(a) Phosphine
P4(s)+3NaOH(aq)+3H2O(l)→3NaH2PO2(aq)Sodiumhypophosphite+PH3(g)Phosphine↑
(b) Helium
(c) The electrode potential of F2 and Cl2 are:
F2+2e−→2F−;E∘=+2.87V
Cl2+2e−→2Cl−;E∘=1.36V
Since E∘ for F2/F− is higher than that of Cl2/Cl−, therefore, F2 is more easily reduced than Cl2. As a result, F2 is stronger oxidising agent than Cl2.
(d) It undergoes disproportionation to form PH3 and H3PO4.
4H3PO3Δ−→PH3Phosphine+3H3PO4Orthophosphoric acid
(e)PbS+4O3→PbSO4+4O2