Oxides, Peroxides and Superoxides of Group 1
Burn a strip of lithium ribbon in air and you get a fine white powder of Li₂O. Burn sodium under the same conditions. You get a pale-yellow solid that is mostly Na₂O₂, sodium peroxide. Drop a chunk of potassium into an oxygen atmosphere. The product is KO₂, potassium superoxide. This is an orange-yellow paramagnetic solid. It will cheerfully react with your exhaled CO₂ to regenerate breathable O₂. That is exactly why KO₂ sits inside submarine and mine-rescue oxygen generators. Three elements, one reagent, three structurally and electronically distinct anions of oxygen. The puzzle is not that alkali metals react with oxygen. Every s-block metal does that eagerly. The real question is why the identity of the oxide ion itself changes systematically as you descend the group. This is a question about electron counting dressed up as a question about ionic radii. Oxygen supports at least three anionic forms in the gas phase and in ionic solids. These are the oxide ion O²⁻, the peroxide ion O₂²⁻ (an O–O single bond, diamagnetic), and the superoxide ion O₂⁻ (a bond order of 1.5, one unpaired electron, paramagnetic). Each of these anions has its own gas-phase stability relative to O₂ and O²⁻. This stability is governed by the molecular orbital energetics of the O–O bond. But whether a given anion actually forms as an isolable solid depends on something else. It is decided by a competition downstream of that gas-phase chemistry: the lattice energy gained by pairing that particular anion with a particular alkali-metal cation. Small, highly charged cations pack a lot of lattice energy around a small, highly charged anion. This is the essence of radius-ratio and charge-density thinking. It is already familiar from the Descriptive Chemistry of Alkali Metals with Water, Oxygen and Halogens discussion. Li⁺ is tiny (ionic radius ≈ 76 pm, six-coordinate). It pairs best with the small, doubly-charged O²⁻ (140 pm). Cs⁺ is huge (167 pm). Its lattice energy is barely affected by shrinking the anion charge from 2− to 1−. So the electronically preferred, lower-charge, larger superoxide ion (O₂⁻, effective radius ≈ 158 pm) wins instead. This size-matching principle is sometimes phrased as "hard cations want hard (small, highly charged) anions." It is one of the cleanest applications of lattice-energy reasoning in the entire periodic table. It also sets up the more general thermodynamic framework developed under Lattice Energy and Thermodynamic Stability of s-Block Salts. This article works through the molecular-orbital picture of the three oxygen anions. It covers the Born–Haber-type energetics that decide which one precipitates for each alkali metal. It traces the group-1 trend Li → Na → K → Rb → Cs. Finally, it covers the practical consequences — from bleaching chemistry to submarine air regeneration. It also explains a seemingly paradoxical fact. Na₂O₂ reacts with water to give H₂O₂, not H₂O.
- Oxygen forms three anions of interest: oxide O²⁻ (no O–O bond), peroxide O₂²⁻ (O–O single bond, bond order 1, diamagnetic), and superoxide O₂⁻ (bond order 1.5, one unpaired electron, paramagnetic).
- Molecular-orbital theory (using O₂'s π* orbitals) predicts the O–O bond lengthens and weakens as electrons are added: O₂ (121 pm) → O₂⁻ (128–135 pm) → O₂²⁻ (149 pm).
- Going down Group 1, the normal combustion product changes: Li → Li₂O (oxide), Na → Na₂O₂ (peroxide), K/Rb/Cs → MO₂ (superoxide).
- The switch is a lattice-energy effect: small cations (Li⁺, Na⁺) stabilize the more highly charged, smaller O²⁻ ion; large cations (K⁺, Rb⁺, Cs⁺) cannot recoup the extra lattice energy needed to favor O²⁻ over O₂⁻, so the singly-charged superoxide becomes competitive.
- Quantitatively this is a "radius-ratio rule" applied to anion selection, formalized by comparing lattice energies (via Kapustinskii or Born–Haber cycles) for the same cation with O²⁻, O₂²⁻, and O₂⁻.
