Most elements react with oxygen to form oxides in which the oxidation number of O is –II. At first s — Physical Chemistry — Kinetics Chemistry Question
Alkali Metal Oxides, Peroxides, and Superoxides
Most elements react with oxygen to form oxides in which the oxidation number of O is –II. At first sight it seems a striking contradiction that some alkali metals, the strongest reducing agents themselves, burn in air to give peroxides or superoxides with fairly strong oxidizing properties. Why is the oxygen only partially reduced in these reactions? To find out, we must start from some basic properties of the elements.
[VISUAL]
We know the energy required to convert the elements into separated ions. How much energy is released when these ions combine to form ionic crystals? The lattice energy of an ionic solid can be estimated by the Kapustinskii equation. In its simplest form,
ΔU(lattice) = –107000 * ν * z+ * z- / (r+ + r-)
where ν is the total number of ions in the empirical formula, z+ and z– are the charges of the individual ions, r+ and r– are the ionic radii in pm, and the result is given in kJ mol–1.
Explain the trend in the ionization energies of the metals.
Model Answer
The atoms have analogous valence electron configurations, therefore the ionization energy decreases with increasing size.
The enthalpy of formation of the free ions dramatically increases in the sequence superoxide peroxide oxide. Why?
Model Answer
The superoxide ion is easily (and exothermically) formed by the addition of an electron to a neutral molecule. To form a peroxide ion, another electron must be added to this negative ion which requires considerable energy. In the oxide ion, the double negative charge must be accommodated by a single atom and, in addition, the O-O bond must also be broken.
Calculate the molar lattice energies of the oxides, peroxides, and superoxides of the three lightest alkali metals. Calculate the amount of energy released in each of the nine possible reactions leading to the oxides, peroxides, or superoxides of the elements. Always assume that 2 moles of a solid alkali metal react with oxygen to form a single product.
Model Answer
The results are summarized in the following table. All values are in kJ mol–1.
Product | Lattice energy | Formation of metal ions | Formation of oxygen ions | total | ΔrH (reaction)
Li2O | 1336 | 904 | 2240 | –2972 | –732
Li2O2 | 1336 | 553 | 1889 | –2578 | –689
2 LiO2 | 1336 | –86 | 1250 | –1829 | –579
Na2O | 1190 | 904 | 2094 | –2653 | –559
Na2O2 | 1190 | 553 | 1743 | –2335 | –592
2 NaO2 | 1190 | –86 | 1104 | –1646 | –542
K2O | 996 | 904 | 1900 | –2309 | –409
K2O2 | 996 | 553 | 1549 | –2064 | –515
2 KO2 | 996 | –86 | 910 | –1446 | –536
For the reaction of each alkali metal with an excess of oxygen, which is the energetically most favoured product?
Model Answer
Li2O, Na2O2, KO2.
Rationalize your results and try to explain, in terms of basic factors, why the composition of the most favoured product changes as we move down Group IA. What products do you expect from the reaction of Rb and Cs with oxygen?
Model Answer
The energy required for the formation of the anion from O2 steeply increases in the sequence O2 – → O2 2– → O 2– and so does the lattice energy released during the formation of the corresponding ionic compound. (Peroxide has a greater charge than superoxide, and oxide is smaller in size than peroxide.) The smaller the cation, the greater are the lattice energies and also the differences between those of oxide, peroxide, and superoxide. In the case of Li, the difference between the lattice energies is large enough to compensate for the higher energy required to form the oxide ion, thus Li2O is the most stable product. For larger cations, the lattice energy differences are smaller and do not cover this energy requirement, so the anions with progressively lower formation energies become preferred. The preferred product in the reaction of Rb or Cs with O2 should be the superoxide, even more so than for K. Indeed, this is actually the case.
Does this mean that peroxides and superoxides, which are powerful oxidizers, cannot be reduced by one of the most powerful reducing agents, metallic potassium?
Model Answer
Of course not. Our calculations have identified the most stable compound when oxygen is in excess. Should we compare the reactions of 1 mol O2 with various amounts of metals, the oxide would always be at the top. Potassium peroxide and superoxide are easily reduced by an excess of metallic potassium, e.g.: KO2 + 3 K = 2 K2O
Let’s turn now to the rest of the periodic system. Most other elements that form ionic oxides have multiply charged cations of relatively smaller size – properties that are generously rewarded in lattice energies. The alkali metals do not have this option. Why?
Model Answer
For the alkali metals to form a cation M 2+ would require the removal of an electron from an inner electron shell which needs disproportionately large amounts of energy (on the order of 3000 - 5000 kJ mol–1).
Consider a metal that forms a cation M 2+ with a radius of 100 pm (the cations of most metals are smaller than this). Compare the lattice energies of its oxide and peroxide. What products do you expect from the reaction of such metals with oxygen?
Model Answer
MO: –3567 kJ/mol, MO2: –3136 kJ/mol. The difference is greater than in the case of Li. Clearly the oxide is the favoured product.
Only one of the non-radioactive Group IIA metals can form a peroxide when heated in air at atmospheric pressure. Which one? Estimate a limit for the size of its cation on the basis of our model.
Model Answer
Barium, which has the largest cation. According to our model, the peroxide is stable if the difference between the lattice energies of the oxide and the peroxide is smaller than the difference between the enthalpies of formation of O 2– and O2 2–, i. e. 351 kJ/mol.
8 * 107000 / (r+ + 140) – 8 * 107000 / (r+ + 173) < 351
Hence we obtain r+ > 127.7 pm, which is in good agreement with the experimental findings: the ionic radius of Sr 2+ is 118 pm, that of Ba 2+ is 135 pm.
In conclusion, you can see that the extremely strong reducing ability of the alkali metals and the fact that some do not completely reduce oxygen when they burn in air have a common underlying reason. What is this?
Model Answer
The relatively small attraction force that the outer electrons experience from the core in the atoms of alkali metals. This results, on the one hand, in low ionization energies and, on the other hand, in cations of large size. As we have shown, peroxides and superoxides are formed if the counter ions are large enough.