/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 629 Consider the formation of an ion... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 629

Consider the formation of an ionic molecule, \(A B\), by the donation of an electron from atom \(\mathrm{A}\) to atom \(\mathrm{B}\). If the ionization potential of \(\mathrm{A}\) is \(5.5 \mathrm{eV}\) and the electron affinity of \(\mathrm{B}\) is \(1.5 \mathrm{eV}\), and the ion \(\mathrm{B}^{-}\) has a radius of \(0.20 \mathrm{~nm}\), what is the maximum size of \(\mathrm{A}^{+}\) that would lend itself to the formation of an energetically stable bond?

Short Answer

Expert verified
Rearranging and calculating for r: \(r = \frac{(1.6 x 10^{-19} C)^{2} (8.99 x 10^9 N m^{2} C^{-2})}{6.4 x 10^{-19} J} = 3.6 \times 10^{-10} m \) Since the radius of B鈦� is given as 0.20 nm (or 2.0 x 10鈦宦光伆 m), we can subtract it from the total distance between the ions to find the maximum size of A鈦�: Max size of A鈦� = r - radius of B鈦� = 3.6 x 10鈦宦光伆 m - 2.0 x 10鈦宦光伆 m = 1.6 x 10鈦宦光伆 m Therefore, the maximum size of A鈦� that would lend itself to the formation of an energetically stable bond is \(1.6 \times 10^{-10} m\).

Step by step solution

01

Find the energy released in the formation of the ionic bond

In order to determine the energy released during the formation of the ionic bond, we need to calculate the difference between the ionization potential of A and the electron affinity of B. This can be done using the following formula: Energy released = Ionization potential of A - Electron affinity of B Plug in the given values: Energy released = 5.5 eV - 1.5 eV = 4.0 eV
02

Convert the energy released to Joules

Since we're dealing with radius in nanometers and energy in electron volts, it would be helpful if we convert the energy released from electron volts (eV) to Joules (J). 1 eV = 1.6 x 10鈦宦光伖 J So, Energy released = 4.0 eV x 1.6 x 10鈦宦光伖 J/eV = 6.4 x 10鈦宦光伖 Joules
03

Use Coulomb's law to find the maximum size of A鈦�

Coulomb's law states that the electrostatic force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them: \(F = \frac{k q_1 q_2}{r^2}\) Where F is the electrostatic force, k is the electrostatic constant (8.99 x 10鈦� N m虏/C虏), q1 and q2 are the charges of the ions, and r is the distance between the ions. The electrostatic energy can be calculated as: Energy = F x r = \(k \frac{q_1 q_2}{r}\) For an energetically stable bond, the electrostatic energy should be equal to the energy released during the formation of the ionic bond: \Formula: 6.4 x 10鈦宦光伖 J = \(k \frac{q_1 q_2}{r}\) Rearrange the formula to solve for r (which will give us the sum of the radii): \(r = \frac{q_1 q_2 k}{6.4 x 10^{-19} J}\) In the case of A鈦� and B鈦�, the charges q1 and q2 are both equal to the elementary charge e (1.6 x 10鈦宦光伖 C): \(r = \frac{(1.6 x 10^{-19} C)^{2} (8.99 x 10^9 N m^{2} C^{-2})}{6.4 x 10^{-19} J}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Ionization Potential
The ionization potential, also known as ionization energy, is a crucial concept in understanding ionic bonding. It refers to the amount of energy required to remove an electron from a neutral atom. In our exercise, we see atom A with an ionization potential of 5.5 eV. This means it takes 5.5 electron volts to remove one electron from atom A.

The higher the ionization potential, the more energy is needed to strip away an electron. This highlights how tightly an atom holds onto its electrons. During the formation of an ionic bond, one atom will lose an electron (ionization) and another will gain (electron affinity), creating oppositely charged ions. This difference in energy levels between atoms contributes to the overall stability of the ionic compound.
  • Ionization potential is measured in electron volts (eV).
  • Energy needed to remove an electron.
  • Depends on how tightly electrons are bound to an atom.
Electron Affinity
Electron affinity is closely related to our understanding of ionic bonding as it represents the energy change when an electron is added to a neutral atom. In the exercise, atom B has an electron affinity of 1.5 eV, signifying the energy released upon gaining an electron.

