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.

  • 2
    $\begingroup$ We can't. We just know it is hypophosphite; that was not obvious at all, nor could it be easily derived from first principles. $\endgroup$ Feb 8, 2017 at 16:53

2 Answers 2


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

  • $\begingroup$ Can sodium phosphate form? $\endgroup$ May 27, 2017 at 5:32
  • $\begingroup$ @Mockingbird if we use some phosphorus compound like phosphorus pentoxide, phosphoric acid or phosphorus trihalides and react with sodium hydroxide, sodium phoshate will form. $\endgroup$ May 27, 2017 at 7:12
  • $\begingroup$ 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? $\endgroup$ Jul 26, 2021 at 11:12

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 $-2$.


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

Solving for $X$ gives $+1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.