier to use these Arabic numbers in calculations):

  (a) If the oxidation state is positive it is indicated by an oxidation number using a plus sign (+) immediately before the arabic number, eg, +1, +2

  (b) If the oxidation state is negative it is indicated by an oxidation number using a minus sign (-) immediately before the arabic number, for example, -1, -2

  (c) If the oxidation state is zero it is indicated by an oxidation number using the arabic number 0

We can use the concept of an oxidation number even when there is no oxygen present in a molecule!
We can apply the concept of oxidation numbers to the atoms in hydrides, fluorides, chlorides, bromides, iodides, nitrides, sulfides, phosphides, etc
But before we can do this, we will need to learn the rules for assigning oxidation numbers ...

How to Apply the Rules to Find the Oxidation Number to Represent the Oxidation State of an Element

First, learn the following 4 rules:

Then, follow the steps given below to calculate the oxidation number of an element in a compound.

Note: Unless stated otherwise, assume that all the atoms of the same element in a molecular formula have the same oxidation number.

The best way of learning how to do this is to do some examples .....

Worked Examples: Calculating Oxidation Numbers for Elements in Electrically Neutral Species

Question 1: Find the oxidation number that represents the oxidation state of fluorine in F₂.

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

   Calculate the oxidation number of fluorine

2. What data (information) have you been given in the question?

   Extract the data from the question:

   formula of a molecule: F_2(g)

3. What is the relationship between what you know and what you need to find out?

   Step 1: Write the formula for the species (molecule, ion or compound).

   F₂

   Step 2: Assign a symbol to the overall charge on the species, and include its value if known.

   ON(F₂) = 0
   The F₂ molecule is electrically neutral, it has no net overall charge.

   Step 3: List the elements making up the species (molecule, ion or compound) and note how many atoms of this element are present.

   2 atoms of fluorine = 2 × F

   Step 4: Assign a symbol to represent the oxidation state of each element in the species.

   Let ON(F) = oxidation number of each atom of fluorine in the molecule

   Step 5: Write in any known values for the oxidation number for the elements above.

   ON(F) = ?

   Step 6: Write an equation using these symbols to show that the sum of the oxidation number for "atoms" of each element present is equal to the overall charge on the species.

   ON(F₂) = 2 × ON(F)
   Note that the F₂ molecule is made up of 2 F atoms and each atom must be assigned an oxidation number.

   Step 7:Substitute the known values into the equation:

   0 = 2 × ON(F)

4. Solve the equation to find the oxidation number of fluorine

   0 = 2 × ON(F)

   Divide both sides of the equation by 2:

   0 ÷ 2 = [2 × ON(F)] ÷ 2
   0 = ON(F)

   Oxidation number for fluorine = 0

5. Is your answer plausible?

   One of the Rules for Assigning Oxidation Numbers, Rule 1, states that the,

   "Oxidation number of the atoms of an element in the free state, including its standard state state, is 0"

   
   F_2(g) represents the free state (and standard state) for fluorine.
   So the oxidation number of fluorine in F₂ must be 0

   Since this agrees with the value we calculated above, we are confident that our answer is plausible.

6. State your solution to the problem "find the oxidation number for fluorine":

   Oxidation number of fluorine in F₂ is 0

Question 2: Find the oxidation number that represents the oxidation state of carbon in CO₂

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

   Calculate oxidation number of carbon (C)

2. What data (information) have you been given in the question?

   Extract the data from the question:

   Formula of a compound: CO₂

3. What is the relationship between what you know and what you need to find out?

   Step 1: Write the formula for the species (molecule, ion or compound).

   CO₂

   Step 2: Assign a symbol to the overall charge on the species, and include its value if known.

   ON(CO₂) = 0
   The CO₂ molecule is electrically neutral, it has no net overall charge.

   Step 3: List the elements making up the species (molecule, ion or compound) and note how many atoms of this element are present.

   1 atom of carbon = 1 × C

   2 atoms of oxygen = 2 × O

   Step 4: Assign a symbol to represent the oxidation number of each element in the species.

   Let the oxidation number of carbon be ON(C)

   Let the oxidation number of oxygen be ON(O)

   Step 5: Write in any known values for the oxidation number for the elements above.

   ON(C) = ?

   ON(O) = -II = -2
   (from the Rules, oxidation number of O is always -II (-2) except in peroxides, superoxides and oxygen fluorides)

   Step 6: Write an equation using these symbols to show that the sum of the oxidation number for "atoms" of each element present is equal to the overall charge on the species.

   ON(CO₂) = ON(C) + [2 × ON(O)]
   Note that the CO₂ molecule is made up of 1 C atom and 2 O atoms and each atom must be assigned an oxidation number.

   Step 7:Substitute the known values into the equation:

   0 = ON(C) + (2 × -2)

4. Solve the equation to calculate the oxidation number (state) of carbon

   ON(C) = 0 - (2 × -2)
   ON(C) = 0 - (-4)
   ON(C) = +4

   The oxidation number of carbon in CO₂ is IV (or +4)

5. Is your answer plausible?

   Oxygen is more electronegative than carbon, that is, oxygen has a greater power to attract the shared negatively charged electrons in the covalent bond towards itself than a carbon atom has, so we expect oxygen to have a negative value for its oxidation number while carbon will therefore have a positive value for its oxidation number.
   In this respect our answer is plausible.

