Respuesta :

kenmyna

 The  formula  of ionic  compound made  from  strontium (sr)  and  fluorine (F)  is SrF2


Explanation

  • The ionic compound    are  compound made up of ions.
  • This ions  are  atoms or ions  that  gains or lose electrons  giving them  a positive  or negative charge   respectively.
  • Metal tends  to loss  electrons while nonmetal  gain  electrons.
  • In SrF2,  Sr( a metal)   lose two  electrons while  two  F (non metal) gain  one electron each to form an ionic compound.
transitionstate

Answer:

             SrF₂

Explanation:

                        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

Conclusion:

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

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