$\text{P}_4+3\text{OH}^-+3\text{H}_2\text{O}\xrightarrow{ \ \ \ \ \ \ \ }\text{PH}_3+3\text{H}_2\text{PO}^-_2$
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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(Consider heat capacity of all solutions as $4.2 J g ^{-1} K ^{-1}$ and density of all solutions as $1.0 \ g mL ^{-1}$ )
$1.$ Enthalpy of dissociation (in $kJ mol ^{-1}$ ) of acetic acid obtained from the Expt. $2$ is
$(A)$ $1.0$ $(B)$ $10.0$ $(C)$ $24.5$ $(D)$ $51.4$
$2.$ The $pH$ of the solution after Expt. $2$ is
$(A)$ $2.8$ $(B)$ $4.7$ $(C)$ $5.0$ $(D)$ $7.0$
Give the answer question $1$ and $2.$
$6-$ amino $-4-$ hydroxycyclohex $-2-$ ene $-1-$ sulphonic acid is
[Use : $\left.\mathrm{h}=6.63 \times 10^{-34}\, \mathrm{Js}, \mathrm{m}_{\mathrm{e}}=9.0 \times 10^{-31}\, \mathrm{~kg}\right]$