Question

What would be the compound formula of strontium Bromide?

Answer 1

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.