Electron Orbital and Getting to "Other Side"
Date: September 2006
I have a question about electron orbitals. Say
you have 4 electrons in a 2P orbital. It is my understanding
that the dumbbell shape of the 2P orbital is based on probability
of where the electron will be at any given moment. My question:
The shape of the 2P orbital tapers down to nothing right at the
nucleus. That would indicate a zero probability of an electron
being there. If there is a zero probability of the electron ever
being there, how does an electron get from one side of the
nucleus to the other? How would it get to the other "cloud"?
Your objection to the electron "getting to the other side" is a
frequent one. While the detailed quantitative "explanation" is a bit
involved, the principle involved in straightforward. The Heisenberg
Uncertainty Principle is at work. The derivation of hydrogen-like
wavefunctions assumes the nucleus is a "point charge", that is, it
has no "size". That is an approximation and the if one does the
math, there is a finite probability of the electron "jumping" to the
other side. However, what is more fundamental is that the
wavefunction is a probability and the uncertainty in the position,
for a typical momentum of a bound electron is of the order of
magnitude of the size of the atom itself. What causes the confusion
is we have the tendency to think in terms of a planetary model, and
this is going to lead to ambiguities like you describe. Instead, one
must "pitch" that model and think in terms of a less intuitive
probability model. Then the question becomes meaningless. There is
no "one side or the other" of the nucleus. The web site:
gives some of the reasoning in more detail.
An excellent question.
The orbital does not tell you the probability
of where the electron is at any given moment,
but rather the probability of *finding* the
electron at a particular place when it has
a specific amount of energy and angular motion.
They sound like they should be the same thing
but in the slippery world of quantum mechanics,
they are not exactly. So, in an electron is in
a p state it can *be* anywhere in space, but
your probability of *catching* it exactly at
the nucleus would be zero. Very weird stuff, I know!
Robert Q. Topper
I am sure you will get quite a few responses from those of us who
will be somewhat annoyed that you have been trapped into learning a
concept that is not properly handled. This is not your fault.
Textbooks (and instructors) have to change and there are some of us
who are working on it.
The problem is that while most textbooks and instructors do say that
the p-orbital shape is a probability map, they also imply that the
p-orbital probability map is a physical reality. It is not. In fact
the probability map that is often drawn in textbooks is not even the
actual solution to the Schrodinger equations but is the psi-square
function. So we really should not think of those pictures as
anything truly reflecting reality - they are just visualization
aids. Moreover, the "lobes" are not even truly lobes as a viewing of
the graph in x-y coordinates indicate that the graph extends well
beyond the "lobes". In effect, what you are being presented with is
just the highest probability surface (or shell).
To understand what I mean by this. Think of an x-y graph of the
function x=y. Anywhere along this line the answer is true, x is
equal to y. Suppose I put an additional constraint and say, graph x
= y except for when x = 3 in which case x is undefined. What I
imagine you would do then is graph the x=y line and put a hollow
circle around the x=3 (and y=3) point indicating that there is a
discontinuity (undefined) in the line. But everywhere else x is
still equal to y and there need not be an explanation on how the
value of x at 2.999999... gets to become 3.00000...1 without passing
through an x=3. We simply say that the math describes that as a
discontinuity in the equation.
This is the same for the Schrodinger equations. The two lobes are
mathematical representations drawn in a 3-dimensional graph where
the value at the node (origin on the graph) is undefined. Anywhere
else on this graph the probability is defined.
Greg (Roberto Gregorius)
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Update: June 2012