Can somebody explain me the disproportionation of Phosphorus (P4) into Phosphine (PH3) and dihydro hypophosphite ion ??? Also I need balanced reaction for it.

disproportionation reaction of P4, i.e., P4 is oxidised as well as reduced in the reaction. P4 is the oxidising as well as the reducing agent. P4 is oxidised to H2PO2 - and reduced to PH3. So the(unbalanced) half-equations are as follows.

P4 => 4PH3, (1) and

P4 => 4 H2PO2- (2)

Following the general procedure, we can write

P4 + 12 H2O + 12 e- => 4 PH3 + 12 OH- (1a)

P4 + 8 OH- => 4 H2PO2 - + 4 e- (2a)

Eq(1a) is the reduction half-equation and Eq(2a) is the oxidation half equation.

Multiplying Eq (2a) by 3 so that the two half equations have equal number of electrons,

we get

3 P4 + 24 OH- => 12 H2PO2 - + 12 e- (2b)

Adding Eq (1a) and Eq(2b) and cancelling the common terms , we get the balanced ionic equation

4 P4 + 12 H2O + 12 OH- => 4 PH3 + 12 H2PO2 -