Verified Solution • 2025–26 Session moderate

QJEEM23Chace5: JEE Main 2023 Chemistry - Class 11 JEEM23Chace5 Solution: Better method for preparation of , among the following is :

Better method for preparation of BeF2\mathrm{BeF}_{2} , among the following is :

Multiple Choice Options

  • A BeH2+F2⟶ΔBeF2\mathrm{BeH}_{2}+\mathrm{F}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2}
  • B (NH4)2BeF4⟶ΔBeF2\left(\mathrm{NH}_{4}\right)_{2} \mathrm{BeF}_{4} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2}
  • C Be+F2⟶ΔBeF2\mathrm{Be}+\mathrm{F}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2}
  • D BeO+C+F2⟶ΔBeF2\mathrm{BeO}+\mathrm{C}+\mathrm{F}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2}

Step-by-Step Verified Solution

According to the NCERT (s block) textbook, the better method for the preparation of BeF₂ is heating (NH₄)₂BeF₄.

Option B :

(NH4)2BeF4⟶ΔBeF2+2NH4F\left(\mathrm{NH}_{4}\right)_{2} \mathrm{BeF}_{4} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2} + 2\mathrm{NH}_{4}\mathrm{F}

This method involves the thermal decomposition of ammonium beryllium fluoride ((NH₄)₂BeF₄) to form beryllium fluoride (BeF₂) and ammonium fluoride (NH₄F). The reaction is more efficient compared to other options and results in a purer product.

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