Question

Use the following data to estimate $\mathbf{∆}{\mathbf{H}}_{\mathbf{f}}^{\mathbf{°}}$for magnesium fluoride.

$$\begin{gathered} \mathbf{Mg}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{F}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }{\mathbf{MgF}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)} \\ \mathbf{Lattice}\mathbf{}\mathbf{energy}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathbf{2913}\mathbf{}\mathbf{KJ}\mathbf{/}\mathbf{mol} \\ \mathbf{First}\mathbf{}\mathbf{inoziation}\mathbf{}\mathbf{energy}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Mg}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{735}\mathbf{}\mathbf{KJ}\mathbf{/}\mathbf{mol} \\ \mathbf{Second}\mathbf{}\mathbf{Ionization}\mathbf{}\mathbf{energy}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Mg}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1445}\mathbf{}\mathbf{KJ}\mathbf{/}\mathbf{mol} \\ \mathbf{Electron}\mathbf{}\mathbf{affinity}\mathbf{}\mathbf{of}\mathbf{}\mathbf{F}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathbf{328}\mathbf{}\mathbf{KJ}\mathbf{/}\mathbf{mol} \\ \mathbf{Bond}\mathbf{}\mathbf{energy}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{F}}_{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{154}\mathbf{}\mathbf{}\mathbf{KJ}\mathbf{/}\mathbf{mol} \\ \mathbf{Enthalpy}\mathbf{}\mathbf{of}\mathbf{}\mathbf{sublimation}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Mg}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{150}\mathbf{}\mathbf{KJ}\mathbf{/}\mathbf{mol} \end{gathered}$$

Short answer

$-1085\mathrm{KJ}/\mathrm{mol}$

Step-by-step solution

The first step will be the sublimation of Mg(s):

The conversion of the solid phase directly to the gaseous phase, with no intermediate liquid phase, is called sublimation.

Enthalpy of sublimation for Mg $=150$kJ/mol

$\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Mg}\left(\mathrm{g}\right)$

The next stage will be the ionization of the gaseous Mg:

The amount of energy required for removing the most loosely bound electron of an isolated gaseous atom is called it's ionization energy.

$\mathrm{Mg}\left(\mathrm{g}\right)\stackrel{{\mathrm{IE}}_1}{\to }{\mathrm{Mg}}^{+}\left(\mathrm{g}\right)$

First ionization energy $=735$kJ/mol

$\mathrm{Mg}\left(\mathrm{g}\right)\stackrel{{\mathrm{IE}}_2}{\to }{\mathrm{Mg}}^{2+}\left(\mathrm{g}\right)$

Second ionization energy$=1445$ kJ/mol

The fluorine bond will break:

Energy is always needed for breaking a bond which is known as bond energy.

${\mathrm{F}}_2\left(\mathrm{g}\right)\to 2\mathrm{F}\left(\mathrm{g}\right)$

The bond energy of ${\mathrm{F}}_2=154$kJ/mol

Fluorine atoms will now ionize:

When an electron is added to a neutral atom for forming a negative ion, the amount of energy released is called electron affinity.

$2\mathrm{F}\left(\mathrm{g}\right)\to 2{\mathrm{F}}^{-}\left(\mathrm{g}\right)$

Electron affinity of F$=-328$kJ/mol

Electron affinity of 2F $=2(-328)=-656$kJ/mol

Now find out lattice energy:

The energy liberated when gaseous ions combine to form an ionic solid is called lattice energy.

${\mathrm{Mg}}^{2+}\left(\mathrm{g}\right)+2{\mathrm{F}}^{-}\left(\mathrm{g}\right)\to {\mathrm{MgF}}_2\left(\mathrm{s}\right)$

Lattice energy$=-2913$ kJ/mol

Add all these values to evaluate the formation enthalpy of $\mathrm{MgF_2}$:

$∆{\mathrm{H}}_{\mathrm{f}}^{°}=150+735+1445+154-656-2913=-1085$kJ/mol

The enthalpy of formation of ${\mathrm{MgF}}_2$is calculated to be $-1085$kJ/mol.