Sciencemadness Discussion Board

Oxygenic acid?

YT2095 - 7-7-2008 at 01:48

why isn`t there an Oxygenic acid?

it`s group 6, same as S, Se and Te...
and they all follow the formula H2xO4, so can H2OO4 (or H2O5) exist?

Edit: I just realised this is in Biochem rather than Beginnings, sorry about that, my eyes need testing again :(


[Edited on 7-7-2008 by YT2095]

ScienceSquirrel - 7-7-2008 at 03:46

Oxygen is a first row element and only has it's s and p shells available for bonding and it's only common valence state is 2. Sulphur and subsequent elements have a d shell which is available for bonding so they can from higher valence compounds eg SO2, SO3.

YT2095 - 7-7-2008 at 03:52

but sulpher doesn`t have a D shell does it?

it`s (Ne)3S23P5

ScienceSquirrel - 7-7-2008 at 04:12

True but it has an empty d shell available to it so it can 'expand it's octet'.

http://sg.answers.yahoo.com/question/index?qid=2008052119365...

ScienceSquirrel - 7-7-2008 at 04:25

That should have read that it does not have a FILLED d shell but it has the 3d shell available to it.
First row transition elements fill this shell, take a look at the electronic configuration of iron for example.

JohnWW - 7-7-2008 at 05:27

That is right, ScienceSquirrel. Oxygen cannot expand beyond an octet of 2s and 2p valence electrons. A consequence of this is that it cannot have an oxidation state higher than +4, and that only by losing an electron to form a cation with three covalent bonds, as in OF3+, the central O atom in O3, oxonium salts as OR3+, protonated water as H3O+, and protonated ethers as OR2H+.

Pyrovus - 7-7-2008 at 06:09

Technically it is possible to draw a valid Lewis structure for H2OO4, however, because you don't have d orbitals available, the central atom will need to form two coordinate covalent bonds in order to bind to all the other oxygens. Now, let's consider the stability of such a monstrosity. Oxygen-oxygen single bonds are quite unstable, and you have four of them packed into a tiny molecule. Not only that, but you have an ugly +2 formal charge on the central oxygen atom. As a result of this, it will be much less stable than the straight chain isomer HO-O-O-O-OH. And hydrogen polyoxides are very unstable; hydrogen peroxide (H2O2) is a powerful oxidiser, hydrogen trioxide (H2O3) has only a transitory existence, and H2O4 and H2O5 if they even exist will be much more unstable still. Oxygenic acid would therefore be a less stable isomer of a molecule which is insanely unstable to begin with; consequently there is no reason to expect it to exist.

ScienceSquirrel - 7-7-2008 at 06:42

Theoretically you can draw a ring isomer of ozone akin to cyclopropane buit it would be like keeping a polecat in a paper bag in stability terms :-D

12AX7 - 7-7-2008 at 07:11

There's hydrooxygenic acid (in analogy to hydrosulfuric acid, H2S, also a weak acid). Nasty stuff, also known as dihydrogen monoxide. :P

Tim

ShadowWarrior4444 - 7-7-2008 at 16:21

Quote:
Originally posted by 12AX7
There's hydrooxygenic acid (in analogy to hydrosulfuric acid, H2S, also a weak acid). Nasty stuff, also known as dihydrogen monoxide. :P

Tim


Have you read the MSDS on this chemical? It is truly a blight upon the Earth. It's not even *organic!* Prone to spontaneous autoprotonation, too.

http://www.youtube.com/watch?v=hzLs60ZaNW4

More on topic: I wonder if higher oxidation states of oxygen can be supported under high pressures.

JohnWW - 7-7-2008 at 18:35

Quote:
Originally posted by ShadowWarrior4444
More on topic: I wonder if higher oxidation states of oxygen can be supported under high pressures.

