How are there different shapes of orbitals?

[questionposter]

If all atoms have the same energy levels, as in the electrons in the ground state electron in lead wouldn't be closer than in hydrogen, and those energy levels can absorb only specific energies of light, why are there many different shaped orbitals?

[Schrödinger's hat]

why are there many different shaped orbitals?

 

 

[math]\left(i\hbar\frac{\partial}{\partial t} + \frac{\hbar^2}{2m}\nabla^2 + \frac{e^2}{4\pi\epsilon_0 r}\right)\Psi(\mathbf{r},\,t) = 0[/math]

Edited December 7, 2011 by Schrödinger's hat

[mississippichem]

[math]\left(i\hbar\frac{\partial}{\partial t} + \frac{\hbar^2}{2m}\nabla^2 + \frac{e^2}{4\pi\epsilon_0 r}\right)\Psi(\mathbf{r},\,t) = 0[/math]

 

Allow me to translate:

 

electrons with the same principle quantum number (the "n" levels) can take on one of several angular momentum states.

 

An electron in an n=m level can take on one of m angular momentum states from 0 to n-1.

 

So for n=2 there are two possible angular momentum states, 0 and 1. There are actually additional quanutm numbers that further define the "orbital shape" as well (the next quantum member distinguishes between orbitals with the same angular momentum that are degenerate in the absence of an applied magnetic field).

 

schroedinger's_hat: you could've just used the time independent equation with a potential .

Edited December 7, 2011 by mississippichem

[Schrödinger's hat]

schroedinger's_hat: you could've just used the time independent equation with a potential .

My brain actually parsed the time derivative bit as E..

 

While I'm actually talking I may as well say a bit about how the wave equation entails shape.

 

The electron behaves almost purely as a wave in this case. And it follows that big scary wave equation I wrote.

 

As mississippichem said, there are different quantum numbers involved, and they define different solutions to the equation.

 

It's a bit hard to explain for an electron (or a quantum system in general). So let's use a classical one.

 

Hopefully you can get access to a guitar/bass guitar. A violin or other stringed instrument also works if you're good, but guitar is best.

 

Just touch the string on the 12th fret, and pluck it near the sound hole (or on your violin play a harmonic, if you can do this you probably already know about harmonics and can just follow the guitar explanation).

 

What you've done is cancel out some of the resonant modes of the string. You can release it with both hands and you'll hear that the pitch stays 1 octave above the root note.

 

The vibration on the string is a wave with one 'quantum number' known as a harmonic. Different harmonics have a different number (not really quantum number, but analogous in the maths), and they have a different shape.

One is the full string vibrating, one is still in the middle with two halves vibrating, then three stationary points, and so on.

If this were a quantum system you'd be more likely to find the phonon/particle/whatever where the string is vibrating most

and unlikely to find it where it isn't vibrating much.

The other difference is you can put energy into all the harmonics at once.

 

The electron is much the same, but it does sort of a funky 3D 'vibration' (the more 'vibration', the more likely you are to find it)

The different quantum numbers define the 'still' (unlikely to find) and 'moving' (likely) points in much the same way as the harmonic defined the number of still points on the guitar string.

[questionposter]

[math]\left(i\hbar\frac{\partial}{\partial t} + \frac{\hbar^2}{2m}\nabla^2 + \frac{e^2}{4\pi\epsilon_0 r}\right)\Psi(\mathbf{r},\,t) = 0[/math]

 

I've seen that wikipedia equation before...

Allow me to translate:

 

electrons with the same principle quantum number (the "n" levels) can take on one of several angular momentum states.

 

An electron in an n=m level can take on one of m angular momentum states from 0 to n-1.

 

So for n=2 there are two possible angular momentum states, 0 and 1. There are actually additional quanutm numbers that further define the "orbital shape" as well (the next quantum member distinguishes between orbitals with the same angular momentum that are degenerate in the absence of an applied magnetic field).

 

schroedinger's_hat: you could've just used the time independent equation with a potential .

 

So really the only reason then is because energy isn't the only thing that determines the wave mechanics of an electron?

[swansont]

So really the only reason then is because energy isn't the only thing that determines the wave mechanics of an electron?

 

Bingo. The equation has three degrees of freedom, so there are three variables.

[questionposter]

Bingo. The equation has three degrees of freedom, so there are three variables.

 

How does "i" play a role in reality like that? Do you need it to cancel out some other "i" that's generated by the equation?

[swansont]

How does "i" play a role in reality like that? Do you need it to cancel out some other "i" that's generated by the equation?

