Find the formula of the ionic compound made from sodium (na) and iodine (i).

             NaI

                        Sodium belongs to group 1A elements and has an atomic number of 11, hence in neutral state it will have 11 electrons and will be having electronic configuration as follow,

              Na  =  11  =  1s², 2s², 2p⁶, 3s¹

As the valence shell  is 3 and there is only 1 electrons in valence shell hence it will loose one electrons to attain noble gas configuration.

              Na⁺  =  10  =  1s², 2s², 2p⁶

Or,

              Na⁺  =  10  =  [Ne]                                ∴ Ne  =  Neon

                          While, Iodine belongs to group 7A elements and has an atomic number of 53, hence in neutral state it will have 53 electrons and will be having electronic configuration as follow,

              I  =  53  =  1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁵

As the valence shell  is 5 and there are seven electrons in valence shell hence it will gain one electrons to attain noble gas configuration.

              I⁻¹  =  54  =  1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶

Or,

              I⁻¹  =  54  =  [Xe]                                ∴ Xe  =  Xenon

In order to make a neutral ionic compound between Na⁺ and I⁻¹ we will take one Na⁺ and one I⁻¹ because one -ve charges is required to neutralize one +1 charge. Therefore,

                                   (1 × Na⁺) + (1 × I⁻¹)  =  NaI

c8h10 is your answerr