Under high pressures, one just might get the partly ionic tetrafluoride, [OF3]+F-. However, it is unlikely, because no-one has succeeded in obtaining nitrogen pentafluoride, [NF4]+F-, in the same manner, although the NF4+ cation has been known for some time as salts with ClO4-, SbF6-, PF6-, BF4-, and NO3-. Otherwise, O(IV) is known only in O3 and just possibly OF3+, although it may be said to be tetravalent in OR3+, H3O+, OR2H+, and CO.

Pyrovus - 8-7-2008 at 00:21

Since when has ozone contained oxygen in the +4 state?

YT2095 - 8-7-2008 at 00:43

so if I understand this correctly it "Borrows" electrons that are fixed in one orbital geometry and forces them into a different geometry.
as opposed to Promoting them in Energy such as n=0 ---> n=1 whereby it would emit a photon when it "fell" back down to it`s normal state.

ScienceSquirrel - 8-7-2008 at 02:44

If you have a look in a good inorganic textbook you should find a chapter on p d hybridisation, there is a bit here to get you going.
http://en.wikipedia.org/wiki/Orbital_hybridisation
It is a bit hand wavy and has been replaced by Molecular Orbital Theory but it is OK for accounting for simple molecules like SF6.

YT2095 - 8-7-2008 at 03:02

Thank you!

it would seem that an apparently Simple question leads to quite a deep rabbit hole! ;)

JohnWW - 8-7-2008 at 15:35

Quote:
Originally posted by Pyrovus
Since when has ozone contained oxygen in the +4 state?
As I explained above, the central O atom in the bent O3 molecule contains O(IV), with three covalent bonds to the other O atoms, and a positive charge.

ScienceSquirrel - 8-7-2008 at 18:11

This is Beginnings so I deliberately kept away from charged species etc in my explanation.
The fact is that hybridisation is an OK way to explain bonding in neutral compunds of p group elements at a high school, early college level.
Just like VSEPR is a good way to predict molecular geometry.

Pyrovus - 8-7-2008 at 22:54

Quote:
Originally posted by JohnWW
Quote:
Originally posted by Pyrovus
Since when has ozone contained oxygen in the +4 state?
As I explained above, the central O atom in the bent O3 molecule contains O(IV), with three covalent bonds to the other O atoms, and a positive charge.


That does not make it +4. The central atom does not lose control of electrons through those three covalent bonds with the other oxygens; as they have the same electronegativity. The central oxygen atom consequently has an oxidation number of +1.

unionised - 9-7-2008 at 10:04

Isn't there a bit of a steric hinderance problem with (formally) 4 O-- ions squashed round the O6+ ion too?

woelen - 9-7-2008 at 10:30

Quote:
Originally posted by ScienceSquirrel
Oxygen is a first row element and only has it's s and p shells available for bonding and it's only common valence state is 2. Sulphur and subsequent elements have a d shell which is available for bonding so they can from higher valence compounds eg SO2, SO3.
This is not an explanation. Nitrogen, which also only has s and p shells can form stuff like HNO3, N2O5, NO2 and even H3NO4 (under rather extreme low temperature conditions). The salt Na3NO4 is more stable and even exists at room temperature.

12AX7 - 9-7-2008 at 13:00

Oxygen and fluorine are capable of losing electrons (forming positively charged ions), but the ionization energy exceeds the electron affinity of any other atoms suitable for making an acid (i.e., oxygen itself). Besides, there is no electronegativity difference between O and O, so the bonds would be highly covalent and any such assembly would quickly decompose to diatomic oxygen (or if you assembled a fluoro-yl species, OF2 as well).

I don't think steric hinderance would matter much; although O(2-) ions are quite large, even the quite small Si(4+) and S(6+) are extremely comfortable surrounded by them. But your point was that O(6+) should be even smaller, in fact quite a bit smaller than the isoelectronic helium due to the extra charge. In that case, (O)O3 or even just (O)O2 might be the most that can fit. The latter does exist, i.e. ozone, but not with a 4+ charged center.

