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drag the tiles to the correct boxes. not all tiles will be used. the ox…

Question

drag the tiles to the correct boxes. not all tiles will be used. the oxidation number of oxygen is -2, and that of chlorine is -1. use this information to match the names of the compounds to their chemical formulas. manganese(ii) chloride manganese(iv) chloride manganese(ii) oxide manganese(iv) oxide mncl₂ → mno₂ → reset next

Explanation:

Response

To solve this, we use the oxidation numbers to determine the correct compound names for the given formulas.

For $\boldsymbol{MnCl_2}$:
  • Chlorine (Cl) has an oxidation number of -1. Let the oxidation number of Mn be $x$.
  • In a compound, the sum of oxidation numbers is zero. So, for $MnCl_2$: $x + 2\times(-1) = 0$.
  • Solving: $x - 2 = 0 \implies x = +2$.
  • So, Mn has an oxidation state of +2, and the anion is chloride (Cl⁻). Thus, the compound is manganese(II) chloride.
For $\boldsymbol{MnO_2}$:
  • Oxygen (O) has an oxidation number of -2. Let the oxidation number of Mn be $y$.
  • For $MnO_2$: $y + 2\times(-2) = 0$.
  • Solving: $y - 4 = 0 \implies y = +4$.
  • So, Mn has an oxidation state of +4, and the anion is oxide (O²⁻). Thus, the compound is manganese(IV) oxide.
Final Matching:
  • $MnCl_2$: manganese(II) chloride
  • $MnO_2$: manganese(IV) oxide

So, drag "manganese(II) chloride" to the box next to $MnCl_2$ and "manganese(IV) oxide" to the box next to $MnO_2$.

Answer:

To solve this, we use the oxidation numbers to determine the correct compound names for the given formulas.

For $\boldsymbol{MnCl_2}$:
  • Chlorine (Cl) has an oxidation number of -1. Let the oxidation number of Mn be $x$.
  • In a compound, the sum of oxidation numbers is zero. So, for $MnCl_2$: $x + 2\times(-1) = 0$.
  • Solving: $x - 2 = 0 \implies x = +2$.
  • So, Mn has an oxidation state of +2, and the anion is chloride (Cl⁻). Thus, the compound is manganese(II) chloride.
For $\boldsymbol{MnO_2}$:
  • Oxygen (O) has an oxidation number of -2. Let the oxidation number of Mn be $y$.
  • For $MnO_2$: $y + 2\times(-2) = 0$.
  • Solving: $y - 4 = 0 \implies y = +4$.
  • So, Mn has an oxidation state of +4, and the anion is oxide (O²⁻). Thus, the compound is manganese(IV) oxide.
Final Matching:
  • $MnCl_2$: manganese(II) chloride
  • $MnO_2$: manganese(IV) oxide

So, drag "manganese(II) chloride" to the box next to $MnCl_2$ and "manganese(IV) oxide" to the box next to $MnO_2$.