Question

eca review/hw
name:
binary ionic compounds with transition metals
transition metals have unpredictable charges.
list the six that are on your oxidation chart with roman numerals
to show they can vary. striped ½ and ½ on coloring sheets too
ions names ions names
ionic compound names and formulas
a. write the ions, and then the correct chemical formula for these compounds by using the criss cross rule.
name ions compound formula

1. lead (ii) chloride
2. copper (ii) iodide
3. manganese fluoride
4. tin (iv) nitride
5. iron (iii) phosphide
6. sodium iodide
7. caesium phosphide

a. find the ions that make up these compounds, then the name. a few are done for you as examples.

1. ag₂s silver sulfide 8 pbf₄ lead (iv) fluoride
2. zno zinc oxide 9. ga₂o₃
3. cuf
4. cubr₂
5. mncl₂
6. snbr₂
7. rb₂o 10 cscl
8. v₂o₅
9. mg₃n₂
10. hg₂o
11. hgo

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
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
$Ga_{2}O_{3}$	$Ga^{3+},O^{2 - }$	-
$CsCl$	$Cs^{+},Cl^{-}$	-
$V_{2}O_{5}$	$V^{5+},O^{2 - }$	-
$Mg_{3}N_{2}$	$Mg^{2+},N^{3 - }$	-
$Hg_{2}O$	$Hg_{2}^{2+},O^{2 - }$	-
$HgO$	$Hg^{2+},O^{2 - }$	-