Total number of species from the following which can undergo disproportionation reaction is ________.
H₂ O₂, ClO₃⁻, P₄, Cl₂, Ag, Cu⁺¹, F₂, NO₂, K⁺

Solution

Chemistry ma’am · 3 steps

Disproportionation needs a middle oxidation state, one with room to go both up and down.

1. The test. In disproportionation, one element in a single species is oxidised and reduced at the same time. So that element must sit in an intermediate oxidation state, with a lower state and a higher state it can reach.

   Species	Oxidation state	Goes down to	Goes up to
   H₂O₂	O: -1	-2 in H₂O	0 in O₂
   ClO₃⁻	Cl: +5	-1 in Cl⁻	+7 in ClO₄⁻
   P₄	P: 0	-3 in PH₃	+1 in H₂PO₂⁻
   Cl₂	Cl: 0	-1 in Cl⁻	+1 in ClO⁻
   Cu⁺	Cu: +1	0 in Cu	+2 in Cu²⁺
   NO₂	N: +4	+3 in NO₂⁻	+5 in NO₃⁻

2. The three that fail. Ag is at 0, its lowest state, so it can only be oxidised. K⁺ is at +1, its highest state, so it can only be reduced. F₂ is the most electronegative element and has no positive oxidation state, so it can only be reduced to F⁻.
3. So. H₂O₂, ClO₃⁻, P₄, Cl₂, Cu⁺ and NO₂ can disproportionate, which makes 6.

The common slip: 7. Counting F₂ along with Cl₂ gives 7. Fluorine has no positive oxidation state, so it can only be reduced, never disproportionated.

The idea to keep. For each species, ask two questions: can this element go to a lower oxidation state, and can it go to a higher one? Only a yes to both allows disproportionation. Fluorine never qualifies.

Answer from the NTA final answer key. Solution by PrepWiser.