Question

which of the following would be expected to have the lowest melting point? ki nabr licl

Explanation:

Step1: Recall ionic melting point factors

Ionic compounds' melting points depend on lattice energy. Lattice energy ($U$) is related to ion charges ($q_+$, $q_-$) and ion radii ($r_+$, $r_-$) by $U \propto \frac{|q_+ q_-|}{r_+ + r_-}$. Smaller ions and higher charges give higher lattice energy (and melting point).

Step2: Analyze ion sizes and charges

All compounds (KI, NaBr, LiCl) have +1 and -1 charges (so $|q_+ q_-|$ is same). Now compare ion radii:

• Cation radii: $K^+ > Na^+ > Li^+$ (down a group, radius increases).
• Anion radii: $I^- > Br^- > Cl^-$ (down a group, radius increases).

Step3: Calculate sum of radii

For KI: $r(K^+) + r(I^-)$ is largest (both ions are largest among the three). For NaBr: $r(Na^+) + r(Br^-)$. For LiCl: $r(Li^+) + r(Cl^-)$ (smallest sum).

Step4: Relate to lattice energy and melting point

Lattice energy is inversely proportional to sum of radii (when charges are same). So KI has the smallest lattice energy (since sum of radii is largest), hence lowest melting point.

Answer:

KI