JEE Main 2023 · 13 April · evening · Single correct · +4 / −1
Better method for preparation of BeF₂, among the following is :
- (a) BeH₂+F₂ →(Δ) BeF₂
- (b) (NH₄)₂ BeF₄ →(Δ) BeF₂
- (c) Be+F₂ →(Δ) BeF₂
- (d) BeO+C+F₂ →(Δ) BeF₂
Answer and solution
Answer (b) (NH₄)₂ BeF₄ →(Δ) BeF₂
Solution
Chemistry ma’am · 3 steps
Skip the fluorine gas: heat a salt that already holds the fluorides, and BeF₂ is left behind.
- Why it is tricky. BeF₂ is needed anhydrous, so it cannot be crystallised from water. Beryllium salts hold on to water strongly, and heating the hydrate tends to hydrolyse it instead of drying it.
- Standard method. Heat ammonium fluoroberyllate, (NH₄)₂BeF₄. It breaks up cleanly, the ammonium fluoride part leaves as vapour, and pure anhydrous BeF₂ stays behind.
(NH₄)₂BeF₄ →(Δ) BeF₂ + 2NH₄F
Route Problem BeH₂ + F₂ BeH₂ is itself hard to make, and F₂ reacts violently (NH₄)₂BeF₄ on heating none: easy to handle, volatile by-product, pure anhydrous BeF₂ Be + F₂ Be has a protective oxide film, and F₂ is hazardous and costly BeO + C + F₂ this is the BeCl₂ method with Cl₂, not a good fluoride route - So. The better method is the thermal decomposition of (NH₄)₂BeF₄, option (b).
The idea to keep. Anhydrous BeF₂ comes from heating (NH₄)₂BeF₄. Anhydrous BeCl₂ comes from BeO + C + Cl₂ at 600 to 800 K. Keep the two routes paired in your head.
Answer from the NTA final answer key. Solution by PrepWiser.