Write phosphorous reacts with aqueous caustic soda to give hypophosphite and phosphine.

Explanation

a) The ionic skeleton equation is
`P_(4)overset(OH^(-))rarrPH_(3)+H_(2)PO_(2)^(-)`
b) Writing oxidation numbers
`overset(0)(P_(4))rarroverset(-3)(P)overset(+1)(H_(3))+overset(+1)(H_(2))overset(+1)(H_(2))overset(+1)(P)overset(-2)(O_(2)^(-))`
c) Locating atoms undergoing change in oxidation numbers
`overset(0)(P_(4))rarroverset(-3)(P)H_(3)+H_(2)overset(+1)(P)O_(2)^(-)`
d) Dividing the reaction into two two halves and balancing in acidic medium, separately
Oxidation half - reaction : `P_(4)rarrH_(2)PO_(2)^(-)`
Step 1 : Balance phosphorous atoms
`P_(4)rarr4H_(2)PO_(2)^(-)`
Step 2 : Balance oxygen atoms
`P_(4)+8H_(2)Orarr4H_(2)PO_(2)^(-)`
Step 3 : Balance hydrogen atoms
`P_(4)+8H_(2)O+8OH^(-)rarr4H_(2)PO_(2)^(-)+8H_(2)O`
Step 4 : Balance charge
`P_(4)+8OH^(-)rarr4H_(2)PO_(2)^(-)+4e^(-)" .....(a)"`
Reduction half - reaction : `P_(4)rarrPH_(3)`
Step 1 : Balance phosphorous atoms
`P_(4)rarr4PH_(3)`
Stpe 2 : Balance hydrogen atoms
`P_(4)+12H_(2)Orarr4PH_(3)+12OH^(-)`
Step 3 : Balance charge
`P_(4)+12H_(2)O+12e^(-)rarr4PH_(3)+12OH^(-)`
e) Equalising the electrons and adding the two halves
`"eq (a)"xx3+"eq (b)"xx"1, we get"`
`{:(3P_(4)+24H_(2)Orarr12H_(2)PO_(2)^(-)+12e^(-)),(P_(4)+12H_(2)O+12e^(-)rarr4PH_(3)+12OH^(-)),(bar(4P_(4)+12OH^(-)+12H_(2)Orarr" ")),(" "4PH_(3)+12H_(2)PO_(2)^(-)),(I_(4)+3OH^(-)+3H_(2)OrarrPH_(3)+3H_(2)PO_(2)^(-)):}`
This is the balanced equation.