Answer :
Because the attractive forces are governed by the relationship [tex]F = \frac{ q_{1} q_{2}}{ r^{2} } [/tex] we know that the bond strength between the ions of opposite charge depends on the charges on the ions and the distance between the centers of the ions when they pack to form a crystal.
In the example of MgO (magnesium oxide) and NaCl, MgO has a much higher lattice energy because the ions are +2 and -2, instead of +1 and -1.
But your problem doesn't deal with the magnitude of the charge; it concerns the ionic radii.
Smaller ions are packed closer together, meaning the attractive forces are working across a smaller distance and are thus stronger. We know based on periodic trends that as you move down a group, the ion radius increases. Therefore, the lattice energy decreases.
Both Mg and Ca are alkaline earth metals (group II on the periodic table). But Mg is one row above Ca, meaning its ionic radius is smaller. Therefore, its lattice energy is larger.
In the example of MgO (magnesium oxide) and NaCl, MgO has a much higher lattice energy because the ions are +2 and -2, instead of +1 and -1.
But your problem doesn't deal with the magnitude of the charge; it concerns the ionic radii.
Smaller ions are packed closer together, meaning the attractive forces are working across a smaller distance and are thus stronger. We know based on periodic trends that as you move down a group, the ion radius increases. Therefore, the lattice energy decreases.
Both Mg and Ca are alkaline earth metals (group II on the periodic table). But Mg is one row above Ca, meaning its ionic radius is smaller. Therefore, its lattice energy is larger.