Tim

franklyn - 9-7-2008 at 23:58

There is apparently a tetraoxide R-O-O-O-O-R known
http://131.104.156.23/Lectures/CHEM_462/462_chapter_5.html

and H-O-O-O-H
http://en.wikipedia.org/wiki/Trioxidane
seen below



Click the image above to download these two pdf files together in a zip file
Length of Ox chain.pdf
Formation of Ozonides.pdf

.

ScienceSquirrel - 10-7-2008 at 02:49

Quote:
Originally posted by woelen
Quote:
Originally posted by ScienceSquirrel
Oxygen is a first row element and only has it's s and p shells available for bonding and it's only common valence state is 2. Sulphur and subsequent elements have a d shell which is available for bonding so they can from higher valence compounds eg SO2, SO3.
This is not an explanation. Nitrogen, which also only has s and p shells can form stuff like HNO3, N2O5, NO2 and even H3NO4 (under rather extreme low temperature conditions). The salt Na3NO4 is more stable and even exists at room temperature.


Yes it is.

Nitrogen is a lot less electronegative than oxygen so it can use it's lone pair to form a bond.
Have a look at the structure here
http://en.wikipedia.org/wiki/Nitric_acid
The fact is that nitric acid is not wildly stable but it is a lot more stable than O3 or RO3+.
The central oxygen in an oxygenic acid would not be able to use it's d orbitals like sulphur etc to bond and forming coordinate bonds would give structures in which it has to carry a formal positive charge.

Jdurg - 10-7-2008 at 17:52

I feel very fortunate (or unfortunate depending on how you view it) to have some VERY beautiful samples of an inorganic ozonide. My lump of potassium metal that I've had for many years now has developed a beautiful coating of the very dark red potassium ozonide. The same lump shows bits of pure metal, a great deal of potassium oxide, some potassium peroxide, potassium superoxide, and potassium ozonide.

The wikipedia entry for Potassium contains a picture of my sample of potassium from the tail-end of 2006. The red color is now more prominant and there are more orange crystals throughout the lump. I can never cut into this chunk ever due to the sheer amount of unstable oxides on there, but still pretty neat to see.

woelen - 11-7-2008 at 02:25

Quote:
Originally posted by ScienceSquirrelYes it is.

Nitrogen is a lot less electronegative than oxygen so it can use it's lone pair to form a bond.
Have a look at the structure here
http://en.wikipedia.org/wiki/Nitric_acid
The fact is that nitric acid is not wildly stable but it is a lot more stable than O3 or RO3+.
The central oxygen in an oxygenic acid would not be able to use it's d orbitals like sulphur etc to bond and forming coordinate bonds would give structures in which it has to carry a formal positive charge.
I see your point. So, with nitrogen it is the difference in electronegativity, and with sulphur it is the additional d-orbitals that allow the compounds to be sufficiently stable that they can exist. Good to learn something new ;).

kmno4 - 19-7-2008 at 07:48

Quote:
Originally posted by 12AX7
(...)
I don't think steric hinderance would matter much; although O(2-) ions are quite large, even the quite small Si(4+) and S(6+) are extremely comfortable surrounded by them. But your point was that O(6+) should be even smaller, in fact quite a bit smaller than the isoelectronic helium due to the extra charge. In that case, (O)O3 or even just (O)O2 might be the most that can fit. The latter does exist, i.e. ozone, but not with a 4+ charged center.
Tim

Firstly - there are no any ions in such covalent bounded particles as H2SO4, HNO3, SiO2, HClO4.... etc. You must not compare O(-2) ion with covalent oxygen.
I used to count distances between oxygens in orthonitrate anion NO4(-3), from crystallographic data (articles from Wiley).
It turn out that this distance is "critical", almost equal to Van der Waals's radius for oxygen. So, similar structures: OO4(2-) and FO4(-) should be far less stable than NO4(3-), because of smaller radius of central atom. This steric hinderance can be very important reason of nonexistence of H2OO4.
Of course, not the only one.

[Edited on 19-7-2008 by kmno4]