Can the 5+ and 4+ oxidation state of Vanadium ion exist independently?

From my class text book:

The oxidation states of Vanadium ions 5+ and 4+ are shown in the compounds $$\ce{VO2^{+}}$$ and $$\ce{VO^{2+}}$$ respectively, whereas $$\ce{V^{3+}}$$ and $$\ce{V^{2+}}$$ are shown like this in the redox reaction (reduction half equation).

Why is this? cannot $$\ce{V^{5+}}$$ and $$\ce{V^{4+}}$$ exist indepepently as ions like $$\ce{V^{3+}}$$ and $$\ce{V^{2+}}$$ (when measuring the reduction potential)?

When an atom $$M$$ looses one or more electrons, it becomes positively charged and becomes the ion $${M^{z+}}$$. In aqueous solution this positive charge attracts the unused doublets of the Oxygen atom from $$H_2O$$, and it repells the proton of $$H^+$$.

If $$Z$$ = $$1$$, the ion $$M^+$$ attracts several molecules water around it, and it makes a Debye layer of oriented $$H_2O$$ molecules around the ion, and no chemical changes.

If $$Z = 2$$, the attraction is strong enough to repell one proton $$H^+$$ from one of the molecules adsorbed around the $$Z^{2+}$$ ion. The solution becomes a little bit acidic, and the $$M^{2+}$$ion is partially hydrolyzed into the ion [$${M(OH)]^+}$$. It is an equilibrium :$$M^{2+} + H_2O \ce{<=>}[M(OH)]^+ + H^+$$

If $$Z = 3$$, the attraction between $$Z^{3+}$$ and water is strong enough to at least transform all $$M^{3+}$$ions into $$[M(OH)]^{2+}$$, and part of them to [$$M(OH)_2]^+$$ with the two reactions : $$M^{3+} + H_2O\ce{->} [M(OH)]^{2+} + H^+$$ $$[M(OH)]^{2+} + H_2O \ce{<=>} [M(OH)_2]^+ + H^+$$ This is the case if $$M^{3+}$$ is $$Al^{3+}$$ or $$Fe^{3+}$$

If $$Z = 4$$, all ions $$M^{4+}$$ are at least hydrolyzed into $$[M(OH)_3]^+$$ and even into $$M(OH)_4$$, according to : $$M^{4+} + 4 H_2O \ce{->}M(OH)_4 + 4 H^+$$It should be mentioned that $$M(OH)_4$$ has a tendency to loose water and produce a structure that has acidic properties. So it is often presented under the formula $$H_4MO_4$$ or $$H_2MO_3$$. That is the case with $$M = Si$$. Silicic acid is often presented as $$H_4SiO_4$$ or $$H_2SiO_3$$

With Vanadium $$V^{4+}$$, it goes another way. The molecule $$V(OH)_4$$ does not exist, as the ion $$V^{4+}$$ is too small to fix $$4$$ groups $$OH^-$$. As a consequence, $$V^{4+}$$ gives another positive ion $$VO^{2+}$$ according to : $$V^{4+} + H_2O \ce{->} [VO]^{2+}+2H^+$$ Strangely enough, it looks as if the non-existent $$V(OH)_4$$ behaves like a basic hydroxide and gets dissociated according to : $$V(OH)_4\ce{->}[VO]^{2+}+ 2 OH^–$$ If $$Z=5$$, the ion $$M^{5+}$$ can be hydrolyzed according to $$M^{5+} + 3 H_2O \ce{->} [MO_3]^{-}+6H^+$$$$M^{5+} + 4 H_2O \ce{->} [MO_4]^{3-}+8H^+$$ But with Vanadium $$V$$, the ion $$V^{5+}$$ may be hydrolyzed according to :$$V^{5+} + 2 H_2O \ce{->} [VO_2]^{+}+4H^+$$$$V^{5+} + 3 H_2O \ce{->} [VO_3]^{-}+6H^+$$The second reaction is favored at high pH.

This theoretical development is taken from General Chemistry, by Linus Pauling, W. F. Freemann and Co., San Francisco, 1953, Chapter XXI.