Assigning Oxidation numbers

  1. The oxidation number for an atom of any free (uncombined) element is ZERO.
    • Examples: Na, Ca have zero oxidation number.
  2. The oxidation number of an element in self-combination is always ZERO.
    • Examples: H2, O2, P4have zero oxidation number.
  3. In most hydrogen containing compounds, oxidation number of hydrogen is + 1.
    • (Exception is when H combines with alkali metals or alkaline earth to form hydrides of metals such as: NaH, LiH, CaH2. Then, the oxidation number of H is-1).
  4. In compounds involving the alkali metals, the elements are assigned oxidation number of +1.
  5. In compounds involving the alkaline earth metals, the elements are asasigned oxidation number of +2.
  6. Oxygen is usually assigned an oxidation number of -2 for oxides. It has an oxidation number of -1 in peroxides (H2O2).
  7. Fluorine always has oxidation number of -1 in compounds. The other elements in that group is usually -1 in compounds with elements of low electronegativity.
  8. The sum of oxidation numbers of all the atoms in the formula for a neutral compound is ZERO.
  9. The sum of oxidation numbers of an ion or complex ion is the same as the charge on that ion.
  10. Negative oxidation number in compounds of two unlike atoms are assigned to the more electronegative atom.

    Examples: Oxidation numbers
    HCl H = +1, Cl = -1
    H2O H = +1, O = -2

  11. In combinations of nonmetals not involving hydrogen and oxygen, the nonmetal that is more electronegative is considered negative.

    Examples: Oxidation numbers
    CCl4 C = +4, Cl = -1
    SF6 S = +6, F = -1
    CS2 C = +4, S = -2

     

     

    Example 1. Assign oxidation numbers to each atom in SO2.

    Solution

    Step 1: Start with atoms which are known. We know oxygen is – 2 in compounds.

    Step 2: Solve for other atoms Since oxygen is –2, the total of the oxidation numbers for oxygen is –2×2 = –4. Remember, oxidation numbers are assigned per atom. The sum for SO2 must be zero, so S = +4

    So, S = +4 , O = –2

    Example 2. Assign oxidation numbers for each atom in K2Cr2O7.

    Solution

    Step 1: Start with atoms which are known.

    Each O is –2.
    Each K is +1.

    Step 2: Solve for other atoms.

    The total for all the oxygen atoms is –2 × 7 = –14.
    The total for the potassium atoms is +1 × 2 = +2.

    The total for the “molecule” must be zero. Therefore the total for all the Cr atoms is 14 – 2 = +12. If there are two chromium atoms, then each chromium atom is +12 ÷ 2 = +6.

    So, K = +1, Cr = +6, O = –2.

    Example 3. Assign oxidation numbers for each atom in Fe(NO3)3.

    Solution

    Step 1: Start with atoms which are known.

    Each O is –2.

    Step 2: Solve for other atoms.

    Both Fe and N are unknown, so let’s break the compound into its ions: Fe3+ and NO3.

    The oxidation number for Fe is equal to its charge, +3.

    The total for all oxygen atoms in the nitrate ion is -2×3 = –6.

    The sum of oxidation numbers for the nitrate ion must be –1.

    The oxidation number for nitrogen is 6 – 1 = +5.

    So, Fe = +3, N = +5, O = –2.