For \(\mathrm{a} \mathrm{K}^{+}-\mathrm{Cl}^{-}\)ion pair, attractive and repulsive energies \(E_{A}\) and \(E_{R}\), respectively, depend on the distance between the ions \(r\), according to $$ \begin{aligned} E_{A} &=-\frac{1.436}{r} \\ E_{R} &=\frac{5.86 \times 10^{-6}}{r^{9}} \end{aligned} $$ For these expressions, energies are expressed in electron volts per \(\mathrm{K}^{+}-\mathrm{Cl}^{-}\)pair, and \(r\) is the distance in nanometers. The net energy \(E_{N}\) is just the sum of the preceding two expressions. (a) Superimpose on a single plot \(E_{N}, E_{R}\), and \(E_{A}\) versus \(r\) up to \(1.0 \mathrm{~nm}\). (b) On the basis of this plot, determine (i) the equilibrium spacing \(r_{0}\) between the \(\mathrm{K}^{+}\)and \(\mathrm{Cl}^{-}\)ions, and (ii) the magnitude of the bonding energy \(E_{0}\) between the two ions. (c) Mathematically determine the \(r_{0}\) and \(E_{0}\) values using the solutions to Problem \(2.14\) and compare these with the graphical results from part (b).

Short Answer

Expert verified
Question: Find the equilibrium spacing between a K+ and Cl- ion pair and the magnitude of the bonding energy based on the given attractive and repulsive energies. Compare the graphical and mathematical methods' results. Answer: To find the equilibrium spacing and bonding energy, first calculate the net energy by summing the attractive and repulsive energies. Plot the energies against the distance between the ions. Find the equilibrium spacing (r_0) at the lowest point of the net energy curve on the graph. Calculate the bonding energy (E_0) using the equilibrium spacing value. Mathematically determine r_0 and E_0 by finding the derivative of the net energy function and setting it to zero. Solve for r to find r_0 and then use this value to find E_0. Compare the values obtained from the graphical and mathematical methods.

Step by step solution

01

Define the given functions

We have two functions for attractive and repulsive energies: $$ \begin{aligned} E_{A} &=-\frac{1.436}{r} \\ E_{R} &=\frac{5.86 \times 10^{-6}}{r^{9}} \end{aligned} $$ Where \(E_A\) and \(E_R\) are the attractive and repulsive energies respectively, and \(r\) is the distance between the ions in nanometers.
02

Calculate the net energy

The net energy, \(E_N\), is the sum of both attractive and repulsive energies. $$ E_N=E_A+E_R=-\frac{1.436}{r}+\frac{5.86 \times 10^{-6}}{r^9} $$
03

Plot the energies

Plot the three energies, \(E_N\), \(E_R\), and \(E_A\) against \(r\) up to \(1.0 nm\). (This step would typically be performed on graph paper or using a graphing software, such as Desmos or Excel.)
04

Find the equilibrium spacing and bonding energy

(i) Observe the plotted graph and find the equilibrium spacing \(r_0\) at the lowest point of the net energy curve (The minimum net energy corresponds to the maximum stability of the ion pair). (ii) The bonding energy \(E_0\) can be found at this equilibrium spacing - use the values of \(r_0\) obtained from the graph and put it into the net energy equation: $$ E_0 = E_N(r_0) $$
05

Mathematically determine \(r_0\) and \(E_0\)

To mathematically find the equilibrium spacing \(r_0\), first find the derivative of the net energy function \(E_N\) with respect to \(r\), then set it to zero and solve for \(r\): $$ \frac{dE_N}{dr} = -\frac{-1.436}{r^2} - \frac{9(5.86 \times 10^{-6})}{r^{10}} = 0 $$ Solve this equation for \(r\) to find the equilibrium spacing \(r_0\). Next, use the value of \(r_0\) to find the bonding energy \(E_0\): $$ E_0 = E_N(r_0) $$
06

Compare the results

Compare the values of \(r_0\) and \(E_0\) obtained from the graphical and mathematical methods. If done correctly, the values should be in close agreement.

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Key Concepts

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

Equilibrium Spacing
Equilibrium spacing, often represented as \( r_0 \), is a significant concept in the study of ionic bonds. In simple terms, it is the optimal distance between two oppositely charged ions at which the energies of attraction and repulsion balance each other out. At this point, the system achieves maximum stability and minimum potential energy.

