Question

Calculate this distance for the potassium bromide molecule, described in Exercise 42.2.

Short answer

The distance for the potassium bromide molecule is 1.80nm.

Step-by-step solution

Step 1

Definition of Ionization energy of an atom

It is the amount of energy required to discharge a electron from isolated gaseous atom in ground state.

Definition of Electron affinity of an atom

It is the amount of energy that released when an electron is added to a neutral atom to form a negatively charged ion.

Energy required to create bond or well separated ions

The net energy$\left(E_n\right)$ required to create bond or well separated ions is difference of energy of ionization$\left(E_i\right)$ and energy of electron affinity$\left(E_a\right)$.

Mathematically it can be given by

$E_n=E_i-E_a$

Potential energy of system of two charges

The potential energy of system of two charges $(q_1,q_2)$separated distance r is given by

$U=\frac{k{q}_1{q}_2}{r}$

Where , Uis the potential energy of the system.

The calculation of energy required to create system of well separated ions of potassium and bromine

Given (From exercise 42.2):

The ionization energy of potassium atom is$E_i=4.3eV$

The energy of electron affinity of bromine is$E_a=3.5eV$

Using

$E_n=E_i-E_a$

Put the values of constants in above equation

$E_n=4.3eV-3.5eV=0.8eV$

Thus, the energy required to create system of well separated ions of potassium and bromine.

The calculation of distance between two charges

Using

$$\begin{gathered} U=\frac{k{q}_1{q}_2}{r} \\ r=\frac{k{q}_1{q}_2}{U} \end{gathered}$$

Now,put the values of constants in above equation

$$\begin{gathered} r=\frac{9\times {10}^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2{\left(1.60\times {10}^{-19}\mathrm{C}\right)}^2}{0.8\mathrm{eV}\left(1.60\times {10}^{-19}\mathrm{J}/\mathrm{eV}\right)} \\ r=1.80\mathrm{nm} \end{gathered}$$

Thus, the distance for the potassium bromide molecule is 1.80nm.