On reaction of Beo with HF in presence of `NH_3` gives [x]. [x] on thermal decomposition gives `BeF_2` & `NH_4F`. [x] is:

Options

A. `(NH_4)_2BeF_4`
B. `(NH_4)_4 BeF_6`
C. `(NH_4)_3BeF_5`
D. `NH_4BeF_3`

Answer

To solve the question, we need to determine the compound [x] formed when beryllium oxide (BeO) reacts with hydrofluoric acid (HF) in the presence of ammonia (NH₃).

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are beryllium oxide (BeO), hydrofluoric acid (HF), and ammonia (NH₃).

2. **Write the Reaction**:
- When BeO reacts with HF in the presence of NH₃, the reaction can be represented as:
\[ \text{BeO} + \text{HF} + \text{NH}_3 \rightarrow [x] \]

3. **Determine the Product [x]**:
- The product formed from this reaction is ammonium beryllium tetrafluoride, which can be represented as:
\[ [x] = \text{NH}_4\text{BeF}_4 \]
- The balanced equation for this reaction is:
\[ \text{BeO} + 4 \text{HF} + 2 \text{NH}_3 \rightarrow \text{NH}_4\text{BeF}_4 + 2 \text{H}_2\text{O} \]

4. **Thermal Decomposition of [x]**:
- Upon thermal decomposition of ammonium beryllium tetrafluoride (NH₄)₂BeF₄, it breaks down into beryllium fluoride (BeF₂) and ammonium fluoride (NH₄F):
\[ \text{NH}_4\text{BeF}_4 \rightarrow \text{BeF}_2 + \text{NH}_4\text{F} \]

5. **Final Answer**:
- Therefore, the compound [x] is:
\[ \text{NH}_4\text{BeF}_4 \]