Question

choose the ionic compound with the higher boiling point.

Explanation:

Step1: Recall Ionic Bonding Theory

Ionic compounds' boiling points depend on lattice energy. Lattice energy ($U$) is related to ion charges ($q_+$, $q_-$) and ion radii ($r_+$, $r_-$) by the formula $U \propto \frac{|q_+ q_-|}{r_+ + r_-}$. Higher lattice energy means higher boiling point (stronger ionic bonds, more energy to break).

Step2: Analyze Ions in CsI and NaCl

• For $\ce{CsI}$: $\ce{Cs^+}$ (larger radius, Group 1, period 6) and $\ce{I^-}$ (larger radius, Group 17, period 5).
• For $\ce{NaCl}$: $\ce{Na^+}$ (smaller radius, Group 1, period 3) and $\ce{Cl^-}$ (smaller radius, Group 17, period 3).

Step3: Compare Lattice Energies

Both compounds have $q_+ = +1$, $q_- = -1$ (same ion charges). So lattice energy depends on ion radii sum. Smaller ion radii (NaCl: $\ce{Na^+}$ and $\ce{Cl^-}$) give smaller $r_+ + r_-$, so higher $U$ (since $U \propto \frac{1}{r_+ + r_-}$). Higher lattice energy means higher boiling point.

Answer:

NaCl (the ionic compound with the higher boiling point is NaCl)