Chapter 7, Problem 84A

Interpretation Introduction

Interpretation:

The formula of the ionic compounds formed when ${\text{Cr}}^{\text{2+}}$ and ${\text{Cr}}^{\text{3+}}$ reacts with Fluorine and oxygen ions needs to be determined.

Concept introduction:

Ionic bonds are formed when there is complete transfer of electrons from one atom to another. These atoms either lose or gain electrons to become negatively or positively charged ions. The forces of attraction between these ions causes the ionic bond formation. Ionic compounds are formed only when there is a balance between their charges.

Answer to Problem 84A

The formula of the ionic compounds formed when ${\text{Cr}}^{\text{2+}}$ and ${\text{Cr}}^{\text{3+}}$ reacts with Fluorine and oxygen ions is ${\text{CrF}}_2$, $\text{CrO}$, ${\text{CrF}}_3$, ${\text{Cr}}_2{\text{O}}_3$.

Explanation of Solution

The element Chromium is a transition metal. Its atomic number is 24 and electronic configuration is ${\text{1s}}^2{\text{2s}}^2{\text{2p}}^6{\text{3s}}^2{\text{3p}}^6{\text{4s}}^1{\text{3d}}^5$

As there are 6 half filled shells, it can donate 2 electrons to become ${\text{Cr}}^{\text{2+}}$ and it can lose three electrons to become ${\text{Cr}}^{\text{3+}}$.

When ${\text{Cr}}^{\text{2+}}$ reacts with Fluorine ion, ${\text{F}}^{\text{-}}$ and oxygen ion, ${\text{O}}^{\text{2-}}$ it will form

When ${\text{Cr}}^{\text{3+}}$ reacts with Fluorine ion, ${\text{F}}^{\text{-}}$ and oxygen ion, ${\text{O}}^{\text{2-}}$ it will form

Conclusion

The formula of the ionic compounds formed when ${\text{Cr}}^{\text{2+}}$ and ${\text{Cr}}^{\text{3+}}$ reacts with Fluorine and oxygen ions is ${\text{CrF}}_2$, $\text{CrO}$, ${\text{CrF}}_3$, ${\text{Cr}}_2{\text{O}}_3$.