# Why oxides with the element in higher oxidation state are more acidic

We see this trend of acidic-basic property of oxides in most cases: $\ce{Mn2O7}$ is acidic, $\ce{MnO}$ is basic, the intermediate oxides are progressively less acidic and more basic, as the oxidation state of $\ce{Mn}$ decreases.

What is the reason of this general trend in almost all cases?

I have seen many explanations of this, most of them involving the proton releasing capacity of $\ce{E-O-H}$ bond, in the hydrolysed oxide. But most oxides of transition metals do not hydrolyze in water. So, I am seeking an alternative explanation without having to involve the hydrolysis of oxide. Also, most explanations on the internet lack any reference, so I prefer a reference.

A simple answer for this would be to say that $$\ce{Mn2O7}$$ has +7 oxidation state. Thus it is electron deficient and behaves as a lewis acid. $$\ce{MnO}$$ has +2 oxidation state and it is comparatively electron rich. Thus it behaves as a lewis base. So compounds in the middle are amphoteric.
In the bonds $$M^{z+}-O-H$$, where $$M$$ can be $$Mn$$ or $$Cr$$, the atom $$M$$ is positively charged. When this charge $$z$$ is large, $$M$$ repells the $$H$$ atom stronger than if z is small. So the molecule containing these bonds is a stronger acid. For example, $$Z$$ = $$6$$ in $$H_2CrO_4$$ and $$Z = 7$$ in $$HMnO_4$$. They both contain at least one $$M-O-H$$ bond. As $$Z$$ is high in these molecules, they are strong acids. Molecules containing $$H, O$$ and the same atoms $$Mn$$ and $$Cr$$ at a lower oxidation state exist. But they are not acidic in water. Examples : $$Mn^{2+}$$ makes $$Mn(OH)_2$$ which is insoluble in water and not acidic. $$Cr^{3+}$$ and $$Cr^{2+}$$ both produces $$Cr(OH)_3$$ and $$Cr(OH)_2$$ which are also insoluble and not acidic.