Question

For each of the following sets of ionic substances, arrange the members in order of increasing lattice energy (smallest first). (b)CaO, K2O, K2S, CaS

(c) ScCl3, TiCl4 , KCl, CaCl2

Answer 1

Answer 2

To arrange the members in order of increasing lattice energy, we need to consider the charges of the ions and the sizes of the ions.

(b) For the set of ionic substances: CaO, K2O, K2S, CaS

In general, the lattice energy increases with increasing charge magnitude and decreasing ion size.

CaO: Calcium (Ca) has a 2+ charge, and oxygen (O) has a 2- charge. Both ions have relatively small sizes. Therefore, the lattice energy is relatively high for CaO.

K2O: Potassium (K) has a 1+ charge, and oxygen (O) has a 2- charge. Both ions have relatively larger sizes compared to Ca and O. Therefore, the lattice energy is lower compared to CaO.

K2S: Potassium (K) has a 1+ charge, and sulfur (S) has a 2- charge. Both ions have relatively larger sizes compared to Ca and O. The lattice energy of K2S would be similar to K2O.

CaS: Calcium (Ca) has a 2+ charge, and sulfur (S) has a 2- charge. Both ions have relatively small sizes. The lattice energy of CaS would be similar to CaO.

Therefore, the arrangement of increasing lattice energy would be: K2O < K2S < CaS < CaO

(c) For the set of ionic substances: ScCl3, TiCl4, KCl, CaCl2

In this set, all the substances have chloride ions (Cl-) with a 1- charge. The difference lies in the cations.

KCl: Potassium (K) has a 1+ charge, and its ion size is larger compared to the other cations in the set. Therefore, KCl would have the lowest lattice energy.

CaCl2: Calcium (Ca) has a 2+ charge, and its ion size is smaller compared to K. Therefore, CaCl2 would have a higher lattice energy than KCl.

ScCl3: Scandium (Sc) has a 3+ charge, and its ion size is smaller compared to Ca. Therefore, ScCl3 would have a higher lattice energy than CaCl2.

TiCl4: Titanium (Ti) has a 4+ charge, and its ion size is smaller compared to Sc. Therefore, TiCl4 would have the highest lattice energy in this set.

Therefore, the arrangement of increasing lattice energy would be: KCl < CaCl2 < ScCl3 < TiCl4