 

Are you referring to [imath]\sqrt{-1}[/imath] ? That will go away whenever you are solving the problem; at some point you multiply by the complex conjugate and all of the final terms are real.

[questionposter]

Are you referring to [imath]\sqrt{-1}[/imath] ? That will go away whenever you are solving the problem; at some point you multiply by the complex conjugate and all of the final terms are real.

 

Ok, well what about different atoms changes the momentum at different energies if a ground-state s orbital is the same distance from the nucleus in every atom? Cause an increasing number of protons doesn't seem to change what electrons do, just how much can be in a neutral system.

Edited December 9, 2011 by questionposter

[Schrödinger's hat]

The size of the various orbitals decreases with increasing number of protons.

The shape stays pretty much the same, because you still have the same symmetries.

[questionposter]

The size of the various orbitals decreases with increasing number of protons.

The shape stays pretty much the same, because you still have the same symmetries.

 

Some chemist told me that isn't right though, that the same orbitals actually are the same distances from every nucleus and the only reason they would seem different is because with bigger atoms you'd measure negative joules for some reason.

Edited December 9, 2011 by questionposter

[mississippichem]

Some chemist told me that isn't right though, that the same orbitals actually are the same distances from every nucleus and the only reason they would seem different is because with bigger atoms you'd measure negative joules for some reason.

 

Well it gets a bit complicated for multi-electron systems. You have to account for electron-electron repulsive forces and that spawns a particularly nasty set of differential equations (far nastier than the regular Schroedinger equation). The atomic radius of beryllium is smaller than that of lithium yet they have the same highest occupied orbital in the ground state, (n=2, l=0). The nuclear charge is larger though. Similar trends are observed throughout the periodic table with a few exceptions.

 

Not sure I understand the part about negative joules. All electrons in atomic orbitals have negative energies (think...you have to add joules to excite an electron to a higher energy state, and the ground state atomic orbitals are stable).

 

Careful with chemist advice about atomic orbitals . Sometimes it's kind of luck of the draw whether or not they really know what they are talking about here. Maybe you just misunderstood what your chemist friend was saying though.

 

(We also have a lot of specialized quantum jargon and conventions in chemistry that can be confusing to an outsider.)

Edited December 9, 2011 by mississippichem

[questionposter]

Well it gets a bit complicated for multi-electron systems. You have to account for electron-electron repulsive forces and that spawns a particularly nasty set of differential equations (far nastier than the regular Schroedinger equation). The atomic radius of beryllium is smaller than that of lithium yet they have the same highest occupied orbital in the ground state, (n=2, l=0). The nuclear charge is larger though. Similar trends are observed throughout the periodic table with a few exceptions.

 

Not sure I understand the part about negative joules. All electrons in atomic orbitals have negative energies (think...you have to add joules to excite an electron to a higher energy state, and the ground state atomic orbitals are stable).

 

Careful with chemist advice about atomic orbitals . Sometimes it's kind of luck of the draw whether or not they really know what they are talking about here. Maybe you just misunderstood what your chemist friend was saying though.

 

(We also have a lot of specialized quantum jargon and conventions in chemistry that can be confusing to an outsider.)

 

Well, I don't know what to tell ya. Classically you would think electrons would be closer to a nucleus with more protons, but atoms don't operate on classical mechanics. Plus, in order to get a high degree in chemistry, you need to know some quantum mechanics. If a ground-state electron is closer to the nucleus then it shouldn't have the same energy, but it does.

 

Actually, I think even swan corrected me on this. I said electrons in bigger nuclei are closer and even he said that's not right.

Edited December 10, 2011 by questionposter

[swansont]

Well, I don't know what to tell ya. Classically you would think electrons would be closer to a nucleus with more protons, but atoms don't operate on classical mechanics. Plus, in order to get a high degree in chemistry, you need to know some quantum mechanics. If a ground-state electron is closer to the nucleus then it shouldn't have the same energy, but it does.

 

Actually, I think even swan corrected me on this. I said electrons in bigger nuclei are closer and even he said that's not right.

 

If I did, it was a different discussion, and we likely weren't discussing the 1S orbital.

 

It's easy to show that this is wrong. The ionization energy for H is 13.6 eV. The ionization energy for He, with two electrons in the 1S state, is around 25 eV. The electrons are in the same S state but don't have the same energy.

[questionposter]

If I did, it was a different discussion, and we likely weren't discussing the 1S orbital.

