Hey guys. When i was studying the molecular orbital diagram for HF, it was said to me that 1s od hydrogen CANNOT overlap with 2py or 2px of fluorine due to the assymetry of these two orbitals, so it overlaps exclusively with 2pz. (And cannot overlap with 2s of fluorine becouse of large energy difference). This was said several times, and even in general for overlaping of orbitals. And it makes sense. But in orbital hybridisation theory, s orbital overlaps with all three p orbitals. How can that be? Why H didn't overlap with other 2p orbitals then? Thanks!
Strictly speaking, the hydrogen 1s orbital overlaps with all orbitals on F. However, due to symmetry, the effect of the overlap cancels out for the px and pz orbitals and is of little importance for the 2s orbital. In valence bond theory, you would also not consider forming hybrid orbitals if the effects of the overlap are insignificant, as in HF. In other molecules, like CH4, the overlap between 1s on H and px, py on C is also insignificant, but as all sp3 hybrids contain some pz and 2s contribution, the overlap has to be taken into account for all 4 orbitals. If you do all the math correctly, a description in terms of s and p orbitals is exactly equivalent to one in terms of hybrid orbitals.
Thanks, I think I got it! But how about this one? http://www.upl.co/uploads/IMAG0075.jpg This is the picture taken from my textbook. The whole point of it was JUST to explain how regular overlap cannot explain 104.5 angle (later hybridisation was taken into account). But how can 1s combine with py? Visually, i can imagine that happening, but doesn't this disobey that rule for assymetric orbitals? Of course, they combine end to end, but then any 1s and any p orbital could do this (why HF couldn't do it?), so where would be the problem of whole assymetry? pz and py - here they seem equally significant. Thanks!
It might be that you are "on the wrong page", I can't tell from your post. The p orbitals (2p) have two signs; one lobe is positive and one is negative, hence any s orbital which interacts (evenly) with both lobes will have zero net interaction energy (to a first approximation). IOW, the only way for an s orbital to bond with a p orbital is by coming in onto the "tip" of one lobe. The x, y, z designation of p orbitals is arbitrary, different authors will assume bonding is s-pz or s-px (s-py, too of course). Which coordinate you pick to specify the orbital is up to you, they are orthogonal and in the absence of a electrical/magnetic field equivalent. So, I really don't understand your problem with one H's s orbital interacting with a px orbital and another H atom's s orbital interacting with that same atoms py orbital...that second H atom can "call" the p orbital anything it wants. Relative to the H atom (pick the one you are talking about) the two p orbitals in the plane perpendicular to the bond axis will have a net interaction of zero due to the cancelation of interaction of the s to the + and s to the - lobes of each orbital. Note that this sign of the orbital's lobes is NOT the spin of the electron(s) in that orbital!! Unrelated and not to be confused! Note the color of the lobes in this table: https://en.wikipedia.org/wiki/Atomic_orbital#Orbitals_table Note that you can treat the two colors as "bonding" and "antibonding" although that is technically wrong, as bonding and antibonding orbitals are molecular, not atomic, but it gives you the idea about their opposite signs. I SHOULD have first made the point that these concepts are geometrical, having to do with the positioning of the orbitals in space. As you imagine a s orbital (a sphere) getting closer and closer to the p orbitals of another atom, you need to picture the placement of those 3 p orbitals ...you could picture the s just touching the tip of one p, or it coming in mid-way between two p orbitals and 'touching' them both, and even coming in between all three and touching all 3. You have to take it on faith that electrons pair and so the predominant touch will be between one single p orbital and the s orbital. If you accept this, then the s orbital will be on the p orbitals axis, and then you can talk about the contributions (or lack of) from the other orbitals.
Water is better described in terms of pure p orbitals than in terms of hybrids as the 2s orbital in oxygen is energeticallly too low to participate equally in bonding. The bond angle is no convincing argument as orbital overlap is not the only criterion that determines bond geometries.