Heating white phosphorus with conc. NaOH solution gives mainly

When white phosphorus ($$P_4$$) is heated with concentrated NaOH solution, it undergoes disproportionation.

The reaction of white phosphorus with concentrated NaOH solution is:

$$P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$$

In this reaction:

- Phosphorus in $$P_4$$ has an oxidation state of 0.

- In $$PH_3$$ (phosphine), phosphorus has an oxidation state of $$-3$$ (reduction).

- In $$NaH_2PO_2$$ (sodium hypophosphite), phosphorus has an oxidation state of $$+1$$ (oxidation).

This is a disproportionation reaction where phosphorus is simultaneously oxidised and reduced.

The main products are phosphine ($$PH_3$$) and sodium hypophosphite ($$NaH_2PO_2$$).

Hence, the correct answer is Option D.