Question

What would be the compound formula of strontium Bromide?

Answer 1

Hint: We know that strontium is a group 2 element. Strontium symbol is Sr. because it is a group 2 element so its valency is $ + 2 $ . And these group 2 elements are ionic. We know that bromine is a p-block element so it has $ - 1 $ valency. Hence the compound formula for strontium bromide will be $ SrB{r_2} $

Complete Step By Step Answer:
As we know that Sr is a group 2 element. Group 2 elements are known as alkaline earth metals. Since these elements' electronic configuration is $ n{s^2} $ so these elements have $ + 2 $ a positive charge. And compounds of these elements are ionic. It means that these elements make an ionic compound. Sr atomic number is $ 38 $ and its atomic mass is $ 87.62gmo{l^{ - 1}} $ .
Now, we know that Br is the p-block element and its atomic number is 35. Since its electronic configuration is $ \left[ {Ar} \right]4{s^2}3{d^{10}}4{p^5} $ so bromine has 7 valence electrons so it has $ - 1 $ charge. Bromine is also ionic. Bromine has a negative charge.
Now we will discuss the reactivity of strontium towards bromine. All the alkaline earth metals react with halogen by the simple reaction at elevated temperature.
 $ M + {X_2} \to M{X_2} $
Here M= metal, X= halogen ( F, Cl, Br, I )
The reaction of strontium with bromine-
 $ S{r_{(s)}} + B{r_{2(l)}} \to SrB{r_{2(s)}} $
Hence the formula of strontium bromide would be $ SrB{r_2} $ .
Since it is an ionic compound. It is odorless.
On flame test, it burns with a bright red color flame. It is used as flares for signaling etc.

Note:
Strontium has $ + 2 $ a positive charge and bromine has $ - 1 $ a negative charge and they combine to form an ionic compound $ SrB{r_2} $ . Since strontium is alkaline earth metal and bromine is non-metal so they form an ionic compound. $ SrB{r_2} $ has some pharmaceutical uses. It is white crystalline salt.