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JEE Main 2023 · Chemistry · s-Block, p-Block Elements and Hydrogen · medium

White phosphorus is warmed with concentrated aqueous NaOH under CO2. The gaseous product and the ratio of P atoms oxidised to P atoms reduced (as per…

Single correct option

White phosphorus is warmed with concentrated aqueous NaOH\mathrm{NaOH} under CO2\mathrm{CO_2}. The gaseous product and the ratio of P atoms oxidised to P atoms reduced (as per P4+3NaOH+3H2O→PH3+3NaH2PO2\mathrm{P_4+3NaOH+3H_2O\to PH_3+3NaH_2PO_2}) are respectively
  1. A. PH3\mathrm{PH_3}; 3:13:1
  2. B. PH3\mathrm{PH_3}; 1:31:3
  3. C. H2\mathrm{H_2}; 3:13:1
  4. D. PH3\mathrm{PH_3}; 1:11:1
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Answer: (A) PH3\mathrm{PH_3}; 3:13:1

This is a disproportionation. In P4\mathrm{P_4}, P is 00. In PH3\mathrm{PH_3} P is −3-3 (reduced, 1 atom) and in NaH2PO2\mathrm{NaH_2PO_2} P is +1+1 (oxidised, 3 atoms). So oxidised : reduced =3:1=3:1 and the gas is the phosphine PH3\mathrm{PH_3}. Electron balance check: 1×31\times3 gained =3×1=3\times1 lost. Option (D) is wrong since electron balance forces the 3:1 ratio.
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