Final answer:
The oxidation state of niobium (Nb) in the compound Nb(AsO₄)₂ is +6, calculated by balancing the oxidation states of the arsenate ions and considering the neutral charge of the compound.
Step-by-step explanation:
To determine the oxidation state of niobium (Nb) in Nb(AsO₄)₂, one must first recognize that oxygen has a typical oxidation state of -2. Arsenate (AsO₄)⁻ is a polyatomic ion where arsenic has an oxidation state of +5. With four oxygen atoms, the total oxidation state for oxygen in one arsenate ion is -8. However, the arsenate ion has a charge of -3, which tells us that arsenic's oxidation contributes +5 to this total (since -8 from oxygen plus +5 from arsenic equals the -3 charge of the ion).
Since we have two arsenate ions in Nb(AsO₄)₂, the total oxidation state for arsenate is -6 because 2 x (-3) = -6. The compound is neutral, so the sum of all oxidation states must be zero. Therefore, the oxidation state of niobium must balance the -6 from the arsenate. This leads us to the equation: x (oxidation state of Nb) + (-6) = 0, solving for x gives us an oxidation state of +6 for niobium in this compound.