Question

In the given sequence of reactions: \[ \text{P}_4 + \text{NaOH} + \text{H}_2\text{O} \;\longrightarrow\; \dots \;\longrightarrow\; X \;+\; 2\,\text{NaH}_2\text{PO}_2 \;\longrightarrow\; \dots \] The final product obtained with copper and \(\text{H}_2\text{SO}_4\) is:

• A. \(\text{Cu}_3(\text{PO}_4)_2\)
• B. \(\text{Cu}_3\text{P}_2\)
• C. \(\text{Cu}(\text{OH})_2\)
• D. \(\text{CuCO}_3\)

Hint:

Phosphine (PH\(_3\)) can reduce certain metal salts to metal phosphides. - In an acidic medium with copper, phosphide formation is common.

Answer 1

The Correct Option is B

Step 1: Reactions of Phosphorus with Alkali - White phosphorus (\(\text{P}_4\)) reacts with \(\text{NaOH}\) and water to give phosphine (\(\text{PH}_3\)) and hypophosphite (\(\text{NaH}_2\text{PO}_2\)). 

Step 2: Reaction with Copper and Acid - Phosphine can further react under specific conditions (involving \(\text{Cu}\) and \(\text{H}_2\text{SO}_4\)) to produce copper phosphide, \(\text{Cu}_3\text{P}_2\). Hence, the final product is (copper phosphide), \(\text{Cu}_3\text{P}_2\).