Which molecule represent by the bold atoms are in their highest oxidation state?

Options

A. `H_(2)S_(2)O_(8)`
B. `P_(4)O_(10)`
C. `F_(2)O`
D. `Mn_(2)O_(7)`

Explanation

To determine which molecule has the bold atoms in their highest oxidation state, we will calculate the oxidation states of the relevant atoms in each molecule mentioned in the question.

### Step-by-Step Solution:

1. **Identify the Molecules**: The molecules we need to analyze are H₂S₂O₈, P₄, F₂O₂, and Mn₂O₇.

2. **Calculate the Oxidation State for H₂S₂O₈**:
- The oxidation state of oxygen (O) is typically -2.
- In H₂S₂O₈, there are 8 oxygen atoms: \( 8 \times (-2) = -16 \).
- Let the oxidation state of sulfur (S) be \( x \). Since there are 2 sulfur atoms, we have:
\[ 2x + (-16) = 0 \implies 2x = 16 \implies x = +8 \]
- The oxidation state of sulfur in H₂S₂O₈ is +6.

3. **Calculate the Oxidation State for P₄**:
- In P₄, phosphorus (P) is in its elemental form, so its oxidation state is 0.

4. **Calculate the Oxidation State for F₂O₂**:
- The oxidation state of fluorine (F) is -1.
- For F₂O₂, we have:
\[ 2(-1) + 2x = 0 \implies -2 + 2x = 0 \implies 2x = 2 \implies x = +1 \]
- The oxidation state of oxygen in F₂O₂ is +1.

5. **Calculate the Oxidation State for Mn₂O₇**:
- The oxidation state of oxygen (O) is -2.
- In Mn₂O₇, there are 7 oxygen atoms: \( 7 \times (-2) = -14 \).
- Let the oxidation state of manganese (Mn) be \( x \). Since there are 2 manganese atoms, we have:
\[ 2x + (-14) = 0 \implies 2x = 14 \implies x = +7 \]
- The oxidation state of manganese in Mn₂O₇ is +7.

6. **Compare the Oxidation States**:
- H₂S₂O₈: Sulfur = +6
- P₄: Phosphorus = 0
- F₂O₂: Oxygen = +1
- Mn₂O₇: Manganese = +7

7. **Determine Highest Oxidation States**:
- The highest oxidation states among the calculated values are:
- Sulfur in H₂S₂O₈: +6
- Phosphorus in P₄: 0 (not the highest)
- Oxygen in F₂O₂: +1 (not the highest)
- Manganese in Mn₂O₇: +7

### Conclusion:
The molecules with the bold atoms in their highest oxidation states are:
- H₂S₂O₈ (Sulfur: +6)
- Mn₂O₇ (Manganese: +7)

Thus, the answer to the question is that both H₂S₂O₈ and Mn₂O₇ contain atoms in their highest oxidation states.