   Work backwards by using your calculated value for the oxidation number of carbon, and your value for the oxidation number of oxygen as given by the "rules" to calculate the overall charge on the CO₂ molecule.
   Let ON(CO₂) = charge on the molecule
   and ON(O) = oxidation number of oxygen = -2
   and ON(C) = oxidation number of carbon = +4
   Then:
   ON(CO₂) = ON(C) + [2 × ON(O)]
   ON(CO₂) = +4 + (2 × -2) = +4 + (-4) = +4 - 4 = 0
   So CO₂ is found to have 0 overall charge.
   We are therefore reasonably confident that our answer is correct.

6. State your solution to the problem "find the oxidation number for carbon":

   Oxidation number of carbon is IV (or alternatively, +4)

Worked Examples: Calculating Oxidation Numbers for Elements in Charged Species (Ions)

Question 1: Determine the oxidation number of sulfur in S^2-.

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

   Calculate oxidation number for sulfur

2. What data (information) have you been given in the question?

   Extract the data from the question:

   formula of the ion: S^2-

3. What is the relationship between what you know and what you need to find out?

   Step 1: Write the formula for the species (molecule, ion or compound).

   S^2-

   Step 2: Assign a symbol to the overall charge on the species, and include its value if known.

   Let ON(S^2-) = charge on ion

   ON(S^2-) = -2

   Step 3: List the elements making up the species (molecule, ion or compound) and note how many atoms of this element are present.

   1 atom of sulfur = 1 × S

   Step 4: Assign a symbol to represent the oxidation number of each element in the species.

   Let ON(S) = oxidation number of sulfur in the ion

   Step 5: Write in any known values for the oxidation number for the elements above.

   ON(S) = ?

   Step 6: Write an equation using these symbols to show that the sum of the oxidation number for "atoms" of each element present is equal to the overall charge on the species.

   ON(S^2-) = ON(S)

   Step 7:Substitute the known values into the equation:

   -2 = ON(S)

4. Solve the equation to calculate the oxidation number

   ON(S) = -2 = -II

5. Is your answer plausible?

   Rule 2 states that "the oxidation number for the ion of an element is equal to the charge on the ion."
   S^2- is the sulfide ion, the ion of the element sulfur.
   So, the oxidation number of S^2- will be equal to the charge on the ion.
   Charge on S^2- is 2-
   Therefore oxidation number of sulfur in S^2- must be -2, or, -II

   Since this value agrees with that which we calculated above, we are reasonably confident that our answer is plausible.

6. State your solution to the problem "determine the oxidation number for sulfur":

   Oxidation number for sulfur in S^2- is -II (or, alternatively, -2)

Question 2: Determine the oxidation number for chromium in Cr₂O₇^2-

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

   Calculate the oxidation number of chromium

2. What data (information) have you been given in the question?

   Extract the data from the question:

   Formula of the polyatomic ion: Cr₂O₇^2-

3. What is the relationship between what you know and what you need to find out?

   Step 1: Write the formula for the species (molecule, ion or compound).

   Cr₂O₇^2-

   Step 2: Assign a symbol to the overall charge on the species, and include its value if known.

   Let ON(Cr₂O₇^2-) be the charge on the polyatomic ion
   ON(Cr₂O₇^2-) = -2

   Step 3: List the elements making up the species (molecule, ion or compound) and note how many atoms of this element are present.

   2 atoms of chromium = 2 × Cr
   7 atoms of oxygen = 7 × O

   Step 4: Assign a symbol to represent the oxidation number of each element in the species.

   Let ON(Cr) = oxidation number of each chromium atom

   Let ON(O) = oxidation number of each oxygen atom

   Step 5: Write in any known values for the oxidation number for the elements above.

   ON(Cr) = ?

   ON(O) = -II = -2
   (from the Rules, oxidation number of O is always -II (-2) except in peroxides, superoxides and oxygen fluorides)

   Step 6: Write an equation using these symbols to show that the sum of the oxidation number for "atoms" of each element present is equal to the overall charge on the species.

   ON(Cr₂O₇^2-) = [2 × ON(Cr)] + [7 × ON(O)]
   remember, each molecule of Cr₂O₇^2- is made up of 2 Cr atoms and 7 O atoms and each atom has an oxidation number

   Step 7:Substitute the known values into the equation:

   -2 = [2 × ON(Cr)] + [7 × -2]

4. Solve the equation to calculate the oxidation number

   -2 = [2 × ON(Cr)] + [7 × -2]
   -2 = 2ON(Cr) -14

   Collect like terms:
   -2 + 14 = 2ON(Cr)
   +12 = 2ON(Cr)

   Divide both sides of the equation by 2:
   +12 ÷ 2 = 2ON(Cr) ÷ 2
   +6 = ON(Cr) = VI

5. Is your answer plausible?

   Chromium is a transition metal so we expect it to be less electronegative than oxygen, therefore we expect chromium to have a positive oxidation number and for oxygen to have a negative oxidation number.
   On this basis, our answer is plausible.

   Work backwards by using our calculated value for the oxidation number of chromium and the known oxidation number for oxygen to see if we arrive at a charge of 2- for the polyatomic ion given in the question:
   ON(Cr) = +6
   ON(O) = -2
   ON(polyatomic ion) = (2 × +6) + (7 × -2)
       = +12 + -14 = -2
   Charge on the polyatomic ion = 2-

   Since this is the charge on the polyatomic ion given in the question, we are confident our answer is plausible.

6. State your solution to the problem "determine the oxidation number for chromium":

   Oxidation number of chromium in Cr₂O₇^2- is VI (or, alternatively, +6)