 

It's easy to show that this is wrong. The ionization energy for H is 13.6 eV. The ionization energy for He, with two electrons in the 1S state, is around 25 eV. The electrons are in the same S state but don't have the same energy.

 

So could the nucleus of an atom be so big that its boundaries exceed the most probably location of its ground state electron? Like say we were able to fuse two uranium atoms together to make a really big nucleus...

Edited December 10, 2011 by questionposter

[swansont]

So could the nucleus of an atom be so big that its boundaries exceed the most probably location of its ground state electron? Like say we were able to fuse two uranium atoms together to make a really big nucleus...

 

In principle, yes. But such nuclei would be incredibly unstable.

[questionposter]

In principle, yes. But such nuclei would be incredibly unstable.

 

Wait, how could the energies of electrons not be continuous then? Couldn't there just be a small and smaller fraction of energy that electrons could form hybrid orbtials with? Because you could have helium who's electrons are 1/2 the distance of hydrogen, and then lithium's who are 1/3 the distance of hydrogen and etc, and since there's no limit to how big a nucleus can actually be, could an electron actually posses any energy?

Edited December 11, 2011 by questionposter

[swansont]

Wait, how could the energies of electrons not be continuous then? Couldn't there just be a small and smaller fraction of energy that electrons could form hybrid orbtials with? Because you could have helium who's electrons are 1/2 the distance of hydrogen, and then lithium's who are 1/3 the distance of hydrogen and etc, and since there's no limit to how big a nucleus can actually be, could an electron actually posses any energy?

 

The energy within an atom is discrete. That doesn't change with the size of the nucleus.

[DrRocket]

If all atoms have the same energy levels, as in the electrons in the ground state electron in lead wouldn't be closer than in hydrogen, and those energy levels can absorb only specific energies of light, why are there many different shaped orbitals?

 

Maybe you ought to read this.

 

http://en.wikipedia.org/wiki/Atomic_orbital

 

Then think a bit more deeply about what the term "orbital" means (and what it does not mean) and how the term applies to atoms with a very large number of protons and electrons.

 

quantum mechanics can get messy.

[mississippichem]

quantum mechanics can get messy.

 

Two electrons is one electron too many.

 

It's really amazing that we can even get approximate solutions for large multi-electron systems.

 

Without assuming the nuclei are still (Born-Oppenheimer) we wouldnt even be able to get that.

[questionposter]

The energy within an atom is discrete. That doesn't change with the size of the nucleus.

 

But how could a ground state electron in an s orbital 1/2 the distance of hydrogen's s orbital from the nucleus have the same energy as one that is 1/3 the distance of hydrogen's ground state electron from the nucleus

 

Or basically, if the atoms in helium are 1/2 the distance of hydrogen, how could they have the same energy as electrons in lithium which would have electrons 1/3 the distance of hydrogen?

Edited December 12, 2011 by questionposter

[swansont]

But how could a ground state electron in an s orbital 1/2 the distance of hydrogen's s orbital from the nucleus have the same energy as one that is 1/3 the distance of hydrogen's ground state electron from the nucleus

 

Or basically, if the atoms in helium are 1/2 the distance of hydrogen, how could they have the same energy as electrons in lithium which would have electrons 1/3 the distance of hydrogen?

 

They aren't half the distance, and we've established that they don't have the same energy.

[questionposter]

They aren't half the distance, and we've established that they don't have the same energy.

 

So if they don't have the same energy, couldn't an electron have any energy by going to different atoms? Also, then how much closer are electrons in helium than hydrogen?

[swansont]

So if they don't have the same energy, couldn't an electron have any energy by going to different atoms? Also, then how much closer are electrons in helium than hydrogen?

 

In principle, yes. That's not something that is restricted — you have quantized energy levels in a bound system, so there are forbidden energy values in each, but there is nothing that keeps that energy from being allowed in another system. That's why you have various materials that are useful in different situations, depending on the application. e.g. semiconductor lasers that emit red light are a different material than those that emit blue light.

[IM Egdall]

In principle, yes. That's not something that is restricted — you have quantized energy levels in a bound system, so there are forbidden energy values in each, but there is nothing that keeps that energy from being allowed in another system. That's why you have various materials that are useful in different situations, depending on the application. e.g. semiconductor lasers that emit red light are a different material than those that emit blue light.

 

Are these forbidden energy values for a given bound system absolute, or are they just the most probable energy values? In other words, can an electron have any energy value in a bound system, but the probability is the highest at the so-called permitted energy values?