Question

writing chemical formulae (ii)

1. iron(ii) chloride
2. copper(i) sulfide
3. lead(iv) iodide
4. tin(ii) fluoride
5. mercury(i) bromide
6. tin(ii) oxide
7. chromium(iii) oxide
8. gold(i) iodide
9. manganese(ii) nitride
10. cobalt(iii) phosphide
11. iron(iii) chloride
12. copper(ii) sulfide
13. lead(ii) bromide
14. tin(iv) iodide
15. mercury(ii) fluoride
16. tin(iv) oxide
17. manganese(iii) chloride
18. chromium(ii) nitride
19. gold(iii) oxide
20. cobalt(ii) phosphide
21. tin(ii) sulfide
22. mercury(i) sulfide
23. gold(iii) bromide
24. manganese(ii) oxide
25. chromium(ii) chloride

Explanation:

Step1: Identify ion charges

Use Roman numerals for cation charge; recall common anion charges:

• Halides (Cl⁻, Br⁻, I⁻, F⁻): $1^-$
• Oxide ($O^{2-}$): $2^-$
• Sulfide ($S^{2-}$): $2^-$
• Nitride ($N^{3-}$): $3^-$
• Phosphide ($P^{3-}$): $3^-$
• Mercury(I) is $Hg_2^{2+}$ (special case)

Step2: Balance charge for neutrality

Cation total charge = Anion total charge; use subscripts to balance.

Step3: Write final formula

Simplify subscripts to smallest whole numbers (except $Hg_2^{2+}$).

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Answer:

1. $\text{FeCl}_2$
2. $\text{Cu}_2\text{S}$
3. $\text{PbI}_4$
4. $\text{SnF}_2$
5. $\text{Hg}_2\text{Br}_2$
6. $\text{SnO}$
7. $\text{Cr}_2\text{O}_3$
8. $\text{AuI}$
9. $\text{Mn}_3\text{N}_2$
10. $\text{CoP}$
11. $\text{FeCl}_3$
12. $\text{CuS}$
13. $\text{PbBr}_2$
14. $\text{SnI}_4$
15. $\text{HgF}_2$
16. $\text{SnO}_2$
17. $\text{MnCl}_3$
18. $\text{Cr}_3\text{N}_2$
19. $\text{Au}_2\text{O}_3$
20. $\text{Co}_3\text{P}_2$
21. $\text{SnS}$
22. $\text{Hg}_2\text{S}$
23. $\text{AuBr}_3$
24. $\text{MnO}$
25. $\text{CrCl}_2$