- A$P_4 + OH^-\to PH_3 + H_2PO_2^-$
- B$Cl_2 + OH^-\to Cl^-+ \mathop O\limits^{\circleddash} Cl$
- C$Br_2 + OH^-\to Br^-+ BrO_3^-$
- ✓$2KClO_3 \to 2KCl + 3O_2$
It is a redox reaction where one element works both ways with part of it being oxidized and another part is reduced. It does not have to be strictly ultramolecular, because the oxidizing and reducing agents need not come from the same molecule.
$(a) \,P _4+\overline{ O } H \longrightarrow PH _3+ H _3 PO _2^{-}$
Phosphorus is oxidized and reduced.
$(b) \,Cl _2+\overline{ O } H \longrightarrow Cl ^{-}+ ClO ^{-}$
Chlorine is oxidized and reduced.
$(c) \,Br _2+\overline{ O } H \longrightarrow Br ^{-}+ BrO _3^{-}$
Bromine is oxidized and reduced
$(d) \,2 KClO _3 \longrightarrow 2 KCl +3 O _2$
Here, chlorine is reduced from $+5$ to $-1$ oxidation state while oxygen is oxidized from $-2$ to $0$ oxidation state.
$\therefore$ It is not a disproportionation reaction.
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$NH_3(g) + O_2(g) \rightarrow NO(g) + H_2O(l)$
$NO(g) + O_2(g) \rightarrow NO_2(g)$
To obtain maximum mass of $NO_2$ from a given mass of mixture $NH_3$ and $O_2$, the ratio of mass of $NH_3$ to $O_2$ should be -