When `F_(2)` gas reacts with `H_(2)O` it forms

Options

A. `O_(2)` gas
B. `O_(2) F_(2)`gas
C. HOF
D. Both (2) & (3)

Answer

To determine the products formed when fluorine gas (F₂) reacts with water (H₂O), we can follow these steps:

### Step 1: Write the unbalanced chemical equation
When fluorine gas reacts with water, the initial unbalanced equation can be written as:
\[ \text{F}_2 + \text{H}_2\text{O} \rightarrow \text{HF} + \text{O}_2 \]

### Step 2: Identify the products
From the reaction, we can see that the products formed are hydrofluoric acid (HF) and oxygen gas (O₂).

### Step 3: Balance the reaction
To balance the reaction, we need to ensure that the number of atoms of each element is the same on both sides of the equation.

1. **Fluorine (F)**: There are 2 fluorine atoms in F₂ and 1 in HF. To balance, we need 2 HF:
\[ \text{F}_2 + \text{H}_2\text{O} \rightarrow 2\text{HF} + \text{O}_2 \]

2. **Hydrogen (H)**: Now, we have 2 hydrogen atoms in 2 HF, which means we need 1 H₂O to balance the hydrogen:
\[ \text{F}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{HF} + \text{O}_2 \]

3. **Oxygen (O)**: There are 2 oxygen atoms in 2 H₂O and 1 in O₂. Thus, we need to adjust the water:
\[ 2\text{F}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{HF} + \text{O}_2 \]

### Final Balanced Equation
The final balanced equation is:
\[ 2\text{F}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{HF} + \text{O}_2 \]

### Step 4: Identify oxidation states
- In H₂O, H is +1 and O is -2.
- In HF, H is +1 and F is -1.
- In O₂, O is 0.

### Step 5: Determine oxidation and reduction
- Oxygen in H₂O goes from -2 to 0 (oxidation).
- Fluorine goes from 0 in F₂ to -1 in HF (reduction).

### Conclusion
When fluorine gas reacts with water, it forms hydrofluoric acid (HF) and oxygen gas (O₂).

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