InterviewSolution
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InterviewSolution
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Given that `""^(r)Ca^(2+) = 114 ` pm and `""^(r)CO_(3)^(2-) = 185` pm , calculate the lattice energy of ` Ca CO_(3)`. |
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Answer» Applying Kapustinskii 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)`. |
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