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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.

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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.

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It does not appear to be always true. According to Wikipedia, copper(I) becomes acidic, forming an anionic hydroxide complex, at about 1.5 pH unit lower than copper (II). The higher oxidation state of copper qualifies as less acidic.

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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.

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