Exercise 29 outlines how energy may be extracted by transferring an electron from an atom that easily loses an electron from an atom that easily loses an electron to one with a large appetite for electrons , then allowing the two to approach , forming an ionic bond.

1. Consider separately the cases of hydrogen bonding with fluorine and sodium bonding with fluorine in each case , how close must the ions approach to reach “break even” where the energy needed to transfer the electron between the separated atoms is balanced by the electrostatic potential energy of attraction? The ionization energy of hydrogen is 13.6 eV , that of sodium is 5.1 eV , and the electron affinity of fluorine is 3.40 Ev.
2. Of HF and NaF , one is considered to be an ionic bond and the other a covalent bond . Which is which and Why?

Ionic bonds, also known as electrovalent bonds, are a type of linkage created in a chemical molecule by the electrostatic attraction of ions with opposing charges. When the valence (outermost) electrons of one atom are permanently transferred to another atom, a bond of this kind is created.

The Energy input=Energy Output

Energy input+liberated Energy by fluorine=Ionisation Energy of hydrogen.

Thus, Energy Input=Ionisation Energy of hydrogen-Liberated Energy by fluorine….

$13.6\text{\hspace{0.17em}eV}-3.40\text{\hspace{0.17em}eV}=10.2\text{\hspace{0.17em}eV}$

Thus The Energy input that need to be balanced by potential energy, $10.2\text{\hspace{0.17em}eV}$.

The potential energy between the hydrogen ion and the fluorine ion is given by

$\mathrm{U}=-\frac{{\mathrm{e}}^2}{4\mathrm{\pi }{\in }_0\mathrm{a}}$

To, Calculate the Separation in order to release $10.2\text{\hspace{0.17em}eV}$

$\mathrm{a}=-\frac{{\mathrm{e}}^2}{4\mathrm{\pi }{\in }_0\mathrm{U}}$

Substitute,$\mathrm{e}=1.6\times {10}^{-19}$,${\in }_0=8.85\times {10}^{-12}{\text{\hspace{0.17em}C}}^{\text{2}}{\text{/N.m}}^2$ , and$\mathrm{U}=10.2\text{\hspace{0.17em}eV}$in the above equation.

Hence, Ion need to be at least as close at $0.14\text{\hspace{0.17em}nm}$ so that it can release as much energy as it spends when forming Ions.

Energy input+liberated Energy by fluorine=Ionisation Energy of sodium..

Thus, Energy Input=Ionisation Energy of sodium-Liberated Energy by fluorine….

$5.1\text{\hspace{0.17em}eV}-3.40\text{\hspace{0.17em}eV}=1.70\text{\hspace{0.17em}eV}$

Thus The Energy input that need to be balanced by potential energy $1.70\text{\hspace{0.17em}eV}$.

The potential energy between the Sodium ion and the fluorine ion is given by

$\mathrm{U}=-\frac{{\mathrm{e}}^2}{4\mathrm{\pi }{\in }_0\mathrm{a}}$

To , Calculate the Separation in order to release$1.70\text{\hspace{0.17em}eV}$

$\mathrm{a}=-\frac{{\mathrm{e}}^2}{4\mathrm{\pi }{\in }_0\mathrm{U}}$

Substitute, $\mathrm{e}=1.6\times {10}^{-19}$, ${\in }_0=8.85\times {10}^{-12}{\text{\hspace{0.17em}C}}^{\text{2}}{\text{/N.m}}^2$, and $\mathrm{U}=1.70\text{\hspace{0.17em}eV}$ in the above equation.

By putting all Values in Formula of U we get

$\mathrm{a}=8.5*{10}^{-10}\mathrm{m}$

= $0.85\mathrm{nm}(1\mathrm{nm}={10}^{-9}\mathrm{m})$

Hence Hydrogen and Fluorine need to Separated by at most $0.14\text{\hspace{0.17em}nm}$while Sodium and fluorine need to be separated by at most $0.85\text{\hspace{0.17em}nm}$.