BeO reacts with HF in presence of ammonia to give (A) which on thermal decomposition produces [B] and ammonium fluoride . Oxidation state of Be in [A] is _____________

Explanation

To solve the problem step by step, we will analyze the reaction of beryllium oxide (BeO) with hydrofluoric acid (HF) in the presence of ammonia (NH3) and determine the products formed, as well as the oxidation state of beryllium in the product.

### Step 1: Identify the Reaction
Beryllium oxide (BeO) reacts with hydrofluoric acid (HF) in the presence of ammonia (NH3).

### Step 2: Write the Reaction
The reaction can be represented as follows:
\[ \text{BeO} + \text{HF} + \text{NH}_3 \rightarrow \text{BeF}_2 + \text{NH}_4\text{F} \]

Here, BeF2 is the product (A) formed from the reaction.

### Step 3: Identify the Products
From the reaction, we can identify:
- Product (A): Beryllium fluoride (BeF2)
- By-products: Ammonium fluoride (NH4F)

### Step 4: Thermal Decomposition of Product A
When BeF2 is subjected to thermal decomposition, it can break down into:
\[ \text{BeF}_2 \rightarrow \text{Be} + \text{F}_2 \]

Here, Be is the product (B) formed from the thermal decomposition of BeF2.

### Step 5: Determine the Oxidation State of Beryllium in Product A
In BeF2, the oxidation state of beryllium (Be) can be calculated as follows:
- Fluorine (F) has an oxidation state of -1.
- Since there are two fluorine atoms, the total contribution from fluorine is -2.
- Let the oxidation state of beryllium be x. The overall charge of the compound is neutral (0).

Thus, we can set up the equation:
\[ x + 2(-1) = 0 \]
\[ x - 2 = 0 \]
\[ x = +2 \]

Therefore, the oxidation state of beryllium in product (A) (BeF2) is +2.

### Final Answer
The oxidation state of Be in [A] is **+2**.

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