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I faced a question where I had to find the amphoteric oxides among some given oxides. The answer key says $\ce{Cr2O3} $ is amphoteric but $\ce{CrO}$ isn't. But why is that?

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Basic oxide are those oxides that dissolve in water to form soluble hydroxide. Basic oxides are very soluble for this reason. Acidic oxides are that oxides that dissolve in water to form strong acids. Generally, covalent oxides form acidic oxides as the element to which oxygen is bonded is electronegative. As charge increases, electronegativity increase and thus oxide become acidic. This also applies for elements having multiple oxides. As oxidation no. increases, charge increases and thus oxides become acidic. This link shares a good information on this topic:

Since the acidity of a cation rises rapidly with charge, d-block elements which exhibit a wide variety of oxidation numbers may have one or more oxides that exhibit only basic properties and one or more oxides that exhibit only acidic properties. The higher the oxidation number the more acidic the corresponding oxide. Chromium is an example of such an element. 

$$ \begin{array}{|c|c|c|}\hline \text{Oxide}&\text{Oxidation number}&\text{Category}\\\hline \ce{CrO}&\ce{Cr^2+}&\text{basic}\\\hline \ce{Cr2O3}&\ce{Cr^3+}&\text{amphoteric}\\\hline \ce{CrO3}&\ce{Cr^6+}&\text{acidic}\\\hline \end{array} $$

The amphoteric character of chromium(III) is stated in its Wikipedia article:

Chromium(III) oxide is amphoteric. Although insoluble in water, it dissolves in acid to produce hydrated chromium ions, $\ce{[Cr(H2O)6]^3+}$ which react with base to give salts of $\ce{[Cr(OH)6]^{3−}}$. It dissolves in concentrated alkali to yield chromite ions $(\ce{[CrO2]^{-}})$.

So, the rule of thumb is as oxidation number increases, charge on metal increases, acidic character increases. Also, basic oxide is soluble in water, acidic oxide is insoluble in water. This can be used to differentiate between acidic and basic oxide.

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    $\begingroup$ I understand the explanation, but find it incomplete. Apparently, it is just circumventing the main underlying logic. You say that increasing oxidation number means increasing acidity. But what is the reason for this? Why does acidity increase with increasing oxidation number? $\endgroup$ – Gaurang Tandon May 11 '18 at 2:32
  • $\begingroup$ Increasing oxidation number on the metal atom leaves more empty orbitals to accept electron density, so shouldn't that increase the Lewis acidity contribution to the behavior of the metal oxide? I agree, if you can still say, "but why?", then the underlying logic (if any) is not sufficient. $\endgroup$ – timaeus222 May 15 '18 at 18:22

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