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.
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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. $\endgroup$– Ivan NeretinCommented Feb 8, 2017 at 16:53
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2 Answers
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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, $\ce{P2H4}$.
answered Feb 8, 2017 at 17:19
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$\begingroup$ @Mockingbird if we use some phosphorus compound like phosphorus pentoxide, phosphoric acid or phosphorus trihalides and react with sodium hydroxide, sodium phoshate will form. $\endgroup$ Commented May 27, 2017 at 7:12
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$\begingroup$ 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? $\endgroup$ Commented Jul 26, 2021 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 $+1$.
