What are some examples of ionic formulas?

Strontium Fluoride is an Ionic compound , with formula #Sr# #F_2#.

Let us see how we got the answer; Look at the electronic arrangement of Sr and F atom.

Sr ( Z= 38) has electrons with following electronic configuration. 1#s^2#2#s^2#2#p^6#3#s^2#3#p^6#4#s^2#3#d^10#4#p^6#5#s^2#

It loses two electron in its 5s subshell to achieve stability and forms ion #Sr^(2+)#.

#Sr^(2+)# 1#s^2#2#s^2#2#p^6#3#s^2#3#p^6#4#s^2#3#d^10#4#p^6#

F ( Z=9) on the other hand has nine electrons and wants to gain one more electron to achieve stable noble gas configuration.Fluorine atom on gaining one electron forms negative fluoride ion, #F^-# ion.

F( Z=9) = 1#s^2#2#s^2#2#p^5#

#F^-# = 1#s^2#2#s^2#2#p^6#

Two Fluorine atoms gains one electrons each ( total of two) from one Sr atoms, each Sr atom loses two electrons ( one each) to two chlorine atoms , in this process each Sr atom becomes #Sr^(2+)# ion and each fluorine atom after gaining one electron becomes #F^-# ion.

so in all we have one #Sr^(2+)# ions and two fluoride #F^-# ions.
so the formula becomes Sr#F_2#.

Look at the electronic arrangement of Ca and N atom.

Ca ( Z= 20) has 20 electrons with following electronic configuration. 1#s^2#2#s^2#2#p^6#3#s^2#3#p^6#4#s^2#.

It loses two electron in its 4s subshell to achieve stability and forms ion #Ca^(2+)#.

#Ca^(2+)# = 1#s^2#2#s^2#2#p^6#3#s^2#3#p^6#

N ( Z=7) on the other hand has seven electrons and wants to gain three electrons to achieve stable noble gas configuration.Nitrogen atom on gaining three electrons forms negative nitride ion, #N^-3# ion.

N ( Z=7) = 1#s^2#2#s^2#2#p^3#

#N^-3# = 1#s^2#2#s^2#2#p^6#

Two Nitrogen atoms gains three electrons each ( total of six) from three Ca atoms, each Ca atom loses two electrons (total of six) to two Nitrogen atom , in this process each Ca atom becomes #Ca^(2+)# ion and nitrogen atom after gaining three electrons becomes #N^-3# ion.

so in all we have three #Ca^(2+)# ions and two nitride #N^-3# ions.
so the formula becomes #Ca_3# #N_2#.