3
$\begingroup$

I realize that nodes (both angular and radial) are areas with zero probablility of finding an electron. I realize that plotting the square of the radial eigenfunction for an orbital will give the probability density of finding an electron at a given distance from the nucleus.

So for the 1s orbital (zero nodes), the graph of the square of the eigenfunction will be one hump. For the 2s orbital (one node) there will be a small hump followed by a large hump with a zero point in between them (talking about the sqaure of the eigenfunction not the eigenfunction itself).

For the 1s orbital, the peak of the hump will represent the most probable radius the electron will be found from the nucleus. The probability will lower as the radius increases or decreases from the most probable radius (similar to gaussian distribution), and the electron will have zero probability of being found within the nucleus of the atom (0 radius).

And from what I understand the radial node(for the 2s orbital) is physically manifested at the point where the 1s orbital and 2s orbital meet (my instructor used the analogy of russian nesting dolls), and on the NON squared eigenfunction plot of the 2s orbital, the node is represented where the wave function crosses the x axis going from +y to -y.

So here is what Im really wondering (please correct me on anything stated previously if I am mistaken). Is the first (smaller) hump on the squared radial eigenfunction (probability density function) for the 2s orbital the probability that the electron is in the 1s orbital (I believe this would be called an excited state if im not mistaken)? This would make the most sense to me, as it would be saying that it is possible for a 2s electron to be in either the 1s orbital or 2s orbital, but not in between (where the wave function goes from +y to -y)

Edit: Perhaps it is incorrect to say an electron from the 2s orbital has the possibility of being in the 1s orbital. What I really mean is the electron may occupy the same space as the 1s orbital, but since orbitals are just 3d areas where an electron is most likely to be found perhaps I am not incorrect in saying that a 2s electron occupies the 1s orbital since the 1s orbital is technically just a region in space. My guess is that the probability of a 2s electron occupying the same space as the 1s orbital is lower than it occupying the space of the 2s orbital due to repulsion by the electrons that already occupy the 1s orbital.

Here is a picture representing everything ive tried to explain so far:

s orbitals

$\endgroup$
  • 1
    $\begingroup$ You have it all backwards. The point where the 1s orbital and 2s orbital meet is not important at all, and in particular has nothing to do with the radial node of 2s. (These are different points, in fact.) When an electron sits on 2s, well, this means it sits totally, 100% on 2s and not on 1s. It does not care at all about 1s. It does not even "know" whether 1s exists. True, that electron spends part of its own existence in the region of the first hump where most of 1s also resides, but then again, every orbital overlaps with every other. It's not like orbitals own the space they occupy. $\endgroup$ – Ivan Neretin Sep 8 '16 at 18:36
  • $\begingroup$ related: chemistry.stackexchange.com/questions/53926/… $\endgroup$ – DavePhD Sep 8 '16 at 18:39
4
$\begingroup$

I realize that plotting the square of the radial eigenfunction for an orbital will give the probability density of finding an electron at a given distance from the nucleus.

This isn't quite right. The square of the radial function will give the probability density for s-orbital, but this is not the probability of finding the electron at a given distance from the nucleus, because the number of points a given distance from the nucleus increase with $r^2$. To get the plots in Fig. c, the radial function squared in multiplied by $r^2$.

For example, for 1s, the probability density is greatest at the nucleus, but by multiplying by $r^2$ when r=0, a radial probability density of zero is obtained.

Furthermore, as you can see from the graphs on pages 187 and 188 here, for 2s and 3s the most probable location (highest probability density) is at the nucleus (r=0). Again, only by multiplying by $r^2$ when r=0, is a radial probability density of zero is obtained.

So "Fig. a" in the OP is misleading, the central region of 2s and 3s should be shaded much more intensely than the outer region(s).

So here is what Im really wondering (please correct me on anything stated previously if I am mistaken). Is the first (smaller) hump on the squared radial eigenfunction (probability density function) for the 2s orbital the probability that the electron is in the 1s orbital (I believe this would be called an excited state if im not mistaken)? This would make the most sense to me, as it would be saying that it is possible for a 2s electron to be in either the 1s orbital or 2s orbital, but not in between (where the wave function goes from +y to -y)

No, the 2s electron is only in the 2s orbital.

$\endgroup$
  • $\begingroup$ So why does the probability distribution curve have a smaller 1st hump (smaller radius) than the 2nd hump (larger radius) for the 2s orbital if the probability that the electron is closer to the nucleus is higher? $\endgroup$ – Keaton Sep 8 '16 at 19:32
  • $\begingroup$ @Ki11akd0g the probability that the electron is in a given infinitesimal volume element (probability density) is highest at the nucleus, but the probability that it is a given distance from the nucleus (radial probability density) is at the outermost hump, primarily because a spherical shell have infinitely more points than a single point, and because a bigger radius shell has a greater area than a small radius shell. . $\endgroup$ – DavePhD Sep 8 '16 at 19:39
  • 1
    $\begingroup$ I actually understand the concept of nodes now, what was misleading is the color patterning depicted on thw images. I was considering the 2s orbital to contain a 1s orbital inside. The 2s orbital and 1s orbital are completely separate, although they may overlap and occupy the some of the same space, they are separate entities. $\endgroup$ – Keaton Sep 8 '16 at 19:42
  • 1
    $\begingroup$ AND I understand what you meant by the probability distribution vs the radial probability distribution. The probability of finding an electron decreases as you move further from the nucleus (on a straight line path away from the nucleus). But there are alot more points at which the radius equals a specific value the further you move away from the nucleus (there is a greater area of points at r=2 than r=1). So its the product of the probability of actually finding an electron at a given distance from the nucleus and the greater area at which that probability applies to. THANKS SO MUCH! $\endgroup$ – Keaton Sep 8 '16 at 20:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.