Answer:
The correct option is;
d) LiF
Explanation:
The Lattice energy of ionic compounds can be calculated using the modified Coulomb's law as follows;
[tex]U = \dfrac{k' \cdot Q_1\cdot Q_2}{r_0}[/tex]
Where;
Q₁ and Q₂ = The ionic charges
k' = A constant depending on the ionic and valence configuration
r₀ = The distance between the two nucleus
Given the four compounds each of the metallic ions are of the same group (group one) have the same (only one) valence electron, and are combined to the same ion (fluoride ion) such that Q₂, is the same for all compounds, the factor that will influence the lattice energy more will be the r₀
The sizes of the atoms Li, Na, K and Cs increases as arranged as such Li has the least r₀, and therefore LiF the least lattice energy
From chemistry test online, we have;
The lattice energy of LiF = -1045 kJ/mole
The lattice energy of NaF = -924 kJ/mole
The lattice energy of KF = -822 kJ/mole
The lattice energy of CsF = -737 kJ/mole.
