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

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

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$1, 3, 1, 3$

Vedclass · Answer

(D) 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,$a=1, b=3, c=1, d=3$.

Column-$I$	Column-$II$
$A$. $C_2O_4^{2-} \rightarrow 2CO_2$	$P$. $2$
$B$. $Cr_2O_7^{2-} + 14H^{+} \rightarrow 2Cr^{3+}$	$Q$. $6$
$C$. $2Na_2S_2O_3 \rightarrow Na_2S_4O_6$	$R$. $3$
$D$. $MnO_4^{-} \rightarrow MnO_2$	$S$. $1$

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: