Question

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

Short answer

In short, the trends in lattice energy are determined by the electrostatic attraction between ions, which depends on their charges and ionic radii. (a) The lattice energy of CaF₂ is greater than BaF₂ because the smaller ionic radius of Ca²⁺ leads to stronger electrostatic attraction between the ions. (b) The lattice energy decreases from NaCl to RbBr to CsBr due to increasing ionic radii, resulting in weaker electrostatic attraction. (c) BaO has a higher lattice energy than KF because Ba²⁺ has a greater charge, creating a stronger electrostatic attraction with O²⁻ compared to K⁺ with F⁻, despite the slightly smaller ionic radius of Ba²⁺.

Step-by-step solution

Case (a): CaF₂ > BaF₂

To explain this trend, we need to examine the ionic radii of the Ca²⁺ and Ba²⁺ ions and the F⁻ ions in both compounds. The ionic radii of Ca²⁺ and Ba²⁺ ions are: - Ca²⁺: \(0.099\,nm\) - Ba²⁺: \(0.135\,nm\) Since the ionic radius of Ba²⁺ is larger than that of Ca²⁺, the distance between the positive and negative ions in BaF₂ is greater than in CaF₂. As a result, the electrostatic attraction between the ions in BaF₂ is weaker than in CaF₂. Therefore, the lattice energy of CaF₂ is greater than BaF₂.

Case (b): NaCl > RbBr > CsBr

Similar to case (a), we will examine the ionic radii of the cations and anions in these compounds. The ionic radii of Na⁺, Rb⁺, and Cs⁺ ions are: - Na⁺: \(0.102\,nm\) - Rb⁺: \(0.148\,nm\) - Cs⁺: \(0.169\,nm\) As we move down the group in the periodic table, the ionic radius increases. Therefore, the distance between the positive and negative ions in CsBr is larger compared to RbBr and NaCl. Consequently, the electrostatic attraction between the ions is weaker in CsBr than in RbBr and NaCl. Similarly, the ionic radii of Cl⁻ and Br⁻ ions are: - Cl⁻: \(0.181\,nm\) - Br⁻: \(0.196\,nm\) The ionic radius of Br⁻ is larger than that of Cl⁻, implying that the distance between the positive and negative ions in RbBr is greater than in NaCl. Hence, the lattice energy of NaCl is greater than RbBr and CsBr.

Case (c): BaO > KF

Comparing the ionic radii of Ba²⁺ and K⁺ along with the charges, we have: - Ba²⁺: \(0.135\,nm\) with charge +2 - K⁺: \(0.138\,nm\) with charge +1 Although the ionic radius of Ba²⁺ is smaller than that of K⁺, the charge of Ba²⁺ is higher, resulting in a stronger electrostatic attraction between the Ba²⁺ and O²⁻ ions compared to the K⁺ and F⁻ ions in KF. This stronger electrostatic attraction results in a higher lattice energy for BaO compared to KF.

Key concepts

Ionic Radius

The ionic radius is the measure of an ion's size, which is crucial in determining the characteristics of ionic compounds like lattice energy. An ion's size can greatly affect how closely it can pack with other ions in a solid structure. For example, when comparing ions like Ca²⁺ and Ba²⁺, we see that the Ba²⁺ ion has a larger ionic radius of 0.135 nm, compared to Ca²⁺ at 0.099 nm.

This difference in size means that in a lattice structure, Ba²⁺ ions are held at a greater distance from neighboring anions than Ca²⁺ ions. The larger ionic radius generally makes the lattice less stable, because ions are not held as tightly together. This results in lower lattice energy. Therefore, CaF₂ has greater lattice energy compared to BaF₂ mainly due to the smaller ionic radius of Ca²⁺ which allows for stronger electrostatic attractions.

Electrostatic Attraction

Electrostatic attraction refers to the force that pulls oppositely charged ions together. It is a key factor in determining lattice energy, which is the energy needed to separate a mole of an ionic solid into gaseous ions. This attraction depends on both the charges of the ions and their distances from each other.

For example, in compounds like NaCl and CsBr, the varying ionic radii lead to differences in the strength of electrostatic attractions. Na⁺ and Cl⁻ ions have a smaller ionic radius compared to Cs⁺ and Br⁻ ions. This closer proximity results in a stronger electrostatic attraction and, therefore, higher lattice energy in NaCl versus CsBr. Similarly, the charge also plays a crucial role, as seen in BaO where the 2+ charge on Ba²⁺ results in stronger attractions and higher lattice energy compared to a compound like KF with a 1+ charge on K⁺.

Ionic Compounds

Ionic compounds are formed from the electrostatic attraction between cations and anions. These compounds are usually characterized by high melting and boiling points due to the strong attractions between ions. The structure of these compounds in a solid state is a repeating pattern known as a lattice, which maximizes the attractive forces while minimizing repulsion between alike charges.

The stability and properties of ionic compounds like CaF₂, BaF₂, NaCl, and BaO largely depend on the lattice energy, which is influenced by the ionic radii and charges of the ions involved. In general, compounds with smaller and more highly charged ions, like in BaO, tend to have higher lattice energy. This makes the compound more stable, solid at room temperature, and often less soluble compared to others with larger ionic radii and lower charges, thus impacting how these compounds behave in different environments.