Identify the true statement.

Options

A. Among the oxides MnO, `Mn_2O_3, MnO_2 and Mn_2O_7` , the last one is most acidic
B. `H^-` is a powerful reductant and a very strong base
C. In the molecule `FNO_3`, two oxygen atoms are in -1 oxidation state each
D. There are four P-P bonds in the `P_4` molecule

Explanation

To solve the question of identifying the true statement among the given options, we will analyze each statement step by step.

### Step 1: Analyze the first statement regarding the oxides of manganese (MnO, Mn2O3, MnO2, Mn2O7).

1. **Determine the oxidation states of manganese (Mn) in each oxide:**
- For **MnO**:
- Let the oxidation state of Mn be \( x \).
- The formula is \( x + (-2) = 0 \) (since oxygen has a charge of -2).
- Solving gives \( x = +2 \).
- For **Mn2O3**:
- The formula is \( 2x + 3(-2) = 0 \).
- Solving gives \( 2x - 6 = 0 \) → \( x = +3 \).
- For **MnO2**:
- The formula is \( x + 2(-2) = 0 \).
- Solving gives \( x - 4 = 0 \) → \( x = +4 \).
- For **Mn2O7**:
- The formula is \( 2x + 7(-2) = 0 \).
- Solving gives \( 2x - 14 = 0 \) → \( x = +7 \).

2. **Determine acidity based on oxidation states:**
- As the oxidation state of Mn increases, its acidity also increases.
- Therefore, Mn2O7 (with Mn in +7 oxidation state) is the most acidic among the listed oxides.

### Conclusion for Step 1:
- The first statement is **true**: Mn2O7 is the most acidic oxide.

### Step 2: Analyze the second statement regarding H- as a reductant and strong base.

1. **H- as a strong base:**
- H- (hydride ion) is indeed a strong base. It can deprotonate carboxylic acids to form carboxylate ions.
- Reaction: \( RCOOH + H^- \rightarrow RCOO^- + H2 \) (where H2 gas is released).

2. **H- as a reductant:**
- H- can reduce aldehydes to alcohols.
- Reaction: \( RCHO + H^- \rightarrow RCH2OH \).

### Conclusion for Step 2:
- The second statement is also **true**: H- is a powerful reductant and a strong base.

### Step 3: Analyze the third statement regarding the oxidation states in FnO3.

1. **Determine oxidation states in FnO3:**
- Fluorine (F) is more electronegative than both nitrogen (N) and oxygen (O).
- Assign oxidation states:
- F = -1
- Let the oxidation state of N be \( x \).
- For the three O atoms, we can assume two have -2 and one has 0 (as per the structure).
- The equation becomes: \( x + (-1) + 2(-2) = 0 \).
- Solving gives \( x - 1 - 4 = 0 \) → \( x = +5 \).

2. **Check the statement:**
- The statement claims that two oxygen atoms are in -1 oxidation state, which is incorrect.

### Conclusion for Step 3:
- The third statement is **false**.

### Step 4: Analyze the fourth statement regarding P4 molecule.

1. **Determine the number of P-P bonds in P4:**
- The structure of P4 shows that there are 6 P-P bonds.
- Each phosphorus atom forms bonds with three other phosphorus atoms.

### Conclusion for Step 4:
- The fourth statement is **false**.

### Final Conclusion:
- The true statements are the first and second statements. The third and fourth statements are false.

### Summary of True Statements:
1. The first statement about Mn2O7 being the most acidic is true.
2. The second statement about H- being a powerful reductant and strong base is also true.