Question

Reaction of $BeO$ with ammonia and hydrogen fluoride gives A which on thermal decomposition gives $BeF _2$ and $NH _4 F$ What is ' $A ^{\prime}$ ?

• A. $\left( NH _4\right) BeF _3$
• B. $H _3 NBeF _3$
• C. $\left( NH _4\right) Be _2 F _5$
• D. $\left( NH _4\right)_2 BeF _4$

Answer 1

The Correct Option is D

Let's analyze the chemical reaction given in the problem: beryllium oxide (\( \text{BeO} \)) reacts with ammonia (\( \text{NH}_3 \)) and hydrogen fluoride (\( \text{HF} \)) to give a compound 'A'. The compound 'A' upon thermal decomposition yields beryllium fluoride (\( \text{BeF}_2 \)) and ammonium fluoride (\( \text{NH}_4\text{F} \)). We need to identify the compound 'A'.

To solve this, let's consider the components involved in the reaction:

1. \(\text{BeO}\) reacts with \(\text{NH}_3\) and \(\text{HF}\) to form a beryllium-ammonium-fluoride compound.
2. The product forms \(\text{BeF}_2\) and \(\text{NH}_4\text{F}\) upon decomposition, indicating that 'A' is likely a complex compound containing these ions.
3. Let's consider the stoichiometry required to balance the creation and decomposition:

Since \(\text{A}\) when decomposed yields two products, \(\text{BeF}_2\) should have a single beryllium for stoichiometric balance. The formula \(\left( \text{NH}_4 \right)_2 \text{BeF}_4\) accommodates this:

1. \(\left( \text{NH}_4 \right)_2 \text{BeF}_4\) decomposes to \(\text{BeF}_2\) and 2 \(\text{NH}_4\text{F}\).
2. Thus, the breakdown is: \(\left( \text{NH}_4 \right)_2 \text{BeF}_4 \rightarrow \text{BeF}_2 + 2 \, \text{NH}_4\text{F}\).
3. This matches the components given: a source for \(\text{BeF}_2\) and \(\text{NH}_4\text{F}\).

Thus, the correct compound 'A' is \(\left( \text{NH}_4 \right)_2 \text{BeF}_4\), because:

• It contains beryllium, ammonium, and fluoride ions adequately.
• On decomposition, it yields the exact products as described: \(\text{BeF}_2\) and \(\text{NH}_4\text{F}\).

 

Let's now evaluate other options:

1. \(\left( \text{NH}_4 \right) \text{BeF}_3\): Contains only one ammonium ion, making it incapable of yielding \(\text{NH}_4\text{F}\) properly.
2. \(\text{H}_3 \text{NBeF}_3\): This doesn't even contain sufficient ammonium for producing \(\text{NH}_4\text{F}\).
3. \(\left( \text{NH}_4 \right) \text{Be}_2 \text{F}_5\): Incorrect ratio of beryllium and fluorine for the decomposition products given.

Therefore, the correct answer is: \(\left( \text{NH}_4 \right)_2 \text{BeF}_4\).