2
$\begingroup$

H2O has a 109.5 degree bond angle, but CO2 has exactly 180 degrees. Is there a qualitative reason for this? It's hard to believe CO2 is exactly 180 degrees unless there were some symmetry, but the same symmetry argument should apply to H2O then. So, is it really exactly 180 degrees?

Correction: H2O has 104.5 degrees and the tetrahedral ball-stick model is therefore a little inaccurate.

$\endgroup$
  • 3
    $\begingroup$ Looks like it belongs to Chemistry.SE. $\endgroup$ – Wildcat Sep 13 '14 at 6:23
  • 3
    $\begingroup$ $H_2O$ has more than just two bonds. The Oxygen atom in the centre also has two lone pairs of electrons. So you're distributing four electron pairs in free space. On the other hand, Carbon in $CO_2$ just has two double bonds, which is why they go exactly opposite each other. $\endgroup$ – mikhailcazi Sep 13 '14 at 6:36
  • $\begingroup$ @mikhailcazi It's amazing that a simple tetrahedral ball-stick model (representing sp3 orbitals) gives nearly the exact bond angles for both molecules though the true multi-electron Schrodinger equation is so complicated! I can't think of a better success in QM simplification. Since you didn't submit an answer, I can't give you answer credit, but your comment helped me the most. $\endgroup$ – bobuhito Sep 16 '14 at 6:41
1
$\begingroup$

$H_2O$ has more than just two bonds. The Oxygen atom in the centre also has two lone pairs of electrons. So you're distributing four electron pairs in free space.

H2O

On the other hand, Carbon in $CO_2$ just has two double bonds, which is why they go exactly opposite each other.

enter image description here

$PS$: As to why the tetrahedral analogy doesn't give a perfect answer: each electron pair repels the other electron pair, but the repulsive forces differ depending on whether the electron pair is a bond pair or a lone pair.
If I remember right, the repulsion between two lone pairs is the most, followed by the repulsion between a lone pair and a bond pair, which is in turn larger than the repulsion between two bond pairs:
$\text{LP-LP > LP-BP > BP-BP}$

This is why $H_2O$ doesn't have a perfect $109.5^\circ$ angle between the bond pairs. The repulsion of the lone pairs being more, the $H-O$ bond pairs are pushed closer together, making it $104^\circ$ instead.

$\endgroup$
2
$\begingroup$

Disclaimer: I will try my best to make this rather long story short.

First of all, we must say that we are talking about the so-called equilibrium geometry - molecular geometry that corresponds to the true minimum on the potential energy surface, a surface which describes the energy of a molecule as a function of nuclear coordinates.

Secondly, potential energy surface (PES) is a mathematical abstraction that appears only in approximate treatments of molecular systems within Born–Oppenheimer approximation. In this approximation the state (or, speaking classically, the motion) of electrons is treated independently from that of nuclei, and each electronic state of a molecule there exist the corresponding PES. This approximation breaks out when two PESes come close to each other or even intersect, but it is generally accurate at least for molecules in their ground electronic state (i.e. electronic state of the lowest energy).

Note, however, that even when PES for a particular electronic state of interest is well separated from PESes corresponding to other electronic states, there might exist more than one minimum on the same PES. Now, clearly, to meaningfully speak about equilibrium geometry for a molecule in a particular electronic state it is required that different minima on the corresponding PES are well separated, or, in other words, that there exist one distinct minimum on the potential energy surface. And this is again not always the case: there exist some non-rigid molecules, for which minima are not well separated, or, in other words, which have few equilibrium geometries. And some relatively small external influences can significantly change the molecular geometry.

But how small is small? And how big should be the barrier between two minima so that we say they are well separated? All this depends on the problem at hands and the corresponding physical conditions. For instance, if we are talking about molecules in a gas phase under usual temperatures (300 K), then, say, ethane is non-rigid molecule, since its conformations are separated by just a few kJ/mol (comparable to energies of thermal motion), and thus, the molecular geometry of ethane constantly changes.


But, aside from this, for rigid (under usual conditions) molecules, like CO2 and H2O, we can meaningfully speak about their equilibrium geometries. And we can calculate equilibrium geometries for, say, ground electronic states, in the Born-Oppenheimer approximation. We can do what is known as the geometry optimization: starting from some initial geometry we try to minimise its electronic energy (including nuclear-nuclear repulsion energy) by varying arrangement of nuclei. We should be careful, since in general (as we have already said) we are not guaranteed that there exist only one minimum, but for small molecules such as CO2 and H2O that is not an issue.

