Rules for Assigning Oxidation Numbers

1. Free Elements:

   • The oxidation number of an atom in its elemental form is 000.

   • Examples: O2\text{O}_2O2​, N2\text{N}_2N2​, S8\text{S}_8S8​, Na\text{Na}Na all have Ox. number=0\text{Ox. number} = 0Ox. number=0.

2. Monatomic Ions:

   • The oxidation number equals the ion's charge.

   • Examples: Na+=+1\text{Na}^+ = +1Na+=+1, Cl−=−1\text{Cl}^- = -1Cl−=−1, Mg2+=+2\text{Mg}^{2+} = +2Mg2+=+2.

3. Hydrogen:

   • In compounds: +1+1+1 (except in metal hydrides where it is −1-1−1, e.g., NaH\text{NaH}NaH).

4. Oxygen:

   • In most compounds: −2-2−2 (except in peroxides like H2O2\text{H}_2\text{O}_2H2​O2​, where it is −1-1−1, and in compounds with fluorine like OF2\text{OF}_2OF2​, where it is +2+2+2).

5. Alkali Metals (Group 1\text{Group 1}Group 1):

   • Always +1+1+1 in compounds.

   • Example: NaCl,K2O\text{NaCl}, \text{K}_2\text{O}NaCl,K2​O.

6. Alkaline Earth Metals (Group 2\text{Group 2}Group 2):

   • Always +2+2+2 in compounds.

   • Example: CaCl2,MgO\text{CaCl}_2, \text{MgO}CaCl2​,MgO.

7. Halogens (Group 17\text{Group 17}Group 17):

   • Usually −1-1−1, except when bonded to oxygen or another halogen with a higher electronegativity.

   • Example: Cl2O=+1\text{Cl}_2O = +1Cl2​O=+1 for Cl\text{Cl}Cl.

8. Neutral Compounds:

   • The sum of oxidation numbers equals 000.

   • Example: In H2O\text{H}_2\text{O}H2​O: 2(+1)+(−2)=02(+1) + (-2) = 02(+1)+(−2)=0.

9. Polyatomic Ions:

   • The sum of oxidation numbers equals the ion's charge.

   • Example: In SO42−\text{SO}_4^{2-}SO42−​: S=+6\text{S} = +6S=+6, 4×(−2)=−84 \times (-2) = -84×(−2)=−8; +6+(−8)=−2+6 + (-8) = -2+6+(−8)=−2.