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As we all know, if we put $\ce{H^+}$ ions in water then water reacts with them and produces $\ce{H_3O^+}$:

$$\ce{H_2O + H^+ <=> H_3O^+}$$

My question is, why does water react with $\ce{H^+}$ in the first place? It doesn't have an electron, then how can the reaction occur?

And again, it's already stable and adding one more $\ce{H^+}$ will, in my understanding, destabilize it as it will have one more proton in the molecule.

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  • $\begingroup$ Do you mean "react"? $\endgroup$ – sixtytrees Jul 19 '16 at 16:15
  • $\begingroup$ There's no free H+ (bare proton!), generated via dissotiation, in water or in any solution. Or maybe you're interested in nuclear reactions? $\endgroup$ – Mithoron Jul 20 '16 at 1:43
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Water has two lone pairs. H$^+$ is a naked nucleus with a positive charge. It is extremely reactive and would bind to any neutral molecule. Water has two lone pairs of electrons and H$^+$ binds one of them to form H$_3$O$^+$.

In absence of water it would bind other things. For example, it would bind methane to form unstable CH$_5^+$ that breaks down to CH$_3^+$ + H$_2$.

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  • $\begingroup$ so the extra proton forces (in lose terms) the water molecule to make bonds? Does the water molecules get forced to become unstable? $\endgroup$ – MartianCactus Jul 19 '16 at 16:53
  • $\begingroup$ So the lone proton doesn't need an electron to form the bond? It can just get attracted towards the molecule and then stick to it? $\endgroup$ – MartianCactus Jul 19 '16 at 16:56
  • $\begingroup$ Yes. Strictly speaking H+ is a very high energy species. Binding to anything brings its energy down, so it is ready to bind whatever. Now there are molecules that can bind H+ at lower energy cost (like water, or, even more so NH3). For other molecules it is more expensive. But the net energy change for H+ +X ->HX+ is favoring the products. Once again it is because H+ is so high in energy. $\endgroup$ – sixtytrees Jul 19 '16 at 16:57
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    $\begingroup$ @Adi, If you just look at the water molecule, it seems like it's a "bad deal". The thing is you have to look at the whole system. As much as water is getting more unstable at the $\ce{H3O+}$ molecule, the high energy $\ce{H+}$ is getting stabilized. The final energy in the system is going to be lower than the initial energy (also the reason why this reaction is exothermic, this energy has to go somewhere..). $\endgroup$ – IanC Jul 19 '16 at 17:24
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    $\begingroup$ @IvanNeretin The accuracy of my answer depends on the way the question was asked. Correct explanation would involve Gibbs energy for H$^+$, H2O and H3O$^+$. But that wouldn't really explain anything. I mean, for someone new to chemistry it is just numbers. The matter of fact is H$^+$ reacts with any neutral molecule (we can discuss HeH$^+$ elsewhere). It even can react with cathions (N2H5$^+$+ H$^+$ -> N2H6$^{2+}$) but there are many neutral molecules and even ions that water wouldn't react with. $\endgroup$ – sixtytrees Jul 19 '16 at 19:13
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The simple answer is that H$_3$O$^+$ is more stable that H$^+$+ H$_2$O. The reason for this is that the H$_3$O$^+$ ion is isoelectronic with ammonia and has the same type of geometry, a triangular based pyramid (trigonal pyramid). It is possible to crystallise some salts with H$_3$O$^+$ as a counter ion, but of course these are no longer in water.

Neutral water normally contains a small concentration of ions, 10$^{-7}$ M H$^+$ and OH$^-$ at pH = 7 due to the dissociation/ionisation 2H$_2$O=H$_3$O$^+$+ OH$^-$. As water is a highly hydrogen bonded liquid there are always H atoms between any two oxygen atoms, and any oxygen has three hydrogens surrounding it, two normal bonds and one Hydrogen bond. Any positive charge now added will form the H$_3$O$^+$ and this has now to be incorporated into the H bonding network. Thus in bulk water H$_3$O$^+$ is probably not a compound in the normal sense of the word because there will only be be a small energy barrier for the charge to move around along the network of hydrogen bonding (all water molecules are the same) as the water molecules jostle due to their thermal energy, i.e. Hydrogen bonds making and breaking. Alternatively you can think of this charge migrating effect as the charge being spread over more than one water molecule, thus stabilising it. It is possible therefore to imagine complexes of various numbers of waters with a single charge, such as H$_5$O$_2^+$ .

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In chemistry course, we are taught at the beginning that a bond is formed between 2 atoms, each of which providing 1 electron to make the bond. But this is more a way to introduce the octet rule and shape of the molecules. In most reactions (as we write them) electrons move by pair.

If a bond is made of 2 electrons, it does not have to be 1+1 (one electron from each atom) but it can be 2+0 (one atom brings 2 electrons, the other one does not).

In the case of $H^+$, it has no electron where it should have 2 when bonded. This empty "slot" is really eager to be filled (octet rule) so it "steals" electrons from water which has 2 big non-bonding pairs of electrons on the oxygen atom. The result is that the positive charge on $H_3 O^+$ is shared by all 3 hydrogen atoms. This dilution of charge is greatly favorable.

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The electron configuration of the hydrogens are approximately

$$\ce{\underset{1s}{[\uparrow]}}$$

although some of the charge is pulled off into the oxygen.

The electron configuration of a neutral oxygen atom is

$$[\ce{He}] \underset{2s}{[\uparrow \downarrow]} \underset{2p}{[\uparrow \downarrow \vert \uparrow \vert \uparrow]}$$

In $\ce{H2O}$ each free electron in a hydrogen pairs up with a free electron in an oxygen (approximately.) So how can a $\ce{H3O^+}$ ion form?

In $\ce{H3O^+}$ there is a missing electron so the oxygen could have an electron configuration of

$$[\ce{He}] \ce{\underset{2s}{[\uparrow \downarrow]}} \underset{2p}{[\uparrow \vert \uparrow \vert \uparrow]}$$

which allows for a hydronium ion to form.

Now in an actual hydronium ion the oxygen would have hybridized orbitals of the form

$$[\ce{He}] \underset{(2s)(2p)^3}{[\uparrow \downarrow \vert \uparrow \vert \uparrow \vert \uparrow]}$$

which is a tetrahedral arrangement.

Now this is only an approximation. A more in depth look would go deeper into molecular orbital theory and bond symmetries but that gets annoying for polyatomic molecules.

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Also, you are right in saying that hydronium ion is unstable than water. But a system containing hydronium ion is thermodynamically more stable (has lower gibbs energy) system containing a proton and water molecule. The following exposition is based on the subjective knowledge, will be glad to be corrected if I am wrong. This due to the following arguments: 1) number of possible configurations of system increased. In the two particle system, charge was only on proton, which is now shared between several hydrogen atoms bonded to oxygen. but one would argue that number of molecules decreased, but entropy is not only about number of molecules, but it is about number of possible configuration of a system for different parameters There must be entropy increase due to increased number of ways a charge can distributed, but I am not able to explain how to count this number of ways a charge can be distributed, Please let me know, if anybody understands where and how this entropy increase is coming from.

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