Question

day 4 hw: binary ionic compounds with transition metal cations
name the compound from just the formula
to figure out the charge of the transition metal, just \uncross the charges\
chemical formula name

1. snf₂

1. pbo₂

1. snf₄

1. cuo

1. cu₂o

1. fecl₃

1. al₂s₃

write the chemical formula for the following compounds.
chemical name cation anion chemical formula

1. copper ii chloride

1. zinc sulfide

1. tin iv bromide

1. mercury ii iodide

1. copper ii sulfide

1. nickel iii sulfide

1. iron iii oxide

1. titanium iii phosphide

Explanation:

Step1: Determine the charge of transition metal

For binary ionic compounds with transition metals, use the charge of the non - metal (anion) to find the charge of the transition metal (cation). For example, in \(SnF_2\), fluoride (\(F^-\)) has a charge of \(- 1\). Let the charge of \(Sn\) be \(x\). Using the formula \(x + 2\times(-1)=0\) (since the compound is neutral), we get \(x = + 2\).

Step2: Name the compound

Name the compound as [transition metal name (charge in Roman numerals)] [non - metal root + -ide]. For \(SnF_2\), it is tin (II) fluoride.
For \(PbO_2\), oxide (\(O^{2 -}\)) has a charge of \(-2\). Let the charge of \(Pb\) be \(y\). Using \(y+2\times(-2) = 0\), we get \(y=+4\). So it is lead (IV) oxide.
For \(SnF_4\), using \(z + 4\times(-1)=0\) (where \(z\) is the charge of \(Sn\)), \(z = + 4\). Name is tin (IV) fluoride.
For \(CuO\), oxide (\(O^{2 -}\)), let charge of \(Cu\) be \(a\), \(a+(-2)=0\), \(a = + 2\). Name is copper (II) oxide.
For \(Cu_2O\), let charge of \(Cu\) be \(b\), \(2b+(-2)=0\), \(b = + 1\). Name is copper (I) oxide.
For \(FeCl_3\), chloride (\(Cl^-\)), let charge of \(Fe\) be \(c\), \(c + 3\times(-1)=0\), \(c=+3\). Name is iron (III) chloride.
For \(Al_2S_3\), aluminum (\(Al\)) has a fixed charge of \(+3\), sulfide (\(S^{2 -}\)). Name is aluminum sulfide (since \(Al\) is not a transition metal, no Roman numeral needed).

For writing formulas:

• Copper (II) chloride: cation \(Cu^{2+}\), anion \(Cl^-\), formula \(CuCl_2\).
• Zinc sulfide: cation \(Zn^{2+}\) (fixed charge), anion \(S^{2 -}\), formula \(ZnS\).
• Tin (IV) bromide: cation \(Sn^{4+}\), anion \(Br^-\), formula \(SnBr_4\).
• Mercury (II) iodide: cation \(Hg^{2+}\), anion \(I^-\), formula \(HgI_2\).
• Copper (II) sulfide: cation \(Cu^{2+}\), anion \(S^{2 -}\), formula \(CuS\).
• Nickel (III) sulfide: cation \(Ni^{3+}\), anion \(S^{2 -}\), formula \(Ni_2S_3\).
• Iron (III) oxide: cation \(Fe^{3+}\), anion \(O^{2 -}\), formula \(Fe_2O_3\).
• Titanium (III) phosphide: cation \(Ti^{3+}\), anion \(P^{3 -}\), formula \(TiP\).

Answer:

1. tin (II) fluoride
2. lead (IV) oxide
3. tin (IV) fluoride
4. copper (II) oxide
5. copper (I) oxide
6. iron (III) chloride
7. aluminum sulfide
8. \(Cu^{2+}\), \(Cl^-\), \(CuCl_2\)
9. \(Zn^{2+}\), \(S^{2 -}\), \(ZnS\)
10. \(Sn^{4+}\), \(Br^-\), \(SnBr_4\)
11. \(Hg^{2+}\), \(I^-\), \(HgI_2\)
12. \(Cu^{2+}\), \(S^{2 -}\), \(CuS\)
13. \(Ni^{3+}\), \(S^{2 -}\), \(Ni_2S_3\)
14. \(Fe^{3+}\), \(O^{2 -}\), \(Fe_2O_3\)
15. \(Ti^{3+}\), \(P^{3 -}\), \(TiP\)