Why does sodium lose its electrons?

I had answered a post about why sodium loses its electron. Borek had also provided an answer that I generally agreed with, but I wanted to add more detail.

As we go across the periodic table the nuclear charge increases and … the elements are more electronegative. Sodium has a larger 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 understand why sodium, which has a larger 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 because sodium is larger than fluorine?

My post:

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.

Dan:

I don’t understand how your discussion about molecules helps in answering the original question about atoms.

Looking back at my post, I must agree with Dan. Although I tried to avoid too much detail, I also did not explain my thought process very well. I also cannot publish an entire chapter on this topic.

In discussing electronegativity theory, I had to explain Pauling’s theory of ionic attraction, which he used to explain why some bonds were stronger than predicted. This is a paradoxical theory as ionic bonds had large energy differences which Pauling concluded corresponded with strong bonds, yet these bond readily dissociated into their ions.

At this point, we must be very careful as we have several different ideas that are affecting the results. They relate to bond strength and how they are measured. It is expected that high melting points of ionic bonds correspond with strong bonds. If true, then the ions of NaOH and HCl should have great affinity for one another. They do not. In the post, “Why is an acid-base equilibrium always shifted toward the weaker acid?”, I explained why ammonia is a stronger base than fluoride ion. The subatomic properties can better support the attraction than the net difference in charge. A net charge does not necessarily result in a strong field or attractive force. The properties of ionic compounds like sodium chloride have strong matrices, but a strong matrix does not make individual bonds strong. You can drive a car on a frozen lake made up of hydrogen bonds. You should not conclude from this that hydrogen bonds are strong bonds.

There are four forces of nature, the strong force, weak force, gravity, and electromagnetic force. Chemical bonds use an electromagnetic force therefore different bonds types describe the magnitude of different bond strengths. We may use a simple analysis for estimating the strength of chemical bonds based upon their charge and inverse square of the distance (Coulomb’s Law).

Atoms are inherently polar due to quantum effects. Protons are positive and electrons are negative. We should not think of atoms as neutral any more than think the north and south poles of a magnet balance the magnetic field. Even though the noble gas helium is electronically neutral, the electrons of helium can be protonated because the electrons create a local charge gradient.

To date, our theory of atomic structure has not caught up with how we should describe these quantum effects or apply them to structures. For example, bond lengths use the sum of the ionic radii. Since anions are negative and cations positive (ignoring the negative field surrounding the nucleus), the anions have been given a different radius so the sum of the radii of cations and anions equal the bond length.

In a different model, the bond length of HF (92 pm) and the radius of a sodium cation (ca. 90 pm) are consistent with the 10 electrons having a similar volume. The bond length of HF reinforces the estimates of sodium’s radius. Logically, replacing a proton with a sodium cation should result in a bond length of 182 pm (92 pm + 90 pm = 182 pm). The bond length of sodium fluoride is much larger, 231 pm (Pauling). The difference is consistent with a gap between the fluoride anion and sodium cation, 231 pm – 90 pm – 92 pm = 49 pm.

We cannot determine where electrons are. If we consider the acidity of HF, we could ask whether the electrons remain with the proton or the fluorine? Although we may be inclined to think they should remain with the fluorine because it has a larger charge and a larger field, that need not be so. The attractive force varies with the inverse square of the separation. If the proton-electron pair distance were small, then the force may be larger and the electrons could remain with the proton. However, we know that HF ionizes into a fluoride anion and water picks up the proton.

We can use this example to determine the relative force between a proton and an electron pair in reactions with water. Water is a useful solvent as it can donate an electron pair from oxygen or a proton. I propose a simple test to determine whether a pair of electrons is more strongly attracted to a proton or a nucleus. If the compounds LiH, BeH2, BH3, CH4, NH3, H2O, and HF were to react with water, to which atom would the electrons remain? In the example below, the products to the left form by hydride abstracting a proton from water. The products to the right form from water abstracting the proton and the electrons remaining with the lithium. Lithium hydride forms the products to the left.

Li(+) + HO(-) + H2 <- LiH + H2O -> Li(-) + H3O(+)

In the reactions of LiH, BeH2, and BH3 with water, the electrons remain with the proton (hydride ion). As the nuclear charge increases, the electrons are pulled closer to the nucleus (and away from the protons). In the reactions of NH3, H2O, and HF with water, the electrons now remain with the nucleus. This series of compounds share the same nuclear kernel, a central nucleus surrounded by a pair of electrons. We should conclude from this simple experiment that even though lithium, beryllium, and boron have more protons than electrons (+1, +2, and +3), the Coulombic attraction of a pair of electrons to a single proton is greater than to the nucleus. When the nuclear charge increases, the Coulombic attraction of a single proton is overcome and the electrons remain with the nucleus.

The Coulombic forces are also revealed in the bond lengths of these hydrides. The bond lengths decrease as the nuclear charge increases.
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How should we interpret atomic structure? If we were to examine a slice of a lithium cation, in the center are three protons surrounded by two electrons. Even though the net field of the electrons are spread over a larger area, their repulsive force to other electrons should grow exponentially as expected by the inverse square of the separation. At close distances, the electron shell of a pair of electrons will be repulsive to other electrons. At greater distances, the net field of the excess protons will become attractive to other electrons. We should expect the laws of physics to apply to all atoms and I argue is consistent with the bond length data shown above.

The attraction of a cation to electrons is consistent with the crystal structures as well. LiF has the same structure as NaCl, namely a face centered cubic structure. A lithium cation is surrounded by six fluoride anions. Therefore, the electrons must be much further from the cation and the electron-electron repulsion is minimized. If the nuclear charge is increased, the attractive field increases and electrons become pulled closer to the nucleus. When the electrons are far away, the e-e gap is larger and a larger number of electrons can surround a nucleus. LiF has six pairs of electrons around each lithium. With methane, the electrons are closer to the nucleus, the e-e gap is smaller and a limit of four pairs is reached.

A similar analysis should apply to a sodium cation. At short distances, the negative field of the electrons should be greater and at larger distances, the nuclear field should become larger as it has a larger net charge. Because the field decreases with the inverse square of the separation, a larger distance will also result in a smaller force. We should expect that a sodium cation would have a low affinity for additional electrons.

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