Electron in a covalent bond: what happens when it moves out? Let's look at a covalent bond in, say, a water molecule:
H : O : H
H is bonded to O by a pair of electrons. 
But electrons are in constant motion. What happens when the electron leaves?
Why does not the bond break? Both nuclei on both sides of it cannot move, obviously.
Also, if the electron has to orbit around it's respective atom, does it mean that it would actually spend only 1/10 of the time in the "sweetspot" position (right in between the two cores), and will spend 9/10 or most of its time "orbiting" (ie. out of that position)?
I am rather puzzled by this. Any ideas will be greatly appreciated.
 A: 
But electrons are in constant motion.

This is wrong, as we can't classically interpret electrons. They are a wave function that describes their probabilities. So you should look at a covalent bond as a situation where it is very likely you will find the electron around those two atoms.

Also, it electron has to orbit around its respective atom, does it mean that it would actually spend only 1/10 of the time in the "sweetspot" position (right in between the two cores), and will spend 9/10 or most of its time "orbiting" (i.e. out of that position)

Indeed... This would be strange. However, calculating the probabilities, we find out that it is most probable to find the electron in its "sweetspot". This does not mean that most of the time the electron is in the mid-section of the nuclei, as I just said, you can't think of an electron as "being around" any spot, at least until you measure it...
I would suggest that you start by studying the much simpler case of the bond in $H_2^{+}$ before going to more complex cases like a water molecule
A: Covalent bonding is a complicated thing. The laws governing how electrons behave in atoms and molecules are the laws of quantum mechanics, and so an attempt to understand these things in terms of 'orbiting electrons' is likely to fail. The reality is that electrons don't have well-defined positions and velocities --- you simply cannot say that your electrons are here at one instant of time and move round to here at a later time.
Instead, the most complete description of our electrons we can give is not their postions and velocities but instead an object called their wavefunction. A wavefunction can be thought of as a probability density function for the position of a particle. That is to say, it is a function (just like $x^3 + 1$ or $2^x$) that describes the relative likelihoods of finding the electron at each point in space.
For an electron involved in a covalent bond, its wavefunction might look something like this
 
where the ticks mark the positions of the nuclei, the horizontal axis corresponds to position, and the vertical axis corresponds to the value of the wavefunction --- i.e., to the probability of finding the electron at that position. As you can see, the electron is most likely to be found somewhere between the nuclei, holding the molecule together. If you like, you can imagine that the electron 'spends most of its time' in between the two nuclei. As I said, the electron doesn't have a well-defined position or velocity. But if you must visualise it classically, imagine that the electron is jittering about between the nuclei, rather than orbiting anything. Electrons don't orbit, and it's silly in my opinion to teach people that they do.
A: 
But electrons are in constant motion. What happens when the electron leaves? Why does not the bond break?

The bond would be weaker, but only for a short time, in which the 1836-times heavier nucleus won't go anywhere far. The negatively charged electron is attracted back to the nuclei, since they are positively charged.
In common situations, water molecules are in large number and they mutually interact, so rogue electron wouldn't get too far - it would end up on the closest molecule and perhaps be part of the separation process 
$$
\text{H}_2 \text{O} + \text{H}_2 \text{O} \rightarrow \text{OH}^- +\text{H}_3\text{O}^+.
$$
