Answer :
Explanation:
When white phosphorus (P4) is boiled in an alkaline medium, it disproportionates to form phosphine (PH3) and hypophosphite ions (H2PO2^-) according to the following redox reaction:
\[ \text{P}_4(s) + 3\text{NaOH}(aq) + 3\text{H}_2\text{O}(l) \rightarrow \text{PH}_3(g) + \text{H}_2\text{PO}_2^-(aq) + 3\text{Na}^+(aq) + 3\text{e}^- \]
In this reaction, phosphine (PH3) and hypophosphite ions (H2PO2^-) are produced.
Now, to determine the molar ratio between phosphine and hypophosphite ions produced, we need to balance the redox reaction first. After balancing, we can see that for every 1 mol of phosphine (PH3) produced, 1 mol of hypophosphite ions (H2PO2^-) is also produced.
So, the molar ratio between phosphine and hypophosphite ions produced is 1:1.