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Chapter 4: Problem 19

Born-Haber cycle is used to determine (1) lattice energy (2) electron affinity (3) tonisation energy (4) any of these

Short Answer

Expert verified
The Born-Haber cycle is used to determine lattice energy, ionization energy, and electron affinity.

Step by step solution

01

- Understand the Born-Haber Cycle

The Born-Haber cycle is a thermochemical cycle used to analyze the formation of an ionic compound from its constituent elements.
02

- Identify the Components Involved

The cycle involves several steps including the sublimation of the metal, the ionization of the metal atoms, the dissociation of the non-metal molecules, and the electron affinity of the non-metal atoms.
03

- Recognize the Lattice Energy

The lattice energy is derived from the Born-Haber cycle as it represents the energy released when the ions come together to form the ionic solid.
04

- Determine What the Cycle Can Calculate

The Born-Haber cycle is used to determine the lattice energy of an ionic compound. It is also crucial in calculating other thermodynamic quantities such as ionization energy and electron affinity.
05

- Answer the Question

Given the components and purposes of the Born-Haber cycle, it can be concluded that it is used to determine lattice energy, ionization energy, and electron affinity.

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Key Concepts

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

Lattice Energy
Lattice energy is the energy released when oppositely charged ions come together to form a solid. This concept is essential in understanding ionic compounds as it shows how strong the bonds are within the compound. To calculate lattice energy using the Born-Haber cycle, you need several steps and intermediate values:
  • Sublimation energy: Energy required to convert a solid metal into gaseous atoms.
  • Bond dissociation energy: Energy to break the bond in a non-metal molecule.
  • Ionization energy: Energy to remove an electron from metal atoms.
  • Electron affinity: Energy change when non-metal atoms gain electrons.
Lattice energy is always a large negative value because it's an exothermic process. High lattice energy means a more stable ionic compound.
The Born-Haber cycle helps us visualize these energy changes, giving a clearer picture of how ionic compounds form and why they are so stable.
Electron Affinity
Electron affinity is the energy change that occurs when an electron is added to a neutral atom. This process is usually exothermic as atoms release energy when gaining an electron to achieve a stable electron configuration.
  • First Electron Affinity: Energy change when one electron is added.
  • Second Electron Affinity: Additional energy change when another electron is added.
For example, when a chlorine atom gains an electron to form a chloride ion, the electron affinity is negative because energy is released. In the Born-Haber cycle, electron affinity is a crucial step as it represents the electron gain by the non-metal part of the ionic compound.
Understanding electron affinity helps you predict how atoms will interact, showing the tendency of atoms to form anions in ionic compounds.
Ionization Energy
Ionization energy is the energy needed to remove an electron from an atom, creating a cation. It is usually endothermic, meaning the process consumes energy. The Born-Haber cycle uses ionization energy to understand the formation of metal cations in ionic compounds.
  • First Ionization Energy: Energy to remove the first electron from a neutral atom.
  • Second Ionization Energy: Energy required to remove a second electron from a +1 cation.
Metals typically have low ionization energies, making them more likely to lose electrons and form cations. This property is essential in forming ionic bonds, as the energy provided to remove the electron is later compensated by forming the ionic compound.
By analyzing ionization energies, you can predict how easily an element will form a positive ion, which is integral in completing the Born-Haber cycle.

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Most popular questions from this chapter

The energy of \(2 \mathrm{~s}\) is greater than \(\sigma^{*} 1 \mathrm{~s}\) orbital because (1) \(\sigma 2 \mathrm{~s}\) -orbital is bigger than \(\sigma\) /s-orbital. (2) \(\sigma 2 \mathrm{~s}\) is a bonding orbital, whereas \(\left.\sigma^{*}\right] \mathrm{s}\) an antibonding orbital. (3) \(\sigma 2 \mathrm{~s}\) orbital has a greater value of \(n\) that \(\sigma * 2 \mathrm{~s}\) -orbital. (4) \(\sigma 2 \mathrm{~s}\) orbital is formed only after \(\sigma \mid \mathrm{s}\).

The bonds between \(\mathrm{P}\) atoms and \(\mathrm{Cl}\) atoms in \(\mathrm{P} \mathrm{Cl}_{5}\) are likely to be (1) ionic with no covalent character (2) covalent with some ionic character (3) covalent with no ionic character (4) ionic with some metallic character

The lone pair of electrons in a molecule spread out (1) more than the bond pair electrons (2) less than the bond pair electrons (3) as much as the bond pair electrons (4) only when they are present in oxygen atoms

The bond angles of \(\mathrm{N} 1 \mathrm{I}_{3}, \mathrm{NlI}_{4}\) and \(\mathrm{NII}_{2}\) are in the order (1) \(\mathrm{NII}_{2}^{-}>\mathrm{NII}_{3}>\mathrm{NII}_{4}^{+}\) (2) \(\mathrm{NII}_{4}^{+}>\mathrm{NII}_{3}>\mathrm{NII}_{2}^{-}\) (3) \(\mathrm{NII}_{3}>\mathrm{NII}_{2}>\mathrm{NII}_{4}^{-}\) (4) \(\mathrm{NII}_{3}>\mathrm{NII}_{4}^{+}>\mathrm{NII}_{2}^{-}\)

Which of the following statement is wrong? (1) Bond length decreases with increase in \(\mathrm{p}\) -character, (2) Bond energy of \(\mathrm{HF}\) is the highest among hydrogen halides. (3) Bond energy decreases with increase in the lone pair electrons on bonded atoms. (4) Bond energy of a covalent bond increases with increase in the difference of electronegativities of bonded atoms.
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