Answer:
Lattice energy of the ionic compounds takes the following decline in trend: [tex]AlCl_{3} > AlBr_{3} < MgBr_{2} < LiBr[/tex] i.e AlCl3, AlBr3, MgBr3, LiBr.
Explanation:
The factors that determine lattice energies of ionic compound are:
Across the period (on the periodic table), Mg has a higher ionic radius (smaller lattice energy) than Al. Conversely from [tex]F^{-}[/tex] to [tex]I^{-}[/tex] there is increase in lattice energies caused by increase in ionic radii down the group.
Hence the order of lattice energies from largest to smallest is:
[tex]AlCl_{3} > AlBr_{3} < MgBr_{2} < LiBr[/tex]
The correct size of decreasing expected lattice energy is; LiBr, MgBr2, AlBr3, AlCl3.
The lattice energy refers to the energy released when two isolated ions form a crystal lattice. The lattice energy depends on the size of the combining ions. The smaller the size of the ions the larger the lattice energy.
Looking at the compounds shown, we can see that in terms of size, Al^3+< Mg^2+ <Li^+. For the anions, Cl^- <Br-. Hence the correct size of decreasing expected lattice energy is; LiBr, MgBr2, AlBr3, AlCl3.
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