- Key idea: the alkali metal cation's size decides which oxygen anion is thermodynamically stabilized enough in the solid to outcompete disproportionation back to O²⁻ and O₂.
Cation size controls anion identity: small cations electrostatically favor small, highly-charged O²⁻; large cations cannot supply enough lattice energy to make the highly-charged anion worthwhile, so the larger, lower-charge, MO-stabilized O₂⁻ (or intermediate O₂²⁻) becomes the thermodynamic product instead.
Where this lives today
Potassium superoxide's ability to simultaneously absorb CO₂ and release O₂ is a live-support workhorse reaction:
This single reaction converts exhaled CO₂ (the problem) directly into breathable O₂ (the solution) without needing any external oxygen supply, which is why KO₂ canisters and related "oxygen candle" and self-contained self-rescuer chemistry underlie submarine atmosphere regeneration, mine-rescue breathing apparatus, and early closed-circuit spacecraft life-support concepts. Sodium peroxide's clean hydrolysis to H₂O₂ (Example 2) made Na₂O₂ historically important as an industrial bleaching and oxidizing agent, and it remains a convenient solid-state source of "active oxygen" in some cleaning and water-treatment formulations, since it avoids transporting and stabilizing liquid hydrogen peroxide directly. The lattice-energy logic developed here — that cation size selects anion charge state by controlling how much electrostatic stabilization is available to "pay for" a more highly charged anion — generalizes far beyond Group 1. It explains why alkaline-earth peroxides (BaO₂ is far more stable/isolable than the analogous MgO₂, mirroring exactly the Li/Cs contrast but shifted up one group), why certain transition-metal and lanthanide oxide/peroxide chemistries show comparable radius sensitivity, and it is the same reasoning that will be generalized quantitatively under Lattice Energy and Thermodynamic Stability of s-Block Salts, where the same Born–Haber/Kapustinskii machinery is used to rationalize solubility trends, thermal-decomposition temperatures of carbonates and nitrates, and the broader "radius-ratio" logic that runs through essentially all of ionic solid-state chemistry. The superoxide radical anion O₂⁻ also has a substantial afterlife in biochemistry: it is the primary reactive oxygen species generated during mitochondrial electron transport leakage and by NADPH oxidase in immune cells, and its controlled detoxification by the metalloenzyme superoxide dismutase (which cycles a metal center between oxidation states to disproportionate 2 O₂⁻ + 2H⁺ → H₂O₂ + O₂, structurally the same disproportionation chemistry seen in Example 2) is essential to aerobic life — a striking case of the same O–O bond-order/MO logic appearing in both a submarine's air scrubber and a living cell's antioxidant defense.
Ask the tutor: How would this lattice-energy argument change if we used a Born–Haber cycle with real experimental lattice enthalpies instead of Kapustinskii estimates — would the Na/K crossover point shift?
The physical picture
Start from O₂ itself. The molecular orbital diagram of the oxygen molecule is the textbook example of why simple Lewis structures fail: the ground state is a triplet, with two unpaired electrons in the degenerate π orbitals (π₂ₚₓ and π₂ₚᵧ). The valence MOs, in increasing energy, are σ₂ₛ, σ₂ₛ, σ₂ₚᵤ, {π₂ₚₓ, π₂ₚᵧ}, {π₂ₚₓ, π₂ₚᵧ}, σ₂ₚᵤ. For neutral O₂ (12 valence electrons) the filling gives (σ₂ₛ)²(σ₂ₛ)²(σ₂ₚ)²(π₂ₚ)⁴(π₂ₚ)², with the last two electrons singly occupying the two degenerate π orbitals (Hund's rule) — this is the paramagnetism you can demonstrate by pouring liquid O₂ between the poles of a magnet. Bond order is computed the usual way,
Adding electrons to make the anions simply continues filling π: one electron gives superoxide, O₂⁻ (bond order 1.5), two electrons gives peroxide, O₂²⁻ (bond order 1, both π orbitals fully occupied, diamagnetic, no unpaired spins). Each additional electron in an antibonding π* orbital weakens and lengthens the O–O bond and lowers the vibrational stretching frequency — a trend you can watch directly in Raman/IR spectra of the isolated solids.
