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An Ionic Bond. (a) Calculate the electric potential energy for a K\(^+\) ion and a Br\(^-\) ion separated by a distance of 0.29 nm, the equilibrium separation in the KBr molecule. Treat the ions as point charges. (b) The ionization energy of the potassium atom is 4.3 eV. Atomic bromine has an electron affinity of 3.5 eV. Use these data and the results of part (a) to estimate the binding energy of the KBr molecule. Do you expect the actual binding energy to be higher or lower than your estimate? Explain your reasoning.

Short Answer

Expert verified
The electric potential energy is approximately -3.09 eV, and the estimated binding energy is 4.71 eV. The actual binding energy is likely higher due to factors not included in this estimate.

Step by step solution

01

Identify the Charges of Ions

The potassium ion K\(^+\) has a charge of \(+1e\) and the bromine ion Br\(^-\) has a charge of \(-1e\) where \(e\) is the elementary charge, approximately \(1.6 \times 10^{-19}\) C.
02

Understand the Coulomb's Law Equation

The electric potential energy \(U\) between two point charges can be calculated with the formula: \[ U = \frac{k \cdot q_1 \cdot q_2}{r} \] where \(k\) is Coulomb's constant \( (8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2) \), \(q_1\) and \(q_2\) are the charges, and \(r\) is the separation distance.
03

Plug Values into Coulomb's Law

For the K\(^+\) and Br\(^-\) ions, we substitute: \( q_1 = +e, q_2 = -e \), and \( r = 0.29 \, \text{nm} = 0.29 \times 10^{-9} \text{m} \). The potential energy is then: \[ U = \frac{8.99 \times 10^9 \times (1.6 \times 10^{-19})^2}{0.29 \times 10^{-9}} \approx -4.94 \times 10^{-19} \text{J} \].
04

Convert Energy to Electronvolts

To find the energy in electronvolts (eV), convert joules to eV using the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\). Thus, \[ U = \frac{-4.94 \times 10^{-19} \text{J}}{1.6 \times 10^{-19} \text{J/eV}} \approx -3.09 \text{eV} \].
05

Calculate the Binding Energy

The estimated binding energy is the sum of the ionization energy of the potassium and the electron affinity of bromine minus the electric potential energy: \[ E_{binding} = 4.3 \, \text{eV} + 3.5 \, \text{eV} - 3.09 \, \text{eV} = 4.71 \, \text{eV} \].
06

Analyze the Expectation for Actual Binding Energy

The actual binding energy is expected to be higher than the estimated 4.71 eV because this estimate does not account for additional forces, such as the covalent interactions and electron correlation effects, which would further stabilize the molecule.

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Key Concepts

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

Electric Potential Energy
Electric potential energy is a crucial concept when it comes to understanding ionic bonds. In simple terms, it is the energy stored due to the position of charged particles in an electric field. For ions like K\(^+\) and Br\(^-\), being oppositely charged, their electric potential energy depends on their separation distance.
To calculate this energy, we use Coulomb's Law. The potential energy, denoted as \( U \), can be expressed in the form of:
  • \( U = \frac{k \cdot q_1 \cdot q_2}{r} \)
where \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2) \), \( q_1 \) and \( q_2 \) are the charges of the ions, and \( r \) is the distance of separation.
In our exercise, substituting the values for a K\(^+\) ion and Br\(^-\) ion at their equilibrium distance of 0.29 nm, we calculate the energy to be around \(-3.09\) eV. This negative value indicates a stable configuration, showing that energy is released as these ions come together.
Ionization Energy
Ionization energy is another fundamental concept related to ionic bonding. It refers to the amount of energy needed to remove an electron from an atom to form a positive ion.
For potassium (K), the ionization energy is 4.3 eV. This signifies that 4.3 eV of energy is required to remove one electron from a neutral potassium atom to create a K\(^+\) ion.
Knowing the ionization energy is essential because it directly affects the potential for an atom to lose an electron and thus engage in ionic bonding. A higher ionization energy means the atom holds onto its electrons more tightly, making it harder to form positive ions.
Electron Affinity
Electron affinity is the flip side of ionization energy. It measures an atom's eagerness to gain an electron.
For bromine (Br), the electron affinity is 3.5 eV. This indicates that when an electron is added to a neutral bromine atom to form a Br\(^-\) ion, 3.5 eV of energy is released.
This released energy contributes to the overall energy dynamics in the formation of an ionic bond. A higher electron affinity means the atom easily accepts electrons, favoring negative ion formation, which is crucial for achieving ionic stability.
Binding Energy
Binding energy is a measure of the strength of an ionic bond; it is the energy needed to separate the ions in a molecule.
The binding energy of KBr was estimated by combining all the critical energies: the ionization energy of potassium, the electron affinity of bromine, and the electric potential energy between them.
  • Calculated Binding Energy: \( 4.3 \, \text{eV} + 3.5 \, \text{eV} - (-3.09 \, \text{eV}) = 4.71 \, \text{eV} \)
This value reflects the ionic strength; however, the real binding energy could be higher. This is because other forces, such as covalent character and electron correlation, contribute to further stabilization, which our straightforward estimation does not cover.
Coulomb's Law
Coulomb's Law is an essential principle for calculating the interactions between charged particles. It shows how two point charges exert force upon each other, directly impacting their potential energy.
  • The formula: \( F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \)
where \( F \) is the force between the charges, \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between them.
For potential energy, the formula transforms slightly, reflecting the linear relationship with inverse distance (\( r \)). Knowing that opposite charges attract and like charges repel, Coulomb's Law helps to predict how ions will behave when brought close together, thereby forming ionic bonds. In the context of KBr, Coulomb's Law helps estimate how close they come to their energetic balance, which defines the ionic bond's stability.

