Oxidation Numbers - Introduction (Edexcel International A Level Chemistry)

• There are three definitions of oxidation and reduction  used in different branches of chemistry
• Oxidation and reduction can be used to describe any of the following processes

Definitions and Examples of Oxidation & Reduction

Use the acronym "Oil Rig" to help you remember the definitions of oxidation and reduction

Oxidation Number

• The oxidation number of an atom is the charge that would exist on an individual atom if the bonding were completely ionic
• It is like the electronic ‘status’ of an element
• Oxidation numbers are used to
  • Tell if oxidation or reduction has taken place
  • Work out what has been oxidised and/or reduced
  • Construct half equations and balance redox equations

Oxidation Numbers of Simple Ions

Answers:

a) 0     b) +3    c) +2   

d) -2    e) 0      f) +3

 

• So, in simple ions, the oxidation numbers of the atom is the charge on the ion:
  • Na⁺, K⁺, H⁺ all have an oxidation number of +1
  • Mg²⁺, Ca²⁺, Pb²⁺ all have an oxidation number of +2
  • Cl^–, Br^–, I^– all have an oxidation number of -1
  • O^2-, S^2- all have an oxidation number of -2

Oxidation Number Rules

• A few simple rules help guide you through the process of determining the oxidation number of any element
• Remember, you are determining the oxidation number of a single atom
• The oxidation number (ox.no.) refers to a single atom in a compound

Oxidation Number Rules Table

Roman Numerals 

• Transition metals are characterised by having variable oxidation numbers.
• Oxidation numbers can be used in the names of compounds to indicate which oxidation number a particular element in the compound is in
• Where the element has a variable oxidation number, the number is written afterwards in Roman numerals.
• This is called the STOCK NOTATION (after the German inorganic chemist Alfred Stock), but is not as widely used for non-metals, so SO₂ is often referred to as sulfur dioxide rather than sulfur(IV) oxide
• For example, iron can be both +2 and +3 so Roman numerals are used to distinguish between them
  • Fe²⁺ in FeO can be written as iron(II) oxide
  • Fe³⁺ in Fe₂O₃ can be written as iron(III) oxide
• More complicated examples include other atoms / ions as part of the formula 
  • Potassium manganate(VII) implies that the manganese is in a +7 oxidation
  • Potassium manganate(VII) contains the potassium ion K⁺ and the manganate ion MnO₄^– 
    • Since the oxygen in the manganate ion is in the -2 oxidation state, there is a total of -8 from the oxygen
    • The manganate ion has an overall -1 charge, which means that the manganese ion must be in the +7 oxidation state

Molecules or Compounds

• In molecules or compounds, the sum of the oxidation numbers on the atoms is zero

Oxidation Number in Molecules or Compounds Table

• Because CO₂ is a neutral molecule, the sum of the oxidation states must be zero
• For this, one element must have a positive oxidation number and the other must be negative

How do you determine which is the positive one?

• The more electronegative species will have the negative value
• Electronegativity increases across a period and decreases down a group
• O is further to the right than C in the periodic table so it has the negative value

How do you determine the value of an element's oxidation number?

• From its position in the periodic table and / or the other element(s) present in the formula

Hydrogen in metal hydrides, H^-

• An example of a metal hydride is sodium hydrogen, NaH which is a neutral compound 
• The hydrogen atom is present as a hydride ion 
  • This has the symbol H^- and therefore the oxidation state -1
• We can also think of this as a sum
  • Since group 1 metals always have an oxidation state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0)

Oxygen in peroxides, O^-

• Peroxides include hydrogen peroxide, H₂O₂, which again is a neutral compound 
• The sum of the oxidation states of the hydrogen and oxygen must be 0 
• Since each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to accommodate the neutral charge of the compound

Ions

• The oxidation number of elements within an ion can be calculated using the rules shown in the Oxidation Number Rules Table (above)

Answers

• • In all of the answers, the O is more electronegative than the other element, so the oxidation number of O will be –2

Answer 1:

• • The nitrite ion contains two oxygen atoms
    • 2 x –2 = –4
  • The overall ion has an oxidation state of –1, which means that the oxidation numbers of the elements in the ion must add up to –1
  • Therefore, the oxidation number of N is +3
    • N + (2 x –2) = –1
      N = +3

Answer 2:

• • The nitrate ion contains three oxygen atoms
    • 3 x –2 = –6
  • The overall ion has an oxidation state of –1, which means that the oxidation numbers of the elements in the ion must add up to –1
  • Therefore, the oxidation number of N is +5
    • N + (3 x –2) = –1
      N = +5

Answer 3:

• • The thiosulfate ion contains three oxygen atoms
    • 3 x –2 = –6
  • The overall ion has an oxidation state of –2, which means that the oxidation numbers of the elements in the ion must add up to –2
  • The oxidation number of both S atoms is +4
    • 2S + (3 x –2) = –2
      2S = +4
  • Therefore, the oxidation number of S is +2
    • 2S = +4
    • S = +2

Answer 4:

• • The tetrathionate ion contains six oxygen atoms
    • 6 x –2 = –12
  • The overall ion has an oxidation state of –2, which means that the oxidation numbers of the elements in the ion must add up to –2
  • The oxidation number of all four S atoms is +10
    • 4S + (6 x –2) = –2
      4S = +10
  • Therefore, the oxidation number of S is +2.5
    • 4S = +10
    • S = 2.5
  • This specific question is a rare example showing a fractional oxidation number for the element in question / sulfur
    • This is because the sulfur atoms are in two different environments 
      
    • The fractional oxidation number shows the average oxidation number of the sulfur environments