JEE Main 2023 · 24 January · morning · Single correct · +4 / −1
Reaction of BeO with ammonia and hydrogen fluoride gives A which on thermal decomposition gives BeF₂ and NH₄F. What is 'A'?
- (a) (NH₄)Be₂F₅
- (b) (NH₄)₂BeF₄
- (c) (NH₄)BeF₃
- (d) H₃NBeF₃
Answer and solution
Answer (b) (NH₄)₂BeF₄
Solution
Chemistry ma’am · 4 steps
Be stops at four bonds, so excess fluoride gives the tetrahedral [BeF₄]²⁻ ion.
- Covalency of Be. Beryllium has only four valence orbitals, one 2s and three 2p, and no d orbitals. So its maximum covalency is 4, a part of its anomalous behaviour, while heavier group 2 metals can reach 6.
- Form the complex. NH₃ and HF give NH₄⁺ and F⁻. Be²⁺ takes four fluoride ions to fill all four orbitals, giving [BeF₄]²⁻. Two NH₄⁺ ions balance the 2- charge.
BeO + 2NH₃ + 4HF → (NH₄)₂BeF₄ + H₂O
- Heat it. On heating, the salt splits into exactly the products given: (NH₄)₂BeF₄ →(Δ) BeF₂ + 2NH₄F. This is the standard way to prepare anhydrous BeF₂.
- So. A is ammonium tetrafluoroberyllate, (NH₄)₂BeF₄. That is option (b).
The idea to keep. Be can never exceed four bonds because it has no d orbitals. With halide ions it forms [BeX₄]²⁻, and it never forms [BeX₆]⁴⁻ the way Al forms [AlF₆]³⁻.
Answer from the NTA final answer key. Solution by PrepWiser.