Practice Problem 1

Determine the oxidation number of each element in the following compounds:

(a) BaO₂     (b) (NH₄)₂MoO₄      (c)Na₃Co(NO₂)₆      (d) CS₂

Solution

(a) If the oxidation number of the oxygen in BaO₂ were -2, the oxidation number of the barium would have to be +4.  But elements in Group IIA can't form +4 ions.  This compound must be barium peroxide, [Ba²⁺][O₂^2-].   Barium therefore is +2 and oxygen is -1.

(b) (NH₄)₂MoO₄ contains the NH⁴⁺ ion, in which hydrogen is +1 and nitrogen is -3.  Because there are two NH⁴⁺ ions, the other half of the compound must be an MoO₄^2- ion, in which molybdenum is -6 and oxygen is -2.

(c) Sodium is in the +1 oxidation state in all of its compounds.  This compound therefore contains the Co(NO₂)₆^3- complex ion.  This complex ion contains six NO^2- ions in which the oxidation number of nitrogen is +3 and oxygen is -2.  The oxidation state of the cobalt atom is therefore +3.

(d) The most elecronegative element in a compound always has a negative oxidation number.  Since sulfur tends to form -2 ions, the oxidation number of the sulfur in CS₂, is -2 and the carbon is +4.