eca review/hw\nname: \nbinary ionic compounds with transition metals\ntransition metals have unpredictable…
eca review/hw\nname: \nbinary ionic compounds with transition metals\ntransition metals have unpredictable charges.\nlist the six that are on your oxidation chart with roman numerals\nto show they can vary. striped ½ and ½ on coloring sheets too\nions names ions names\nionic compound names and formulas\na. write the ions, and then the correct chemical formula for these compounds by using the criss cross rule.\nname ions compound formula\n1. lead (ii) chloride\n2. copper (ii) iodide\n3. manganese fluoride\n4. tin (iv) nitride\n5. iron (iii) phosphide\n6. sodium iodide\n7. caesium phosphide\na. find the ions that make up these compounds, then the name. a few are done for you as examples.\n1. ag₂s silver sulfide 8 pbf₄ lead (iv) fluoride\n2. zno zinc oxide 9. ga₂o₃\n3. cuf\n4. cubr₂\n5. mncl₂\n6. snbr₂\n7. rb₂o 10 cscl\n11. v₂o₅\n12. mg₃n₂\n13. hg₂o\n14. hgo
Answer
Explanation:
Step1: Identify ions for lead (II) chloride
Lead (II) is $Pb^{2 + }$ and chloride is $Cl^{-}$.
Step2: Apply criss - cross rule
The charges cross over to give the formula $PbCl_{2}$.
Step3: Identify ions for copper (II) iodide
Copper (II) is $Cu^{2+}$ and iodide is $I^{-}$.
Step4: Apply criss - cross rule
The formula is $CuI_{2}$.
Step5: For Manganese fluoride
Assume manganese in +2 state (common), $Mn^{2+}$ and fluoride is $F^{-}$, formula is $MnF_{2}$.
Step6: For Tin (IV) nitride
Tin (IV) is $Sn^{4+}$ and nitride is $N^{3 - }$, formula is $Sn_{3}N_{4}$ after criss - cross.
Step7: For Iron (III) phosphide
Iron (III) is $Fe^{3+}$ and phosphide is $P^{3 - }$, formula is $FeP$.
Step8: For Sodium iodide
Sodium is $Na^{+}$ and iodide is $I^{-}$, formula is $NaI$.
Step9: For Caesium phosphide
Caesium is $Cs^{+}$ and phosphide is $P^{3 - }$, formula is $Cs_{3}P$.
Step10: For $Ag_{2}S$
Ions are $Ag^{+}$ and $S^{2 - }$.
Step11: For $ZnO$
Ions are $Zn^{2+}$ and $O^{2 - }$.
Step12: For $CuF$
Ions are $Cu^{+}$ and $F^{-}$.
Step13: For $CuBr_{2}$
Ions are $Cu^{2+}$ and $Br^{-}$.
Step14: For $MnCl_{2}$
Ions are $Mn^{2+}$ and $Cl^{-}$.
Step15: For $SnBr_{2}$
Ions are $Sn^{2+}$ and $Br^{-}$.
Step16: For $Rb_{2}O$
Ions are $Rb^{+}$ and $O^{2 - }$.
Step17: For $PbF_{4}$
Ions are $Pb^{4+}$ and $F^{-}$.
Step18: For $Ga_{2}O_{3}$
Ions are $Ga^{3+}$ and $O^{2 - }$.
Step19: For $CsCl$
Ions are $Cs^{+}$ and $Cl^{-}$.
Step20: For $V_{2}O_{5}$
Ions are $V^{5+}$ and $O^{2 - }$.
Step21: For $Mg_{3}N_{2}$
Ions are $Mg^{2+}$ and $N^{3 - }$.
Step22: For $Hg_{2}O$
Ions are $Hg_{2}^{2+}$ and $O^{2 - }$.
Step23: For $HgO$
Ions are $Hg^{2+}$ and $O^{2 - }$.
Answer:
| Name | Ions | Compound Formula |
|---|---|---|
| lead (II) chloride | $Pb^{2 + },Cl^{-}$ | $PbCl_{2}$ |
| copper (II) iodide | $Cu^{2+},I^{-}$ | $CuI_{2}$ |
| Manganese fluoride | $Mn^{2+},F^{-}$ | $MnF_{2}$ |
| Tin (IV) nitride | $Sn^{4+},N^{3 - }$ | $Sn_{3}N_{4}$ |
| Iron (III) phosphide | $Fe^{3+},P^{3 - }$ | $FeP$ |
| Sodium iodide | $Na^{+},I^{-}$ | $NaI$ |
| Caesium phosphide | $Cs^{+},P^{3 - }$ | $Cs_{3}P$ |
| $Ag_{2}S$ | $Ag^{+},S^{2 - }$ | Silver sulfide |
| $ZnO$ | $Zn^{2+},O^{2 - }$ | Zinc oxide |
| $CuF$ | $Cu^{+},F^{-}$ | - |
| $CuBr_{2}$ | $Cu^{2+},Br^{-}$ | - |
| $MnCl_{2}$ | $Mn^{2+},Cl^{-}$ | - |
| $SnBr_{2}$ | $Sn^{2+},Br^{-}$ | - |
| $Rb_{2}O$ | $Rb^{+},O^{2 - }$ | - |
| $PbF_{4}$ | $Pb^{4+},F^{-}$ | lead (IV) fluoride |
| $Ga_{2}O_{3}$ | $Ga^{3+},O^{2 - }$ | - |
| $CsCl$ | $Cs^{+},Cl^{-}$ | - |
| $V_{2}O_{5}$ | $V^{5+},O^{2 - }$ | - |
| $Mg_{3}N_{2}$ | $Mg^{2+},N^{3 - }$ | - |
| $Hg_{2}O$ | $Hg_{2}^{2+},O^{2 - }$ | - |
| $HgO$ | $Hg^{2+},O^{2 - }$ | - |