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 ______

We need to find the oxidation state of Be in compound [A], which is formed when BeO reacts with HF in the presence of ammonia. When BeO reacts with HF in the presence of $$NH_3$$, it forms ammonium tetrafluoroberyllate: $$BeO + 4HF + 2NH_3 \rightarrow (NH_4)_2[BeF_4] + H_2O$$ So compound [A] is $$(NH_4)_2[BeF_4]$$.

On thermal decomposition, compound [A] gives compound [B] and ammonium fluoride: $$(NH_4)_2[BeF_4] \xrightarrow{\Delta} BeF_2 + 2NH_4F$$ So compound [B] is $$BeF_2$$. This is consistent with the information given.

In $$(NH_4)_2[BeF_4]$$, $$NH_4^+$$ has a charge of +1 each (two of them contribute +2) and $$[BeF_4]^{2-}$$ has a charge of -2. In the $$[BeF_4]^{2-}$$ ion, each F has an oxidation state of -1, so 4 fluorine atoms contribute $$4 \times (-1) = -4$$. Let the oxidation state of Be be $$x$$; then $$x + (-4) = -2$$, giving $$x = +2$$.

The oxidation state of Be in compound [A] is +2.