Find the formula of the ionic compound made from strontium (sr) and fluorine (f).

             SrF₂

                        Strontium belongs to group 2A elements and has an atomic number of 38, hence in neutral state it will have 38 electrons and will be having electronic configuration as follow,

              Sr  =  38  =  1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s²

As the valence shell  is 5 and there are only 2 electrons in valence shell hence it will loose two electrons to attain noble gas configuration.

              Sr²⁺  =  36  =  1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶

Or,

              Sr²⁺  =  36  =  [Kr]                                ∴ Kr  =  Krypton

                          While, Fluorine belongs to group 7A elements and has an atomic number of 9, hence in neutral state it will have 9 electrons and will be having electronic configuration as follow,

              F  =  9  =  1s², 2s², 2p⁵

As the valence shell  is 2 and there are seven electrons in valence shell hence it will gain one electron to attain noble gas configuration.

              F⁻¹  =  10  =  1s², 2s², 2p⁶

Or,

              F⁻¹  =  10  =  [Ne]                                ∴ Ne  =  Neon

In order to make a neutral ionic compound between Sr²⁺ and F⁻¹ we will take one Sr²⁺ and two F⁻¹ because two -ve charges are required to neutralize one +2 charge. Therefore,

                                   (1 × Sr²⁺) + (2 × F⁻¹)  =  SrF₂

1 is the answer to this problem