We know from experimental data that $\ce{H-O-H}$ bond angle in water is approximately 104.5 degrees. If its two lone pairs were bonds (which is unfortunately impossible) also $\ce{O-H}$ bonds and a perfect tetrahedron resulted, then VSEPR theory would predict that the bond angle would be 109.5 degrees - this number can be easily derived using the geometry of a tetrahedron. However, how would people give estimates of the actual bonding angle of water, which is caused by a slightly greater repulsion by the lone pairs than there would be if they were bonds? What physics would be involved in the calculation? Thanks!
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2 Answers
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how would people give estimates of the actual bonding angle of water
What physics would be involved in the calculation
Background
That's a very good question. In many cases Coulson's Theorem can be used to relate bond angles to the hybridization indices of the bonds involved.
$$\ce{1+\lambda_{i} \lambda_{j} cos(\theta_{ij})=0}$$
where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.
For example, in the case of methane, each $\ce{C-H}$ bond is $\ce{sp^3}$ hybridized and the hybridization index of each $\ce{C-H}$ bond is $\sqrt3$. Using Coulson's theorem we find that the $\ce{H-C-H}$ bond angle is 109.5 degrees.
How can we use this approach to answer your question?
Unlike methane, water is not a perfectly tetrahedral molecule, so the oxygen bonding orbitals and the oxygen lone pair orbitals will not be exactly $\ce{sp^3}$ hybridized. Since addition of s-character to an orbital stabilizes electrons in that orbital (because the s orbital is lower in energy than the p orbital) and since the electron density is higher in the lone pair orbital than the $\ce{O-H}$ orbital (because the electrons in the lone pair orbital are not being shared with another atom), we might expect that oxygen lone pair orbital will have more s-character and the oxygen $\ce{O-H}$ orbital will have less s-character.
If we examine the case where the $\ce{O-H}$ bond is $\ce{sp^4}$ hybridized, we find from Coulson's Theorem that the $\ce{H-O-H}$ angle is predicted to be around 104.5°. So indeed, in agreement with our prediction, removing s-character from the oxygen $\ce{O-H}$ orbital gives rise to the observed bond angle.
Note on reality
For over 50 years students have been told that water is roughly $\ce{sp^3}$ hybridized. The general description is that there are two equivalent O-H sigma bonds and two equivalent lone pairs orbitals. The lone pair - lone pair repulsion is greater than the sigma bond - sigma bond repulsion, so the hybridization changes as described above and the lone pair-O-lone pair angle opens up slightly and the $\ce{H-O-H}$ angle closes down to the observed 104.5 degrees.
With the advent of photoelectron spectroscopy it was found that the two lone pairs in water were not equivalent (2 signals were observed for the lone pairs). Now, the hybridization of water is described as follows:
- 2 $\ce{sp^4}$ O-H sigma bonds
- one lone pair in a p orbital
- and the second lone pair in an $\ce{sp}$ orbital
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As you said, lone pairs result in slightly more repulsion than bonds, because the lone pairs themselves expand a little around the atom. Common practice that I've learned is just to subtract about 2.5 degrees from the bond angle for every lone pair. For instance, a molecule such as NCl3 has 4 negative charge centers, giving it a tetrahedral group geometry, but it has one lone pair, giving it bond angles of about 107 degrees. Following this pattern, water would have bond angles of about 104.5 degrees.
Here's a picture that sort of shows how the lone pairs expand, reducing the bond angles slightly.
