Observe the following reaction:aP_4{_{text{(s | vedclass

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(A) The given reaction is a disproportionation reaction where $P_4$ is both oxidized and reduced.Step $1$: Write the half-reactions.Reduction: $P_4 \r

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$1, 3, 3, 1$

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(A) The given reaction is a disproportionation reaction where $P_4$ is both oxidized and reduced.Step $1$: Write the half-reactions.Reduction: $P_4 \rightarrow PH_3$Oxidation: $P_4 \rightarrow H_2 PO_2^-$Step $2$: Balance the atoms and charges.Reduction: $P_4 + 3 H_2 O + 3 e^- \rightarrow PH_3 + 3 OH^-$Oxidation: $P_4 + 8 OH^- \rightarrow 4 H_2 PO_2^- + 4 e^-$Step $3$: Multiply to equalize electrons and add the equations.$4(P_4 + 3 H_2 O + 3 e^- \rightarrow PH_3 + 3 OH^-) \Rightarrow 4 P_4 + 12 H_2 O + 12 e^- \rightarrow 4 PH_3 + 12 OH^-$$3(P_4 + 8 OH^- \rightarrow 4 H_2 PO_2^- + 4 e^-) \Rightarrow 3 P_4 + 24 OH^- \rightarrow 12 H_2 PO_2^- + 12 e^-$Summing them: $7 P_4 + 12 H_2 O + 12 OH^- \rightarrow 4 PH_3 + 12 H_2 PO_2^-$Dividing by $4$: $1 P_4 + 3 OH^- + 3 H_2 O \rightarrow 1 PH_3 + 3 H_2 PO_2^-$Thus, $a=1, b=3, c=3, d=1$.

Observe the following reaction:
$aP_4{_{\text{(s)}}} + bOH^{-}{_{\text{(aq)}}} + cH_2O_{\text{(l)}} \rightarrow dPH_3{_{\text{(g)}}} + eH_2PO_2^{-}{_{\text{(aq)}}}$
The values of $a, b, c,$ and $d$ are respectively:

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Observe the following reaction:$aP_4{_{\text{(s)}}} + bOH^{-}{_{\text{(aq)}}} + cH_2O_{\text{(l)}} \rightarrow dPH_3{_{\text{(g)}}} + eH_2PO_2^{-}{_{\text{(aq)}}}$The values of $a, b, c,$ and $d$ are respectively: