[galpinj]

Hey guys,

So basically I'm curious about the reasoning behind this question. I understand the principal quantum number, azimuthal quantum number, etc. and how this shows that, mathematically, only one orbital is possible for n=1. However, I haven't found a clear reason as to why it is this way. Is it because at n=1 the "space" is too small to accommodate more than 2 electrons? Please provide a lay mans explanation. I've been told it deals with schrodinger's equation, the uncertainty principle, and a whole lot of other complex math, but I just want to know the basic reason why you cants have 8 electrons at n=1.

Thanks everyone!

[AWK]

Quantum numbers result from If you are powerful in math look into textbook of quantum chemistry solution of the Schrodinger equation (linear partial differential equation). It is pure mathematics.  If you are powerful in math look into textbook of quantum chemistry, otherwise see Wikipedia: Schrödinger_equation.

[Irlanur]

Is it because at n=1 the "space" is too small to accommodate more than 2 electrons? Please provide a lay mans explanation

I think you are looking for an explanation which simply does not exist. it definitely has nothing to do with 'enough space' or anything like that. It is a mathematical necessity.

1) First of all you have to remember (or learn) that "the orbitals" are only exact solutions in the case of the Schroedinger H-Atom. As soon as we are dealing with more than 1 electron, orbitals are only a helpful concept, but definitely not physical solutions, which means that discussing basic questions with them is dangerous.

2)Why do you only ask about n=1? if that bothers you, it should also bother you that there is no 2d orbital...


maybe we can get closer to an answer that you are happy with, but i highly doubt it.

[mjc123]

Maybe this will be helpful. You remember how, for a particle in a box, the lowest energy wavefunction has no nodes (except at the box walls, which don't exist in the H atom case) - the half-wavelength equals the length of the box. The next lowest has one node - it is a full wavelength. The next has two, and so on. The more nodes, the greater the curvature of the wavefunction and the higher the energy.
Now consider the H atom. The lowest energy wavefunction has no nodes. There is only one way to achieve this, a spherically symmetric wavefunction with no radial nodes - a 1s orbital.
The next lowest level has one node. There are two ways of doing this. The 2s orbital has a radial node (there is a surface, at a certain distance from the nucleus, where the value of the wavefunction is zero). The 2p orbitals have an angular node, giving rise to the familiar two lobes + and -. The next level has 2 nodes, which can be done in 3 ways, giving 3 types of orbital - 3s with 2 radial nodes, 3p with one radial and one angular node, and 3d with 2 angular nodes.
As with linear momentum in the box case, the orbital angular momentum increases with the number of angular nodes; in fact this number is equal to the orbital quantum number l.

[galpinj]

Is it because at n=1 the "space" is too small to accommodate more than 2 electrons? Please provide a lay mans explanation

I think you are looking for an explanation which simply does not exist. it definitely has nothing to do with 'enough space' or anything like that. It is a mathematical necessity.

1) First of all you have to remember (or learn) that "the orbitals" are only exact solutions in the case of the Schroedinger H-Atom. As soon as we are dealing with more than 1 electron, orbitals are only a helpful concept, but definitely not physical solutions, which means that discussing basic questions with them is dangerous.

2)Why do you only ask about n=1? if that bothers you, it should also bother you that there is no 2d orbital...

Thank you for the response!

I am also bothered about why there is no 2d, 3f, etc; I didn't include it in the question for simplification. I can understand that it must be related to a very complicated mathematical equation; however, math is also grounded in reality. I'm sure that there is reason behind the limitations! I just don't know what those reasons are...


maybe we can get closer to an answer that you are happy with, but i highly doubt it.

[galpinj]

Maybe this will be helpful. You remember how, for a particle in a box, the lowest energy wavefunction has no nodes (except at the box walls, which don't exist in the H atom case) - the half-wavelength equals the length of the box. The next lowest has one node - it is a full wavelength. The next has two, and so on. The more nodes, the greater the curvature of the wavefunction and the higher the energy.
Now consider the H atom. The lowest energy wavefunction has no nodes. There is only one way to achieve this, a spherically symmetric wavefunction with no radial nodes - a 1s orbital.
The next lowest level has one node. There are two ways of doing this. The 2s orbital has a radial node (there is a surface, at a certain distance from the nucleus, where the value of the wavefunction is zero). The 2p orbitals have an angular node, giving rise to the familiar two lobes + and -. The next level has 2 nodes, which can be done in 3 ways, giving 3 types of orbital - 3s with 2 radial nodes, 3p with one radial and one angular node, and 3d with 2 angular nodes.
As with linear momentum in the box case, the orbital angular momentum increases with the number of angular nodes; in fact this number is equal to the orbital quantum number l.

Thank you so much! That makes some sense. I have looked at the wavelength/node example in a different post, but didn't quite grasp what they were saying. I'll spend some more time on this topic to see if I can get a better handle on what you've said. Thank you again!

[Enthalpy]

It's definitely not a matter of available room. An electron can fit in arbitrarily small room, according to any experiment humans have been able to conduct. Nor is 1s so much smaller than the others.

But little room implies a bigger kinetic energy of the electron. More generally, a big change in the wavefunction over a short distance, for instance a change of sign, or a big fading, implies much kinetic energy; if you like Fourier transforms, the short distance needs components with a big wave number k, which contribute much kinetic energy (ħk)²/(2m).

