Simulation · Chemistry · Class 11
Up and down the oxidation-state ladder
From the lesson Disproportionation and fractional oxidation numbers in Redox Reactions. Change the values and watch what happens.
Up and down the oxidation-state ladderChemistry · Class 11
The idea behind it
NCERT §7.3.1
- In a disproportionation reaction one element in one oxidation state is oxidised and reduced at the same time. The reacting substance must contain that element in a state from which it can both rise and fall.
- Hydrogen peroxide decomposes as 2H₂O₂ → 2H₂O + O₂: oxygen at −1 goes both to −2 (in water) and to 0 (in O₂).
- White phosphorus in alkali gives phosphine and hypophosphite (P 0 → −3 and 0 → +1); sulphur in alkali gives sulphide and thiosulphate (S 0 → −2 and 0 → +2).
- Chlorine in cold dilute alkali: Cl₂ + 2OH⁻ → ClO⁻ + Cl⁻ + H₂O, with Cl going from 0 to +1 and to −1. The hypochlorite formed is the active part of household bleach.
- Fluorine cannot disproportionate. It is the most electronegative element, cannot take a positive state, and in alkali gives F⁻ and OF₂ in a reaction where oxygen, not fluorine, is oxidised.
- Among the chlorine oxoanions, ClO⁻, ClO₂⁻ and ClO₃⁻ can disproportionate. ClO₄⁻ cannot, because chlorine is already at its highest state, +7, and can only be reduced.
- An averaged oxidation number can be fractional when atoms of one element sit in different environments: C₃O₂ gives +4/3, Br₃O₈ gives +16/3 and S₄O₆²⁻ gives +2.5. The real structures hold whole-number states, for example +5, 0, 0, +5 for the four S in tetrathionate.
- Mixed oxides such as Fe₃O₄, Mn₃O₄ and Pb₃O₄ also show fractional averages. Pb₃O₄ behaves as 2PbO + PbO₂, and the O₂⁺ and O₂⁻ ions give oxygen +½ and −½.