To determine which of the given compounds (BF3, SnCl2, SnCl4) behave as Lewis acids, we need to understand the definition of a Lewis acid. A Lewis acid is a species that can accept an electron pair due to being electron deficient or having an incomplete octet.
### Step-by-Step Solution:
1. **Identify the Electron Configuration of Each Compound:**
- **BF3 (Boron Trifluoride):** Boron (B) has 3 valence electrons and forms three bonds with fluorine (F), resulting in only 6 electrons around boron. This means boron has an incomplete octet and is electron deficient.
- **SnCl2 (Stannous Chloride):** Tin (Sn) has 4 valence electrons. In SnCl2, tin forms two bonds with chlorine atoms, which means it has 6 electrons around it (4 from itself and 2 from the two Cl atoms). Although it has a complete octet, the presence of electronegative chlorine atoms can withdraw electron density, making it behave as a Lewis acid.
- **SnCl4 (Stannic Chloride):** In SnCl4, tin forms four bonds with chlorine atoms, resulting in 8 electrons around it. However, like SnCl2, the electronegative chlorine atoms can withdraw electron density, allowing it to also act as a Lewis acid.
2. **Determine the Lewis Acid Behavior:**
- **BF3:** Since boron has an incomplete octet, it readily accepts an electron pair, thus behaving as a Lewis acid.
- **SnCl2:** Although it has a complete octet, the electron-withdrawing effect of chlorine makes it capable of accepting electron pairs, qualifying it as a Lewis acid.
- **SnCl4:** Similar to SnCl2, it can also accept electron pairs due to the influence of the electronegative chlorine atoms, making it a Lewis acid as well.
3. **Conclusion:**
All three compounds (BF3, SnCl2, SnCl4) behave as Lewis acids.
### Final Answer:
**BF3, SnCl2, and SnCl4 all behave as Lewis acids.**