| Species | Valence e⁻ (O₂ core) | π* occupancy | Bond order | O–O length (pm) | Magnetism |
|---|---|---|---|---|---|
| O₂²⁺ (dioxygenyl) | 10 | 1 | 2.5 | ~112 | paramagnetic |
| O₂ | 12 | 2 | 2 | 121 | paramagnetic (triplet) |
| O₂⁻ (superoxide) | 13 | 3 | 1.5 | 128–135 | paramagnetic (1 e⁻) |
| O₂²⁻ (peroxide) | 14 | 4 | 1 | 149 | diamagnetic |
Now bring in the metal. Oxide, O²⁻, is not simply "O₂²⁻ with another electron and no O–O bond" — it is atomic oxygen fully reduced, with no O–O linkage at all; it is the anion of the binary oxide M₂O. Whether a solid alkali-metal oxide contains isolated O²⁻ ions, O₂²⁻ dimers, or O₂⁻ radicals is not decided by which anion is intrinsically "most stable" in the gas phase (in fact O²⁻ is unstable as a free gas-phase ion — it is only the surrounding lattice of cations that stabilizes it enough to exist at all). It is decided by which anion, paired with the cation actually present, gives the lowest-energy ionic solid.
The lattice-energy competition can be pictured with a simple electrostatic scaling argument before ever writing a full Born–Haber cycle. Coulombic lattice energy scales, to a first approximation via the Born–Landé/Kapustinskii-type relation, as
For a fixed cation radius , replacing O²⁻ (, pm) with O₂⁻ (, pm, treating the whole ion as a single quasi-spherical unit) roughly halves the charge product and increases the denominator — both effects lower . This loss of lattice energy is a bigger relative penalty when is small (because is dominated by 's change) and a smaller relative penalty when is already large (because dominates the sum and the anion swap barely moves it). That asymmetry — small cations "feel" the anion-charge change strongly, large cations barely feel it — is the entire content of the Li → Cs trend, and it is worked through quantitatively below.
The core ideas
The governing relationship is the lattice-energy dependence on ionic charge and radius, captured compactly by the Kapustinskii equation (a generalized, structure-independent version of the Born–Landé/Born–Mayer treatments):
where:
- = number of ions per formula unit,
- = magnitudes of cation and anion charges,
- = sum of ionic radii in picometres,
- the pm term is an empirical compressibility correction (Kapustinskii's constant).
Reading: lattice energy grows linearly with the product of the charges but only inversely with the sum of the radii. Because the charge product term is multiplicative while the radius term is additive, doubling an anion's charge (2 vs 1) is a much bigger lever on when the radii are small (so the fractional change in from switching anions is large) than when the cation is already big. The second governing relationship is simply an energy-cycle statement of competitive stability: for two candidate solids (containing O²⁻) and (containing O₂²⁻) formed from the same metal and O₂, the more stable product is the one with the more negative overall formation enthalpy,
where:
- = sublimation enthalpy of the metal (atomization),
- = first ionization energy of M,
- the O-based term bundles the gas-phase energy cost of making O²⁻, O₂⁻ or O₂²⁻ from O₂ (dissociation plus one or two electron affinities, with the second electron affinity of oxygen being strongly endothermic),
- = lattice energy of the specific M/anion pair, released on solid formation (a stabilizing, negative contribution to ).
Reading: everything on the right-hand side except is essentially fixed by the choice of anion (it doesn't care which alkali metal you use) — the electron affinity ladder for O₂ → O₂⁻ → O₂²⁻ is a fixed gas-phase cost. The only term that depends on which metal you paired the anion with is . So the entire Li → Cs trend in product identity is a trend in how much lattice energy each cation can supply to offset the increasingly unfavorable gas-phase cost of building up more negative charge on oxygen.
Goal: show that forming O²⁻ from O₂ is strongly endothermic in the gas phase, so that only a large lattice energy can make the solid oxide thermodynamically preferred over its peroxide/superoxide alternatives.
