Question

The lattice energy of KF is 794 kJ/mol, and the interionic distance is 269 pm. The Na-F distance in NaF, which has the same structure as KF, is 231 pm. Which of the following values is the closest approximation of the lattice energy of NaF?

(a) 682 kJ/mol
(b) 794 kJ/mol
(c) 924 kJ/mol
(d) 1588 kJ/mol
(e) 3175 kJ/mol

Answer 1

Given the inversely proportional relationship between lattice energy and internuclear distance, the closest lattice energy value of NaF, based on the provided data, is approximately 924 kJ/mol. The correct option is C.

Step-by-step explanation:

The question is asking to estimate the lattice energy of NaF given the known lattice energy of KF and the corresponding interionic distances. Lattice energy is directly proportional to the charge on the ions (which are the same for KF and NaF) and inversely proportional to the internuclear distance (the distance between the ions).

Since the question provides the distance of Na-F in NaF is shorter compared to the distance of K-F in KF (231 pm compared to 269 pm respectively), we can infer that the lattice energy of NaF will be higher because of the shorter internuclear distance.

Using the Coulomb's Law approximation for the lattice energies, which suggests that they are inversely proportional to the distance between ions, we can directly compare the two given distances and energies.

Given that the ratio of the distances (269/231) is roughly 1.16 and assuming all other factors are equal, multiplying the given lattice energy of KF by this ratio gives us an estimate for the lattice energy of NaF, which would be around 924 kJ/mol (794 kJ/mol * 1.16).