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Why H2O readily picks up H+ in acid and Carbon does not?


BurningKrome

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OK…for as much chemistry as I’ve taken, my High Schooler asked me a question I can’t actually explain…so I’m hoping for help.

 

 

We were discussing basic chem: the number of bonds for atoms and how they form bonds based on the number of needed electrons in the valence shell. HOWEVER we also discussed how they can transiently form lesser or more bonds than predicted by the periodic table.

 

 

I.e. H2O regularly picks up transient H+ ions to form H30+ in acidic solutions. And carbon will form a very brief carbocations (carbon with only three bonds and a pos charge) during Sn1 and E1 reactions.

 

 

Since H20 can relatively readily form transient H3O+ in an acidic solution, she assumed that carbon could also readily form a transient additional bond (I.e. CH5+) in an acidic solution.

 

 

When I said it doesn’t she asked the horrifying question “why not?”

 

 

So, I guess I don’t actually know. So my question is:

 

 

Why does a molecule like water readily pickup a transient excess hydrogen bond and carbon does not? Another good example for us would be perchlorate (ClO4-)…where the Chloride has four oxygen even though it only has one unsatisfied valence electron.

 

 

Thanks!

 

 

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Well, generally things with less total electrons "behave" themselves better. Carbon only has 6 electrons, and two of them are pretty much dedicated to the 1s shell (ask Helium). So it won't be making any permanent 5th bonds. Transient I dunno, it would have to be quite transient.

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To add to and elaborate on Mr. Skeptic, oxygen, carbon and chlorine all have the capacity to fill eight slots in it's most exposed and lowest energy electron orbitals. Thus by geometric fit and by attainable bond energy levels these atoms all have the reasonable ability to form up to four bonds. They will readily form bonds where electrons are surplus or slots are unfilled (these bonds have very low formation energies and high dissociation energies) and they will more reluctantly form bonds within the outer valance in total (which generally have high formation energies and modest to low dissociation energies).

 

Practically, due to energy differentials between the bonds of H2O H3O+ and H4O+2 only the first two will be observed because an H2O molecule will strip off any H from an H4O +2 that might happen to form long before it is observed. CH5 is nearly out of the question for the same reason because the bonding energy difference is so high in attempting to share an inner valance electrons as Mr. Skeptic indicated.

 

In short it is all about energy balance. Bonds with energy requirements that are balanced and in alignment with available energy can coexist will be observed, those that are wildly out of place will not.

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Thanks for the info. Gives me more to work with. However, when discussing the bonding energy for ADDITIONAL bonds (ionizing bonds or those bonds beyond that which forms a completed valence)...what is the controlling factor for these (electron affinity, electron negativity, resonance, ETC.?)

 

Sticking with hydrogen for ease...if HCl was added to a solvent containing 50% H2O and 50% methane...the H20 would have observable amounts of H30 to absorb the excess H+, however the Methane would have non-observable amounts.

 

For lack of a better description, the Kb of H2O >>> Kb of CH4. (The K base of Water is much much greater than the K base of Methane).

 

WHY? :) What controls the affinity of additional ionizing hydrogen bonds causing Water to have such a significantly higher affinity than Ch4?

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Well, there are two reasons. Oxygen (along with nitrogen and fluorine) is one of the most electronegative elements around. In addition, it has extra electron pairs. So the electron pairs provide a place to bond, whereas the electronegativity means it steals most of the electron from the nearby hydrogens. Because of this, the oxygen has a partial negative charge as it likes, and with three hydrogens the hydrogens all share the positive charge (which they kind of like).

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Of course :) That makes sense. Somewhat obvious, now that it is pointed out :)

 

In the case of Oxygen (or more specifically H2O) does the third hydrogen actually form a covalent bond with a lone pair...or is the H+ held mostly by the charge (like a bastardized ionic bond)? If a covalent bond, am I correct in inferring that resonance plays a part in keeping the third hydrogen in place?

 

Thanks!

 

Oh...and to satisfy her curiosity...

 

I know about carbocation intermediaries (loss of a bond) ...but are there any instances where carbon will PICK UP an ADDITIONAL (5th) bond...even transitorily?

 

Thanks!

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  • 2 months later...

As far as I'm aware the addition hydrogen adds on as a dative (or co-ordinate) covalent bond just like in NH4. As for carbon forming five bonds I'd doubt it very much because carbon forms four bonds via sp3 hybridism. I couldn't imagine a 1s electron functioning as a bonding electron...

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My apologises. I've just checked a textbook and in mass spectrometry CH4 loses an electron to form the radical .CH4+ this reacts with CH4 to form the CH3 radical and CH5+ This is unusual as it only has eight electrons but still forms 5 bonds. These are distributed between the five bonds (hence the + charge) and the structure is thought to be trigonal bipyramidal. This structure has not been determined as it's too unstable. It's merely proposed from theoretical calculation.

 

CH5+ is an unstable compound which functions as a powerful acid and can protonate pretty much everything when it protonates something a proton is removed but no electrons.

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  • 3 weeks later...

I'm not sure if this has been said already, but since non-ionic carbon has a full set of bonds and no lone pairs of electrons, it is incapable of exceeding its valency by virtue of the orbitals it has available to it. Many atoms are able to exceed their valency on account of having empty, low-lying (in terms of energy) orbitals, such as d orbitals.

 

You mention CH5+ in the last post. The ONLY place this is ever, ever shown in text books is in chapters dealing with mass spec. Even then, there is no proof to say it is actually a species that exists. Pentavalent carbon is not something that can exist, especially CH5+.

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I'm not sure if this has been said already, but since non-ionic carbon has a full set of bonds and no lone pairs of electrons, it is incapable of exceeding its valency by virtue of the orbitals it has available to it. Many atoms are able to exceed their valency on account of having empty, low-lying (in terms of energy) orbitals, such as d orbitals.

 

You mention CH5+ in the last post. The ONLY place this is ever, ever shown in text books is in chapters dealing with mass spec. Even then, there is no proof to say it is actually a species that exists. Pentavalent carbon is not something that can exist, especially CH5+.

 

If it has an existence at all, I imagine that the extra [ce]H^+[/ce] would not have a full order sigma bond due to carbon's lack of available p-orbitals or lack d-orbitals to expand the hybrid, as you stated. It might exist very quickly as a van-der Waals complex under low molecular KE conditions. Van der Waal's complexes can seem to violate standard valency rules. Another, more extreme example is the oddity [ce][PbHe_{21}]^{2+}[/ce] ion. It is only metastable but can exist for long enough to conduct IR in a solid Argon matrix (cold as hell). And yes, thats [ce] He [/ce] not [ce] H [/ce].

 

This is unusual as it only has eight electrons but still forms 5 bonds. These are distributed between the five bonds (hence the + charge) and the structure is thought to be trigonal bipyramidal. This structure has not been determined as it's too unstable. It's merely proposed from theoretical calculation.

 

Jahn-Teller distortion of said trigonal bi-pyramid would determine whether the [ce] CH_{5}^{+}[/ce] was an excited state bonded complex or a Van der Waals complex. if it undergoes Jahn-Teller distortion than we're dealing with partial order bonds here. Problem is, I doubt it lasts long enough to get a good IR and there is no way to get a good electron-diffraction pattern which would be ideal for discovering Jahn-Teller distortion.

Edited by mississippichem
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