So, if you do these calculation, you will indeed found out that (at least in their ground electronic states) CO2 and H2O molecules have the equilibrium geometry mentioned in books.

But even for rigid (under specified conditions) molecules, one should remember, that even at the absolute zero of temperature nuclei are constantly vibrating near the equilibrium. So if you perform some physical measurement (say, GED) you want necessarily "catch" each and every molecule in the equilibrium geometry, rather you obtain the average picture.

$\endgroup$
  • $\begingroup$ In your terms, my question is how do we know theoretically that CO2 belongs to D∞h? Within the "filled-shells model", one could position 3 nuclei (with charges +6, +4, and +6) and apply the many-bodied Schrodinger equation to 16 electrons...then find the nuclei positions which have the lowest energy...but that problem is too difficult for current computers. mikhailcazi's comment actually answers my question for why CO2 should be more linear than H2O, but I suspect that perfect linearity has never been theoretically proven. Has it? $\endgroup$ – bobuhito Sep 14 '14 at 17:18
  • $\begingroup$ @bobuhito: I"m sure someone's done some X-ray imaging or something, but you don't need to -- what direction would be preferred in the CO${}_{2}$ molecule? In the H${}_2$O molecule, they bend away from the non-binding pair. In the carbon dioxide molecule, all of the pairs are bonding to oxygen atoms, symmetrically. There is no, and can be no, preferred direction, so the angle is 180 degrees. Also, the fact that carbon dioxide is nonpolar is evidence of this, since the fact that oxygen has a different electronegativity would create a net moment if the angle was not $180^{\circ}$ $\endgroup$ – Jerry Schirmer Sep 15 '14 at 19:10
  • $\begingroup$ @bobuhito If you accept a quantum mechanical calculation as proof then the answer is yes! For example, the water dimer has been studied > 20 years ago with exactly the ingredients you outline. CO2 has more valence electrons but a search with "CO2" and "ab initio" or "first principles" will give you plenty of references. $\endgroup$ – Felix Sep 15 '14 at 19:24
  • $\begingroup$ @bobuhito, I'm pretty sure someone did high-level quantum chemistry calculation of CO2. And I'm pretty sure the equilibrium geometry they obtained was indeed linear. $\endgroup$ – Wildcat Sep 15 '14 at 20:30
  • $\begingroup$ @Wildcat In the Born-Oppenheimer (or Car-Parrinello) approximation, does a single SiO2 molecule come out as linear? And, is a single SiO2 molecule truly linear? I know that most SiO2 is bulk silica which changes everything, but wondering if the ball-stick model still works for a single SiO2 molecule (which would be equivalent to a CO2 molecule though Si's ball is bigger). $\endgroup$ – bobuhito Sep 16 '14 at 12:53
1
$\begingroup$

In H2O the center atom has four valence electrons, in CO2 the center atom has six valence electrons. H2O forms two simple bonds, while CO2 forms two double bonds. Why should it behave the same?

Simple qualitative answer: Think about the ball and stick model (Not sure, how these are really called. I mean the ones with plastic spheres and soft plastic bonds). If you model H2O, you end up with the bent geometry (104 deg). If you do the double bonds on CO2, you'll end up with 180 deg symmetry.

A thorough quantum-mechanical description requires somewhat more effort as you are dealing with multi-electron systems.

Just saw, that somebody else was faster (and more exact).

$\endgroup$
1
$\begingroup$

There is plenty of theoretical and experimental evidence that CO2 is linear and H2O has a tetrahedral geometry. For instance, these geometries have been calculated ab initio (from quantum mechanics) several times in Car-Parrinello studies: they indeed converge to a linear O=C=O configuration, or an approximately tetrahedral structure for H2O, for the ground state.

Infrared spectroscopy confirms that CO2 is linear since linear molecules have an additional vibrational mode compared to non-linear ones (actually the story is slightly more complicated for CO2 because of its symmetry - see Example 2 in the Chemwiki - Vibrational Modes).

H2O is a highly polar molecule, whereas the linear CO2 is nonpolar. As a consequence, water - in contrast to the much heavier CO2 - is liquid at room temperature because due to its polarity it can easily associate and form clusters. Finally, the crystal structure of both molecules depend on their molecule geometry, and this has also been verified by ab initio calculations.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.