If two alkali metal atoms join with an oxygen atom, an ionic bond forms. Since hydrogen has the same number of valence electrons as alkali metals, why can't water be ionic?
This is what I'm thinking:
$$\ce{(H+)2O^2-}$$
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1$\begingroup$ In other words.. Is water ionic or molecular/covalent? $\endgroup$– Geoff HutchisonFeb 6, 2015 at 3:01
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2$\begingroup$ It seems water can be even "superionic" under high pressure and temperature: chemistry.stackexchange.com/questions/26863/… $\endgroup$– MithoronMar 4, 2015 at 15:09
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1$\begingroup$ chemistry.stackexchange.com/questions/17064/… $\endgroup$– MithoronMay 10, 2015 at 14:13
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2 Answers
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First of all, the difference between ionic and covalent bonds is not sharp. As electronegativity differences increase, you move away from covalent and towards ionic bonds. There are "in between" states like polar covalent, where one side of the bond is stronger but not fully ionic. And this I think is the main reason: hydrogen has fairly high Pauling electronegativity (2.20), rather close to oxygen (3.44), which seems polar covalent overall (and why we get hydrogen bonding with water). In contrast, the alkali metals all have electronegativity less than 1.00, a much bigger difference versus oxygen and thus a more ionic bond.
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$\begingroup$ What do you mean that one size of a polar bond is stronger? $\endgroup$ Dec 10, 2014 at 6:34
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1$\begingroup$ Side, not size. One atom is more electronegative and has the electron more often. $\endgroup$– user467Dec 10, 2014 at 12:00
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1$\begingroup$ This video from Khan Academy explains this gradual move from ionic to covalent bonding due to differences in electronegativity: youtube.com/watch?v=126N4hox9YA $\endgroup$ Feb 21, 2015 at 18:17
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No compound is purely ionic. Water is about 33% ionic.
Linus Pauling discusses the ionic character of water in his famous "The Nature of the Chemical Bond" [1] at pages 100-101.
Pauling says as a first approximation water should be considered as having contributions from four resonance structures:
\begin{array}{lc} &37\% \qquad & \ce{H-O-H} \\ &24\% \qquad & \ce{(H-O)^-~H+} \\ &24\% \qquad & \ce{H+~(O-H)-} \\ &15\% \qquad & \ce{H+~O^{2-}~H+} \end{array}
However, he goes on to explain that this is based upon considering each $\ce{O-H}$ bond independently, and the fact that the two $\ce{O-H}$ bonds are not independent reduces the contribution of $\ce{H+~O^{2-}~H+}$ somewhat and increases the other three contributions.
In any case, the purely convalent form $\ce{H-O-H}$ should be considered as contributing only somewhere between 37% and 44% according to Pauling.
References
- Pauling, L. The Nature of the Chemical Bond and the Structure of Molecules and Crystals: An Introduction to Modern Structural Chemistry, 3rd ed.; Cornell University Press, 1960. ISBN 978-0-8014-0333-0.
