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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1$\begingroup$ 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 Neretin Feb 8 '17 at 16:53
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Actually, there are two reaction involving:
- $$\ce{P4 + 3NaOH + 3H2O -> PH3 + 3NaH2PO2}$$
White phosphorus react with sodium hydroxide and water to produce phosphine and sodium hypophosphite . Sodium hydroxide - cold, concentrated solution. This reaction takes place slowly. (Chemiday 1)
The O.S. of phosphorus in sodium hypophosphite is calculated in @Sayantan.G's answer.
- $$\ce{P4 + 8NaOH + 4H2O ->[\Delta] 4Na2HPO3 + 6H2}$$
White phosphorus react with sodium hydroxide and water to produce disodium hydrogen phosphite and hydrogen. Sodium hydroxide - concentrated solution. The reaction takes place in a boiling solution.(Chemiday 2)
The O.S. of phosphorus in disodium hydrogen phosphite is:
$$\ce{ +2*1 + 1 + x + 3*(-2) = 0}$$
$$\ce{ x = +3}$$
answered Feb 8 '17 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. $\endgroup$ – Nilay Ghosh May 27 '17 at 7:12
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When Phosphorous reacts with $\ce{NaOH}$ . It produces Phosphine and Sodium Hypophosphite.
The reaction is as follows:-
$$\ce{4P + 3NaOH +3H2O -> PH3 + 3NaH2PO2}$$
Oxidation State of Phosphorous in Sodium Hypophosphite is $+1$
As $Na$ is $+1$, $H$ is $+1$ and $O$ is $-2$.
Solution:-
$1+1*2+X-2*2=0$ (As net charge is zero)
Solving $X$ gives $+1$.
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1$\begingroup$ I think @Palash ,U should go through Nilay's answer . O.S. Must be +1 or +3 ,I think so😇 $\endgroup$ – Sayantan Ghosh Feb 9 '17 at 13:07
