Trioxalatoaluminate (III) and tetrafluoroborate(III) ions are :

Options

A. `[Al(C_(2)O_(4))_(3)], [BF_(4)]^(3-)`
B. `[Al(C_(2)O_(4))_(3)]^(3+), [BF_(4)]^(3+)`
C. `[Al (C_(2)O_(4))_(3)]^(3-), [BF_(4)]^(-)`
D. `[Al(C_(2)O_(4))_(3)]^(2-), [BF_(4)]^(2-)`

Explanation

To determine the formulas for the trioxalatoaluminate(III) and tetrafluoroborate(III) ions, we will break down the components of each complex ion step by step.

### Step 1: Analyze Trioxalatoaluminate(III)
1. **Identify the components**:
- "Trioxalato" indicates there are three oxalate ions (C2O4).
- "Aluminate(III)" indicates the presence of aluminum (Al) with a +3 oxidation state.

2. **Write the formula for oxalate**:
- The oxalate ion (C2O4) has a charge of -2.

3. **Combine the components**:
- Since there are three oxalate ions, the total contribution from the oxalate ions is 3 × (-2) = -6.
- The aluminum ion contributes +3.

4. **Calculate the total charge**:
- Let the overall charge of the complex ion be X.
- Therefore, we have:
\[ +3 + (-6) = X \]
- This simplifies to:
\[ X = -3 \]

5. **Final formula for trioxalatoaluminate(III)**:
- The formula is Al(C2O4)3, and the charge is -3.

### Step 2: Analyze Tetrafluoroborate(III)
1. **Identify the components**:
- "Tetrafluoro" indicates there are four fluoride ions (F).
- "Borate(III)" indicates the presence of boron (B) with a +3 oxidation state.

2. **Write the formula for fluoride**:
- The fluoride ion (F) has a charge of -1.

3. **Combine the components**:
- Since there are four fluoride ions, the total contribution from the fluoride ions is 4 × (-1) = -4.
- The boron ion contributes +3.

4. **Calculate the total charge**:
- Let the overall charge of the complex ion be Y.
- Therefore, we have:
\[ +3 + (-4) = Y \]
- This simplifies to:
\[ Y = -1 \]

5. **Final formula for tetrafluoroborate(III)**:
- The formula is BF4, and the charge is -1.

### Conclusion:
- The final formulas for the complex ions are:
- Trioxalatoaluminate(III): Al(C2O4)3^3-
- Tetrafluoroborate(III): BF4^-