Which of the following given below can liberate `Br_(2)` from `KBr`?
`F_(2),Cl_(2),I_(2) and " Conc. "H_(2)SO_(4)`

Options

A. `F_(2), Cl_(2), I_(2) and "Conc. "H_(2)SO_(4)`
B. `Cl_(2), I_(2) and "Conc. "H_(2)SO_(4)`
C. `F_(2),Cl_(2), I_(2)`
D. `F_(2),Cl_(2) and "Conc. "H_(2)SO_(4)`

Answer

To determine which of the given substances can liberate \( \text{Br}_2 \) from \( \text{KBr} \), we will analyze each option one by one.

### Step 1: Analyze Fluorine (\( \text{F}_2 \))
Fluorine is the most powerful oxidizing agent among the halogens. It can oxidize bromide ions (\( \text{Br}^- \)) to bromine (\( \text{Br}_2 \)).

**Reaction:**
\[ \text{F}_2 + 2 \text{KBr} \rightarrow 2 \text{KF} + \text{Br}_2 \]

**Conclusion:** Fluorine can liberate \( \text{Br}_2 \) from \( \text{KBr} \).

### Step 2: Analyze Chlorine (\( \text{Cl}_2 \))
Chlorine is also a strong oxidizing agent, but not as strong as fluorine. It can oxidize bromide ions to bromine.

**Reaction:**
\[ \text{Cl}_2 + 2 \text{KBr} \rightarrow 2 \text{KCl} + \text{Br}_2 \]

**Conclusion:** Chlorine can liberate \( \text{Br}_2 \) from \( \text{KBr} \).

### Step 3: Analyze Iodine (\( \text{I}_2 \))
Iodine is a weaker oxidizing agent compared to both fluorine and chlorine. It cannot oxidize bromide ions to bromine.

**Conclusion:** Iodine cannot liberate \( \text{Br}_2 \) from \( \text{KBr} \).

### Step 4: Analyze Concentrated Sulfuric Acid (\( \text{H}_2\text{SO}_4 \))
Concentrated sulfuric acid is a strong oxidizing agent and can oxidize bromide ions to bromine.

**Reaction:**
\[ \text{H}_2\text{SO}_4 + 2 \text{KBr} \rightarrow \text{K}_2\text{SO}_4 + \text{KHSO}_4 + \text{Br}_2 + \text{SO}_2 + 2 \text{H}_2\text{O} \]

**Conclusion:** Concentrated sulfuric acid can liberate \( \text{Br}_2 \) from \( \text{KBr} \).

### Final Conclusion:
The substances that can liberate \( \text{Br}_2 \) from \( \text{KBr} \) are:
- Fluorine (\( \text{F}_2 \))
- Chlorine (\( \text{Cl}_2 \))
- Concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \))

Thus, the answer is that **Fluorine, Chlorine, and Concentrated Sulfuric Acid** can liberate bromine from potassium bromide.