Question

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

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⁻ is given as 0.20 nm (or 2.0 x 10⁻¹⁰ m), we can subtract it from the total distance between the ions to find the maximum size of A⁺: Max size of A⁺ = r - radius of B⁻ = 3.6 x 10⁻¹⁰ m - 2.0 x 10⁻¹⁰ m = 1.6 x 10⁻¹⁰ m Therefore, the maximum size of A⁺ that would lend itself to the formation of an energetically stable bond is \(1.6 \times 10^{-10} m\).

Step-by-step solution

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

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 10⁻¹⁹ J So, Energy released = 4.0 eV x 1.6 x 10⁻¹⁹ J/eV = 6.4 x 10⁻¹⁹ Joules

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

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 10⁹ N m²/C²), 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 10⁻¹⁹ 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⁺ and B⁻, the charges q1 and q2 are both equal to the elementary charge e (1.6 x 10⁻¹⁹ C): \(r = \frac{(1.6 x 10^{-19} C)^{2} (8.99 x 10^9 N m^{2} C^{-2})}{6.4 x 10^{-19} J}\)

Key concepts

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’s 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.