Why is $\ce{HNO3}$ a stronger oxidising agent than $\ce{H3PO4}$? $\ce{N}$ and $\ce{P}$ have the same oxidation number. Is it the electronegativity difference between $\ce{N}$ and $\ce{P}$?

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    $\begingroup$ Do you think the electronegativity of N helps? N: 3.04 while P: 2.19. $\endgroup$
    – Huy Nguyen
    Jun 6, 2015 at 22:15
  • $\begingroup$ Yes, may be. N atoms accept electrons more easily than P atoms, and this can affect his reduction potential. But I'm not sure, the difference in electronegativity is not that big to explain the huge difference in reduction potentials. $\endgroup$
    – Ragnar
    Jun 6, 2015 at 23:32
  • $\begingroup$ related chemistry.stackexchange.com/questions/30863/… $\endgroup$
    – Mithoron
    Jun 7, 2015 at 13:46

4 Answers 4


Let me create background first. Oxidizing agents are the chemical that helps something else oxidize and itself gets reduced.(reduction in charge)

$\ce{N}$ in $\ce{HNO3}$ is in the +5 oxidation state - how do we know that? $\ce{H}$ is +1, $\ce{O}$ is -2 and the overall $\ce{HNO3}$ has a zero net charge. The same goes for $\ce{H3PO4}$ resulting in +5 oxidation state of $\ce{P}$.

That +5 is the highest common oxidation state for $\ce{N}$. If $\ce{N}$ is reduced to say +4 as in $\ce{NO2}$ or +2 in $\ce{NO}$, it's charge is reduced and $\ce{N}$ is therefore an oxidizing agent for that particular reaction

So, $\ce{HNO3}$ is an strong oxidizing agent because the $\ce{N}$ can be readily reduced.

But $\ce{H3PO4}$ is a poor oxidizing agent. Here are two basic reasons:

  1. Nitrogen does not possess $d$-orbitals in valence shell and so its covalency is limited to 4. $\ce{N}$ can however achieve a formal oxidation state of +5 as in the $\ce{NO3-}$ ion. The inability of $\ce{N}$ to unpair and promote its $2s$ electron results in that $\ce{N(+5)}$ is less stable than $\ce{N(+3)}$. However, $\ce{P(+5)}$ is more stable than $\ce{P(+3)}$ and phosphoric acid shows less oxidising properties.

  2. The affinity of phosphorus for oxygen is greater than that of nitrogen; as a result, phosphonic acid ($\ce{H3PO3}$) is good reducing agent.

    $\ce{H3PO4 + 2H+ + 2e-> H3PO3 + H2O}$; $E=-0.276 \ce{V}$

  • $\begingroup$ Thank you so much, now I get it. I didn't think of the contribution of d-orbitals in P. $\endgroup$
    – Ragnar
    Jun 7, 2015 at 23:20
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    $\begingroup$ No, there is no need to bring hypothetical d-orbitals of phosphorus into the game. Electronegativity of the two elements, compared with that of oxygen is enough. $\endgroup$
    – Jan
    Jun 8, 2015 at 1:06

Frost Diagrams explain it very nicely as well. Look at the G/F = zE/V, very positive for HNO3, which makes it a potent oxidizing agent (much less stable). H3PO4 is basically the complete opposite.

As you can see, the G/F=zE/V for HNO3 is very positive, making it a potent oxidizing agent, and much less stable.

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The simplest explanation lies in electronegativity. N is more electronegative than P. And a general trend is that the more electronegative an element is, the more unstable it is in high oxidation states. And since strong oxidizing agents typically are unstable, one can suspect that nitric acid is a stronger oxidizing agent than phosphoric acid.

However, to truly answer your question, one would need to consider the relative stability of the oxidized state and reduced state of the agent. And then lots of different considerations may come into play. For instance, nitric acid develops NO gas, giving an entropic contribution that phosphoric acid does not have.

Even though arguing with electronegativity is simple, it can be useful in a lot of different cases. For instance, high voltage cathode materials for Li-ion batteries depend on electronegative d-metals like Ni and Co. This is because a good Li-ion cathode needs to be a strong oxidizer.


To put it simply, in the lewis structure of nitric acid the nitrogen atom has a formal charge of +1 whereas, in the case of orthophosphoric acid, phosphorus bears no formal charge.

Nitrogen being an electronegative element couldn't bear the positive charge and thus attracts electron from the other substance with which it reacts.

Also notice that ,nitrogen of nitric acid is sp2 hybridized and phosphorous of ortho phosphoric acid is sp3 hybridized meaning that nitrogen is more electronegative ( electron -pulling charge density increases with more s orbital involvement) thus, due to positive charge & sp2 hybrid nature, HNO3 is stronger oxidizing agent than H3PO4 despite the fact that, they both have same oxidation number .in my personal opinion, if one just goes about saying that HNO3 is stronger oxidizing agent, just because Nitrogen is more electronegative than phosphorous- it is not a very good answer on grounds that, nitrous acid (HNO2) is not a stronger oxidizing agent than H3PO4 - despite the fact that Nitrogen is more electronegative than phosphorous ; here, taking the lewis structure of HNO2 into account ,we find that N bears no formal charge with sp2 hybridization more over, it shows smaller oxidation number than P of H3PO4. also the presence of lone pair of electrons makes it a sort of reducing agent ( to a little extent) these all facts are drawn out to say HNO2 is weaker oxidizing agent than H3PO4; in a nutshell, oxidization and electro negativity are bulk properties - one can't easily predict without giving thoughts to the bulk.

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    $\begingroup$ To put it simple is to put it wrong, at least in this case. $\endgroup$ Jun 6, 2021 at 20:07

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