Which of the following trends in lattice energy is due to differences in ionic radii? (a) $\mathrm{LiF}>\mathrm{NaF}>\mathrm{CsF},(\mathbf{b}) \mathrm{CaO}>\mathrm{KCl}\( (c) \)\mathrm{PbS}>\mathrm{Li}_{2} \mathrm{O}$

Short Answer

Expert verified
The correct answer is (a): LiF > NaF > CsF, as this trend in lattice energy is predominantly determined by the differences in ionic radii, with smaller radii leading to higher lattice energy.

Step by step solution

01

Understanding ionic radii trends

In order to find which trend is based on the differences in ionic radii, it is essential to know the general trends of the ionic radii in the periodic table. Typically, the atomic radii or ionic radii increase as one moves down a group and decrease as one moves from left to right across a period.
02

Analyze option (a)

In option (a), the compounds given are LiF, NaF, and CsF. They are all from Group 1 and have the same charge of +1 for the cations and -1 for the F anions. As Li, Na, and Cs are from the same group, we can see that the ionic radii will increase as we move down the group. However, when the ionic radii increase, the lattice energy will decrease. Therefore, the trend in option (a) is determined by the differences in ionic radii. Due to smaller radii, LiF has the highest lattice energy, followed by NaF and CsF.
03

Analyze option (b)

In option (b), the compounds given are CaO and KCl. Ca and K are from different groups and their corresponding anions (O and Cl) also belong to different groups. So, we cannot confidently determine the trend based solely on the ionic radii.
04

Analyze option (c)

In option (c), the compounds given are PbS and Li2O. Pb and Li belong to different groups, and their corresponding anions (S and O) also belong to different groups. So, similar to option (b), we cannot confidently determine the trend based solely on the ionic radii.
05

Determine the correct answer

Based on the analysis of each option, option (a) is the only trend that is predominantly determined by the differences in ionic radii. So, the correct answer is (a): LiF > NaF > CsF.

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