BeO reacts with HF in presence of ammonia to give [A] which on thermal decomposition produces [B] an — s Block Elements Chemistry Question
Question
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 & Explanation
**Step 1: Identify compound [A]** BeO reacts with HF in the presence of ammonia. The product [A] is an ammonium fluoroberyllate complex: (NH₄)₂[BeF₄] This can be confirmed by the thermal decomposition reaction: (NH₄)₂[BeF₄] → BeF₂ [B] + 2NH₃ + 2HF **Step 2: Assign oxidation states in [A]** In (NH₄)₂[BeF₄]: - NH₄⁺ ions: N is -3, H is +1 - [BeF₄]²⁻ complex anion: F is -1 **Step 3: Apply oxidation state rules** For the fluoroberyllate anion [BeF₄]²⁻: - Each F atom has oxidation state = -1 - There are 4 F atoms: 4 × (-1) = -4 - Total charge of complex = -2 **Step 4: Calculate Be oxidation state** Using the equation: Oxidation state of Be + (4 × oxidation state of F) = charge of complex Let oxidation state of Be = x x + 4(-1) = -2 x - 4 = -2 x = +2 **Step 5: Verify with charge balance** (NH₄)₂[BeF₄]: 2(+1) + 1(+2) + 4(-1) = +2 - 2 + 2 - 4 = 0 ✓ Therefore, the answer is **2.00**.