The group 13 has the valence electron configuration of ns2p1, their common oxidation state is +3. Comparatively it is the most stable oxidation state. As atoms of these elements, the 0 is the oxidation state. However, it is not stable, for instance, Al is reactive at normal temperature. It can be oxidized by oxygen in the air. The oxidation numbers of B and Al are +3 in almost all their compounds. However, the heavier elements in the group are more likely to keep their s-electrons (the inter-pair effect), so the oxidation number +1 becomes increasingly important down the group, and thallium (I) compounds are as common as thallium (III) compounds.
B(OH)3OH2(aq) + H2O = H3O+(aq) + B(OH)4-(aq)
Al(H2O)63+(aq) + H2O = H3O+(aq) + [Al(OH)(H2O)5]2+(aq)
The group 14: the carbon family.
Carbon is a very important feature in the organic chemistry, and has 4 lone pair electrons. They are easy to form four covalent bonds. Carbon and silicon usually are +4 oxidation state in their compounds. However, they are not the most stable forms. Carbon’s 0 oxidation state is the most stable form at a particular condition that if the substance is diamond ( It is formed by carbons). Oppositely, if the condition is that the carbon forms the graphite, it will not be the most stable form. Because it is dependent of the forms of crystals. Diamond is an atomic crystal, whereas the graphite is the molecular crystal. Silicon reacts with oxygen then generates the SiO2. Here Si has the +4 oxidation state and this compound is absolutely stable as it is an atomic crystal.
On descending the group, however, the energy separation between the s- and p-orbitals increases and the s-electrons become progressively less available for bonding; in fact, the most common oxidation state for lead is +2.
E(s) + 2H3O+(aq) E2+(aq) + 2H2O(l) + H2(g), E = Sn, Pb.
The common oxidation states of tin are Sn(IV) (or stannic) and Sn(II) (or stannous). Due to the bond strength and ease of electronic promotion related to the IV oxidation state for tin, the relative stabilities of Sn(IV) and Sn(II) are approximately equal. The reason for this can be qualitatively illustrated by the relationship of tin to the other group 14elements. Carbon, silicon, and germanium are typically found in the IV oxidation state, whereas lead is normally found in the II oxidation state. This trend is not unique to the group 14 elements either, as the elements in groups 13-16 also demonstrate a decrease (by two) of the primary oxidation state as the elements of the group increase in size.
Though it is useful to qualitatively rationalize the two stable oxidation states of tin-it is, no doubt, necessary to reinforce these ideas with more tangible evidence-which can be found through analysis of the electron configuration of the elements in Group 14. Typically, Group 14 (IVA) elements follow the electron configuration ns2 np2 in which the s orbital has two electrons and the px and py orbitals contain one electron each. The pz orbital remains empty. In this case, the electrons in the s orbital are an inert pair and are, therefore, not used for bonding. The electrons of the p orbitals are used to form two covalent bonds. This is, of course, the configuration related to the elements which “prefer” a (II) oxidation state.
In the case of the elements that prefer a (IV) oxidation state, one of the s electrons in the inert pair (from the first example) is promoted to the pz orbital that was initially empty. In this case, four covalent bonds can be formed. Though the promotion of the s electron to the pz orbital requires some energy, this configuration is still desirable for the lighter, smaller elements of Group 14. This is because (as stated earlier, though in less detail) the energy required to promote the s electron to the pz orbital is negated by the release of energy from the additional two bonds that form. As for lead, it is more efficient to simply form two covalent bonds than it is to promote the s electron to the pz orbital in exchange for the energy released by the two extra covalent bonds. The reason that tin is stable in both the (II) and (IV) oxidation states is that the energy “costs” are similar in both situations for tin. In short, it requires similar “effort” on the part of the tin to both form two covalent bonds, or to promote an s electron and form four covalent bonds.
(The oxidation states of tin, Mike ward, 1997, USA)
Sn + 2I2 SnI4, Sn2+(aq) + 2I- SnI2(s)
Nitrogen has one of the widest ranges of oxidation states of any element: compounds are known for each whole-number oxidation number from -3 (in NH3) to +5 (in nitric acid and the nitrates). It also occurs with fractional oxidation numbers, such as -1/3 in the azide ion, N3-. Usually the N2(gas) is the most stable (0 oxidation state). The oxidation state is at -3, which is strong reductive. +5 of N is strong oxidative. Those Ns are not stable when those compounds reactive with some Lewis acids and Lewis bases.
Similarly phosphorus has several oxidation states, but its compound are very reactive, because of the fact that its electronegativity is lower, and it has 3d-orbitals available. Thus it is more active.
For the groups of 16 and 17, the common oxidation states of them are -2 and -1 respectively.
In the group 16, O has -1, -2 and 0 oxidation states. i.e. Na2O2, there the oxidation state of oxygen is -1. but it is not stable, and can be reduced to -2, or can be oxidized to 0 when treat with strong reducing reagent.