When discussing the ionic bond between a cation, such as potassium (K+), and an anion, such as chloride (Cl-), the equilibrium spacing can be visualized on a graph where net energy is plotted against the distance between ions. The curve for net energy will typically have a distinct minimum point. This lowest point on the graph indicates the equilibrium spacing \( r_0 \), signifying the most stable arrangement of the ions within the compound.

In calculations and exercises like the given example, finding the equilibrium spacing involves determining where the derivative of the net energy with respect to the inter-ionic distance equals zero. This mathematical approach complements the graphical method and often serves to verify the accuracy of the plotted data.
Attractive and Repulsive Energies
When ions form a bond, there are two primary forces at play: attractive and repulsive energies. Attractive energies arise from the electrostatic force pulling ions with opposite charges together, such as the potassium cation and chloride anion. This energy is inversely proportional to the distance between the ions, meaning that as the ions draw closer, the attraction strengthens.

Conversely, repulsive energies work against this attraction. They emerge from the repulsion between electron clouds when ions are too close. Like attractive energies, repulsive energies are also dependent on the distance between ions, but they follow a considerably faster rate of change as ions approach each other, as illustrated in the given functions.

Understanding these concepts is critical for comprehending how the net energy of a system is calculated by summing the attractive and repulsive energies. By plotting these energies against various distances, students can visually grasp how these forces dictate the behavior of ionic systems and determine the point of equilibrium spacing where the energy is lowest and stability is the highest.
Electron Volts
The electron volt (eV) is a unit of energy that's especially useful in the context of atomic and molecular scales. One electron volt is defined as the amount of kinetic energy gained or lost by a single electron as it moves through an electric potential difference of one volt. It's a practical unit because it relates to the energies typically involved in chemical bonds and changes in electron configurations.

In our exercise with \( \mathrm{K}^{+}-\mathrm{Cl}^{-} \) ion pairs, energies are expressed in electron volts per pair. This is convenient because it gives a direct sense of the scale of energies when ions interact. The electron volt is a much smaller unit than the Joule, the standard SI unit for energy, which makes it all the more suitable for describing the subtle shifts in energy as ions move closer or further apart. For reference, 1 eV equals approximately \( 1.602 \times 10^{-19} \) Joules.
Nanometers
The nanometer (nm) is a unit of length in the metric system, equivalent to one billionth of a meter (\( 10^{-9} \) meters). It's a preferred unit of measurement when dealing with the extremely small scales of atoms and molecules.

As seen in the provided exercise, distances between ions in ionic bonds are often expressed in nanometers. Utilizing nanometers allows scientists and students alike to easily communicate and calculate properties and behaviors of particles at the atomic level. Given that bonds in molecules and solids range on the scale of tenth to few nanometers, using such a precise unit ensures accuracy in descriptions and calculations of chemical structures, particularly when evaluating the effects of interatomic distances on bonding energy and force interactions.

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Most popular questions from this chapter

6 Allowed values for the quantum numbers of electrons are as follows: $$ \begin{aligned} n &=1,2,3, \ldots \\ l &=0,1,2,3, \ldots, n-1 \\ m_{l} &=0, \pm 1, \pm 2, \pm 3, \ldots, \pm l \\ m_{s} &=\pm \frac{1}{2} \end{aligned} $$ The relationships between \(n\) and the shell designations are noted in Table 2.1. Relative to the subshells, \(l=0\) corresponds to an \(s\) subshell \(l=1\) corresponds to a \(p\) subshell \(l=2\) corresponds to a \(d\) subshell \(l=3\) corresponds to an \(f\) subshell For the \(K\) shell, the four quantum numbers for each of the two electrons in the \(1 s\) state, in the order of \(n l m_{i} m_{s}\), are \(100 \frac{1}{2}\) and \(100\left(-\frac{1}{2}\right)\) Write the four quantum numbers for all of the electrons in the \(L\) and \(M\) shells, and note which correspond to the \(s, p\), and \(d\) subshells.

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To what group in the periodic table would an element with atomic number 114 belong?

Compute the percent ionic character of the interatomic bonds for the following compounds: \(\mathrm{TiO}_{2}, \mathrm{ZnTe}, \mathrm{CsCl}, \mathrm{InSb}\), and \(\mathrm{MgCl}_{2}\).

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