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Question: In the reaction, \[{{P}_{4}}+NaOH\to P{{H}_{3}}+Na{{H}_{2}}P{{O}_{2}}\] mole ratio of \[Na{{H}_{2}}P...

In the reaction, P4+NaOHPH3+NaH2PO2{{P}_{4}}+NaOH\to P{{H}_{3}}+Na{{H}_{2}}P{{O}_{2}} mole ratio of NaH2PO2Na{{H}_{2}}P{{O}_{2}} and PH3P{{H}_{3}} is:

Explanation

Solution

Hint: Start this question by writing the complete equation and balance it. Also, keep in mind the oxidation states and atomic numbers while solving it.

Complete step by step answer:
The reaction given above is an example of disproportionation reaction. It is because P4{{P}_{4}}acts both as an oxidant and reductant.
White phosphorus reacts with a cold and concentrated solution of sodium hydroxide and water to produce phosphine and sodium hypophosphite. This reaction takes place slowly.
We can balance the given equation as –
P4+3NaOH+3H2OPH3+3NaH2PO2{{P}_{4}}+3NaOH+3{{H}_{2}}O\to P{{H}_{3}}+3Na{{H}_{2}}P{{O}_{2}}
From the reaction we can see that there are 3 moles of NaH2PO2Na{{H}_{2}}P{{O}_{2}} (sodium hypophosphite) and 1 mole of PH3P{{H}_{3}}.
Therefore, the answer is – the mole ratio of NaH2PO2Na{{H}_{2}}P{{O}_{2}} and PH3P{{H}_{3}}in the given reaction is 3.

Additional information:
Reactants –
White phosphorus - P4{{P}_{4}}
White phosphorus or simply tetraphosphorus (P4{{P}_{4}}) exists as molecules which is made up of four atoms in a tetrahedral structure. The tetrahedral arrangement results in ring strain and instability.
Sodium hydroxide - NaOHNaOH
Water - H2O{{H}_{2}}O

Products –
Phosphine - PH3P{{H}_{3}}
It is a colorless, flammable, toxic gas and is classified as a pnictogen hydride. Phosphine burns and produces a dense white cloud of phosphoric acid.
Sodium hypophosphite - NaH2PO2Na{{H}_{2}}P{{O}_{2}}
Sodium hypophosphite is also known as sodium phosphinate. It is the sodium salt of hypophosphorous acid. It is a solid at room temperature and appears as odorless white crystals.

Note: The given reaction is a disproportionation reaction. It is also known as redox reaction in which a compound of intermediate oxidation state converts to two different compounds, one of higher and one of lower oxidation states, and hence oxidation and reduction takes place simultaneously.