One consequence it that there is one optimum size for the orbital of lowest energy, 1s: a smaller orbital puts the electron nearer to the nucleus and improves the electrostatic energy, but the kinetic energy worsens brutally. You can write simplistic models for the sum of both energies and find an optimum orbital size that is numerically meaningful. An other consequence is that the 1s orbital shrinks when the nucleus contains more protons, because the stronger electrostatic attraction can afford a bigger kinetic energy.

An other consequence is that a wavefunction that changes its sign more times in a given volume (this is simplified thinking) implies a bigger kinetic energy while the electrostatic energy isn't better, so orbitals with more nodes have a higher energy, that is, the electron is less bound to the nucleus.

So an orbital with a radial node (2s) or an azimutal node (2p) binds the electron less than without a node (1s).

Now, we call conventionally "1" the orbital with zero nodes, "2" the orbitals with one node, and so on. So the impossibility of an azimutal node in the "1" orbital results from the naming convention, because the first such orbital is called "2".

It's the very nature of electrons, and of all fermions as opposed to bosons like the photon, that they can't be in the same state. It's a mathematical impossibility, not a repulsion force. Maybe someone else can explain this. And because electrons can have two states of spin, in 1s state you can find 2 electrons, with opposite spins. Same for 2s: 2 electrons. And same for 2p_x, 2p_y and 2p_z, which are called collectively 2p but exist with three different orientations. Two electrons in each 2s, 2p_x, 2p_y and 2p_z make 8 electrons in all "2" orbitals.

In case you wonder if an orbital could have the simple decaying electron density of 1s and at the same time the azimutal behaviour of a p orbital, the answer is no. "p" means that the wavefunction makes one phase turn (well, for some p orbitals at least) for each geometric turn around the nucleus, and since at the nucleus the wavefunction must have every phase from 0° to 360° its only possible value is zero. More generally, only the "s" orbitals have nonzero value at the nucleus.

It happens that each radial node has the same effect on the orbital's energy as each azimutal node for the hydrogen atom.
http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html third frame there
That's why we add the numbers of radial and azimutal nodes and group the orbitals in 1, 2, 3, 4... Though, this is inaccurate for more electrons, and higher orbitals interleave
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/sodium.html
and this is just for one outer electron at sodium. Transition elements do it more.

[galpinj]

It's definitely not a matter of available room. An electron can fit in arbitrarily small room, according to any experiment humans have been able to conduct. Nor is 1s so much smaller than the others.

But little room implies a bigger kinetic energy of the electron. More generally, a big change in the wavefunction over a short distance, for instance a change of sign, or a big fading, implies much kinetic energy; if you like Fourier transforms, the short distance needs components with a big wave number k, which contribute much kinetic energy (ħk)²/(2m).

One consequence it that there is one optimum size for the orbital of lowest energy, 1s: a smaller orbital puts the electron nearer to the nucleus and improves the electrostatic energy, but the kinetic energy worsens brutally. You can write simplistic models for the sum of both energies and find an optimum orbital size that is numerically meaningful. An other consequence is that the 1s orbital shrinks when the nucleus contains more protons, because the stronger electrostatic attraction can afford a bigger kinetic energy.

An other consequence is that a wavefunction that changes its sign more times in a given volume (this is simplified thinking) implies a bigger kinetic energy while the electrostatic energy isn't better, so orbitals with more nodes have a higher energy, that is, the electron is less bound to the nucleus.

So an orbital with a radial node (2s) or an azimutal node (2p) binds the electron less than without a node (1s).

Now, we call conventionally "1" the orbital with zero nodes, "2" the orbitals with one node, and so on. So the impossibility of an azimutal node in the "1" orbital results from the naming convention, because the first such orbital is called "2".

It's the very nature of electrons, and of all fermions as opposed to bosons like the photon, that they can't be in the same state. It's a mathematical impossibility, not a repulsion force. Maybe someone else can explain this. And because electrons can have two states of spin, in 1s state you can find 2 electrons, with opposite spins. Same for 2s: 2 electrons. And same for 2p_x, 2p_y and 2p_z, which are called collectively 2p but exist with three different orientations. Two electrons in each 2s, 2p_x, 2p_y and 2p_z make 8 electrons in all "2" orbitals.

In case you wonder if an orbital could have the simple decaying electron density of 1s and at the same time the azimutal behaviour of a p orbital, the answer is no. "p" means that the wavefunction makes one phase turn (well, for some p orbitals at least) for each geometric turn around the nucleus, and since at the nucleus the wavefunction must have every phase from 0° to 360° its only possible value is zero. More generally, only the "s" orbitals have nonzero value at the nucleus.

It happens that each radial node has the same effect on the orbital's energy as each azimutal node for the hydrogen atom.
http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html third frame there
That's why we add the numbers of radial and azimutal nodes and group the orbitals in 1, 2, 3, 4... Though, this is inaccurate for more electrons, and higher orbitals interleave
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/sodium.html
and this is just for one outer electron at sodium. Transition elements do it more.

Thank you for the response Enthalpy! What you have said is kind of going over my head, so I'm going to read over it a few times to get a better grasp. I appreciate the input!

[Enthalpy]

Forgot the link to the nice gallery:
http://winter.group.shef.ac.uk/orbitron/
Just choose 1s, 2s, 2p... on the left side. The drawings displays the sign of the wavefunction and the nodal surfaces where it passes by zero.

[InnocentRealist]

@ mjc123;

"Maybe this will be helpful. ..."

Finally! An answer that makes sense. If there are more nodes that means the frequency is greater which means more energy. Information like this makes science rewarding (and I haven't seen this anywhere else)! I spend my time trying to make sense of these things, so thank you so much. : )UU