And of course, the other limiting (non-equilibrium) geometry would be a linear geometry with a H-O-H angle of 180 degrees as follows:
So now the question is, why is the former geometry adopted by water molecule? Phrasing this question another way: is it such that the energy of the water molecule is lower when it has a H-O-H bond angle of 104.5 degrees?
Well, the easiest way to find out whether this is true or not (not so much why it is true!) is to plot a graph of the energy of the water molecule against the H-O-H bond angle.
Now usually you wouldn’t be able to calculate the total electronic energy of the water molecule by hand, but since I’m bundled up with this molecular calculation suite called Gaussian03 (yes, I know the latest version is G09!), I’ve performed the calculations using my trusty laptop.
So here’s the graph, or like how they call it, the Walsh Diagram for the water molecule (click to enlarge):
This graph actually shows that the 104.5 degrees H-O-H bond angle gives rise to a lower total electronic energy of water than the linear geometry! So be convinced! That the Singaporean education system isn't wrong! :)
So the next step comes in explaining why that's the case.
Another time. :p
And all calculations were clumsily performed with a minimal STO-3G basis set using the Hartree Fock level of theory, which merely took 5 seconds or so for each single point energy calculation.
Yay! Time to sleep. :)
