Final answer:
To determine the order of ionic compounds from greatest to least lattice energy, consider the charges and sizes of the ions involved. All compounds in the question have a Br⁻ ion and a metal ion from Group 2 with a +2 charge. The order is BeBr₂, MgBr₂, CaBr₂, SrBr₂, BaBr₂, from strongest to weakest lattice energy, based on the increasing ionic size down the group.
Step-by-step explanation:
The question regarding the ordering of compounds from greatest to least lattice energy is based on the principles of lattice energies of ionic compounds. Lattice energy is influenced by the charge on the ions and the size of the ions. A high product of ionic charges will result in a high lattice energy, while a larger internuclear distance will result in a lower lattice energy.
For the compounds given: BeBr₂, CaBr₂, BaBr₂, SrBr₂, and MgBr₂, we can predict the order of lattice energy by considering the charges on the metal ions and the size of the ions. All these compounds contain the Br⁻ ion, so the difference in lattice energy will mainly be due to the metal ions Be²⁺, Ca²⁺, Ba²⁺, Sr²⁺, and Mg²⁺. Since all metal ions carry a +2 charge, the deciding factor will be the ionic radius. Smaller ions will result in higher lattice energies.
The metal ions in the given compounds are all from Group 2 of the periodic table, which means as we go down the group, the size of the ions increases. Therefore, we can expect that the compound with Be²⁺ will have the highest lattice energy since beryllium is the smallest ion, followed by Mg²⁺, Ca²⁺, Sr²⁺, and Ba²⁺ having the least since barium is the largest ion. Hence, the order from greatest to least lattice energy is:
- BeBr₂
- MgBr₂
- CaBr₂
- SrBr₂
- BaBr₂