`BF_3 & C Cl_4` on reaction with water produce _____ & _______ respectively:

Options

A. `H_3BO_3+HF , H_2CO_3` + HCl
B. `H_3BO_3+HBF_4` , No reaction
C. `H_3BO_3+HBF_4 , H_2CO_3 + HCl`
D. `H_3BO_3+HF` No reaction

Explanation

To solve the question, we need to analyze the reactions of BF3 (Boron Trifluoride) and CCl4 (Carbon Tetrachloride) with water.

### Step-by-Step Solution:

1. **Identify the Reactants**: We have two compounds, BF3 and CCl4, that we will react with water (H2O).

2. **Analyze CCl4**:
- CCl4 is a non-polar molecule and does not have any vacant orbitals.
- When CCl4 is mixed with water, there is no reaction because water cannot attack CCl4.
- Therefore, CCl4 does not produce any products when reacted with water.

3. **Analyze BF3**:
- BF3 is a Lewis acid and has an empty p orbital on the boron atom.
- When BF3 reacts with water, the water molecule can donate a pair of electrons to the boron atom, forming a bond.
- The reaction can be represented as:
\[ BF_3 + H_2O \rightarrow HBF_4 \]
- Additionally, if we consider the complete hydrolysis of BF3, it can also produce boric acid (H3BO3):
\[ 4 BF_3 + 3 H_2O \rightarrow H_3BO_3 + 3 HBF_4 \]

4. **Conclusion**:
- From the above analysis, we conclude that when BF3 reacts with water, it produces HBF4 (Hydrofluoric acid) and potentially H3BO3 (Boric acid) as well.
- CCl4 does not produce any products as it does not react with water.

### Final Answer:
- BF3 reacts with water to produce **HBF4**.
- CCl4 does not react with water and produces **no product**.