Question

Order the following compounds according to their expected lattice energies:

Answer 1

Answer:

The order of this compound based on increasing lattice energy is

$LiBr , MgBr_2 ,AlBr_3,AlCl_3$

Explanation:

For this compounds

     $MgBr_2$ contains $Mg^{2+} \ ion \ and \ Br^{-} \ ion$

      $LiBr \ contains \ Li^{+} \ ion\ Br^- \ ion$

      $AlCl_3 \ ion \ contains \ Al^{3+} ion \ and \ Cl^- \ ion$

     $AlBr_3 \ contain \ Al^{3+} \ ion \ and \ Br^- \ ion$

Now lattice energy is a parameter that is dependent on

•      Size of the ion this implies that the higher the size of the ion the lower the lattice energy and vise versa
•   The charge on the ion this implies that the lower the charge on the ion the lower the lattice energy and vise versa

So from this the order of this compound based on increasing lattice energy is

$LiBr , MgBr_2 ,AlBr_3,AlCl_3$

Answer 2

Answer:

Lattice energy of the ionic compounds takes the following decline in trend: $AlCl_{3} > AlBr_{3} < MgBr_{2} < LiBr$ i.e AlCl3, AlBr3, MgBr3, LiBr.

Explanation:

The factors that determine lattice energies of ionic compound are:

• Product of the ionic charges (cations and anions). The higher the product of the charge, the higher the lattice energy
• Size/ionic radii. Lattice energy is inversely proportional to the size of the ions. a decrease in ionic radii results in increase in lattice energy.

Across the period (on the periodic table), Mg has a higher ionic radius (smaller lattice energy) than Al. Conversely from $F^{-}$ to $I^{-}$ 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:

$AlCl_{3} > AlBr_{3} < MgBr_{2} < LiBr$