Final answer:
The compound KBr is expected to have the lowest lattice energy among the options given because it has ions with a +1 and -1 charge and a larger anion compared to KCl, leading to a greater internuclear distance.
Step-by-step explanation:
To identify the compound with the lowest magnitude of lattice energy among KCl, KBr, SrO, and CaO, we need to understand that lattice energy is directly proportional to the product of the ionic charges and inversely proportional to the internuclear distance between ions. Stronger charges and smaller distances typically mean higher lattice energy.
Among the given options, KCl and KBr consist of a +1 cation (K+) with -1 anions (Cl- and Br-, respectively), SrO contains a +2 cation (Sr2+) and a -2 anion (O2-), and CaO also has a +2 cation (Ca2+) and -2 anion (O2-). Given that K+ has a +1 charge and Cl- and Br- have a -1 charge, these compounds will have lower lattice energies than SrO or CaO, which have ions with +2 and -2 charges.
Between KCl and KBr, the bromide ion (Br-) is larger than the chloride ion (Cl-), leading to a greater internuclear distance in KBr. Hence, KBr is expected to have the lowest lattice energy because its ions have both a smaller charge and a larger distance between them compared to SrO and CaO.