Question

Consider the following energy changes: $$\begin{array}{ll} \text {} & \quad { \Delta H} \\ \text {} & {(k J / m o l)} \\ \hline \\ {\mathrm{Mg}(g) \rightarrow \mathrm{Mg}^{+}(g)+\mathrm{e}^{-}} & {735} \\ {\mathrm{Mg}^{+}(g) \rightarrow \mathrm{Mg}^{2+}(g)+\mathrm{e}^{-}} & {1445} \\ {\mathrm{O}(g)+\mathrm{e}^{-} \rightarrow \mathrm{O}^{-}(g)} & {-141} \\ {\mathrm{O}^{-}(g)+\mathrm{e}^{-} \rightarrow 0^{2-}(g)} & {878}\end{array}$$ Magnesium oxide exists as \(\mathrm{Mg}^{2+} \mathrm{O}^{2-}\) and not as \(\mathrm{Mg}^{+} \mathrm{O}^{-}\) Explain.

Short answer

The given energy changes for the formation of ions from gaseous elements show that the formation of \(\mathrm{Mg}^{2+} \mathrm{O}^{2-}\) has a total energy change of \(2917 kJ/mol\), while the formation of \(\mathrm{Mg}^{+} \mathrm{O}^{-}\) has a total energy change of \(594 kJ/mol\). Even though both species involve positive energy changes, the formation of \(\mathrm{Mg}^{+} \mathrm{O}^{-}\) is more energetically favorable due to its lower energy change. Nonetheless, magnesium oxide exists as \(\mathrm{Mg}^{2+} \mathrm{O}^{2-}\) rather than \(\mathrm{Mg}^{+} \mathrm{O}^{-}\) as the given energy changes do not account for other factors that may affect stability in real systems.

Step-by-step solution

Identify the energy changes to form Mg2+ and O2-

First, analyze the two given energy changes required to create \(\mathrm{Mg}^{2+}\) and \(\mathrm{O}^{2-}\) ions. For \(\mathrm{Mg}^{2+}\) 1. \(\mathrm{Mg}(g) \rightarrow \mathrm{Mg}^{+}(g) + \mathrm{e}^-\) , \(\Delta H = 735 kJ/mol\) 2. \(\mathrm{Mg}^{+}(g) \rightarrow \mathrm{Mg}^{2+}(g) + \mathrm{e}^-\) , \(\Delta H = 1445 kJ/mol\) For \(\mathrm{O}^{2-}\) 3. \(\mathrm{O}(g) + \mathrm{e}^- \rightarrow \mathrm{O}^{-}(g)\) , \(\Delta H = -141 kJ/mol\) 4. \(\mathrm{O}^{-}(g) + \mathrm{e}^- \rightarrow \mathrm{O}^{2-}(g)\) , \(\Delta H = 878 kJ/mol\)

Calculate the total energy change to form Mg2+ O2-

Now, calculate the total energy changes required to create the \(\mathrm{Mg}^{2+} \mathrm{O}^{2-}\) ion pair from the gaseous elements: $$\Delta H_{\mathrm{Mg}^{2+} \mathrm{O}^{2-}} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4 = 735 + 1445 - 141 + 878 = 2917 kJ/mol$$

Identify the energy changes to form Mg+ and O-

Analyze the energy changes required to create \(\mathrm{Mg}^{+}\) and \(\mathrm{O}^{-}\) ions. For \(\mathrm{Mg}^{+}\) 1. \(\mathrm{Mg}(g) \rightarrow \mathrm{Mg}^{+}(g) + \mathrm{e}^-\) , \(\Delta H = 735 kJ/mol\) For \(\mathrm{O}^{-}\) 3. \(\mathrm{O}(g) + \mathrm{e}^- \rightarrow \mathrm{O}^{-}(g)\) , \(\Delta H = -141 kJ/mol\)

