91Ó°ÊÓ

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${\mathbf{Cal}}_{\mathbf{2}}$.

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.

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

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

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