Atoms with high electron affinity have a strong tendency to gain electrons and thus form negative ions. This is crucial in the formation of ionic bonds, as one atom donates and another accepts the electron. The energy released during this process aids in stabilizing the newly formed ions. The combination of ionization energy and electron affinity helps predict whether the ionic bonding process is energetically favorable.
  • Measured in electron volts (eV).
  • Energy change when an atom gains an electron.
  • Indicates an atom's propensity to become negatively charged.
Coulomb鈥檚 Law
Coulomb's law is fundamental in understanding the forces that operate within ionic bonds. It describes the electrostatic interaction between charged particles, which in this case are ions. As described in the step-by-step solution, the force between atoms A and B is governed by Coulomb's law:

\[ F = \frac{k q_1 q_2}{r^2} \]

Here, \(F\) represents the electrostatic force, \(k\) is the Coulomb's constant \( (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \), \( q_1 \) and \( q_2 \) are the charges on the ions, and \( r \) is the distance between them.
  • Helps calculate the strength of the electrostatic force.
  • Directly impacts the stability of an ionic bond.
  • Key to determining the optimal distance between ions for bond stability.
Electrostatic Force
Electrostatic force is the glue holding ionic compounds together. It's the attraction between charged ions, arising naturally once an electron is transferred from one atom to another. This force is central in determining the stability of ionic bonds.

The calculated energy via Coulomb's law reflects the electrostatic force's role in maintaining bond stability between oppositely charged ions. A stable ionic compound results when the electrostatic force balances out the energies associated with electron transfer (ionization potential and electron affinity). It tells us how close the ions need to be for a stable bond, an aspect highlighted in the exercise when determining the maximum size of \( A^{+} \).
  • A force that attracts opposite charges.
  • Crucial for the structural integrity of ionic compounds.
  • Determines the required proximity of ions for stability.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Using electron-dot notation, show for each of the following the outer shell electrons for the uncombined atoms and for the molecules or ions that result: (a) \(\mathrm{H}+\mathrm{H} \rightarrow\) Hydrogen molecule (b) \(\mathrm{Br}+\mathrm{Br} \rightarrow\) bromine molecule (c) \(\mathrm{Br}+\mathrm{Cl} \rightarrow\) bromine chloride (d) \(\mathrm{Si}+\mathrm{F} \rightarrow\) silicon fluoride (e) \(\mathrm{Se}+\mathrm{H} \rightarrow\) hydrogen selenide (f) \(\mathrm{Ca}+\mathrm{O} \rightarrow\) calcium oxide

What is the meaning and significance of the Pauling electronegativity scale?

Distinguish a metallic bond from an ionic bond and from a covalent bond.

What is the explanation for the following trends in lattice energies? \(\mathrm{NaF} \quad-260 \mathrm{Kcal} / \mathrm{mole} \quad \mathrm{NaCl}-186 \mathrm{Kcal} / \mathrm{mole}\) \(\mathrm{NaCl}-186 \mathrm{Kcal} / \mathrm{mole} \quad \mathrm{KCl}-169 \mathrm{Kcal} / \mathrm{mole}\) \(\mathrm{NaBr}-177 \mathrm{Kcal} / \mathrm{mole} \quad \mathrm{CsCl}-156 \mathrm{Kcal} / \mathrm{mole}\)

Write the Born-Haber cycle for the formation of crystalline sodium fluoride \(\left(\mathrm{Na}^{+} \mathrm{F}^{-}\right)\), starting with solid \(\mathrm{Na}\) and gaseous F. Then, using the thermochemical data supplied below, determine its heat of formation: (1) \(\mathrm{Na}(\mathrm{s}) \quad \rightarrow \mathrm{Na}(\mathrm{g}) \quad \Delta \mathrm{H}=+26.0 \mathrm{Kcal}:\) sublimation (2) \(\mathrm{F}_{2}(\mathrm{~g}) \quad \rightarrow 2 \mathrm{~F}(\mathrm{~g})\) \(\Delta \mathrm{H}=+36.6 \mathrm{~K}\) cal : dissociation (3) \(\mathrm{Na}(\mathrm{g}) \rightarrow \mathrm{Na}^{+}(\mathrm{g})+\mathrm{e}^{-} \quad \Delta \mathrm{H}=+120.0 \mathrm{~K}\) cal : ionization (4) \(\mathrm{F}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{F}^{-}(\mathrm{g})\) \(\Delta \mathrm{H}=-83.5 \mathrm{Kcal}:\) electron addition (5) \(\mathrm{Na}^{+}(\mathrm{g})+\mathrm{F}^{-}(\mathrm{g}) \rightarrow \mathrm{Na}^{+}, \mathrm{F}^{-}(\mathrm{s}) \Delta \mathrm{H}=-216.7 \mathrm{Kcal}\) : lattice formation.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.