The oxidation states of sulfur in sulfur dioxide and the sulfites is +4, an intermediate value in sulfur’s range from -2 to +6. Hence these compounds can act as either oxidizing agents or reducing agents. The stability of sulfur is increasing by increasing the oxidation states.
SO3(g) + H2SO4(l) = H2S2O7(l)
Sulfur hexafluoride is a thermally stable nontoxic gas that is a good insulator.
All the halogens have the -1 and 0 oxidation states. Fluorine is the most electronegative element of all and has an oxidation state of -1 in all its compounds. Its high electronegativity and small size allow it to oxidize other elements to their highest oxidation states. Chlorine is a strong oxidizing agent and oxidizes metals to high oxidation states; for example, anhydrous iron(III) chloride, not iron(II) chloride, is formed when chlorine reacts with iron:
2Fe(s) + 3Cl2(g) = 2FeCl3(s).
Chlorine has a range of oxidation states, like their compounds, such as HClO4, HClO3, HClO2 and HClO. This property is similar as P.
Noble gases, here xenon is the only noble gas known to form an extensive series of compounds with fluorine and oxygen. Its oxidation states are from +2 to +8(XeO64-). Xenon fluorides are powerful fluorinating agents and xenon oxides are powerful oxidizing agents.
- Transition Metals (the d-block):
Generally speaking, every transition metal has multifarious oxidation states that can form many compounds. Further more, most of them are apply to 18 electrons’ rule. In spite of the fact that some metals have various formal oxidation states, their compounds’ electron counting is a constant of 18 e-.
The oxidation states for the period 4’s transition metals:
For these transition metals, the stability of their compounds depends on their oxidation states, their structures and so on. But the oxidation states are the important feature. When we consider their relative stability of oxidation states, we must count if they are apply to the 18e’ rule. (there is only one except the one that is V.) In the organometallic chemistry, the compound [V(CO)6] has the electron count 17 e-s. Here the v is at the oxidation state 0. But it is still stable at a certain temperature.
Thermodynamically stable transition metal organometallic complexes are formed when the sum of the metal’s d-electrons and the electrons donated by the ligands add up to 18.
Fe and Cu both have two formal oxidation states, and for the low oxidation state i.e. +2 Fe and +1 Cu are less stabile than the high oxidation state compound. After all, Fe (II) and Cu(I) both can be oxidized and reduced, whereas the Fe(III) and Cu(II) just can be reduced. On the other hand, Mn has 6 formal oxidation states, they have various compounds and organometallic complexes. KMnO4 may be reduced to oxidation of +6, +5, or the lowest oxidation state.
Altogether, the middle of oxidation states of a transition metal compound are not very stable, due to the fact that the metal cation is easy to donate or accept electron(s), then generating a new oxidation state metal by the oxidizing or reducing.
(4) Stability Index Diagrams: Pictorial Representations of the Relative Stabilities of Oxidation States for Metallic Elements
Diagrams that plot the free energies of formation of aqueous species, or enthalpies of formation of ionic solids, against the oxidation number n for metallic elements have been exploited by a number of authors in order to display the relative stabilities of oxidation states. The lack of experimental data for unstable oxidation states, and difficulties in estimating required enthalpies/free energies of formation, have limited the utility of such diagrams.
The stability index An of a metallic element M in the +n oxidation state is defined by the equation: An = DeltaHof(Mn+, g) - an(n + 1)/(r + b) where a is an empirical constant (taking the recommended value 1150 kJ mol-1 A), and r is the metallic radius of M in A. b is a constant which reflects the covalent radius of the element bonded to M in a binary compound, and is allocated a representative value of 1.0 A, although this can be varied in order to compare, e.g., fluorides with iodides. Stability index diagrams plot An against n. Examples given for elements of the 3d and 4f series, and for the Group 13 elements, demonstrate the value of such diagrams, and their construction helps students and teachers to rationalize the relative stabilities of oxidation states.
(Smith, Derek W. Stability Index Diagrams: Pictorial Representations of the Relative Stabilities of Oxidation States for Metallic Elements J. Chem. Educ. 1996 73 1099.)
Here is a new theory that Derek W. Smith developed. He found that the stability of oxidation state for metallic elements are apply to a energy equation: An = DeltaHof(Mn+, g) - an(n + 1)/(r + b). This equation is easier to see a metal element’s stability. It made the microcosmic stability of oxidation state macroscopically.
Conclusion
The main factor is energy theory for the stability of a compound. But the relative oxidation states theory is a microcosmic factor. It is essential that the variability and relative stability of oxidation states is able to insight into a compound and their relative elements.
Reference:
-
Chemical principles, Peter Atkins,2nd edition, 2000, Main groups.
- Organometallic chemistry, A Unified Approach, R.C. Mehrotra, 1998.
- Oxidation states of Tin, Mike ward, www.webelements.com.
- Chemistry summarizing, Lan xinzhong, 2000, China Dalian.