Which of the following reactions does not take place?
`BF_(3)^(-)+F^(-)rarrBF_(4)^(-)" ....(I)"`
`BF_(3)+3F^(-)rarrBF_(6)^(3-)" ...(II)"`
`AlF_(3)+3F^(-)rarrAlF_(6)^(3-)" ....(III)"`

Options

A. Only (I)
B. Only (II)
C. Only (III)
D. Only (I) and (III)

Answer

To determine which of the following reactions does not take place, we will analyze each reaction step by step.

### Step 1: Analyze Reaction (I)
**Reaction:** \( BF_3^- + F^- \rightarrow BF_4^- \)

- **Boron (B)** has an electronic configuration of \( 1s^2 2s^2 2p^1 \).
- In the excited state, one electron from the \( 2s \) orbital can be promoted to the \( 2p \) orbital, giving \( 2s^1 2p^2 \).
- This creates a vacant \( p \) orbital, allowing boron to accept a fluoride ion (\( F^- \)).
- The coordination number increases from 3 to 4, and thus this reaction is valid.

### Step 2: Analyze Reaction (II)
**Reaction:** \( BF_3 + 3F^- \rightarrow BF_6^{3-} \)

- Boron can only have a maximum coordination number of 4 due to its electronic configuration.
- It lacks vacant \( d \) orbitals to accommodate a coordination number of 6.
- Therefore, this reaction cannot occur as boron cannot expand its coordination number beyond 4.

### Step 3: Analyze Reaction (III)
**Reaction:** \( AlF_3 + 3F^- \rightarrow AlF_6^{3-} \)

- Aluminum (Al) has an electronic configuration of \( 1s^2 2s^2 2p^6 3s^2 3p^1 \).
- Aluminum has vacant \( d \) orbitals available for bonding.
- This allows aluminum to expand its coordination number from 3 to 6 when reacting with fluoride ions.
- Therefore, this reaction is valid.

### Conclusion
From the analysis:
- Reaction (I) is valid.
- Reaction (II) does not take place.
- Reaction (III) is valid.

Thus, the reaction that does not take place is **Reaction (II)**.

### Final Answer
**The reaction that does not take place is:** \( BF_3 + 3F^- \rightarrow BF_6^{3-} \) (Option B).

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