The set of coefficients of P_4,OH^{Theta} and | vedclass

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(A) The given reaction is a disproportionation reaction: $P_4 + OH^{\Theta} 	o PH_3 + H_2PO_2^{\Theta}$.Step $1$: Oxidation half-reaction: $P_4 	o H

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$1, 3, 3$

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(A) The given reaction is a disproportionation reaction: $P_4 + OH^{\Theta} 	o PH_3 + H_2PO_2^{\Theta}$.Step $1$: Oxidation half-reaction: $P_4 	o H_2PO_2^{\Theta}$. Balancing $P$ atoms: $P_4 	o 4 H_2PO_2^{\Theta}$. Balancing $O$ and $H$ atoms using $OH^{\Theta}$ and $H_2O$: $P_4 + 8 OH^{\Theta} 	o 4 H_2PO_2^{\Theta} + 4 e^-$.Step $2$: Reduction half-reaction: $P_4 	o PH_3$. Balancing $P$ atoms: $P_4 	o 4 PH_3$. Balancing $H$ atoms: $P_4 + 12 H_2O + 12 e^- 	o 4 PH_3 + 12 OH^{\Theta}$.Step $3$: Multiply the oxidation half-reaction by $3$ to equalize electrons: $3 P_4 + 24 OH^{\Theta} 	o 12 H_2PO_2^{\Theta} + 12 e^-$.Step $4$: Add the two half-reactions: $(3 P_4 + 24 OH^{\Theta}) + (P_4 + 12 H_2O + 12 e^-) 	o (12 H_2PO_2^{\Theta} + 12 e^-) + (4 PH_3 + 12 OH^{\Theta})$.Simplifying: $4 P_4 + 12 OH^{\Theta} + 12 H_2O 	o 4 PH_3 + 12 H_2PO_2^{\Theta}$.Dividing by $4$: $P_4 + 3 OH^{\Theta} + 3 H_2O 	o PH_3 + 3 H_2PO_2^{\Theta}$.The coefficients for $P_4$,$OH^{\Theta}$,and $H_2PO_2^{\Theta}$ are $1, 3, 3$ respectively.

The set of coefficients of $P_4$,$OH^{\Theta}$ and $H_2PO_2^{\Theta}$ in the balanced reaction $P_4 + OH^{\Theta} \to PH_3 + H_2PO_2^{\Theta}$ is

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