BeO reacts with HF in presence of ammonia to give [A] which on...

$BeO + 2{NH}_{3} + 4HF \to {\left({NH}_{4}\right)}_{2}{BeF}_{4} + {H}_{2}O$

Compound $A$, is ${\left({NH}_{4}\right)}_{2}{BeF}_{4}$

${\left({NH}_{4}\right)}_{2}{BeF}_{4}\overset{heat}{\to}{BeF}_{2} + 2{NH}_{4}F$

Compound B is ${BeF}_{2}$

In ${\left({NH}_{4}\right)}_{2}{BeF}_{4}$ the oxidation state of $Be$ is $+ 2$.