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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

Expert verified
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 CaF鈧 is greater than 叠补贵鈧 because the smaller ionic radius of Ca虏鈦 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 Ba虏鈦 has a greater charge, creating a stronger electrostatic attraction with O虏鈦 compared to K鈦 with F鈦, despite the slightly smaller ionic radius of Ba虏鈦.

Step by step solution

01

Case (a): CaF鈧 > 叠补贵鈧

To explain this trend, we need to examine the ionic radii of the Ca虏鈦 and Ba虏鈦 ions and the F鈦 ions in both compounds. The ionic radii of Ca虏鈦 and Ba虏鈦 ions are: - Ca虏鈦: \(0.099\,nm\) - Ba虏鈦: \(0.135\,nm\) Since the ionic radius of Ba虏鈦 is larger than that of Ca虏鈦, the distance between the positive and negative ions in 叠补贵鈧 is greater than in CaF鈧. As a result, the electrostatic attraction between the ions in 叠补贵鈧 is weaker than in CaF鈧. Therefore, the lattice energy of CaF鈧 is greater than 叠补贵鈧.
02

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鈦, Rb鈦, and Cs鈦 ions are: - Na鈦: \(0.102\,nm\) - Rb鈦: \(0.148\,nm\) - Cs鈦: \(0.169\,nm\) 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鈦 and Br鈦 ions are: - Cl鈦: \(0.181\,nm\) - Br鈦: \(0.196\,nm\) The ionic radius of Br鈦 is larger than that of Cl鈦, 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.
03

Case (c): BaO > KF

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

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Key Concepts

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

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 Ca虏鈦 and Ba虏鈦, we see that the Ba虏鈦 ion has a larger ionic radius of 0.135 nm, compared to Ca虏鈦 at 0.099 nm.

This difference in size means that in a lattice structure, Ba虏鈦 ions are held at a greater distance from neighboring anions than Ca虏鈦 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, CaF鈧 has greater lattice energy compared to 叠补贵鈧 mainly due to the smaller ionic radius of Ca虏鈦 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鈦 and Cl鈦 ions have a smaller ionic radius compared to Cs鈦 and Br鈦 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 Ba虏鈦 results in stronger attractions and higher lattice energy compared to a compound like KF with a 1+ charge on K鈦.
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 CaF鈧, 叠补贵鈧, 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.

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