# How can I write the formula for strontium fluoride?

Jul 3, 2014

The correct answer is $S r$ ${F}_{2}$.

Let us see how we got the answer; Look at the electronic arrangement of Sr and F atom.

Sr ( Z= 38) has 38 electrons with following electronic configuration. 1${s}^{2}$2${s}^{2}$2${p}^{6}$3${s}^{2}$3${p}^{6}$4${s}^{2}$3${d}^{10}$4${p}^{6}$5${s}^{2}$

It loses two electron in its 5s subshell to achieve stability and forms ion $S {r}^{2 +}$.

$S {r}^{2 +}$ 1${s}^{2}$2${s}^{2}$2${p}^{6}$3${s}^{2}$3${p}^{6}$4${s}^{2}$3${d}^{10}$4${p}^{6}$

F ( Z=9) on the other hand has nine electrons and wants to gain one more electron to achieve stable noble gas configuration.Fluorine atom on gaining one electron forms negative fluoride ion, ${F}^{-}$ ion.

F( Z=9) = 1${s}^{2}$2${s}^{2}$2${p}^{5}$

${F}^{-}$ = 1${s}^{2}$2${s}^{2}$2${p}^{6}$

Two Fluorine atoms gains one electrons each ( total of two) from one Sr atoms, each Sr atom loses two electrons ( one each) to two chlorine atoms , in this process each Sr atom becomes $S {r}^{2 +}$ ion and each fluorine atom after gaining one electron becomes ${F}^{-}$ ion.

so in all we have one $S {r}^{2 +}$ ions and two fluoride ${F}^{-}$ ions.
so the formula becomes Sr${F}_{2}$.

Simple to the point here: http://brainly.com/question/419669