How can I write the formula for strontium fluoride?

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How can I write the formula for strontium fluoride?

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Answer 1 · 2014-07-03T06:31:57

The correct answer is $S r$ $Sr$ $F_{2}$ $F_2$ .

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 $s^{2}$ $s^2$ 2 $s^{2}$ $s^2$ 2 $p^{6}$ $p^6$ 3 $s^{2}$ $s^2$ 3 $p^{6}$ $p^6$ 4 $s^{2}$ $s^2$ 3 $d^{10}$ $d^10$ 4 $p^{6}$ $p^6$ 5 $s^{2}$ $s^2$

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

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

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, $F^{-}$ $F^-$ ion.

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

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

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 $S r^{2 +}$ $Sr^(2+)$ ion and each fluorine atom after gaining one electron becomes $F^{-}$ $F^-$ ion.

so in all we have one $S r^{2 +}$ $Sr^(2+)$ ions and two fluoride $F^{-}$ $F^-$ ions.
so the formula becomes Sr $F_{2}$ $F_2$ .

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How can I write the formula for strontium fluoride?

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