Step 1 — break the O=O bond. Converting ½O₂ into one O atom costs half the O=O bond dissociation energy:
Step 2 — add the first electron (favorable). The first electron affinity of atomic oxygen is exothermic:
Step 3 — add the second electron (strongly unfavorable). Adding a second electron to an already-anionic, electron-rich O⁻ ion means pushing negative charge onto negative charge; the electron–electron repulsion overwhelms any orbital stabilization, so the second electron affinity is strongly endothermic:
Step 4 — total gas-phase cost of making one mole of "free" O²⁻. Summing the three steps,
This is a huge endothermic penalty — free gaseous O²⁻ does not exist as a bound species at all in isolation; it is thermodynamically unstable with respect to autodetachment back to O⁻ + e⁻. The only reason ionic oxides like Li₂O or MgO exist is that the lattice energy released when O²⁻ is surrounded by cations vastly exceeds 852 kJ mol⁻¹ and more than repays this cost. Contrast this with the analogous number for the superoxide route: forming O₂⁻ directly by adding one electron to O₂ has
which is already exothermic without any lattice assistance at all. This single contrast — O²⁻ formation costing +852 kJ mol⁻¹ versus O₂⁻ formation being spontaneously favorable at −42 kJ mol⁻¹ — is the quantitative root of why only cations that can generate enormous lattice energies (small, highly charged: Li⁺, Mg²⁺, Al³⁺-type ions) can force the oxide route, while everything else defaults toward peroxide or superoxide.
Go deeper · Derivation B · 2 minComparing lattice energies of a fixed metal with O²⁻ versus O₂⁻: the Kapustinskii estimate
Goal: quantify, for a single representative cation, how much lattice energy is lost by swapping O²⁻ for O₂⁻, and show this loss shrinks as the cation radius grows — the mechanistic core of the Li→Cs trend. Use the Kapustinskii equation with ions per formula unit for M₂O (M⁺, M⁺, O²⁻) and for MO₂ (M⁺, O₂⁻), taking (O²⁻) = 140 pm and (O₂⁻) = 158 pm (an effective spherical radius reflecting its larger, delocalized charge cloud).
Step 1 — small cation case, Li⁺ (r = 76 pm). For Li₂O ( pm):
For hypothetical LiO₂ ( pm):
Note the lattice energy is quoted per formula unit; to compare fairly per mole of metal we should also examine 2 LiO₂ vs Li₂O (i.e. double the LiO₂ value to put it on a "per 2 Li" basis): kJ mol⁻¹, still far below 2806 kJ mol⁻¹ for Li₂O. The lattice-energy gap favoring the oxide is
— comfortably larger than the ~852 kJ mol⁻¹ gas-phase penalty for making O²⁻ instead of O₂⁻ (which itself is already exothermic), so lithium's small cation easily tips the balance toward the oxide.
Step 2 — large cation case, Cs⁺ (r = 167 pm). For Cs₂O ( pm):
For CsO₂ ( pm):
Doubling to compare on a per-2-Cs basis, kJ mol⁻¹, so the gap is
Step 3 — compare the gaps. The lattice-energy advantage of choosing O²⁻ over O₂⁻ shrinks from 1054 kJ mol⁻¹ (Li) to 763 kJ mol⁻¹ (Cs) — a decrease of almost 300 kJ mol⁻¹ simply from increasing the cation radius. Combined with the essentially fixed +852 kJ mol⁻¹ gas-phase cost of building O²⁻ (Step 4 of the first derivation) and the already-favorable −42 kJ mol⁻¹ cost of building O₂⁻, the net preference for oxide over superoxide, roughly
is strongly positive (favoring oxide) for Li (1054 − 894 ≈ +160 kJ mol⁻¹ net preference for oxide) but has essentially vanished or reversed for Cs (763 − 894 ≈ −131 kJ mol⁻¹, i.e. net preference swings toward superoxide). This crossover — occurring somewhere around Na/K — is exactly where the experimentally observed switchover from peroxide (Na) to superoxide (K, Rb, Cs) sits. The peroxide is the intermediate case: with an anion charge of 2 like oxide but a much larger, more diffuse anion (149 pm vs 140 pm) and a milder, near-thermoneutral gas-phase formation cost, Na₂O₂ occupies the energetic middle ground between the Li₂O and KO₂/RbO₂/CsO₂ extremes.
