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 ____________.

Solution

Chemistry ma’am · 4 steps

Name [A] first: it is ammonium tetrafluoridoberyllate, and the charge balance gives Be its value.

1. Identify A. With HF and ammonia, BeO forms the complex salt (NH₄)₂[BeF₄]. Be is small and highly charged, so it readily forms the tetrahedral [BeF₄]²⁻ ion.

   BeO + 2NH₃ + 4HF → (NH₄)₂[BeF₄] + H₂O

2. Check with B. On heating, (NH₄)₂[BeF₄] → BeF₂ + 2NH₄F. So [B] is BeF₂ and the by-product is ammonium fluoride, which matches the question.
3. Balance charges. Each NH₄⁺ is +1 and each F⁻ is -1. So 2(+1) + x + 4(-1) = 0, which gives x = +2.
4. So. The oxidation state of Be in [A] is 2, the same as in BeO, because no electrons are transferred in this reaction.

The common slip: 4. Confusing the coordination number with the oxidation state gives 4. Four fluorides surround Be, but its oxidation state is still +2.

The idea to keep. Beryllium always stays at +2. Its small size lets it form complexes like [BeF₄]²⁻, but forming a complex changes the coordination number, not the oxidation state.

Answer from the NTA final answer key. Solution by PrepWiser.