Given that ""^(r)Ca^(2+) = 114pm and""^(r)CO_(3)^(2-) = 185pm , calculate the lattice energy ofCa CO_(3).

Given that ""^(r)Ca^(2+) = 114pm and""^(r)CO_(3)^(2-) = 185pm , calcul

1.	Given that ""^(r)Ca^(2+) = 114pm and""^(r)CO_(3)^(2-) = 185pm , calculate the lattice energy ofCa CO_(3).
Answer» Solution :ApplyingKapustinskii equation `U_(L) = 120250 ((v\|z_(+)\|\|z_(-) \|)/(d)) (1 - (34.5)/(d))` Where v = No. of ions per formula UNIT ` Z_(+), Z_(-) = `CHARGE on the cation and anion respectively `d = r_(+) + r_(-)`, i.e., SUM of ionic radii in pm `U_(L)` = LATTICE ENERGY in kJ `mol^(-1)` Here,` v = 2 , z_(+) = + 2, z_(-) = - 2, z_(-) = 2 , i.e., \|z_(+)\| = 2 , \|z_(-)\| = 2 ` ` d = r_(+) + r_(-) = 144 + 185 = 299 pm` `U_(L) = 120250 ((2xx2xx2)/(299))(1 - (34.5)/(299)) = 2846 " kJ mol "^(-1)`.