Chapter 7.2, Problem 9PP

Interpretation Introduction

Interpretation:

The formation of ionic compound from elements Strontium and fluorine needs to be explained.

Concept introduction:

Chemical compounds are formed when the atoms, ions or molecules of two substances are attracted towards each other forming a chemical bond. This bond can be electrostatic when two oppositely charged ions are attracted towards each other and this bond is said to be ionic bond.

Answer to Problem 9PP

1 Strontium will lose 2 electrons which is gained by 2 of Fluorine atoms to form ionic compound, ${\text{SrF}}_2$

Explanation of Solution

The formation of Strontium Fluoride is when Strontium atoms interact with fluorine atom.

The electronic configuration of Strontium is ${\text{1s}}^{\text{2}} {\text{2s}}^{\text{2}} {\text{2p}}^{\text{6}} {\text{3s}}^{\text{2}} {\text{3p}}^{\text{6}} {\text{4s}}^{\text{2}} {\text{3d}}^{\text{1}0}{\text{4p}}^{\text{6}} {\text{5s}}^{\text{2}}$ or $\left[\text{Kr}\right]{\text{5s}}^2$

Strontium atom has two outermost electron which can be lost to become a positive ion.

Figure 1

The electronic configuration of fluorine is ${\text{1s}}^2{\text{ 2s}}^2{\text{ 2p}}_{\text{x}}{}^2{\text{ 2p}}_{\text{y}}{}^2{\text{ 2p}}_{\text{z}}{}^1$

Fluorine atom has 1 half filled shells of p orbital. To fill up the half filled shell, it can take up 1 electron to become a fluoride anion.

Figure 2

The 2 electrons lost by Strontium can be taken up by 2 of Fluorine atoms and form ionic compound. The Strontium is positively charged and Fluorine ion is negatively charged which gets attracted to form ionic compound.

Hence, we can say that one Strontium atom bonds with 2 Fluorine atoms to form ionic compound. The 2 electrons of Strontium gets gained by Fluorine to get an octet structure and form ionic compound, ${\text{SrF}}_2$

Conclusion

1 Strontium will lose 2 electrons which is gained by 2 of Fluorine atoms to form ionic compound ${\text{SrF}}_2$