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.

So, is it true that the $\ce{VO4+}$ ion dost not exist? Or have I missed something?

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 $\ce{VO4+}$?