Assigning Oxidation Numbers to Chemical Formulae

backyardchem Aug 15, 2007 Hobbies
This tutorial is for introductory chemistry students. Learn the five easy steps to assigning oxidation numbers to atoms in chemical formulae. Also, I will give you some problems that teachers love to give on tests because they are "tricky".

What are oxidation numbers?

 

If you are just trying to pass your chemistry class, then the truth is it doesn't matter much. You can do all of the calculations without understanding the general concept. But for those purists out there, here you go.

 

Definition: An oxidation number is the net charge an atom in a chemical species would have if it was completely ionic.

 

So if each atom completely "owned" the electrons closest to them and no electron sharing went on at all, then the atom's charge would be equal to its oxidaiton number. Since covalent bonding (electron sharing) does frequently occur, oxidation numbers often do not reflect that actual nature of bonding and hence are mainly useful for chemistry "bookeeping".

 

The Five Rules

 

The likelihood that your teacher will place this concept on a test is colored in parentheses.

  1. Fluorine always is -1. (very likely)
  2. Oxygen usually is -2. (very likely)
    • In peroxides, it is -1. (likely)
      • Common test problems: H2O2, Na2O2, K2O2
      • Uncommon test problems: Li2O2, Rb2O2, Cs2O2, O2-2
    • In superoxides, it is -1/2. (unlikely)
      • Possible test problems: NaO2, KO2, RbO2, CsO2, O2-1
    • Two oxy-fluorine compounds are exceptions. (unlikely)
      • +1: O2F2
      • +2: OF2
  3. Hydrogen usually is +1. (very likely)
    • In hydrides, it is -1. Hydrides occure when hydrogen is written at the end of the forumla. (likely)
      • Common test problems: LiH, NaH, NH3, PH3, CH4, LiAlH4, NaBH4, SiH4
      • Uncommon test problems: KH, RbH, CsH, AlH3, B2H6, AsH3
  4. Elements by themselves are +0.
  5. The total sum of the oxidation numbers in an ion equals the charge of the ion. The total sum of the oxidation numbers in a neutral compound equals 0.

Some Practice (Also Very Common Test Problems)

 

  • KMnO4
  • Use rules 2 and 5.
  • K = +1, Mn = +7, O = -2
  • K = +1 because of the K+ ion
  • O = -2 because of rule 2.
  • Solve for Mn using rule 5. You know the entire compound must equal 0. So you have 4 Os so that is equal to -8. Then you have 1 K which is +1. That makes -7. Mn therefore must = +7 so that the whole compound equals 0.

  • S8
  • Use rule 8.
  • S = 0
  • S = 0 because it is an element by itself

  • K2Cr2O7
  • Use rules 2, 3, and 5.
  • K = +1
  • O = -2
  • Cr = +6
  • K = +1 because of the K+ ion
  • O = -2 because of rule 2.
  • Solve for Cr using rule 5. 2 Ks make +2.  7 Os make -14

  • NaHSO4
  • Use rules 2, 3, and 5.
  • Na = +1
  • H = +1
  • O = -2
  • S = +6
  • Na = +1 because of the Na+ ion.
  • H = +1 because of rule 3.
  • O = -2 because of rule 2.
  • Use rule 5 to solve for S. +1 from Na plus the +1 from H plus the four -2s from O equals -6. So S must be equal to +6 so that the whole compound can be equal to 0.
What did you think of this tutorial?
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2 CommentsAdd a Comment
N2VPI on Jan 11, 2008
 
I found this tutorial under "Hobbies". Chemistry and Physics and Mathematics are the Essences of the Universe.
ChemKid on Jan 5, 2008
 
Pretty sure the oxidation number for hydrogen in ammonia (NH3) is +1, due to the larger electronegativity of Nitrogen. Same for Methane (CH4) correct me if I'm wrong, but that is how I learned it. p.s. good tutorial :)
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