Chemistry · Redox Equilibrium · NEET
It needs the SAME element, starting from ONE middle oxidation state, to be oxidised (go up) AND reduced (go down) in the same reaction. Example: Cu(+1) splits into Cu(0) and Cu(+2). So the element must have at least one higher state available and one lower state available. If either is missing, disproportionation is impossible.
To disproportionate, part of the element must be oxidised, meaning its oxidation number must increase. But if it is already at the maximum possible value, there is no higher state to move into. So it cannot be oxidised at all. It can only act as an oxidising agent (get reduced). Example: Mn is +7 in MnO4-. Mn cannot go above +7, so MnO4- alone cannot disproportionate.
No. To disproportionate, part of the element must also be reduced, meaning its oxidation number must decrease. If the element is already at its lowest value, it cannot go lower, so it cannot be reduced. It can only act as a reducing agent. So both extremes (highest AND lowest state) cannot disproportionate. Only middle (intermediate) oxidation states can.
Fluorine is the most electronegative element. In all its compounds F is -1 (its lowest state), and F2 is 0 (its highest state). F- is already at the bottom, so it cannot be reduced further; F2 has no positive state, so as an element it has almost no middle state to split from. NCERT states there is no way to convert F- to F2 by ordinary chemical means; it needs electrolysis. So fluorine species do not disproportionate like Cl or Br do.
Find the oxidation number of the key element. Then check: does the element have a higher stable state AND a lower stable state available? If yes, it can disproportionate (like Cl in +1, Mn in +6, P in 0). If it is at the top (Mn +7, Cl +7, S +6, N +5) it cannot. If it is at the bottom (F -1, Cl -1, S -2) it cannot. Middle state = possible; extreme state = impossible.
Yes. In comproportionation, two DIFFERENT oxidation states of the same element combine to give ONE middle state. Example: MnO4- (+7) plus Mn2+ (+2) give MnO2 (+4). This is the reverse of disproportionation. NEET often puts a comproportionation reaction as a wrong option in a disproportionation question, so read carefully.
Which of the following reactions are disproportionation reactions? (a) 2Cu+ -> Cu2+ + Cu (b) 3MnO4^2- + 4H+ -> 2MnO4- + MnO2 + 2H2O (c) 2KMnO4 -> K2MnO4 + MnO2 + O2 (d) 2MnO4- + 3Mn2+ + 2H2O -> 5MnO2 + 4H+
From the Latimer diagram BrO4- (1.82 V) BrO3- (1.5 V) HBrO (1.595 V) Br2 (1.0652 V) Br-, the species undergoing disproportionation is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
MnO4- (permanganate). Mn is +7, the highest possible oxidation state of manganese. It cannot be oxidised further, so it cannot disproportionate; it only acts as a strong oxidising agent. Other examples: KClO4 (Cl +7), H2SO4 (S +6), HNO3 (N +5), K2Cr2O7 (Cr +6).
For many main-group elements the highest oxidation state equals the group number (for example S in group 16 has +6, Cl in group 17 has +7, N in group 15 has +5). Once an element sits at this top value, it has no higher state, so that species cannot be oxidised and cannot disproportionate.
Yes. +6 is an intermediate state of manganese (it has +7 above and +4 below available). So MnO4^2- (manganate, Mn +6) disproportionates into MnO4- (+7) and MnO2 (+4). But MnO4- (Mn +7) is at the top and cannot. This is why the state, not just the element, decides.
NEET regularly asks 'which of these is/are disproportionation reactions' and mixes in traps like comproportionation and thermal decomposition. Knowing the simple rule (only intermediate states disproportionate; highest and lowest states cannot) lets you eliminate wrong options in seconds without balancing anything.
