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Author Topic: Sodium has more nuclear charge than fluorine then why...  (Read 1874 times)

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Nescafe

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Sodium has more nuclear charge than fluorine then why...
« on: June 15, 2012, 06:29:10 PM »

As we go across the periodic table the nuclear charge increases and we see a trend because of this where the elements are more electronegative. Sodium has more nuclear charge than fluorine. Wouldn't this mean that the one electron in its valence shell is going to feel the attractive force of the nucleus more so than electrons on fluorine? I just don't get why sodium which has more nuclear charge than fluorine gives up its electron so easily. I know that atoms want to have a configuration like that of noble gasses, so does stability explain this? Is it cause sodium is larger than fluorine...



Any help would be appreciated,

Nescafé.
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Borek

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Re: Sodium has more nuclear charge than fluorine then why...
« Reply #1 on: June 15, 2012, 10:06:36 PM »

Not only nucleus charge grows, also number of electrons shielding the charge.

But there is something much more important. Electrons don't form an amorphous cloud around the nucleus, they are ordered in a way and occupy orbitals. That basically means that new electrons go further from the nucleus (oversimplification and not exactly true), so the attraction is not higher for each next element.

Once you will learn a little bit about quantum chemistry, quantum numbers, orbital energies and orbital arrangement in space it will become more clear.
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orgopete

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Re: Sodium has more nuclear charge than fluorine then why...
« Reply #2 on: June 16, 2012, 05:00:41 AM »

Not only nucleus charge grows, also number of electrons shielding the charge.

But there is something much more important. Electrons don't form an amorphous cloud around the nucleus, they are ordered in a way and occupy orbitals. That basically means that new electrons go further from the nucleus (oversimplification and not exactly true), so the attraction is not higher for each next element.

Once you will learn a little bit about quantum chemistry, quantum numbers, orbital energies and orbital arrangement in space it will become more clear.

Although Borek refers to an oversimplification (and not exactly true), it is probably a lot more useful than any other treatment I can think of. I don't want to go over my head in a discussion here, but even quantum mechanical calculations must remain consistent with the principles of physics. (I think that could have been construed as a point Einstein was making in the Bohr-Einstein debates, but that would be another day.) Even with a simplistic model of atomic structure, it will come reasonably close to the physicochemical properties.

"Electrons don't form an amorphous cloud around the nucleus, they are ordered in a way and occupy orbitals." Agreed! Ice is tetrahedral, therefore the electrons or electron pairs are concentrated in defined regions. If you compare the bond lengths of LiH, BeH2, BH3, CH4, NH3, H2O, and HF, the bonds become shorter and consistent with an increase in the nuclear field. In reactions, LiH, BeH2(?), and BH3 are hydride donors and NH3, H2O, and HF are proton donors. I interpret this as an indication of the relative shielding effect of the inner electrons. If you compare dilithium (267 pm) and difluorine (142 pm), their distances appear consistent with the inner electron shielding, a nuclear charge effect, and the inverse square law.

Even if a simplistic model does not result in a correct quantum mechanical calculation, it does seem much easier to grasp and would correctly predict the nuclear field of sodium would be much smaller than elements to the right of it in the periodic table. The inverse square law also predicts the field would be much smaller and thus reinforcing its low affinity for additional electrons.

I think the difficulty in understanding the effects centers on how to treat cations. At large distances, we can treat them as Gaussian shells and only consider their net charge. At short distances, we must be cognizant of their microscopic properties, specifically, the electrons are negative and repel other electrons. A net positive charge does not convert electrons into positive charges.

I do not wish to highjack the posters question, but I urge caution in the accepting some tenets of quantum theory. Even orbitals have unanswered questions. Pauling introduced hybridization to achieve tetrahedral structures while s, p, d, and f orbitals describe the atomic emissions. I am not aware of sp3 emissions nor how strict adherence to s and p orbitals should give a symmetrical methane. Even Gilbert Lewis was skeptical of the Bohr interpretation of the emissions of hydrogen.
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Dan

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Re: Sodium has more nuclear charge than fluorine then why...
« Reply #3 on: June 16, 2012, 12:16:45 PM »

Orgopete, while I agree that shielding/penetration effects do manifest in molecular orbitals, I don't understand how your discussion about molecules helps in answering the original question about atoms.

Nescafe, I suggest reading about effective nuclear charge in an inorganic chemistry textbook. Wikipedia page is http://en.wikipedia.org/wiki/Effective_nuclear_charge but is very brief.

