Final answer:
The lattice energy of rubidium iodide (RbI) can be calculated using the Born-Haber cycle and is influenced by the charges on the ions and internuclear distance. However, actual numeric calculation requires specific constants that are not provided. RbI is expected to have a lower lattice energy compared to compounds with higher ionic charges.
Step-by-step explanation:
To calculate the lattice energy of rubidium iodide (RbI), we must first understand the structure of its cubic unit cell, which consists of iodide ions at the corners and a rubidium ion in the center. The formula of the compound, derived from this configuration, is RbI. Using the Born-Haber cycle, we can calculate the lattice energy by considering the relevant enthalpies such as sublimation, ionization energy, bond dissociation energy, and the standard enthalpy of formation.
The formula for lattice energy (U) can be given as:
U = rac{k' * Q1 * Q2}{r0}
Where Q1 and Q2 are the charges on the rubidium and iodide ions, respectively, and r0 represents the internuclear distance.
Rubidium iodide should have charges of +1 and -1. Given that the charge product for RbI is 1*(-1), it's expected to have a relatively lower lattice energy compared to other compounds with higher ionic charges, such as GaP with a charge product of +3*(-3).
Calculating the actual lattice energy would require specific numerical values for the constants involved, which are not provided.
However, comparing with compounds of known lattice energies such as cesium fluoride or barium oxide can give us a conceptual understanding.