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 ____
Answer: 2
Solution
BeO reacts with HF in the presence of ammonia to form ammonium fluoroberyllate:
BeO + 2HF + 2NH₃ → (NH₄)₂[BeF₄] [A]
On thermal decomposition:
(NH₄)₂[BeF₄] → BeF₂ [B] + 2NH₄F
In compound [A], which is (NH₄)₂[BeF₄], beryllium is in the [BeF₄]²⁻ complex ion. Since fluorine has oxidation state -1, and the overall charge is -2:
Let oxidation state of Be = x
x + 4(-1) = -2
x = +2
Answer: 2
BeO + 2HF + 2NH₃ → (NH₄)₂[BeF₄] [A]
On thermal decomposition:
(NH₄)₂[BeF₄] → BeF₂ [B] + 2NH₄F
In compound [A], which is (NH₄)₂[BeF₄], beryllium is in the [BeF₄]²⁻ complex ion. Since fluorine has oxidation state -1, and the overall charge is -2:
Let oxidation state of Be = x
x + 4(-1) = -2
x = +2
Answer: 2
JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Integer answer