Using orbital notation, diagram the formation of an ionic bond between aluminum and fluorine.

In order to make the ionic compound between two given elements, one should first decide to select which element an an anion (containing negative charge) and which element as cation (containing positive charge).

In given statement Fluorine being more electronegative (4.0) than Aluminum (1.61) will bear negative charge while, Aluminium will attain positive charge.

Formation of F⁻ Ion:

                                The atomic number of Fluorine is 9. Hence in neutral state it will carry 9 electrons with following electronic configuration,

                                       1s², 2s², 2px², 2py², 2pz¹

Therefore, in order to complete its octet, Fluorine will accept one electron as,

                                       1s², 2s², 2px², 2py², 2pz²

Or,

                                       1s², 2s², 2⁶

Or,

                                      F⁻  =  [He], 2s², 2⁶

Formation of Al³⁺ Ion:

                                   The atomic number of Aluminium is 13. Hence in neutral state it will carry 13 electrons with following electronic configuration,

                                       1s², 2s², 2⁶, 3s². 3p¹

Therefore, in order to complete its octet, Aluminium will loose three electron as,

                                       1s², 2s², 2⁶

Or,

                                      Al⁺³  =  [He], 2s², 2⁶

Formation of AlF₃:

                                 In order to have a neutral ionic compound, three ions of Fluoride containing -1 charge each will bond with a single cation of Aluminium containing +3 charge. Hence,

                                            3 F⁻ +  Al⁺³  =  AlF₃

it's c. elements are placed in rows based on how may valencies electrons have.