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Chapter 2: Problem 22

Which shows the highest lattice energy? (a) \(\mathrm{RbF}\) \(\square\) (b) CsF (c) \(\mathrm{NaF}\) \(\square\) (d) \(\mathrm{KF}\)

Short Answer

Expert verified
(c) NaF has the highest lattice energy.

Step by step solution

01

Understand Lattice Energy

Lattice energy is the energy released when one mole of an ionic crystalline compound is formed from gaseous ions. It depends on the charge density of the ions: higher charges and smaller ionic radii result in higher lattice energies.
02

Compare Ion Sizes

Identify the cations in each compound: Rb in RbF, Cs in CsF, Na in NaF, and K in KF. Recall that the smaller the ion, the higher the charge density. Arranging these based on size from smallest to largest, we have: Na鈦�, K鈦�, Rb鈦�, Cs鈦�.
03

Analyze Ionic Charges

All the compounds have the same anion, \(F^-\), with a charge of -1, and the cations \(Na^+\), \(K^+\), \(Rb^+\), and \(Cs^+\) all have a charge of +1. Since charge is the same, size becomes the main determining factor of lattice energy.
04

Determine Highest Lattice Energy

Higher lattice energy correlates with smaller ions since they can pack more closely together, increasing electrostatic attraction. Therefore, among options with the same charge, Na鈦�, having the smallest radius, means NaF has the highest lattice energy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ionic Radii
Ionic radii refer to the size of an ion, which is crucial in understanding many properties of ionic compounds. In general, when considering the cations involved like Na鈦�, K鈦�, Rb鈦�, and Cs鈦�, these ions increase in size as you move down the periodic table. A key principle is that smaller ions can get closer to one another, thereby increasing the electrostatic attractions between them.

This increase in proximity means that a compound with smaller ions will typically have a higher lattice energy. When comparing ionic compounds such as NaF, KF, RbF, and CsF, the radii of the alkali metal cations can be arranged in order of increasing size: Na鈦� < K鈦� < Rb鈦� < Cs鈦�. Na鈦�, being the smallest ion, results in stronger attractions and, consequently, a higher lattice energy for NaF.
Charge Density
Charge density is a measure of how much electric charge is present per unit volume of an ion. This concept is important when evaluating the strength of interactions in ionic compounds. Higher charge density typically implies stronger electrostatic attractions because more charge is concentrated within a smaller area.

With ions like Na鈦�, K鈦�, Rb鈦�, and Cs鈦�, even though they all carry a +1 charge, the smaller ionic radius of Na鈦� means it has a higher charge density. Thus, in a series of compounds like NaF, KF, RbF, and CsF, Na鈦� will exert a stronger attractive force on the F鈦� ions due to its higher charge density, leading to a higher lattice energy in NaF compared to the others.
Electrostatic Attraction
Electrostatic attraction is the force that holds oppositely charged ions together in an ionic lattice. This force is influenced by both the charges of the ions and the distance between them. A classic example is the attraction between Na鈦� and F鈦� in sodium fluoride (NaF).
  • Charge: The higher the charge of the ions, the stronger the attraction.
  • Distance: The smaller the distance between ions, the stronger the attraction.
For ions with the same charge, closer proximity due to smaller ionic radii results in greater electrostatic forces. In NaF, the small size of the Na鈦� ion allows it to get closer to the F鈦� ion, resulting in strong electrostatic attraction and, therefore, high lattice energy. This principle is central to understanding why NaF has a higher lattice energy than compounds with larger cations like CsF.

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Element \(X\) is strongly electropositive and \(Y\) is strongly electronegative. Both are univalent. The compound formed would be: (a) \(X^{+} Y^{-}\) \(\square\) (b) \(X-Y\) (c) \(X^{-} Y^{+}\) \(\square\) (d) \(X \rightarrow Y\)
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