semiconductive
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Ionization energy.
I've been looking at ionization charts and noticed I've probably got some kind of misunderstanding about ionization energy.
When doing specstrocopy (for example on hydrogen), we generally compute the energy based on the shell an electron is in. With the pauli exclusion
principle, there can be up to two electrons in any given shell with opposite spins.
For example, the 1s shell of hydrogen has 13.6eV.
It can have one or two electrons put into it.
So, whether there is one electron or two ... in our basic chem class, there was no change to the formulas; we would simply expect that the shell has
a specific energy and compute transistion changes for spectrum based on the energy of the shells involved in a transistion. No reference to different
energies existing because of multiple electrons or only in a shell was made.
But, I'm thinking this has to be incorrect.
If the charge of the nuclus can affect the energy of a shell ... then the energy of a shell also has to change depending on how many electrons it
contains.
I came across a website which seems to support his hypothesis:
https://www.sjsu.edu/faculty/watkins/helium.htm
In physics experiments where gaseous atoms are ionized, the energy differs for each electron removed from a shell.
Eg: Sulfur, we have 4 valence electrons in three 3p orbitals.
S -> S1+ = 999.6 kJ/mol = 239.14 Kcal/mol
S1+ -> S2+ = 2251 kJ/mol = 538.52 Kcal/mol
S2+ -> S3+ = 3361 kJ/mol = 804.07 Kcal/mol
S3+ -> S4+ = 4564 kJ/mol = 1091.87 Kcal/mol
I assume one p orbital will have 2 electrons, but otherwise all three P orbiitals have 1 electron each and are indistinguishable.
So, I'm thinking the spectra of a sulfur atom that is in the S+ state, ought to be different than the spectra of a neutral sulfer atom where only one
electron is being excited between states. Am I correct?
Which is more correct ... assuming that the energy of an electron in a shell changes depending on the number of electrons in that shell; or that only
the energy to completely remove an electron from a shell to infinity changes as a shell becomes depleted ? ( I'm thinking that it's the former. )
when I look at charts for ionization trends, I notice that the shape of the energy of a shell vs. atomic number is approximately parabolic or
hyperbolic (especially for low orbital shells).
I'm thinking that ionization is very similar to changing the atominc number (aka: effective atomic number) that the remaining electrons experience by
1 for each electron removed.
So, if I know the shape of a curve and several points along it ... I should be able to extrapolate the curve using some kind of formula; although the
formula will break down when the shell is filled, or emptied.
A parabola or hyperbola, has a 1st difference that changes linearly with each step taken along it. It shold be uniform.
So, for the sulfur chart ... I'm taking the first difference in KJ energies and I get:
2251 - 999.6 = 1252.4
3361 - 2251 = 1110.
4564 - 3361 = 1203.
And the variace from one step to the next differ by around 12% (+-70kJ/mole) which isn't excellent although it's still useful.
The change in energy of ionization is around 1188kJ per mole for each successive electron, and it's expected to decrease slightly as each subsequent
electron is removed. But, the trend ought to be stable so long as we stay in the same shell.
Sulfur only has 4 electrons; so the trend cant be tested for more removals ... but it is possible to estimate the energy level for Sulfur trapping
an electron and becoming a negative ion; S-.
From the periodic table, S commonly will take on the S^-2 ion state.
so, If a an electron comes near a neutral sulfur atom; I'd estimate that since 1188 > 999.6, that sulfur would not trap the electron and the
removal energy would be negative: -188 kJ. Hower, this has to be wrong.
When I look up the electron affinity of sulfur ... it IS able to trap a free electron, and the estimate is 200kJ/mole. So, I'm almost 100% wrong in
the estimate and it's much larger than 10% off.
What am I doing wrong, logically, and does anyone have a formula / methodology that is simple enough to be done on a hand calculator and accurate
enough to predict what is the maximum negative ionization level that a particular shell will accept?
[Edited on 16-3-2021 by semiconductive]
[Edited on 16-3-2021 by semiconductive]
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DraconicAcid
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Yes, the emission spectrum of an ion will be different from that of an atom. If you look up the emission spectrum of an element, you'll see different
values tabulated for X(I), X(II), X(III), etc. X(I) is for neutral atoms of X, X(II) is the emission spectrum for X+, etc.
