Why is the Lattice energy of #MgO# greater than #MgF_2# even though the size of Oxygen is greater than Fluorine?

1 Answer

The major factors are ionic charge and empirical formula, not ionic size. That is, the influence of ion charge magnitude is larger than the influence of ionic radius.

Explanation:

The Kaspertinskii equation gives a good estimate of the lattice energy #E_text(L)# in kilojoules per mole (usually within 5 % of the actual value).

#E_text(L) = (121400nz_text(+)z_text(-))/(r_text(+) + r_text(-))(1 - 34.5/(r_text(+) + r_text(-)))#

where

#n# is the number of ions in the empirical formula of the solid
#z_text(+)# and #z_text(-)# are the numbers of charges on the cation and anion, and
#r_text(+)# and #r_text(-)# are the radii of the cation and anion in picometres

The lattice energy of #"MgO"#

#n = 2#
#z_text(+) = 2#
#z_text(-) = 2#
#r_text(+) = "72 pm"#
#r_text(-) = "140 pm"#

#E_text(L) = (121400 × 2 × 2 × 2)/(72 + 140) (1 -34.5/(72 + 140)) = 971200/212(1 - 34.5/212) = 4581(1- 0.1627) = 4581 × 0.8373 = "3840 kJ·mol"^"-1"#

The observed value is #"3791 kJ·mol"^"-1"#.

The lattice energy of #"MgF"_2#

#n = 2#
#z_text(+) = 2#
#z_text(-) = 1#
#r_text(+) = "72 pm"#
#r_text(-) = "133 pm"#

#E_text(L) = (121400 × 3 × 2 × 1)/(72 + 133) (1 - 34.5/(72 + 133)) = 728400/205(1 - 34.5/205) = 3817(1- 0.1683) = 3817 × 0.8317 = "3170 kJ·mol"^"-1"#

The observed value is #"2957 kJ·mol"^"-1"#.