Limiting-case check. As (an arbitrarily huge, hypothetical monocation), the Kapustinskii denominators and become nearly equal (both dominated by ), so and differ almost only by the charge-product factor of 2 vs 1 — the lattice energy of the oxide never quite catches down to that of the superoxide relative to what the fixed +852 kJ mol⁻¹ gas-phase penalty demands, so extremely large cations should always favor the superoxide (or even the neutral O₂ dioxygenyl adduct route) over oxide. This matches the empirical fact that only the very largest alkali metal, Cs, gives essentially pure superoxide on combustion, while Rb sits close to the same regime and K is borderline (often isolated as a K₂O₂/KO₂ mixture depending on O₂ pressure and conditions).
Worked examples
Example 1 (foundational): identify the product and its magnetism for Na + O₂. Sodium metal burned in excess dry O₂ gives predominantly sodium peroxide, Na₂O₂, not Na₂O. Using the MO count: Na₂O₂ contains O₂²⁻, bond order 1 (both π* orbitals doubly occupied), so it is diamagnetic — confirmed experimentally by the absence of any measurable magnetic susceptibility beyond normal diamagnetic core contributions. The O–O bond length in Na₂O₂ is about 149 pm, matching the peroxide entry in the bond-order table. If instead Na₂O (containing isolated O²⁻, no O–O bond at all) were the product, it would show no O–O stretching band in the Raman spectrum — but Na₂O₂ shows a clear O–O stretch near 730–790 cm⁻¹, confirming the peroxide assignment structurally as well as thermodynamically. Sodium sits at the crossover point predicted by the lattice-energy argument: too large a cation to make Li-type oxide favorable, but still small/charge-dense enough (compared to K, Rb, Cs) to stabilize the doubly-charged peroxide rather than defaulting fully to superoxide.
Example 2 (applied): why does Na₂O₂ react with water to liberate H₂O₂, and what does KO₂ do instead? Peroxide hydrolysis is simply protonation of the O₂²⁻ ion without breaking the O–O bond:
Each oxygen atom in O₂²⁻ simply picks up an H⁺; the O–O single bond, being a genuine σ bond between two anionic oxygens, survives protonation intact, and the product is hydrogen peroxide. Contrast this with superoxide hydrolysis. Superoxide is a radical anion (bond order 1.5, one unpaired electron); protonation of O₂⁻ is more complex because the resulting neutral HO₂• radical is unstable and disproportionates:
Here only half the oxygen ends up as peroxide; the other half is released as O₂ gas, because superoxide disproportionates (2 O₂⁻ → O₂²⁻ + O₂) far more readily than peroxide does — a direct consequence of the odd-electron, radical character of O₂⁻ making a second O₂⁻ + O₂⁻ combination (forming one O₂²⁻ and releasing one O₂) both spin-allowed and thermodynamically easy. This O₂-releasing hydrolysis (and analogously, CO₂ absorption, discussed below) is the reaction exploited industrially.
Example 3 (exam/competition level): estimate the crossover cation radius at which oxide and superoxide become equally favorable, using the derivation's own numbers. From the Kapustinskii-based derivation, for a generic monovalent cation of radius (in pm), the lattice-energy gap between the "per 2 M" oxide and superoxide routes can be approximated by interpolating the two computed data points. At pm (Li), kJ mol⁻¹; at pm (Cs), kJ mol⁻¹. Treating this dependence as approximately linear over this range (a reasonable local approximation since the Kapustinskii function varies smoothly),
The net thermodynamic preference for oxide over superoxide is kJ mol⁻¹ (the fixed gas-phase cost difference derived earlier). Setting this to zero to find the crossover radius:
This crude linear estimate places the crossover cation radius at roughly 126 pm — squarely between Na⁺ (102 pm, favoring peroxide, an intermediate case not captured by this two-anion-only model) and K⁺ (138 pm, favoring superoxide). Given how coarse the model is (it ignores peroxide entirely and uses only two calibration points), landing within about 12 pm of the K⁺ radius, and correctly bracketing the experimentally observed Na/K changeover, is a reasonable success for a back-of-envelope electrostatic argument — precisely the kind of order-of-magnitude consistency check expected of a Kapustinskii-level estimate rather than a claim of quantitative precision.
