CALCULATION ON OXIDATION NUMBER OF ELEMENTS?: RULES OF OXIDATION NUMBER

Oxidation number is the charge an element appears to carry. The oxidation number of an element in an ionic compound is the charge on its ion in that compound. The oxidation number of an element in a covalent compound is the charge the atom would have if the bonding were to be ionic taking into account the relative electronegative ties of the species present.

                  General Rule For Assigning Oxidation Numbers

1. The algebraic sum of the oxidation number in a covalent compound is zero. The algebraic sum of the oxidation numbers in an ion equals the charge on the ion. e.g.: H2O, sum of oxidation number=2 (+1)+(-2)=0. NH₄⁺, sum of oxidation number=(-3)+4 (+1) =+1.
2. Elements in their uncombined state (I. e standard states) have an oxidation number of zero. For example, the atoms of molecule (e.g. F2, Cl2, O2, H2, S8) or crystals  (e.g. Li, Ca, Au) are assigned oxidation number of zero.
3. oxygen has an oxidation number of -2 except in (a) F2O where it has a value of +2, (b) peroxides, such as H202 and Na202 where it has a value of -1. While superoxides like K02 have oxidation number of -1/2
4. Hydrogen has an oxidation number of +1 except in the metallic hydride such as NaH, where it has a value of -1. other examples include LiAlH4 and NaBH4
5. In all compounds, alkali metal is +1, alkali earth metal is +2, B, Al is +3.
6. Halogen is -1.
7. In a compound not covered by the above rules, the element of high electronegativity has a negative oxidation number and the element of low electronegative type has a positive oxidation number, for example,

                  ClF3; F=-1 while Cl=+3

                  CCl4; Cl=-1 while C=+4

                         Worked examples

calculate the oxidation number of the underlined atoms in the following compounds.

(a) Na2S4O6     (b) V0₂⁺ (c) CH3OH     (d) 0₂^2-      (e) [Al (H₂O)₆]3+

Solutions 

• Na2S4O6
• let x= oxidation number of S                            
•  2 (+1) +4x +6(-2)=0  
•  2 + 4x -12=0 
•   4x= 12-2    
•  4x = 10  
•  x = 10/4 =2.5

• V0₂⁺
•   let x = oxidation number of V                          
•  x+ 2 (-2)=+1                                                          
•    x-4=+1                                                                    
•   x =1+4                                                                   
•   x=+5

• CH3OH                                                                    
•  Let x= oxidation number of C                           
•  x+1 (+1) +3(-1)=0                                                 
•    x+1-3=0                                                                 
•   x-2=0                                                                       
•  x=+2

• 0₂^2-                                                                        
•   Let x= oxidation number of 0                            
•  2x =-2                                                                   
•   x= -2/2 = -1.

• [Al (H₂O)₆]3+
• Let x= oxidation number of Al
• following rule one, H2O=0, since it is a covalent compound
• x + 6(0) = +3
• x + 0 = 3
• x = +3

Flex Questions :

what is the oxidation number of chromium in the following compounds :

1.     Cr (H₂0)₆^3-

2.     Cr₂0₇^2-

^{3.   NaH (find the oxidation number of H)}

                            

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