$\text{P}_4+3\text{OH}^-+3\text{H}_2\text{O}\xrightarrow{ \ \ \ \ \ \ \ }\text{PH}_3+3\text{H}_2\text{PO}^-_2$
Explanation:
$\stackrel{0}{\text{P}_4}+\stackrel{-2+1}{\text{3OH}}+\stackrel{+3-2}{\text{3H}_2\text{O }}\xrightarrow{ \ \ \ \ \ \ \ }\stackrel{-3+1}{\text{PH}_3}+\stackrel{-2+1-2}{\text{3H}_2\text{PO}^-_2}$
Because O.N. of P increases from 0 (P4) to +1 (H2P0–2) and decreases from 0 (P4) to -3 (PH3), therefore, P has undergone both oxidation as well as reduction. So, option (c) is correct. Option (d) is also correct because O.N. of H remains +1 in all the compounds and hence hydrogen is undergoing neither oxidation nor reduction.
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(Given $R=2.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ )
$S{O_{3(g)}} \rightleftharpoons S{O_{2(g)}} + 1/2\,{O_{2(g)}}$ is $4.9 \times 10^{-2}$ then find equilibrium constant for the reaction
$2S{O_{2(g)}} + {O_{2(g)}} \rightleftharpoons 2SO_3(g)$