Limiting cases, scope & common traps
| Situation | What actually happens | Why |
|---|---|---|
| Li + O₂ (excess, dry) | Li₂O (some Li₂O₂ as minor product) | Small Li⁺ gives huge , easily repaying O²⁻'s +852 kJ mol⁻¹ gas-phase cost |
| Na + O₂ (excess) | Na₂O₂ (peroxide) | Na⁺ intermediate size; peroxide's charge (2−) still worth stabilizing, but not quite enough lattice energy to also favor full disproportionation-resistant O²⁻ over O₂²⁻ |
| K, Rb, Cs + O₂ (excess) | MO₂ (superoxide) | Large cations cannot repay the O²⁻ (or even fully the O₂²⁻) charge penalty; the nearly-thermoneutral/exothermic O₂⁻ route wins |
| Li + limited O₂, or controlled synthesis | Li₂O still dominant (Li₂O₂ known but less stable, decomposes on heating) | Li's small radius makes O²⁻ pairing so favorable it dominates regardless of O₂ stoichiometry |
| Heating Na₂O₂ or MO₂ | Peroxides/superoxides can be thermally converted toward the simple oxide with loss of O₂ | Entropy of releasing O₂ gas eventually favors the lower-oxidation-state oxide at high T |
| M₂O + O₂ (all alkalis, forcing conditions) | Higher alkalis convert to peroxide/superoxide; Li resists | Same lattice-energy competition in reverse; consistent with same trend |
The chemist behind it
The systematic recognition that alkali-metal combustion products follow a size-dependent trend across oxide/peroxide/superoxide emerged from mid-20th-century solid-state and structural inorganic chemistry, particularly X-ray and magnetic-susceptibility studies that definitively distinguished O₂²⁻ from O₂⁻ species (settling long arguments about whether "potassium peroxide" was really a distinct paramagnetic superoxide). The lattice-energy/radius-ratio framework itself descends directly from the ionic-model machinery built by Born, Landé, Haber, and later Kapustinskii for computing lattice energies from purely electrostatic and structural data — the same machinery used throughout main-group solid-state chemistry.
Origins · 1 minAnatoli Fedorovich Kapustinskii and how this was discovered
Kapustinskii generalized the Born–Landé lattice-energy formula into a structure-independent equation depending only on ionic charges, radii, and the number of ions per formula unit — removing the need to know a crystal's exact Madelung constant. This made it possible to estimate lattice energies for salts of hypothetical or unstable ions (such as gas-phase O²⁻, which does not exist free but must be assigned a lattice energy inside a real crystal), which is precisely the tool required to compare oxide, peroxide, and superoxide lattice energies for the same cation on a common footing, as done in the derivations above.
Why doesn't lithium ever form a stable superoxide, LiO₂?
Li⁺ is so small and charge-dense that pairing it with the much larger, singly-charged O₂⁻ wastes a huge amount of potential lattice energy compared to pairing it with the smaller, doubly-charged O²⁻. The Kapustinskii-based comparison above shows Li₂O's lattice energy advantage over the hypothetical LiO₂ route is roughly 1054 kJ mol⁻¹ per two lithiums, far exceeding the ~852 kJ mol⁻¹ gas-phase cost of forming O²⁻ instead of O₂⁻. LiO₂ is known only as a metastable, low-temperature or matrix-isolated species; it readily disproportionates toward Li₂O₂ or Li₂O under normal conditions.
