1. The oxidation number of Phosphorus in white phosphorus ($P_4$) is 0.
2. Let the oxidation number of Phosphorus in phosphine ($PH_3$) be x.
The oxidation number of H is +1 and there are 3 H.
$x + 1(3) = 0 $
$\Rightarrow x = -3$
3. Let the oxidation number of Phosphorus in Sodium Hypophosphite $NaH_2PO_2$ be y.
The oxidation number of Na is +1 and there is 1 Na.
The oxidation number of H is +1 and there are 2 H.
The oxidation number of O is -2 and there are 2 O.
$+1+1(2)+y-2(2) = 0$
$\Rightarrow y = +1$
When the oxidation number increases, the element is said to be oxidised.
When the oxidation number decreases, the element is said to be reduced.
In the same reaction Phosphorous is getting both oxidised and reduced.
$\therefore$ This is a disproportionation reaction.
Disproportionation reaction is a reaction in which the same species is both oxidised and reduced.
Note: Such problems can be solved by finding the oxidation number of the required species in different reactants and products. One should be thorough with his concepts to find the oxidation numbers and should have knowledge about the different reactions.
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