Your statement about S only having four electrons is incorrect. https://www.webelements.com/sulfur/atoms.html has all sixteen ionization energies.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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semiconductive
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Quote: Originally posted by DraconicAcid  | Yes, the emission spectrum of an ion will be different from that of an atom. If you look up the emission spectrum of an element, you'll see different
values tabulated for X(I), X(II), X(III), etc. X(I) is for neutral atoms of X, X(II) is the emission spectrum for X+, etc.
Your statement about S only having four electrons is incorrect. https://www.webelements.com/sulfur/atoms.html has all sixteen ionization energies. |
Those electons are not all in the valence P shell.
There are only four electrons in the 3P shell, according to the site you showed me.
It has three up arrows, and one down ... which valildates the assumption I made.
Filling order is one electron per orbit (first), therefore pairing tends to be delayed until there are no empty orbits in the 3rd P shell.
The formula/method I gave is not expected to be reliable for the fifth ionizaation energy when swtiching shells. See the graph .... S and P do not
follow the same line, but are separated by an energy level jump that isn't part of the parabola/hyperbola trend.
And, notice also ... that the energy level diagram for the sulfur atom shows all three P orbitals at the same energy level, regardless of whether
there is one or two electrons in the orbit.
A 3P orbital with two electrons is drawn at exactly the same height as the P orbital with only one electron. That's curious from the standpoint that
electrons fill each orbital before pairing up in order to keep lowest energy configuration; that seems to suggest (to me) that the electron orbital
with two electrons in it should be ever so slightly higher in energy than an orbital with only one electron in it. ( The drawings not to scale
anyway, .... but it's curious the effect of pairing is not included schematically on the energy scale. )
There seems to be a range of energies that a P orbital can have, depending on the number of electrons in the P orbitals .... So, the orbital is not
fixed by the nuclear charge alone ... Rather than a flat line for energy level, shouldn't a P orbital be represented by a "box" in the diagram ?
[Edited on 16-3-2021 by semiconductive]
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semiconductive
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Well, I spent all day looking ... and the only thing I can find that's at all helpful is slater's rules.
To do this as a physics problem, I would need to know some statistics about orbitals and how they shrink with increasing Z.
Then I could compute an approximate electric field around an atom, and estimate the energy that way.
Orbitals are defined for graphing purposes, as the volumes of space where an electron is 90% of the time.
But, I'm not finding any tables that list basic properties of these volumes ...
I mean, I know that an S orbital is spherically symmetric, so it's volume is 4/3 pi r^3 and it's surface area goes as 4pir^2. This would allow me to
compute a spherical volume based on the radius where an S orbital reaches 10% or less likelihood of finding the electron.
It's trivial to compute the volume of the outermost node of an s orbital by just taking the last node radius, and the 10% radius and subtracting the
volumes.
But other orbital shapes are not simple geometric shapes with well known formulas.
I don't see tables that list formulas for the volume of a P orbital, D orbital, etc.
I'd like to be able to know the volume of the outer-most node of these orbitals is given the 10% boundary radius customarily used in orbital
definitions; as these outermost nodes tend to be outside the filled shell AKA the noble gas shell and are the only complicated part of the
calculation; I think the noble gas inner shell can probably be treated as spherically symmetric.
Does anyone know of a table listing of formulas for the volumes of orbitals? This seems like such a basic statistic that someone should have already
worked it out.
[Edited on 17-3-2021 by semiconductive]
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DraconicAcid
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I can't offer you any advice on this beyond "don't expect it to work out neatly".
I remember one calculating the effective nuclear charge of various atoms and ions, thinking that Zeff for K and Ca+ should differ by exactly one
(since they have the same number and arrangements of electrons, and calcium simply has one more proton). It didn't work.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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semiconductive
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It'll be hard for me to quell my perfectionist tendencies.... but, yes, I'm going to have to settle for approximate.
Thanks, I'll keep the advice in mind.
I'll attempt some theoretical calculations, and I'd appreciate you reviewing my steps and seeing if I make any obvious mistakes.
I'm hoping to get qualitiative results without trying to solve schrodinger's equation fully....
Slater's rules seem like a decent start ... if I could improve them.
When I watch videos on shielding, there a note about *partial* shielding. It corroborates your point, zeff usually doesn't change by 1,
Neighboring electrons in the valence shell aid in ejecting an electron that is being ionizedl; so I have multiple reasons to believe in fractional
z_eff changes.
