`NH_(4)HF_(2)` ionises in aqueous medium as

Options

A. `NH_(4)HF^(+)F^(-)`
B. `NH_(4)^(+)HF_(2)^(-)`
C. `NH_(3)HF^(-)H^(+)`
D. `NH_(4)F_(2)^(-)H^(+)`

Explanation

To solve the problem of how \( NH_4HF_2 \) ionizes in an aqueous medium, we can follow these steps:

### Step 1: Identify the compound
The compound in question is \( NH_4HF_2 \), which is ammonium difluoride.

### Step 2: Understand the components
This compound consists of two parts:
- \( NH_4^+ \) (ammonium ion)
- \( HF_2^- \) (difluoride ion)

### Step 3: Write the ionization equation
When \( NH_4HF_2 \) is dissolved in water, it dissociates into its constituent ions. The ionization can be represented as:
\[ NH_4HF_2 (s) \rightarrow NH_4^+ (aq) + HF_2^- (aq) \]

### Step 4: Identify the ions produced
From the equation above, we can see that the ions produced are:
- \( NH_4^+ \) (ammonium ion)
- \( HF_2^- \) (difluoride ion)

### Step 5: Conclusion
Thus, \( NH_4HF_2 \) ionizes in aqueous medium to produce \( NH_4^+ \) and \( HF_2^- \).

### Final Answer
The correct answer is that \( NH_4HF_2 \) ionizes in aqueous medium to form \( NH_4^+ \) and \( HF_2^- \).

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