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Most popular questions from this chapter

To determine the equilibrium separation of the atoms in the HCl molecule, you measure the rotational spectrum of HCl. You find that the spectrum contains these wavelengths (among others): \(60.4 \mu m\), \(69.0 \mu m\), \(80.4 \mu m\), \(96.4 \mu m\), and \(120.4 \mu m\). (a) Use your measured wavelengths to find the moment of inertia of the HCl molecule about an axis through the center of mass and perpendicular to the line joining the two nuclei. (b) The value of \(l\) changes by \(\pm 1\) in rotational transitions. What value of \(l\) for the upper level of the transition gives rise to each of these wavelengths? (c) Use your result of part (a) to calculate the equilibrium separation of the atoms in the HCl molecule. The mass of a chlorine atom is \(5.81 \times 10^{-26}\) kg, and the mass of a hydrogen atom is \(1.67 \times 10^{-27}\) kg. (d) What is the longest-wavelength line in the rotational spectrum of HCl?

The hydrogen iodide (HI) molecule has equilibrium separation 0.160 nm and vibrational frequency \(6.93 \times 10^{13}\) Hz. The mass of a hydrogen atom is \(1.67 \times 10^{-27}\) kg, and the mass of an iodine atom is 2.11 \(\times\) 10\(^-$$^2$$^5\) kg. (a) Calculate the moment of inertia of HI about a perpendicular axis through its center of mass. (b) Calculate the wavelength of the photon emitted in each of the following vibrationrotation transitions: (i) \(n = 1\), \(l = 1 \rightarrow n = 0\), \(l = 0\); (ii) \(n = 1\), \(l = 2\rightarrow n = 0\), \(l = 1\); (iii) \(n = 2\), \(l = 2\rightarrow n = 1\), \(l = 3\).

A hypothetical NH molecule makes a rotational-level transition from \(l\) = 3 to \(l\) = 1 and gives off a photon of wavelength 1.780 nm in doing so. What is the separation between the two atoms in this molecule if we model them as point masses? The mass of hydrogen is 1.67 \(\times\) 10\(^-$$^2$$^7\) kg, and the mass of nitrogen is 2.33 \(\times\) 10\(^-$$^2$$^6\) kg.

If the energy of the \(H_{2}\) covalent bond is -4.48 eV, what wavelength of light is needed to break that molecule apart? In what part of the electromagnetic spectrum does this light lie?

The gap between valence and conduction bands in silicon is 1.12 eV. A nickel nucleus in an excited state emits a gammaray photon with wavelength 9.31 \(\times\) 10\(^-$$^4\) nm. How many electrons can be excited from the top of the valence band to the bottom of the conduction band by the absorption of this gamma ray?

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