Give the formulas of compounds in which (a) the cation is \(\mathrm{Ba}^{2+}\), the anion is \(\mathrm{I}^{-}\) or \(\mathrm{N}^{3-}\). (b) the anion is \(\mathrm{O}^{2-}\), the cation is \(\mathrm{Fe}^{2+}\) or \(\mathrm{Fe}^{3+}\)

Short Answer

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a) Ba2+ with I- and N3- b) O2- with Fe2+ and Fe3+ Answer: The formulas of the compounds are: a) BaI2 and Ba3N2 b) FeO and Fe2O3

Step by step solution

01

Forming compounds with Ba2+

For the first compound, we have \(\mathrm{Ba}^{2+}\) as the cation and \(\mathrm{I}^{-}\) as the anion. In order to balance the charge, we need 2 iodide anions for each barium cation. Therefore, the formula for this compound is \(\mathrm{BaI}_{2}\). For the second compound, we have \(\mathrm{Ba}^{2+}\) as the cation and \(\mathrm{N}^{3-}\) as the anion. In order to balance the charge, we need 3 barium cations for every 2 nitride anions. Therefore, the formula for this compound is \(\mathrm{Ba}_{3}\mathrm{N}_{2}\).
02

Forming compounds with O2-

For the first compound, we have \(\mathrm{O}^{2-}\) as the anion and \(\mathrm{Fe}^{2+}\) as the cation. In order to balance the charge, we need one of each ion type. Therefore, the formula for this compound is \(\mathrm{FeO}\). For the second compound, we have \(\mathrm{O}^{2-}\) as the anion and \(\mathrm{Fe}^{3+}\) as the cation. In order to balance the charge, we need 2 oxide anions for each iron(III) cation. Therefore, the formula for this compound is \(\mathrm{Fe}_{2}\mathrm{O}_{3}\). In conclusion, the formulas of the compounds are: (a) \(\mathrm{BaI}_{2}\) and \(\mathrm{Ba}_{3}\mathrm{N}_{2}\) (b) \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2}\mathrm{O}_{3}\)

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