practice problems - type ii binary compounds
write the correct name for:
1. cus
2. pbbr₄
3. pb₃n₂
4. fe₂o₃
5. fei₂
6. sn₃p₄
7. cu₂s
8. sncl₂
9. cdo
10. zno
11. cucl₂
12. cubr
13. pbo
14. fe₂s₃
15. pbcl₂
16. sno
17. cu₂o
18. pbo₂
19. feo
20. sno₂
21. agbr
22. ag₂s
23. aucl₃
24. mno
25. crcl₃
26. coo
27. mn₂o₃
28. co₂s₃
29. auf
30. crbr₂

Question

Accepted Answer

# Explanation:
## Step1: Recall naming rules for Type II binary compounds
For Type II binary compounds (where the metal can have multiple oxidation states), we use Roman numerals to indicate the oxidation state of the metal. The non - metal part is named as an anion (e.g., sulfide for $S^{2 -}$, bromide for $Br^{-}$, etc.).

## Step2: Determine oxidation states
1. For $CuS$:
   - The sulfide ion ($S^{2 -}$) has a charge of $- 2$. Let the oxidation state of $Cu$ be $x$. Using the charge balance in the compound $CuS$ ($x+( - 2)=0$), we get $x = + 2$. So, it is copper(II) sulfide.
2. For $PbBr_{4}$:
   - The bromide ion ($Br^{-}$) has a charge of $-1$. Let the oxidation state of $Pb$ be $x$. In $PbBr_{4}$, $x + 4\times(-1)=0$, so $x=+4$. It is lead(IV) bromide.
3. For $Pb_{3}N_{2}$:
   - The nitride ion ($N^{3 -}$) has a charge of $-3$. Let the oxidation state of $Pb$ be $x$. In $Pb_{3}N_{2}$, $3x+2\times(-3)=0$, $3x = 6$, $x = + 2$. It is lead(II) nitride.
4. For $Fe_{2}O_{3}$:
   - The oxide ion ($O^{2 -}$) has a charge of $-2$. Let the oxidation state of $Fe$ be $x$. In $Fe_{2}O_{3}$, $2x+3\times(-2)=0$, $2x=6$, $x = + 3$. It is iron(III) oxide.
5. For $FeI_{2}$:
   - The iodide ion ($I^{-}$) has a charge of $-1$. Let the oxidation state of $Fe$ be $x$. In $FeI_{2}$, $x + 2\times(-1)=0$, $x=+2$. It is iron(II) iodide.
6. For $Sn_{3}P_{4}$:
   - The phosphide ion ($P^{3 -}$) has a charge of $-3$. Let the oxidation state of $Sn$ be $x$. In $Sn_{3}P_{4}$, $3x+4\times(-3)=0$, $3x = 12$, $x = + 4$. It is tin(IV) phosphide.
7. For $Cu_{2}S$:
   - The sulfide ion ($S^{2 -}$) has a charge of $-2$. Let the oxidation state of $Cu$ be $x$. In $Cu_{2}S$, $2x+( - 2)=0$, $2x=2$, $x = + 1$. It is copper(I) sulfide.
8. For $SnCl_{2}$:
   - The chloride ion ($Cl^{-}$) has a charge of $-1$. Let the oxidation state of $Sn$ be $x$. In $SnCl_{2}$, $x+2\times(-1)=0$, $x = + 2$. It is tin(II) chloride.
9. For $CdO$:
   - The oxide ion ($O^{2 -}$) has a charge of $-2$. Let the oxidation state of $Cd$ be $x$. In $CdO$, $x+( - 2)=0$, $x = + 2$. It is cadmium(II) oxide.
10. For $ZnO$:
   - The oxide ion ($O^{2 -}$) has a charge of $-2$. Let the oxidation state of $Zn$ be $x$. In $ZnO$, $x+( - 2)=0$, $x = + 2$. It is zinc(II) oxide.
11. For $CuCl_{2}$:
   - The chloride ion ($Cl^{-}$) has a charge of $-1$. Let the oxidation state of $Cu$ be $x$. In $CuCl_{2}$, $x+2\times(-1)=0$, $x = + 2$. It is copper(II) chloride.
12. For $CuBr$:
   - The bromide ion ($Br^{-}$) has a charge of $-1$. Let the oxidation state of $Cu$ be $x$. In $CuBr$, $x+( - 1)=0$, $x = + 1$. It is copper(I) bromide.
13. For $PbO$:
   - The oxide ion ($O^{2 -}$) has a charge of $-2$. Let the oxidation state of $Pb$ be $x$. In $PbO$, $x+( - 2)=0$, $x = + 2$. It is lead(II) oxide.
14. For $Fe_{2}S_{3}$:
   - The sulfide ion ($S^{2 -}$) has a charge of $-2$. Let the oxidation state of $Fe$ be $x$. In $Fe_{2}S_{3}$, $2x+3\times(-2)=0$, $2x=6$, $x = + 3$. It is iron(III) sulfide.
15. For $PbCl_{2}$:
   - The chloride ion ($Cl^{-}$) has a charge of $-1$. Let the oxidation state of $Pb$ be $x$. In $PbCl_{2}$, $x+2\times(-1)=0$, $x = + 2$. It is lead(II) chloride.
