I've come across the reaction of $\ce{P4O10}$ with $\ce{HClO4}$ (perchloric acid) which gives the products as metaphosphoric acid and $\ce{Cl2O7}$ but I'm unable to understand how does perchloric acid form and what causes the trimer of $\ce{HPO3}$ to be formed, also $\ce{P4O10}$ being an acidic oxide how does it react with a strong acid like perchloric acid?

$$\ce{12HClO4 + 3P4O10->6Cl2O7 + 4(HPO3)3}$$

Most importantly how exactly does this

enter image description here

Get converted to a structure like this

enter image description here

Is there some sort of mechanism behind this like we see in T3P coupling reaction (to be clear I'm not saying this reaction is exactly that, I'm just asking if this involves similar mechanism) where oxyphosphorous linkage follows an elimination? (Like the oxygen of $\ce{HClO4}$ form a linkage with phosphorus of $\ce{P4O10}$)

Even though phosphorus pentoxide acts as dehydrating agent how does HPO3 form as a product how does the cage structure there get broken down to $\ce{(HPO3)3}$?

  • 1
    $\begingroup$ I think it's double substitution by $\ce{ClO4-}$ - first at $\ce{P}$ then at $\ce{Cl}$, catalysed by the acid. $\endgroup$
    – Mithoron
    Feb 23, 2022 at 22:25
  • $\begingroup$ The result of the reaction can be any polymer of $\ce{HPO3}$, and not only $\ce{(HPO3)3}$. $\endgroup$
    – Maurice
    Feb 24, 2022 at 10:20
  • $\begingroup$ @Maurice true but considering this case the product mentioned in the book J.D Lee , I want to know how that is formed? $\endgroup$
    – kafka yash
    Feb 24, 2022 at 11:24

1 Answer 1


Take any phosphorus atom. It is linked to three oxygen atoms $\ce{O}$ by three simple $\ce{P-O}$ bonds. These three bonds will be modified in a similar way, according to the following steps.

Step $1$. Approach one $\ce{HClO4}$ molecule. It provides one $\ce{H+}$ ion which will be attached to one of these bonding $\ce{O}$ atoms. The remaining $\ce{ClO4-}$ ion does not play a role presently.

Step $2$. After this attachment, the $\ce{P-O}$ bond coming from the same initial $\ce{P}$ atom is broken. The oxygen atom remains attached to the $\ce{H}$ atom and to the rest of the molecule. A positive charge appears on this initially chosen $\ce{P}$ atom.

Step $3$. Later on a water molecule will be attracted by this positive $\ce{P}$ atom, and will attach one $\ce{OH}$ group on this $\ce{P}$ atom, (plus release a free $\ce{H+}$ atom, that repeats the same game on the next bonding $\ce{O}$ atom). The whole is equivalent to providing one $ \ce{OH-}$ ion on the positive $\ce{P}$ atom.

Step $4$. Repeat the operations $1$ to $3$ on each $\ce{P-O}$ bonds leaving the initially chosen $\ce{P}$ atom. At the end, the $3$ $\ce{P-O}$ bonds (coming from the same $\ce{P}$ atom) are broken, and replaced by one $\ce{H}$ atom attached on each former bonding $\ce{O}$ atom, and $3$ $\ce{OH}$ bonds on the so modified $\ce{P}$ atom. This produces a lonely $\ce{H3PO4}$ molecule, plus the wanted cycle $\ce{(HPO3)3}$. The result is well described by $$\ce{P4O_{10} + 3 H2O -> H3PO4 + (HPO3)3}$$

Now the wanted substance $\ce{(HPO3)3}$ is obtained. It is easy to check that it has the same structure as drawn by Kafka yash in his original post.

Later on, both $\ce{H3PO4}$ and $\ce{HClO4}$ can be dehydrated to produce $\ce{HPO3}$ then $\ce{(HPO3)3}$ and $\ce{Cl2O7}$. The mechanism of these transformations may be studied by the same reasoning, but it is not required by the author.


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