I should be able to compute Z_eff from experimental iionization tables ...
From how I read slater's rules, the assumption is made that shielding due to core electrons is the same all the way across any horizontal period. Only
small (and fractional) changes to Z_eff occur as electrons are added to the valence shell.
Hmmm ... I'm sure the assumption is a little wrong. S orbitals, for example, penetrate slightly into the core; but i'll try it anyway.
From physics, the Rydberg formula is:
Energy of transition / h c = z^2 * R * ( 1/n_2^2 - 1/n_1^2 )
I can use that formula to solve for ionization energy, by making n_1 go away.
That gives me:
Energy of ionization = (z/n_2)^2 * R * h c
The Rydberg constant (R) multiplied by h*c has to be the energy of the hydrogen atom's ground state. R*h*c = 1312 kJ/mol in thermodynamic units.
So, that means:
Z_eff = n sqrt( E_Ion/ E_0 )
I can treat all group 1 elements as if they were hydrogen (according to slater's rules), and simply compute the empirical Z_eff required to get the
ionization energy of each element.
I think experimentally computed values of Z_eff should differ from 1 only by the amount caused by a valence electron's penetration into the core.
Penetration can only increase the effective nuclear charge seen by a valence electron, therefore:
Z_eff is always >= 1.0.
For each orbital type, in any period, I think the penetration should be constant across the period .
So, the computed Z_eff should differ by 1 if I compare the ionization of the last electron from the S orbital in group 1 vs. group 2 elements.
So, I'll test group1 against group 2 and see how close I get to theory ....
Here's my source of ionization energies to use in the computation:
https://en.wikipedia.org/wiki/Ionization_energies_of_the_ele...
E_I( H+ ) = 1312 kJ/mol. Zeff = sqrt( 1312 / 1312 ) * 1 ~= 1.0
E_I( Li +) = 520.2 kJ/mol Zeff = sqrt( 520.2 / 1312 ) * 2 ~= 1.25
E_I(Na+) = 495.8 kJ/mol Zeff= sqrt( 495.8 / 1312 ) * 3 ~= 1.84
E_I(K+) = 418.8 kJ/mol Zeff= sqrt( 418.8 / 1312 ) * 4 ~= 2.26
E_I(Rb+) = 403.0 kJ/mol Zeff= sqrt( 403.0 / 1312 ) * 5 ~= 2.77
E_I(Cs+) = 375.5 kJ/mol Zeff= sqrt( 375.5 / 1312 ) * 6 ~= 3.21
E_I(Fr+) = 380.0 kJ/mol Zeff= sqrt( 380.0./ 1312 ) * 7 ~= 3.76
So far, so good ... Z_eff is monotonically increasing (periodic).
Next, I'll compute Z_eff for the second ionization of group 2 elements.
This ionization energy has to be from an S orbital, just like the previous table; and we're only comparing S orbitals having the same 'n' number.
On the basis of how schrondinger's equation's solutions are computed ... I'd expect these errors to be quite small.
He++ 5250.5 kJ/mol zeff 2.0 err: 0%
Be++ 1757.1 kJ/mol zeff 2.31 err: 3%
Mg++ 1450.7 kJ/mol zeff 3.15 err: 10%
Ca++ 1145.4 kJ/mol zeff 3.74 err: 13%
Sr++ 1064.2 kJ/mol zeff 4.50 err: 16%
Ba++ 965.2 kJ/mol zeff 5.15 err: 18%
Ra++ 979.0 kJ/mol zeff 5.18 err: 8%
These errors are still quite a bit bigger than I'd expect, although they are in the correct ballpark for a useful approximation.
I'm not sure what's going on ...
But since the expeimental ionization energies aren't off by 18% ... I need to look for a theoretical error.
There's only two ideas I can think of right now:
1) The effective charge does not actu like a point source nucleus, but is more like a smeared out charge; so I need to re-solve (or find) the
Schrodinger equation for a nucleus of finite radius rather than a point charge approximation.
2) As each new proton is added, the shielding of each subshell is affected separately ... so that there is a cascade of expansions in sizes that will
become more pronounced as more protons are added to the nucleus.
[Edited on 17-3-2021 by semiconductive]
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