Answer :
Answer:
The lattice energy decreases down the group and increases across the period, therefore, the lattice energy of LiF > lattice energy of LiI, and the lattice energy of NaCl < the lattice energy of AlCl₃
Explanation:
The given parameters are;
Compound [tex]{}[/tex] Lattice Energy (kJ/mol)
LiF [tex]{}[/tex] 1030
NaCl [tex]{}[/tex] 788
MgCl₂ [tex]{}[/tex] 2326
LiI [tex]{}[/tex] 730
AlCl₃ [tex]{}[/tex] 5376
Lattice is the energy that must be added in order to split one mole of a ionic solid compound into gaseous ions
Given that the force of attraction between charges is inversely proportional to the square of the distance separating the charges, we have that as the atomic radius increases progressively down the group, the lattice energy decreases, therefore, LiF which has a shorter inter nuclear distance than LiI due to fluorine, F, being higher up on the periodic table than iodine, I, has a greater lattice energy than LiI
Similarly, given that the atomic radius decreases across the period, and that sodium, Na, comes before aluminum, Al, as we move across period 3, of the periodic table, the inter nuclear distance between atoms of in NaCl is larger than the internuclear distances between the atoms of AlCl₃, as well as the presence of more chlorine atoms in AlCl₃ than in NaCl, the lattice energy of NaCl is smaller than the lattice energy of AlCl₃.