Reaction of BeO with ammonia and hydrogen fluoride gives 'A' which on thermal decomposition gives $$BeF_2$$ and $$NH_4F$$. What is 'A' ?

We need to identify compound 'A' formed when BeO reacts with ammonia and hydrogen fluoride, which on thermal decomposition gives $$BeF_2$$ and $$NH_4F$$.

Analyse the decomposition products.

The thermal decomposition gives $$BeF_2$$ and $$NH_4F$$. Let us work backwards to identify 'A'.

If compound 'A' decomposes to give $$BeF_2 + 2NH_4F$$, then 'A' must contain Be, N, H, and F atoms in the combined ratio. Adding the atoms:

$$BeF_2 + 2NH_4F = Be + 2N + 8H + 4F$$

This corresponds to the formula $$(NH_4)_2BeF_4$$ (ammonium tetrafluoroberyllate).

Verify the formation reaction.

The reaction of BeO with $$NH_3$$ and HF can be written as:

$$BeO + 2NH_3 + 4HF \rightarrow (NH_4)_2BeF_4 + H_2O$$

This is consistent because:

- BeO provides the beryllium

- HF provides the fluorine

- $$NH_3$$ and HF together form $$NH_4^+$$ ions

- The product $$(NH_4)_2BeF_4$$ contains the tetrafluoroberyllate ion $$[BeF_4]^{2-}$$ with two ammonium counterions

Verify the decomposition.

$$(NH_4)_2BeF_4 \xrightarrow{\Delta} BeF_2 + 2NH_4F$$

Atoms balance: Be: 1=1, N: 2=2, H: 8=8, F: 4=2+2=4. Correct.

The correct answer is Option (2): $$(NH_4)_2BeF_4$$.