Knowing the electron gain enthalpy values for O → O- and O → O2- as –141 and 702 kJ mol-1 respectively, how can you account for the formation of a large number of oxides having O2- species and not O-? (Hint: Consider lattice energy factor in the formation of compounds).

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Accepted Answer

**Step 1: The two steps of electron gain by oxygen.**
$$O(g) + e^- ightarrow O^-(g) \qquad \Delta H = -141\ kJ\,mol^{-1} \; (	ext{exothermic})$$
$$O^-(g) + e^- ightarrow O^{2-}(g) \qquad \Delta H = +702\ kJ\,mol^{-1} \; (	ext{endothermic})$$
Adding a second electron to the already negatively-charged $O^-$ ion requires energy to overcome the strong electrostatic repulsion between the incoming electron and the existing negative charge, so this second step is highly endothermic.

**Step 2: Why $O^{2-}$-containing compounds still form.**
$O^{2-}$ ions are not formed in isolation — they are formed in the process of building an ionic crystal lattice together with cations (e.g. $Mg^{2+}, Ca^{2+}$). The formation of the ionic solid from gaseous ions releases a very large amount of lattice energy, because the doubly-charged $O^{2-}$ ion has a much stronger electrostatic attraction to nearby cations than a singly-charged $O^-$ ion would.

**Step 3: Overall energy balance (Born–Haber-type argument).**
The overall enthalpy of formation of a metal oxide (e.g. MgO) sums the endothermic step of forming $O^{2-}$ from $O^-$ together with the highly exothermic lattice enthalpy of the solid. Since lattice enthalpy scales strongly with the product of the ionic charges ($\propto q_1 q_2$), the lattice energy released on forming the $M^{2+}O^{2-}$ lattice is large enough to more than offset the +702 kJ/mol penalty, making the overall formation of the oxide exothermic and thermodynamically favourable.

**Step 4: Why $O^-$-based oxides are not commonly formed instead.** …

Q.Knowing the electron gain enthalpy values for O → O- and O → O2- as –141 and 702 kJ mol-1 respectively, how can you account for the formation of a large number of oxides having O2- species and not O-? (Hint: Consider lattice energy factor in the formation of compounds).

Concept understanding — Dioxygen and Simple Oxides

Q.Knowing the electron gain enthalpy values for O → O- and O → O2- as –141 and 702 kJ mol-1 respectively, how can you account for the formation of a large number of oxides having O2- species and not O-? (Hint: Consider lattice energy factor in the formation of compounds).

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