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We need to identify the best method for preparation of $$BeF_2$$ from the given options.
$$BeO + C + F_2 \xrightarrow{\Delta} BeF_2$$
This method involves the use of carbon as a reducing agent alongside fluorine gas. The reaction is not commonly used because handling fluorine gas at high temperatures with carbon is hazardous and impractical. The conditions are extreme and the reaction is difficult to control.
$$(NH_4)_2BeF_4 \xrightarrow{\Delta} BeF_2 + 2NH_4F$$
This is the thermal decomposition of ammonium tetrafluoroberyllate. On heating, the ammonium salt decomposes cleanly to give pure $$BeF_2$$ and $$NH_4F$$ (which sublimes away). This is the preferred industrial method because it avoids handling dangerous fluorine gas directly, and it produces high-purity $$BeF_2$$.
$$Be + F_2 \xrightarrow{\Delta} BeF_2$$
While this is a straightforward direct synthesis, the reaction between beryllium metal and fluorine gas is extremely vigorous, exothermic, and dangerous. It is difficult to control, making it impractical for preparative purposes.
$$BeH_2 + F_2 \xrightarrow{\Delta} BeF_2 + H_2$$
Beryllium hydride itself is unstable and difficult to prepare and handle. Using it as a starting material adds unnecessary complexity.
The thermal decomposition of ammonium tetrafluoroberyllate (Option 2) is the preferred method because:
(i) It avoids direct handling of reactive fluorine gas with metals.
(ii) The decomposition is clean and controllable.
(iii) The product obtained is of high purity.
The correct answer is Option (2).
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