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

(A) $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{BeF}_{4} \xrightarrow{\Delta} \mathrm{BeF}_{2}$

(B) $\mathrm{BeH}_{2}+\mathrm{F}_{2} \xrightarrow{\Delta} \mathrm{BeF}_{2}$

(C) $\mathrm{Be}+\mathrm{F}_{2} \xrightarrow{\Delta} \mathrm{BeF}_{2}$

(D) $\mathrm{BeO}+\mathrm{C}+\mathrm{F}_{2} \xrightarrow{\Delta} \mathrm{BeF}_{2}$

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · s-Block Elements

Better method for preparation of

BeF_{2}

, among the following is

(A) $(NH_{4})_{2} BeF_{4} Δ BeF_{2}$ ✓ Correct
(B) $BeH_{2} + F_{2} Δ BeF_{2}$
(C) $Be + F_{2} Δ BeF_{2}$
(D) $BeO + C + F_{2} Δ BeF_{2}$

Answer: (A)

Solution

Beryllium fluoride is best obtained by the thermal decomposition of ammonium tetrafluoroberyllate: on heating,

(NH X_{4}) X_{2} BeF X_{4}

loses

NH X_{4} F

and leaves

BeF X_{2}

. This route is preferred because it proceeds smoothly and gives pure, anhydrous

BeF X_{2}

, whereas the direct action of fluorine on beryllium metal or on its hydride is violent and difficult to control. The better preparation is therefore the thermal decomposition of

(NH X_{4}) X_{2} BeF X_{4}

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Single correct option

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JEE Main · 2023 · 13 Apr 2023 · Shift 2 · s-Block Elements

Solution

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