91影视

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鈥�.

$13.6\text{鈥塭痴}鈭�3.40\text{鈥塭痴}=10.2\text{鈥塭痴}$

Thus The Energy input that need to be balanced by potential energy, $10.2\text{鈥塭痴}$.

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

$\mathrm{U}=鈭�\frac{{\mathrm{e}}^2}{4\mathrm{蟺}{鈭�}_0\mathrm{a}}$

To, Calculate the Separation in order to release $10.2\text{鈥塭痴}$

$\mathrm{a}=鈭�\frac{{\mathrm{e}}^2}{4\mathrm{蟺}{鈭�}_0\mathrm{U}}$

Substitute,$\mathrm{e}=1.6脳{10}^{鈭�19}$,${鈭�}_0=8.85脳{10}^{鈭�12}{\text{鈥塁}}^{\text{2}}{\text{/N.m}}^2$ , and$\mathrm{U}=10.2\text{鈥塭痴}$in the above equation.

Hence, Ion need to be at least as close at $0.14\text{鈥塶尘}$ 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鈥�.

$5.1\text{鈥塭痴}鈭�3.40\text{鈥塭痴}=1.70\text{鈥塭痴}$

Thus The Energy input that need to be balanced by potential energy $1.70\text{鈥塭痴}$.

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

$\mathrm{U}=鈭�\frac{{\mathrm{e}}^2}{4\mathrm{蟺}{鈭�}_0\mathrm{a}}$

To , Calculate the Separation in order to release$1.70\text{鈥塭痴}$

$\mathrm{a}=鈭�\frac{{\mathrm{e}}^2}{4\mathrm{蟺}{鈭�}_0\mathrm{U}}$

Substitute, $\mathrm{e}=1.6脳{10}^{鈭�19}$, ${鈭�}_0=8.85脳{10}^{鈭�12}{\text{鈥塁}}^{\text{2}}{\text{/N.m}}^2$, and $\mathrm{U}=1.70\text{鈥塭痴}$ in the above equation.

By putting all Values in Formula of U we get

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

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

Hence Hydrogen and Fluorine need to Separated by at most $0.14\text{鈥塶尘}$while Sodium and fluorine need to be separated by at most $0.85\text{鈥塶尘}$.