Question

question 3. drag and drop the correct formula for the ionic compounds formed between the given cations and anions.
a) copper(i) ion and sulfide ion

b) iron(ii) ion and nitride ion

c) zinc ion and fluoride ion

d) aluminum ion and sulfide ion

:: cus :: cu₂s₂ :: cus₂
:: cu₂s :: fe₂n₃
:: fe₃n₂ :: fen₂ :: fe₃n
:: znf₂ :: zn₂f :: znf
:: zn₂f₂ :: al₂s₃
:: al₃s₂ :: als₃ :: al₂s

Explanation:

Step 1: Analyze Copper(I) and Sulfide Ion

Copper(I) has a charge of \( +1 \) (\( \text{Cu}^+ \)) and sulfide ion has a charge of \( -2 \) (\( \text{S}^{2-} \)). To balance charges, we need 2 \( \text{Cu}^+ \) for 1 \( \text{S}^{2-} \), so the formula is \( \text{Cu}_2\text{S} \).

Step 2: Analyze Iron(II) and Nitride Ion

Iron(II) has a charge of \( +2 \) (\( \text{Fe}^{2+} \)) and nitride ion has a charge of \( -3 \) (\( \text{N}^{3-} \)). The least common multiple of 2 and 3 is 6. So we need 3 \( \text{Fe}^{2+} \) (total \( +6 \)) and 2 \( \text{N}^{3-} \) (total \( -6 \)), giving the formula \( \text{Fe}_3\text{N}_2 \).

Step 3: Analyze Zinc Ion and Fluoride Ion

Zinc ion has a charge of \( +2 \) (\( \text{Zn}^{2+} \)) and fluoride ion has a charge of \( -1 \) (\( \text{F}^- \)). We need 2 \( \text{F}^- \) to balance \( +2 \) from \( \text{Zn}^{2+} \), so the formula is \( \text{ZnF}_2 \).

Step 4: Analyze Aluminum Ion and Sulfide Ion

Aluminum ion has a charge of \( +3 \) (\( \text{Al}^{3+} \)) and sulfide ion has a charge of \( -2 \) (\( \text{S}^{2-} \)). The least common multiple of 3 and 2 is 6. So we need 2 \( \text{Al}^{3+} \) (total \( +6 \)) and 3 \( \text{S}^{2-} \) (total \( -6 \)), giving the formula \( \text{Al}_2\text{S}_3 \).

Answer:

a) \( \boldsymbol{\text{Cu}_2\text{S}} \)
b) \( \boldsymbol{\text{Fe}_3\text{N}_2} \)
c) \( \boldsymbol{\text{ZnF}_2} \)
d) \( \boldsymbol{\text{Al}_2\text{S}_3} \)