Question

An ionic compound forms when calcium (Z = 20) reacts with iodide (Z = 53). If a sample of the compound contains $\mathbf{7}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}$calcium ions, how many iodide ions does it contain?

Short answer

The amount of iodide ions in the sample is$\mathbf{14}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}$

Step-by-step solution

 Step 1: Identify the chemical formula of the ionic compound

The atomic number of calcium (Ca) is 20, so the electronic configuration of Cais 2, 8, 8, 2. So, for attaining a more stable state, Ca easily loses two electrons and form a positive ion, that is,${\mathbf{Ca}}^{\mathbf{+}\mathbf{2}}$

The atomic number of iodine (I) is 53, so the electronic configuration ofI is 2, 8, 18, 8, and 7. Thus, for attaining stability I will gain one electron and form a negative ion, that is,${\mathbf{I}}^{\mathbf{-}}$

Hence, the chemical formula of the ionic compound is${\mathbf{Cal}}_{\mathbf{2}}$.

Relation between the calcium ion and the iodide ions present in the ionic compound

In ${\mathbf{CaI}}_{\mathbf{2}}$the compound, 1 mole of calcium ion is equal to the 2 moles of iodide ions.

So,the amount of iodide ions is twice the amount of calcium ions.

Calculate the amount of iodide present in the ionic compound

The number of calcium ions present in the compound is$\mathbf{7}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}$

$$\begin{gathered} \mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{I}}^{\mathbf{-}}\mathbf{}\mathbf{ions}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{2}\mathbf{\times }\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{Ca}}^{\mathbf{+}\mathbf{2}}\mathbf{}\mathbf{ions} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathbf{\times }\mathbf{7}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{14}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}} \end{gathered}$$

Therefore, the sample contains $\mathbf{14}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}$iodide ions.