Question

Explain the following trends in lattice energy: (a) \(\mathrm{NaCl}>\mathrm{RbBr}>\mathrm{CsBr} ;\) (b) \(\mathrm{BaO}>\mathrm{KF}\) (c) \(\mathrm{SrO}>\mathrm{SrCl}_{2}\)

Short answer

The trends in lattice energy for the given ionic compounds can be explained as follows: (a) NaCl > RbBr > CsBr: Lattice energy decreases as the size of the cations and anions increases, which is the case in this series. (b) BaO > KF: Lattice energy is higher for BaO due to the greater positive charge of Ba²⁺ and the greater negative charge of O²⁻ compared to K⁺ and F⁻. (c) SrO > SrCl₂: SrO has a higher lattice energy than SrCl₂ because the greater negative charge of O²⁻ ion leads to stronger electrostatic attraction with Sr²⁺.

Step-by-step solution

Explaining the trend in the lattice energy of NaCl, RbBr, and CsBr

In this series, NaCl has the highest lattice energy, and CsBr has the lowest. This can be attributed to the difference in the size of the cations and anions involved. Both Na⁺ and Cl⁻ ions are smaller than Rb⁺ and Br⁻ ions, which in turn are smaller than Cs⁺ and Br⁻ ions. Since lattice energy is inversely proportional to ion size, the lattice energy of NaCl will be higher than that of RbBr, which in turn will have a higher lattice energy than CsBr.

Explaining the trend in the lattice energy of BaO and KF

In this case, BaO has a higher lattice energy than KF. This can be explained by looking at the charges of the ions. The Ba²⁺ ion has a greater positive charge than K⁺, and the O²⁻ ion has a greater negative charge than F⁻. As a result, the electrostatic attraction between Ba²⁺ and O²⁻ is stronger than that between K⁺ and F⁻, leading to a higher lattice energy in BaO compared to KF.

Explaining the trend in the lattice energy of SrO and SrCl₂

In this series, SrO has a higher lattice energy than SrCl₂. Since Sr²⁺ is a common cation in both compounds, the difference in lattice energy must be attributed to the anions involved. O²⁻ ion has a higher negative charge than Cl⁻ ions, and due to the greater electrostatic attraction between Sr²⁺ and O²⁻, SrO has a higher lattice energy compared to SrCl₂. In conclusion, the trends in lattice energy covered in this exercise can be explained by considering the charge and size of the ions involved in the ionic compounds.

Key concepts

Ionic Compounds

Ionic compounds are substances composed of positive and negative ions held together by the strong electrostatic forces between oppositely charged particles. These compounds typically form when a metal reacts with a non-metal, transferring electrons from the metal to the non-metal.

Consider sodium chloride (NaCl) as an example. Sodium (Na) loses one electron to become Na⁺, while chlorine (Cl) gains an electron to become Cl⁻. They're then bonded together to form NaCl, an ionic compound. The structure of these compounds is often a repeating 3D lattice where each ion is surrounded by ions of opposite charge, optimizing the ionic bonds and overall stability.

When comparing ionic compounds like NaCl, RbBr, and CsBr, their differences in lattice energy – the energy released when the ions come together to form a solid – can be explained by specific properties of the ions involved. These include ionic size and the magnitude of charges on the ions, both of which are crucial in determining the strength of the attraction within the compound.

Ion Size and Charge

Ion size and charge play a pivotal role in determining the lattice energy of ionic compounds. Generally, smaller ions can get closer together, resulting in a stronger attraction and higher lattice energy. Conversely, larger ions result in weaker attractions and lower lattice energies.

Here's how we apply this concept to explain variances in lattice energies:

• NaCl vs. RbBr vs. CsBr: Sodium has a smaller ionic radius compared to rubidium and cesium, leading to a stronger electrostatic attraction between Na⁺ and Cl⁻ ions, resulting in a higher lattice energy for NaCl. Additionally, the smaller size of Cl⁻ compared to Br⁻ contributes further to NaCl's higher lattice energy.
• BaO vs. KF: In BaO, the ions involved have higher charges compared to those in KF. Since Ba²⁺ is smaller than K⁺ and has a double positive charge, it can attract O²⁻, which has a double negative charge, more strongly than K⁺ can attract F⁻. This results in a stronger electrostatic force and a higher lattice energy for BaO compared to KF.

Electrostatic Attraction

Electrostatic attraction is the force that draws together positively charged cations and negatively charged anions in an ionic compound. It explains how ions with opposite charges can form stable, solid structures despite their individual charge-based tendencies to repel same-charged species.

The strength of this electrostatic attraction is governed by Coulomb's law, which states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between their centers. This means that ions with higher charges will attract each other more strongly, while increased distance caused by larger ionic sizes would weaken this attraction.

Considering SrO vs. SrCl₂ as an example, the higher lattice energy of SrO is due to the electrostatic attraction between Sr²⁺ and O²⁻ being stronger than that between Sr²⁺ and two Cl⁻ ions. This is because while Sr²⁺ remains constant in both compounds, the doubly charged O²⁻ creates a stronger attraction than the singly charged Cl⁻, following Coulomb's law.