Is Na2O2 truly "the" product of sodium combustion, or does some Na2O also form?
Under carefully controlled, oxygen-limited conditions, Na₂O can be prepared, and small amounts may form alongside Na₂O₂ during ordinary combustion in air. But the thermodynamically dominant, easily isolated bulk product when sodium burns in excess dry oxygen is the peroxide, Na₂O₂, consistent with sodium's intermediate cation size sitting near, but on the peroxide side of, the oxide/peroxide-superoxide crossover discussed in the worked examples.
Why is the second electron affinity of oxygen so unfavorable, and how does an oxide ion survive at all in a crystal?
Adding a second electron to O⁻ means forcing extra negative charge onto an ion that is already anionic; the electron–electron repulsion this creates outweighs the modest orbital stabilization gained, making the process endothermic by roughly +744 kJ mol⁻¹. Combined with the bond-breaking cost of atomizing O₂, the overall gas-phase formation of O²⁻ costs about +852 kJ mol⁻¹. Free O²⁻ is thermodynamically unstable in isolation and would simply eject an electron. It survives only inside a crystal lattice, where the surrounding shell of cations supplies enough electrostatic (Madelung) stabilization — often several thousand kJ mol⁻¹ — to more than repay this cost, which is exactly what the lattice-energy side of the Born–Haber cycle is doing physically.
How does the peroxide/superoxide trend connect to the broader idea of lattice energy controlling s-block salt stability?
This trend is a specific, vivid illustration of a completely general principle: whenever two possible anions (or oxidation states, or coordination geometries) compete for the same cation, the winner is decided by which combination maximizes total lattice stabilization relative to the intrinsic gas-phase cost of forming that anion. The same logic explains carbonate versus bicarbonate stability, why small cations stabilize small polyatomic anions preferentially, and why thermal decomposition temperatures of alkali/alkaline-earth carbonates and nitrates increase down each group — all cases developed further under the more general lattice-energy/thermodynamic-stability framework for s-block salts.
Does the same oxide/peroxide/superoxide competition happen in Group 2 (the alkaline earths)?
Yes, and the trend is shifted appropriately by the higher cationic charge. Beryllium and magnesium essentially only form the simple oxide (BeO, MgO); calcium and strontium form oxides with peroxides accessible under forcing conditions; and barium, the largest and least charge-dense Group 2 cation used in practice, forms a genuinely stable peroxide, BaO₂, industrially useful historically as an oxygen source (the Brin process). The 2+ charge on Group 2 cations means even the largest, Ba²⁺, still favors peroxide over superoxide — Group 2 never reaches the true superoxide regime that Group 1's K/Rb/Cs achieve, because the higher cation charge keeps pulling the balance back toward more highly charged anions.
Why is KO₂ paramagnetic, and how is this demonstrated experimentally?
KO₂ contains the O₂⁻ radical anion, whose highest-energy π* orbital holds a single unpaired electron (bond order 1.5). This unpaired electron gives the solid a permanent magnetic moment, measurable directly by magnetic susceptibility (Gouy balance or SQUID magnetometry), which shows KO₂ is attracted into a magnetic field, unlike diamagnetic Na₂O₂ which is weakly repelled. The measured effective magnetic moment is close to the spin-only value for one unpaired electron, corroborating the MO-based bond-order assignment rather than any alternative structural model.
Can these anions interconvert once formed, and does temperature matter?
Yes — peroxide and superoxide solids are thermodynamically metastable relative to full conversion to the simple oxide plus O₂ gas at sufficiently high temperature, because releasing gaseous O₂ carries a large favorable entropy term that eventually overcomes the lattice-energy preference for the higher-oxygen-content anion. Heating KO₂ or Na₂O₂ strongly enough drives loss of O₂ and conversion toward the corresponding oxide/peroxide. This is why storage and handling temperatures matter for KO₂-based life-support canisters, and why the "preferred product" identified by room-temperature combustion is really a kinetically accessible, but not always the absolute highest-temperature thermodynamic, sink.