Final answer:
MgF2 (E) requires the greatest energy input to separate the ions due to its high lattice energy, resulting from the strong electrostatic attraction between Mg2+ and F- ions.
Step-by-step explanation:
Out of the given options, the compound that will require the greatest energy input to separate the ions is MgF2 (E). This is because lattice energy is generally higher for compounds with ions that have higher charges and smaller sizes. Since Mg has a +2 charge and F has a -1 charge, MgF2 has Mg2+ and F- ions. F- is smaller than Cl-, Br-, and I- (which are found in the other compounds listed), and Mg2+ has a higher charge than Na+ in NaCl.
Therefore, due to the greater electrostatic attraction between Mg2+ and F- ions in MgF2, more energy would be required to overcome this attraction and separate the ions. This is supported by data that indicates compounds with a metal cation with a +2 charge and a nonmetal anion with a -2 charge have significantly higher lattice energies than those with +1 and -1 charged ions.