How can we theoretically decide the acidic or basic nature of an oxide? What are the reasons/ factors which make an oxide acidic, basic, amphoteric or netural?
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3$\begingroup$ You might find the answers given to an older question helpful. $\endgroup$– Klaus-Dieter WarzechaCommented Dec 30, 2014 at 10:07
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3 Answers
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In general, the electropositive character of the oxide's central atom will determine whether the oxide will be acidic or basic. The more electropositive the central atom, the more basic the oxide. The more electronegative the central atom, the more acidic the oxide. Electropositive character increases from right to left across the periodic table and increases down the column. The trend of acid-base behaviour is from strongly basic oxides on the left-hand side to strongly acidic ones on the right, via an amphoteric oxide (aluminium oxide) in the middle. An amphoteric oxide is one that shows both acidic and basic properties. This trend applies only to the oxides of the individual elements in the highest oxidation states for those elements. The pattern is less clear for other oxides.
We define non-metal oxide acidity in terms of the acidic solutions formed in reactions with water. For example, sulfur trioxide reacts with water to forms sulfuric acid.
In sum, acidic oxides are oxides of non-metals, and basic oxides are oxides of metals.
There are three non-metal oxides from the upper right portion of the periodic table, $\ce{CO}$, $\ce{NO}$, and $\ce{N2O}$, which have such low oxidation numbers for the central atom that they give neutral aqueous solutions.
Since the acidity of a cation rises rapidly with charge, d-block elements that 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. $\ce{CrO}$ is basic, $\ce{Cr2O3}$ is amphoteric and $\ce{CrO3}$ is acidic.
answered Dec 30, 2014 at 8:42
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Consider an electropositive atom’s oxide and an electronegative one’s. And also consider them hydrolysed (i.e. $\ce{E=O + H2O -> E-(OH)2}$ or $\ce{E-O-E + H2O -> E-OH + HO-E}$).
Oxygen is very electronegative, so it should always be $\delta -$. But the electronegative atom will allow less electron density to be drawn away by the oxygen, so there will be less of a negative charge on the oxygen. This means that displacing the proton to give the oxygen more of a negative charge becomes more favourable. Thus, the oxide is acidic.
Now for the electropositive atom, oxygen now almost gets that atom’s entire electron denisty. This means that oxygen is just a tad too negative to feel well, so it will draw protons out of the surrounding solution to protonate itself. Thus, the oxide is basic.
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1$\begingroup$ It depends on what you mean by be acid or basic. Look at Lux-Flood theory ;) $\endgroup$– ParaH2Commented Jun 10, 2016 at 17:35
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1$\begingroup$ @Shadock Brønsted-Lowry acid/base definition. $\endgroup$– JanCommented Jun 10, 2016 at 17:39
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1$\begingroup$ Never thought of acids and bases in this light ,really liked this answer. $\endgroup$– parth_07Commented Jan 25, 2018 at 22:51
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There is a theory developed by Hermann Lux and Håkon Flood, named Lux-Flood theory to explain the basic or acid character of an oxide.
The rules are very simple.
An acid of Lux-Flood is an acceptor of $\ce{O^{2-}}$
A base of Lux-Flood is a donor of $\ce{O^{2-}}$
Examples
$\ce{CaO}$ is a base of L-F because $\ce{CaO}=\ce{Ca^{2+}}+\ce{O^{2-}}$
$\ce{SiO2}$ is an acid of L-F because $\ce{Si}$ has unoccupied d orbital then it can have a valence upper than two and then accepts $\ce{O^{2-}}$ ions.
Then they can react to give you $\ce{CaSiO3}$
EDIT
If you use Fajans' Rules you will find that $\ce{CaO}$ is more ionic than $\ce{SiO2}$ that's why this reasoning is correct.
To measure the strength of them it is common to use the scale of $\ce{pO^{2-}}=-\log(\ce{O^{2-}})$ like we do for the $\ce{pH}$.
It may exists amphoteric Lux-Flood compound but I have not examples in head right-now.
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1$\begingroup$ Your examples don’t really make sense. $\ce{SiO2}$ is also $\ce{Si^4+ + O^2-}$. Also, while silicon does have d-orbitals somewhere above in the sky, so does calcium. In neither do they take part in bonding in any significant way. I think, I understand where you want to go, but it is badly phrased. $\endgroup$– JanCommented Jun 10, 2016 at 17:38
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$\begingroup$ @Jan I have a better explaination I will edit my post. $\endgroup$– ParaH2Commented Jun 15, 2016 at 11:42
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2$\begingroup$ Very interesting point of view. I would like you to add examples with amphoteric oxides. Also, explanations that include the involvement of 3d orbitals in the chemistry of the 3rd row elements are highly discouraged. $\endgroup$– EJCCommented Jun 15, 2016 at 12:48
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$\begingroup$ @Marko for amphoteric coumpound there is the answer of Yomen I just saw. :) $\endgroup$– ParaH2Commented Jun 15, 2016 at 13:01
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$\begingroup$ I know, but I want wanted you to expand your answer, and threat the amphoteric oxides with the Lux-Flood theory. $\endgroup$– EJCCommented Jun 15, 2016 at 17:01