Question

Find the formula of the ionic compound made from strontium (Sr) and fluorine (F).

Answer 1

SrF₂

Step-by-step 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

Answer 2

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.