A sample of a metal oxide has the formula M _ 0.75 O . In this oxide, the metal M exists in three different oxidation states: +2 , +3 , and +4 . If the number of M ⁴⁺ ions is twice the number of M ³⁺ ions in the sample, the percentage of metal ions existing in the +2 oxidation state is

Step-by-step solution

Let the number of M ³⁺ ions be y . Given that the number of M ⁴⁺ ions is twice the number of M ³⁺ ions, the number of M ⁴⁺ ions is 2y . The total amount of metal M per oxygen atom is 0.75 . Thus, the number of M ²⁺ ions = 0.75 - (y + 2y) = 0.75 - 3y . Applying the principle of electrical neutrality (total positive charge equals the total negative charge of one O ²⁻ ion): 2(0.75 - 3y) + 3(y) + 4(2y) = 2 1.5 - 6y + 3y + 8y = 2 1.5 + 5y = 2 5y = 0.5 y = 0.1 The number of M ²⁺ ions = 0.75 - 3(0.1) = 0.45 . The percenta