[HappyFace]

Hi

I'm new to this and saw that this question was asked back in Oct 2009. However I cannot see how it works out to be
P= -4  when I work it out to be P= -3

Please can it be explained to me as I have it on a current assignement paper.

Many thanks

[Arctic-Nation]

That's because the oxidation number of the phosphorus atoms in pyrophosphoric acid is +5. Each of the phosphorus atoms is connected to 4 oxygen atoms (which gives +4), one of which with a double bond (an extra +1).

Could you link to thread in question? It seems strange that such an error would go uncorrected here.

[HappyFace]

Many thanks for the reply.

I'm not sure how to link to the tread, so I have copied it to my reply..

MakoEyes
Regular Member


Posts: 28


   Oxidation Number
« on: October 02, 2009, 03:50:26 PM » Quote 

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Give the oxidation number of phosphorus in each of the following compounds or ions:
(a) Ca3P2 (b) H3PO3 (c) P2O74– (d) P4O8 (e) Ca(H2PO4)2

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Borek
Mr. pH
Administrator
Deity Member

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Posts: 12201

    Re: Oxidation Number
« Reply #5 on: October 06, 2009, 10:03:32 PM » Quote 

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Quote from: MakoEyes on October 06, 2009, 04:55:30 PM
(c) P2O74– = Still unsure about this one.

2*ONP + 7*(-2) = -4.

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Many thanks once again.
 

[sjb]

In this thread (http://www.chemicalforums.com/index.php?topic=36196.0) the OP seemed to leave after being pointed at an equation, so I'm not sure whether s/he got the right answer or not. In any event, if the equation is correct, then the oxidation state of P is neither 4 or 3.

The equation as written says 2O_P + 7x^-2 = -4, so what is O_P?

edit missing tag

[Arctic-Nation]

Ah yes, now I see the confusion. Borek's formula is of course correct, but reading it as the answer to the original question will send you off-track.

To clarify for HappyFace: the summation of the oxidation numbers of all atoms in a molecule will give you the total charge of that molecule. As the total charge is known in this case, and since oxygen usually has an ON of -2, you can solve the formula towards the ON of phosphorus.