Question

What is the formula of the compound made from beryllium and bromine?

Answer 1

Hint: We are asked about the formula of the compound which is formed by the reaction of beryllium with bromine so we have to check what happens in this reaction. Also it should be known to us that when solid beryllium reacts with liquid bromine, they yield beryllium bromide as the product.

Complete answer:
Let us first discuss a bit about beryllium and bromine followed by their reaction and find its formula:-
Beryllium: It is a divalent element with the symbol Be and has atomic number 4. It is comparatively a rare element in the universe which usually occurs as a product of the spallation of larger atomic nuclei that have collided with cosmic rays.
Bromine: Bromine belongs to the halogen group with the symbol Br and has atomic number 35. It is a fuming red-brown liquid at room temperature that vaporizes readily to form a similarly colored gas. Its properties are between those of chlorine and iodine which also belongs to the halogen group as well.
-When beryllium metal reacts with liquid bromine, they form beryllium bromide. Since beryllium is a divalent atom and bromine is a monovalent atom, to complete their octet the formula would be:$BeB{{r}_{2}}$. The Chemical equation form of this reaction is shown below:-
$Be(s)+B{{r}_{2}}(l)\to BeB{{r}_{2}}$
-Hence the formula of the compound made from beryllium and bromine is: $BeB{{r}_{2}}$.

Note:
-The above reaction takes place at a temperature of about 500${}^\circ C$to 700${}^\circ C$.
-It is also formed when beryllium oxide is treated with hydrobromic acid as shown below:-
$Beo+2HBr\to BeB{{r}_{2}}+{{H}_{2}}O$
-Also remember that beryllium bromide is very hygroscopic and dissolves well in water.