Even though so much energy is required to form a metal cation with a \(2+\)
charge, the alkaline earth metals form halides with the general formula
\(\mathrm{MX}_{2}\), rather than \(\mathrm{MX}\). (a) Use the following data to
calculate \(\Delta H_{i}^{\circ}\) of \(\mathrm{MgCl}\) :
\(\begin{array}{lr}\mathrm{Mg}(s) \longrightarrow \mathrm{Mg}(g) & \Delta
H^{\circ}=148 \mathrm{~kJ} \\ \mathrm{Cl}_{2}(g) \longrightarrow 2
\mathrm{Cl}(g) & \Delta H^{\circ}=243 \mathrm{~kJ} \\ \mathrm{Mg}(g)
\longrightarrow \mathrm{Mg}^{+}(g)+\mathrm{e}^{-} & \Delta H^{\circ}=738
\mathrm{~kJ} \\ \mathrm{Cl}(g)+\mathrm{e}^{-} \longrightarrow
\mathrm{Cl}^{-}(g) & \Delta H^{\circ}=-349 \mathrm{~kJ} \\ & \Delta H_{\text
{lattice }}^{\circ} \text { of } \mathrm{MgCl}= & 783.5 \mathrm{~kJ} /
\mathrm{mol}\end{array}\)
(b) Is \(\mathrm{MgCl}\) favored energetically relative to \(\mathrm{Mg}\) and
\(\mathrm{Cl}_{2} ?\) Explain.
(c) Use Hess's law to calculate \(\Delta H^{\circ}\) for the conversion of
\(\mathrm{MgCl}\) to \(\mathrm{MgCl}_{2}\) and \(\mathrm{Mg}\left(\Delta
H_{\mathrm{f}}^{\circ}\right.\) of \(\left.\mathrm{MgCl}_{2}=-641.6 \mathrm{~kJ}
/ \mathrm{mol}\right)\)
(d) Is \(\mathrm{MgCl}\) favored energetically relative to \(\mathrm{MgCl}_{2}\) ?
Explain.