Final answer:
The ionic compound formed between gold(III) and fluoride consists of one gold(III) ion and three fluoride ions to balance the charges, making option b (1 gold(III), 3 fluoride) the correct answer.
Step-by-step explanation:
When gold(III) forms an ionic compound with fluoride, it does so in a way that the charges of the ions are balanced. Gold(III), indicated by the Roman numeral III, has a charge of +3. Each fluoride ion has a charge of -1. To achieve electrical neutrality, we need three fluoride ions to balance the +3 charge of one gold(III) ion. Therefore, the correct ratio is 1 gold(III) to 3 fluoride ions, making option b the correct answer.
The resulting compound is gold(III) fluoride, which has the formula AuF3. In an ionic crystal structure, such as iron(III) chloride (FeCl3), there are three times as many chloride ions as iron(III) ions. Similarly, in gold(III) fluoride, there would be three fluoride ions for every gold(III) ion. Option b