Calculate the total energy change to form Mg+ O-

Now, calculate the total energy changes required to create the \(\mathrm{Mg}^{+} \mathrm{O}^{-}\) ion pair from the gaseous elements: $$\Delta H_{\mathrm{Mg}^{+} \mathrm{O}^{-}} = \Delta H_1 + \Delta H_3 = 735 - 141 = 594 kJ/mol$$

Compare the energy changes and define stability

Magnesium oxide in the form of \(\mathrm{Mg}^{2+} \mathrm{O}^{2-}\) has a total energy change of \(2917 kJ/mol\), while the formation of \(\mathrm{Mg}^{+} \mathrm{O}^{-}\) has a total energy change of \(594 kJ/mol\). Since both cases involve positive energy changes, neither species is energetically preferred over the constituent atoms. However, as ΔH(Mg2+O2-) > ΔH(Mg+O-), the formation of the latter species requires less energy and is somewhat more energetically favored compared to the former. Keep in mind that these results are based solely on the given energy changes and do not account for any other factors that may influence the overall stability of these species in real systems. In conclusion, although \(\mathrm{Mg}^{2+} \mathrm{O}^{2-}\) is not energetically preferred over the constituent atoms, its formation is more energetically favorable compared to \(\mathrm{Mg}^{+} \mathrm{O}^{-}\). Therefore, magnesium oxide exists as \(\mathrm{Mg}^{2+} \mathrm{O}^{2-}\) rather than \(\mathrm{Mg}^{+} \mathrm{O}^{-}\) based on the given energy changes.

Key concepts

Ionization energy

When atoms lose electrons, it requires a certain amount of energy called ionization energy. Ionization energy indicates how strongly an atom holds onto its electrons. In simpler terms, it's the energy needed to remove an electron from an atom or ion.

Factors Influencing Ionization Energy:

• Atomic Size: Larger atoms have lower ionization energies because their outer electrons are further from the nucleus and are less tightly bound.
• Nuclear Charge: More positively charged nuclei hold their electrons more tightly, leading to higher ionization energies.

For magnesium, the first ionization involves removing one electron to form \(\mathrm{Mg}^{+}(g)\), requiring 735 kJ/mol. The second ionization energy, to form \(\mathrm{Mg}^{2+}(g)\), is even higher at 1445 kJ/mol because removing an electron from a positively charged ion is harder. This is a crucial concept because it helps explain why certain ion configurations like \(\mathrm{Mg}^{2+}\) are more common than others.

Lattice energy

Lattice energy measures the strength of the forces holding ions together in an ionic solid. It's the energy released when ions come together to form a crystal lattice from gaseous ions.

Importance in Stability:

• The greater the lattice energy, the more stable the compound. This energy makes ionic compounds like magnesium oxide (\(\mathrm{MgO}\)) very stable as it compensates for the high ionization energy required to form cations.
• Lattice energy plays a central role in determining the structure and stability of ionic compounds by influencing melting points and solubility.

Understanding lattice energy explains why \(\mathrm{Mg}^{2+} \mathrm{O}^{2-}\), despite its high energy changes, forms a stable compound due to strong electrostatic attractions between ions.

Electron affinity

Electron affinity is the energy change when an electron is added to a neutral atom to form a negative ion. Unlike ionization energy, electron affinity can be negative if energy is released.

How it Works:

• Atoms with high electron affinity accept electrons more easily, usually forming more stable anions.
• Nonmetals typically have high electron affinities, as they are more inclined to complete their octet by gaining electrons.

In the case of oxygen, the first electron addition to form \(\mathrm{O}^{-}(g)\) releases energy (-141 kJ/mol). However, adding a second electron to form \(\mathrm{O}^{2-}(g)\) requires 878 kJ/mol due to repulsion between the negatively charged ion and the incoming electron. This behavior helps us understand why \(\mathrm{O}^{2-}\) forms less readily than \(\mathrm{O}^{-}\) but is stabilized in the ionic compound by lattice energy.