When an electron revolves around the nucleus in P or d-orbitals why does not it collide with the nucleus.
I mean to say that the shape of the orbital narrows near the nucleus , so shouldn't it collide with the nucleus?
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The shape you see associated with the d-orbital is not the actual shape of the d-orbital. That is a probability cloud of the location of the electron. Outside of the shape, or at least in the not shaded-in regions, the probability of finding the electron drops significantly. This means that the narrowing in the shape of the d-orbital indicates that the region where the electron could be found is shrinking. This can be extended to mean that at the nucleus, the probability cloud effectively doesn't exist. This means not that the electron should collide with it, but that the electron cannot be found there ever.
If you are now wondering how the electron gets from one region of the point cloud to another if they it cannot travel through the nucleus, it is simply a matter of treating the electron as a wave. The waveform exists at one location or another. Like a sound wave, it is possible to have regions of destructive interference where the wave doesn't exist but still have the wave exist on either side. Similarly, the probability cloud shows where you can find the electron. It does not say that the electron is never outside of this region nor does it comment on the velocity or motion of the electron. It can move from one region to another non-connected one; the cloud only implies that you will never (never meaning high statistical improbability) measure the electrons position outside of the region.
Having now read Chris' answer, it excellently fills in my loose description of the important mathematics behind the concepts I was attempting to convey. Anyone reading this answer should definitely read that one as well; they are complimentary.
Jim
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First let's take a step back and try to explain what exactly is being shown in "pictures" of orbitals. When you solve the Schrödinger equation for an electron around a nucleus, you can describe the resulting wavefunction any number of ways. One of the most common is in the position basis, so that the (square magnitude of) the wavefunction gives the probability density for the electron to be "found" at that location if you were to instantly measure its position (i.e. force it into an eigenfunction of the position operator).1
So all we have is this "cloud" - a nonnegative function of $\mathbb{R}^3$ describing where the electron might be. Since it is hard to draw functions of 3D space, what people often do is draw a single surface, usually a surface of constant probability density that contains, say, 90% of the total probability inside of it.2 If you think of the cloud as having varying mass density, we want to depict a natural region wherein we can find 90% of the total "mass."
Just because this surface comes to a point doesn't mean the electron is forced to that point. In fact, such simple diagrams don't really say anything about how the probability density is distributed inside the orbital. Moreover, if you believe the probability density doesn't do anything crazy (e.g. go off to infinity), the shrinking of the surface to a point tells us the electron is actually unlikely to be found near the center.3
As I finish writing this I realize it's something of a loose, mathy explanation for what those shapes are supposed to be. On the other hand, Jim's answer gives a far better explanation of physically what's going on.
1 For more explicit formulas, see an answer I wrote here.
2 The reason I specify a surface of constant density is because there are infinitely many surfaces that contain 90% of the probability. Start with one, expand it a little over here, shrink it a little over there, and you're left with a different 90% surface.
3 In fact, the process whereby a proton captures an electron really only happens with s-orbital electrons. All other orbitals have a node at the very center, so the electrons almost never "come close" to the nucleus.
