Final answer:
The charge (oxidation state) of the Niobium ion in Nb(AsO4)2 is +5. This is because each AsO4 ion contributes an oxidation number of -3, totaling -6 for two ions, and the charge must balance out to zero for the neutral compound.
Step-by-step explanation:
To determine the charge (oxidation state) of the Niobium ion in Nb(AsO4)2, we need to consider the oxidation states of the other elements in the compound. We know that the oxidation state of Oxygen is typically -2. Since there are four Oxygen atoms in one Arsenate ion (AsO4), it contributes a total of -8 in oxidation states.
Arsenic (As) typically has an oxidation state of +5 to balance the charge of the oxygen atoms in the Arsenate ion. As Nb(AsO4)2 is a neutral compound, the sum of all oxidation states must equal zero.
Each Arsenate ion (AsO4) contributes -3 to the total charge (5 from As and -8 from the four Os), so two Arsenate ions contribute -6 in total. Therefore, to balance the charge, the niobium (Nb) must have an oxidation state of +5:
+5 (Nb) + 2(-3) (AsO4) = 0
Nb + 2(-3) = 0
Nb = +6 (-1)
Nb = +5
Thus, the niobium ion has an oxidation state of +5 in Nb(AsO4)2.