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In my book (NCERT India), it is mentioned that: (page 224)

...$\ce{Mn2O7}$ gives $\ce{HMnO4}$ and $\ce{CrO3}$ gives $\ce{H2CrO4}$ and $\ce{H2Cr2O7}$. $\ce{V2O5}$ is, however, amphoteric though mainly acidic and it gives $\ce{VO4^3–}$ as well as $\ce{VO2+}$ salts. In vanadium there is gradual change from the basic $\ce{V2O3}$ to less basic $\ce{V2O4}$ and to amphoteric $\ce{V2O5}$. $\ce{V2O4}$ dissolves in acids to give $\ce{VO^2+}$ salts. Similarly, $\ce{V2O5}$ reacts with alkalies as well as acids to give $\ce{VO4^3-}$ and $\ce{VO4+}$ respectively. The well characterised $\ce{CrO}$ is basic but $\ce{Cr2O3}$ is amphoteric.

The sentence under scrutiny has been boldened. I wonder how the $\ce{VO4+}$ ion can exist.

Going by the usual oxidation numbers, four oxide anions should contribute a total of -8 charge. To balance it, vanadium needs to be in +9 oxidation state. But, vanadium has only five valence electrons $(\ce{[Ar] 3d^3 4s^2})$, so how can it even attain the +9 oxidation state? Though, it may be possible that this ion has a different structure (possibly a peroxide linkage?) that allows it to exist with a +5 oxidation state on vanadium, but I am not sure.

I have read relevant documents on the internet, including the Wikipedia page for the vanadate ion and a Scholar search, but neither yield any relevant results. I can't search it on ChemSpider Structure Search either because I don't know its structure, and a textual search yields only unrelated partial matches.

So, is it true that the $\ce{VO4+}$ ion dost not exist, or have I missed something? Or is it that there is a misprint in the book, and they intended to write another ion in place of $\ce{VO4+}$?

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Since the comment by Ivan pointing to a possible misprint got four votes and no answer yet, I'll post it as an answer. In short: this is a misprint in the book, and the book probably meant $\ce{VO2+}$.

Here's a citation to back it up: Bertrand, G. L.; Stapleton, G. W.; Wulff, C. A.; Hepler, L. G. Thermochemistry of Aqueous Pervanadyl and Vanadyl Ions. Inorg. Chem. 1966, 5 (7), 1283-1284. DOI: 10.1021/ic50041a048.

...it does appear that below $\mathrm{pH} \approx2$ the principal species are the pervanadyl and vanadyl ions represented by $\ce{VO2+(aq)}$ and $\ce{VO^2+(aq)}$ (respectively).

Also worth noting that the formation of the pervanadyl ion is more favourable with a free energy of formation of $\pu{-140.3kcal/mol}$, as opposed to $\pu{-106.7kcal/mole}$ for the vanadyl ion (as mentioned in the source above)

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