Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula \(\mathrm{M}_{3} \mathrm{O}_{4}\) are examples of mixed valence compounds in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format \(\mathrm{MO} \cdot \mathrm{M}_{2} \mathrm{O}_{3},\) to permit estimation of the metal's two oxidation states.) (a) \(\mathrm{Sc}_{2} \mathrm{O}_{3}\) (b) \(\mathrm{TiO}_{2}\) (c) \(\mathrm{V}_{2} \mathrm{O}_{5}\) (d) \(\mathrm{CrO}_{3}\) (e) \(\mathrm{MnO}_{2}\) (f) \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) (g) \(\operatorname{Co}_{3} \mathrm{O}_{4}\) (h) NiO (i) \(\mathrm{Cu}_{2} \mathrm{O}\)

Transition metal oxides are compounds that feature at least one oxygen atom bound to a transition metal. These metals have a characteristic ability to exist in multiple oxidation states, meaning they can lose different numbers of electrons when forming compounds. For example, iron can be found in the form of Fe²⁺ or Fe³⁺, as seen in the compound Fe₃O₄, which displays mixed valence. This duality within a single compound illustrates the concept of mixed valency, which is prevalent in several transition metal oxides. Understanding the valence and the ability of transition metals to exhibit various oxidation states is critical for interpreting the chemical and physical properties of these oxides.

In the context of the exercise, the metal ions are present in different oxidation states, and this is especially demonstrated in compounds with the formula M₃O₄. This kind of formula indicates the presence of both divalent and trivalent metal ions. By expressing this formula as MO · M₂O₃, you can more easily assign and calculate the individual oxidation states for each metal ion within the compound.

The valence of a compound refers to the combining power of its elements, based on the number of electrons that can be donated or accepted in a chemical bond. In general, the valence corresponds to the oxidation state of the element when forming compounds. For example, an oxidation state of +2 (often referred to as a valence of 2) implies that the element can give up two electrons.

To find the oxidation state, particularly in transition metal oxides, you typically count the number of electrons lost or gained by the metal to form a stable compound. Oxygen most often has an oxidation state of -2, which is taken as a starting point for calculations. Using the overall charge neutrality of compounds, which must be zero, the oxidation states of the metal ions can be determined through a simple algebraic process, as demonstrated by the step-by-step solution in our exercise.

Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is followed. This law states that the mass of the reactants in a chemical reaction must be equal to the mass of the products.

To balance a chemical equation, you must make certain that the number of atoms of each element on the reactants' side is equal to the number of atoms of that element on the products' side. In the context of oxidation states, balancing involves ensuring that the sum of all oxidation states in a chemical formula equals zero for neutral compounds or equals the overall charge for polyatomic ions. This balancing act helps in determining the correct stoichiometry of the compounds involved in the reaction. As highlighted in the exercise, by knowing that oxygen has a -2 oxidation state, we can deduce the oxidation state of transition metals in their oxide forms by balancing the equation so that the sum of oxidation states equals zero—a technique crucial for understanding complex chemical reactions.