16. For $SnO$:
   - The oxide ion ($O^{2 -}$) has a charge of $-2$. Let the oxidation state of $Sn$ be $x$. In $SnO$, $x+( - 2)=0$, $x = + 2$. It is tin(II) oxide.
17. For $Cu_{2}O$:
   - The oxide ion ($O^{2 -}$) has a charge of $-2$. Let the oxidation state of $Cu$ be $x$. In $Cu_{2}O$, $2x+( - 2)=0$, $2x=2$, $x = + 1$. It is copper(I) oxide.
18. For $PbO_{2}$:
   - The oxide ion ($O^{2 -}$) has a charge of $-2$. Let the oxidation state of $Pb$ be $x$. In $PbO_{2}$, $x+2\times(-2)=0$, $x = + 4$. It is lead(IV) oxide.
19. For $FeO$:
   - The oxide ion ($O^{2 -}$) has a charge of $-2$. Let the oxidation state of $Fe$ be $x$. In $FeO$, $x+( - 2)=0$, $x = + 2$. It is iron(II) oxide.
20. For $SnO_{2}$:
   - The oxide ion ($O^{2 -}$) has a charge of $-2$. Let the oxidation state of $Sn$ be $x$. In $SnO_{2}$, $x+2\times(-2)=0$, $x = + 4$. It is tin(IV) oxide.
21. For $AgBr$:
   - Silver ($Ag$) in this case has an oxidation state of $+1$ (since $Br^{-}$ has a charge of $-1$). It is silver(I) bromide.
22. For $Ag_{2}S$:
   - The sulfide ion ($S^{2 -}$) has a charge of $-2$. Let the oxidation state of $Ag$ be $x$. In $Ag_{2}S$, $2x+( - 2)=0$, $x = + 1$. It is silver(I) sulfide.
23. For $AuCl_{3}$:
   - The chloride ion ($Cl^{-}$) has a charge of $-1$. Let the oxidation state of $Au$ be $x$. In $AuCl_{3}$, $x+3\times(-1)=0$, $x = + 3$. It is gold(III) chloride.
24. For $MnO$:
   - The oxide ion ($O^{2 -}$) has a charge of $-2$. Let the oxidation state of $Mn$ be $x$. In $MnO$, $x+( - 2)=0$, $x = + 2$. It is manganese(II) oxide.
25. For $CrCl_{3}$:
   - The chloride ion ($Cl^{-}$) has a charge of $-1$. Let the oxidation state of $Cr$ be $x$. In $CrCl_{3}$, $x+3\times(-1)=0$, $x = + 3$. It is chromium(III) chloride.
26. For $CoO$:
   - The oxide ion ($O^{2 -}$) has a charge of $-2$. Let the oxidation state of $Co$ be $x$. In $CoO$, $x+( - 2)=0$, $x = + 2$. It is cobalt(II) oxide.
27. For $Mn_{2}O_{3}$:
   - The oxide ion ($O^{2 -}$) has a charge of $-2$. Let the oxidation state of $Mn$ be $x$. In $Mn_{2}O_{3}$, $2x+3\times(-2)=0$, $2x=6$, $x = + 3$. It is manganese(III) oxide.
28. For $Co_{2}S_{3}$:
   - The sulfide ion ($S^{2 -}$) has a charge of $-2$. Let the oxidation state of $Co$ be $x$. In $Co_{2}S_{3}$, $2x+3\times(-2)=0$, $2x=6$, $x = + 3$. It is cobalt(III) sulfide.
29. For $AuF$:
   - The fluoride ion ($F^{-}$) has a charge of $-1$. Let the oxidation state of $Au$ be $x$. In $AuF$, $x+( - 1)=0$, $x = + 1$. It is gold(I) fluoride.
30. For $CrBr_{2}$:
   - The bromide ion ($Br^{-}$) has a charge of $-1$. Let the oxidation state of $Cr$ be $x$. In $CrBr_{2}$, $x+2\times(-1)=0$, $x = + 2$. It is chromium(II) bromide.

# Answer:
1. copper(II) sulfide
2. lead(IV) bromide
3. lead(II) nitride
4. iron(III) oxide
5. iron(II) iodide
6. tin(IV) phosphide
7. copper(I) sulfide
8. tin(II) chloride
9. cadmium(II) oxide
10. zinc(II) oxide
11. copper(II) chloride
12. copper(I) bromide
13. lead(II) oxide
14. iron(III) sulfide
15. lead(II) chloride
16. tin(II) oxide
17. copper(I) oxide
18. lead(IV) oxide
19. iron(II) oxide
20. tin(IV) oxide
21. silver(I) bromide
22. silver(I) sulfide
23. gold(III) chloride
24. manganese(II) oxide
25. chromium(III) chloride
26. cobalt(II) oxide
27. manganese(III) oxide
28. cobalt(III) sulfide
29. gold(I) fluoride
30. chromium(II) bromide

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QUESTION IMAGE

Question

Explanation:

Step1: Recall naming rules for Type II binary compounds

Step2: Determine oxidation states

Answer: