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What are the products obtained in the reaction of white phosphorus with aqueous sodium hydroxide?

I know one will be phosphine ($\ce{PH3}$).

How can we find the oxidation state of phosphorus in the other product?

Thanks for help in advance.

Sayantan Ghosh
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Palash gupta
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    We can't. We just know it is hypophosphite; that was not obvious at all, nor could it be easily derived from first principles. – Ivan Neretin Feb 08 '17 at 16:53

2 Answers2

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Phosphorus reacts with sodium hydroxide to give phosphine and alkali hypophosphite:

$$\ce{P4 + 3NaOH + 3H2O -> 3NaH2PO2 + PH3 ^}$$

The O.S. of phosphorus in disodium hydrogen phosphite is:

$$\ce{ +2*1 + 1 + x + 3*(-2) = 0}$$

$$\ce{ x = +3}$$

Mechanism: The water that is involved is actually moisture that is accounting for the hydrolysis of phosphorus. The reactions can be dissected into three semi-reactions:

$\ce{3P + 3H2O -> 3OPH + 3H}$
$\ce{P +3H -> PH3}$
$\ce{30PH + 3H2O -> 3H3PO2}$

The phosphine so obtained usually inflames spontaneously on coming into contact with the air; each bubble as it escapes forms a vortex ring of smoke. This is due to presence of some amount of $\ce{(PH2)_x}$, the simplest of them being the dimer, $\ce{P2H4}$.

Source: http://phosphorus.atomistry.com/phosphine.html

Nilay Ghosh
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  • Can sodium phosphate form? – Mockingbird May 27 '17 at 05:32
  • @Mockingbird if we use some phosphorus compound like phosphorus pentoxide, phosphoric acid or phosphorus trihalides and react with sodium hydroxide, sodium phoshate will form. – Nilay Ghosh May 27 '17 at 07:12
  • You used Chemiday. That is a well-regarded and reliable source, as surely as my running a sub-3 hour marathon (when I tried one it was closer to seven). Can you review and identify primary references? – Oscar Lanzi Jul 26 '21 at 11:12
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When Phosphorous reacts with $\ce{NaOH}$, it produces Phosphine and Sodium Hypophosphite. The reaction is as follows:

$$\ce{P4 + 3NaOH + 3H2O -> PH3 + 3NaH2PO2}$$

Oxidation State of Phosphorous in Sodium Hypophosphite is $+1$, as $\ce{Na}$ is $+1$, $\ce{H}$ is $+1$, and $\ce{O}$ is $-2$.

Solution:

$1+1 \times 2+X-2 \times 2=0$ (As net charge is zero)

Solving for $X$ gives $+1$.

Mathew Mahindaratne
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Sayantan Ghosh
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