Final answer:
The compounds MgO, KI, and BaO are placed in order of increasing lattice energy based on ionic charges and sizes: KI has the lowest, followed by BaO, and MgO has the highest due to its ions having higher charges and smaller sizes.
Step-by-step explanation:
To put the compounds MgO, KI, and BaO in order of increasing magnitude of lattice energy, we need to consider both the charge of the ions and the size of the ions within the lattice structure. Lattice energy is directly proportional to the product of the charges of the ions and inversely proportional to the distance between the ions, which relates to the ionic sizes.
KI contains K+ and I- ions, with a +1 and -1 charge respectively, giving it the lowest magnitude of lattice energy due to the singly charged ions. BaO contains Ba2+ and O2- ions. Due to the larger size of the Ba2+ ion relative to Mg2+ in MgO, we expect BaO to have a lower lattice energy compared to MgO. MgO contains Mg2+ and O2- ions, which are both highly charged (+2 and -2) and smaller in size than Ba2+ and O2- in BaO, resulting in the highest magnitude of lattice energy among the three compounds.
The correct order, from lowest to highest lattice energy, is therefore: KI < BaO < MgO. This is because the higher the charge and the smaller the ionic sizes, the greater the lattice energy. Since KI has the lowest charge and MgO has ions with higher charges and smaller sizes, MgO will have the highest lattice energy.