All the boron trihalides except `BI_(3)` may be prepared by direction between the elements. Boron trihalides consist of trigonal-planar `BX_(3)` molecules. Unlike the halides of the other elements in the group they are monomeric in the gas, liquid and solid states, `BF_(3)` and `BCl_(3)` are gases, `BBr_(3)` is a volatile liquid and `BI_(3)` is a solid. Boron trihalides are Lewis acids because they form simple Lewis complexes with suitable bases, as in the reaction:
`BF_(3)(g)+NH_(3)(g)toF_(3)B-NH_(3)(s)`
However, boron chlorides, bromides and iodides are susceptible (sensitive) to protolysis by mild proton sources such as water, alcohols and even amines, for example `BCl_(3)` undergoes rapid hydrolysis:
`BCl_(3)(g)+3H_(2)O(l)toB(OH)_(3)(aq)+3HCl (aq)`
It is supposed that the first step in the above reaction is the formation of the complex `Cl_(3)B larr OH_(2)` which then eliminates `HCl` and reacts with water.
Which of the following reactions is incorrect ?

Options

A. `BF_(3)(g) +F^(-)(aq)rarr [BF_(4)]^(-)(aq)`
B. `BCl_(3)(g)+3EtOH(l)toB(OEt)_(3)(l)+3HCl (g)`
C. `BBr_(3)(l)+F_(3)BN(CH_(3))_(3)(s)toBF_(3) (g)+Br_(3)BN(CH_(3))_(3)(s)`
D. `BCl_(3) (g)+2 C_(5)H_(5)N (l)toCl_(3)B(C_(5)H_(5)N)_(2) (s)`

Explanation

To determine which of the given reactions involving boron trihalides is incorrect, we will analyze each option step by step.

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

- **Explanation:**
- Boron trifluoride (BF3) has an empty p-orbital and can accept a pair of electrons from the fluoride ion (F^-).
- This results in the formation of the tetrafluoroborate ion \( BF_4^- \), which is a well-known reaction.

- **Conclusion:** This reaction is correct.

### Step 2: Analyze Option B
**Reaction:** \( BCl_3 + C_2H_5OH \rightarrow B(OEt)_3 + 3HCl \)

- **Explanation:**
- Boron trichloride (BCl3) acts as a Lewis acid and can react with the lone pair on the oxygen of ethanol (C2H5OH).
- The reaction forms a complex where the three chlorine atoms are replaced by ethoxy groups (OEt).
- The reaction produces three moles of hydrochloric acid (HCl) as a byproduct.

- **Conclusion:** This reaction is correct.

### Step 3: Analyze Option C
**Reaction:** \( BBr_3 + F^- \rightarrow BF_3 + Br^- \)

- **Explanation:**
- In this reaction, bromine from \( BBr_3 \) is replaced by fluoride to form \( BF_3 \) and release a bromide ion (Br^-).
- This type of halide exchange is plausible due to the relative electronegativities of bromine and fluorine.

- **Conclusion:** This reaction is correct.

### Step 4: Analyze Option D
**Reaction:** \( BCl_3 + 2C_5H_5N \rightarrow B(C_5H_5N)_2Cl_3 \)

- **Explanation:**
- Here, \( BCl_3 \) is reacting with two moles of pyridine (C5H5N).
- While pyridine can coordinate to boron, the formation of a complex with two pyridine molecules and three chloride ions would imply that boron is forming five bonds.
- Boron can only form a maximum of four bonds due to its tetravalent nature, making this reaction impossible.

- **Conclusion:** This reaction is incorrect.

### Final Answer
The incorrect reaction is **Option D**.

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