Chemistry · Ch 7 — Redox Reactions

The Paradox of Fractional Oxidation Number

You already know from §7.3 that oxidation numbers count whole electrons — an atom gains,

loses or shares whole electrons, never a fraction of one. So it looks like a contradiction

when a handful of real compounds are printed with a fractional oxidation number for

one of their elements: carbon in C₃O₂ works out to $+\frac{4}{3}$ , bromine in Br₃O₈ to

$+\frac{16}{3}$ , and sulphur in Na₂S₄O₆ to $+2.5$ — none of these are whole numbers.

The resolution is that a fractional oxidation number is never the true state of any single

atom. It is an average taken across several atoms of the same element that sit at

different positions in the molecule, each in its own whole-number state. The formula

alone can't see this — only the actual bonded structure can, because the structure shows

which atom is bonded to what.

Carbon suboxide, C₃O₂ is linear: O=C=C=C=O. The two terminal carbons are each in the

$+2$ state and the middle carbon (starred in the structure) is in the $0$ state — the

average is $\frac{(+2) + 0 + (+2)}{3} = \frac{4}{3}$ .

Figure box-C3O2-structureStructure of C3O2 (carbon suboxide), showing the fractional oxidation number of carbon as an average of whole-number states.

$Fig. box-C3O2-structure — Structure of C3O2 (carbon suboxide), showing the fractional oxidation number of carbon as an average of whole-number states.$

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

$O = \overset{+2}{C} = \overset{0}{C^*} = \overset{+2}{C} = O$ …

Tribromooctaoxide, Br₃O₈ has two end bromines each carrying three oxygens (the $+6$

state) around a middle bromine in the $+4$ state (starred) — the average is

$\frac{(+6) + (+4) + (+6)}{3} = \frac{16}{3}$ .

Figure box-Br3O8-structureStructure of Br3O8 (tribromooctaoxide), showing the fractional oxidation number of bromine as an average of whole-number states.

$Fig. box-Br3O8-structure — Structure of Br3O8 (tribromooctaoxide), showing the fractional oxidation number of bromine as an average of whole-number states.$

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

Br $_3$ O $_8$ has two terminal bromine atoms, each bonded to three oxygen atoms (one bridging to the central bromine, two terminal — like a perbromate-type BrO $_3$ group), and one central bromine atom bonded to the two bridging oxygens and one additional terminal oxygen. Assigning oxidation numbers atom by atom: each terminal bromine is in the $+6$ state, while the central bromine (marked with an asterisk) is in the $+4$ state. The bare molecular formula, Br $_3$ O $_8$ , would suggest a single fractional oxidation number of $+\frac{16}{3}$ for bromine — but that number is really just the ave …

Tetrathionate ion, S₄O₆²⁻ is a four-sulphur chain: the two outer sulphurs (each

bonded to three oxygens) are in the $+5$ state, the two inner sulphurs (starred, bonded

only to sulphur) are in the $0$ state — the average is $\frac{5 + 0 + 0 + 5}{4} = 2.5$ .

Figure box-S4O6-structureStructure of the tetrathionate ion S4O6(2-), showing the fractional oxidation number of sulphur as an average of whole-number states.

$Fig. box-S4O6-structure — Structure of the tetrathionate ion S4O6(2-), showing the fractional oxidation number of sulphur as an average of whole-number states.$

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

The tetrathionate ion, S $_4$ O $_6^{2-}$ , has a chain of four sulphur atoms: two terminal sulphur atoms, each bonded to three oxygen atoms (an -SO $_3^-$ -type group), and two central sulphur atoms bonded only to each other and to the terminal sulphurs (an S–S–S–S backbone, no oxygen on the two middle atoms). Assigning oxidation numbers atom by atom: each terminal sulphur is in the $+5$ state, while the two middle sulphur atoms (marked with asterisks) are each in the $0$ state. The bare molecular formula would suggest a single fractional oxidation number of $+2.5$ for sulphur — but that is really just t …

This isn't a one-off oddity of these three species. Several mixed oxides show the

identical pattern — Fe₃O₄, Mn₃O₄ and Pb₃O₄ are the standard examples, each one really a

combination of the metal in two different whole-number oxidation states packed into one

formula (Pb₃O₄, for instance, behaves as if it were 2 parts PbO — lead at $+2$ — fused

with 1 part PbO₂ — lead at $+4$ ; see Problem 7.7 in §7.3.1 for exactly this reaction

chemistry playing out).

The one genuine exception: in O₂⁺ (dioxygenyl) and O₂⁻ (superoxide), the oxidation

numbers really are $+\frac{1}{2}$ and $-\frac{1}{2}$ — not an average of two …

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