Also look up shielding and penetration. Also a brief wiki page on this http://en.wikipedia.org/wiki/Shielding_effect
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cheese (MSW)

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Re: Sodium has more nuclear charge than fluorine then why...
« Reply #4 on: June 18, 2012, 06:24:06 AM »

http://en.wikipedia.org/wiki/Slater%27s_rules
Zeff = Z –S
F: 1s^2 2s^2 2p^5  S = 2×0.85 (1s e⁻s) +6×0.35(2s^2 2p^4 e⁻s) = 3.8
Zeff = 9 – 3.8 = 5.2  (Note that e⁻s in the same valence shell do a poor job of screening each other)
IP α  Zeff/r
Atomic radius (r) = 71.7 pm, IP = 1681 kJ mol^-1
Na: 1s^2 2s^2 2p^6 3s^1
S =   2 (1s e⁻s) + 8×0.85(2s^2 2p^6 e⁻s) = 8.8
The 2s/2p e⁻s are now core e⁻s.
Zeff = 11-8.8 = 2.2
Na: atomic radius = 153.7 IP = 495.8 kJ mol^-1

The electron configurations come from the Schrödinger Wave Equation that was proposed early in 1926 and has,
 except for relativistic corrections (Dirac), agreed with every test scientists can throw at it (kitchen sinks are
somewhat large to probe the subatomic world).

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orgopete

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Re: Sodium has more nuclear charge than fluorine then why...
« Reply #5 on: June 18, 2012, 12:57:43 PM »

Orgopete, while I agree that shielding/penetration effects do manifest in molecular orbitals, I don't understand how your discussion about molecules helps in answering the original question about atoms.

I concede this is accurate. I have been working on this topic for some time and it is easy to overlook a perspective that only I possess. This is an instance. I have more detail in "Why does sodium lose its electrons?", but it is long.

A graph of a point I was attempting to make is shown a lot more clearly here. A problem with distances is I don't know where the electrons are located. Bond lengths show the distance between nuclei. The graph virtually shows the attraction of electrons to a nucleus or you can react the compounds with water and compare the electron pair-proton cohesion.

I understand the point being made, but I don't agree with the logical connection between the calculation and the results in the lab. I felt this was precisely the question the poster was making. Subsequently, cheese (MSW) has added the effective nuclear charge for sodium, 2.2.

If I were a student, I would not find the effective nuclear charge any more helpful to understand why sodium gives up its electron. The point I was attempting to make is that by the inverse square law, the inner shell of electrons are more repulsive at short distances than using the Slater rules or a simple vector calculation of the net charge of a Gaussian shell seems to indicate. Although this may take some reflection, if the negative charge of the inner electrons has a greater repulsive effect than anticipated, they may increase the distance to any electrons being attracted to the net positive field. That is, electrons can be repulsed at short distances and attracted at large distances. As a consequence, the attraction will be weak, if the distance has increased. I argue this analysis is more helpful to predicting the properties of the alkali metals than using an effective nuclear charge.


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cheese (MSW)

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Re: Sodium has more nuclear charge than fluorine then why...
« Reply #6 on: June 20, 2012, 07:50:57 AM »

Slater’s rules were first proposed in the 30s and are still useful in order to give a quick estimation of the effective nuclear charge on a (valence) e⁻ (ve⁻).  Let’s ignore the case where the atom has d or f  e⁻s.
If an e⁻ is in the same sp shell as the ve⁻ the screening factor is 0.35.  As stated previously, valence e⁻s do a poor job of screening each other (consider the screening of a px e⁻ on a py e⁻)
If the sp e⁻s are in the shell that has a principal quantum number 1 less than the ve⁻ then the screening factor for each of these e⁻s is 0.85.  This recognizes that the ve⁻ can penetrate these core e⁻s to some extent (as indicated by the SWE) and as such these core electrons do not screen with a factor of 1.0.
For e⁻s in sp shells that have a principal quantum of two (or lower) less than the n of the ve⁻ the screening factor is the full 1.0.   Penetration of the ve⁻ of these AOs is negligible for screening purposes.
The Zeff given by Slater’s rules (SR) are for simple atoms in remarkable agreement with those calculated by sophisticated self-consistent field (SCF) methods.  The gold standard for these calculations are still the Clementi Raimondi values (CR) published in the 60s.   Hence Zeff (F) = 5.20 (SR) 5.10 (CR); Zeff (Na) = 2.20 (SR) 2.51 (CR). (Values from J. Emsley The Elements.)
http://www.knowledgedoor.com/2/elements_handbook/clementi-raimondi_effective_nuclear_charge.html
As previously stated, IP is proportional to Zeff/r (Coulombic attraction).  It is a problem what to use for r given the nature of e⁻s.  The atomic radii (again from Emsley) give an indication of the average r.  What I could have used is valence shell orbital radii, Rmax. These are F 2s = 38.1; 2p = 40.6 pm
http://www.webelements.com/fluorine/orbital_properties.html
Na 3s = 179.4 pm (!)
http://www.webelements.com/sodium/orbital_properties.html
From the SWE after the 2s2p AOs are filled the next e⁻ enters the 3s AO that is further from the nucleus and experiences more screening because the 2s^2 2p^6 are now core e⁻s.

The fundamental laws necessary for the mathematical treatment of a large part of physics and the whole of chemistry are thus completely known, and the difficulty lies only in the fact that application of these laws leads to equations that are too complex to be solved.  Paul Dirac (But now we have computers!)

I do not like [quantum mechanics], and I am sorry I ever had anything to do with it. Erwin Schrödinger.
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