Which of the following is stable (inert) to fire ?

Options

A. `"CC"l_(4)`
B. `C_(2)H_(5)OH`
C. `CH_(4)`
D. `C_(4)H_(10)`

Explanation

To determine which of the given compounds is stable (inert) to fire, we need to analyze the reactivity of each compound with oxygen, as combustion reactions typically involve the reaction of a substance with oxygen to produce carbon dioxide (CO2) and water (H2O).

### Step-by-Step Solution:

1. **Identify the Compounds**: The question likely presents several compounds, including hydrocarbons and other organic compounds. For this solution, we will consider common examples such as CH4 (methane), C2H5OH (ethanol), and CCl4 (carbon tetrachloride).

2. **Analyze CH4 (Methane)**:
- Methane is a simple hydrocarbon that readily reacts with oxygen in the air.
- The combustion reaction is:
\[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \]
- Conclusion: Methane is not stable to fire.

3. **Analyze C2H5OH (Ethanol)**:
- Ethanol is an alcohol that also reacts readily with oxygen.
- The combustion reaction is:
\[ C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O \]
- Conclusion: Ethanol is not stable to fire.

4. **Analyze C4H10 (Butane)**:
- Butane is another hydrocarbon that combusts in the presence of oxygen.
- The combustion reaction is:
\[ 2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O \]
- Conclusion: Butane is not stable to fire.

5. **Analyze CCl4 (Carbon Tetrachloride)**:
- Carbon tetrachloride is a compound where carbon is bonded to four chlorine atoms. The bond between carbon and chlorine is very strong.
- It requires a significant amount of energy to break these bonds, making it less likely to react with oxygen under normal conditions.
- Conclusion: Carbon tetrachloride is stable (inert) to fire.

6. **Final Answer**: Among the compounds analyzed, the one that is stable (inert) to fire is **CCl4 (Carbon Tetrachloride)**.