# How do valence shell configurations affect attraction and repulsion forces between atoms? [closed]

It seems clear that the electrons of one atom are attracted to the protons of another and that this is the source of attractive forces between atoms. However, it is not clear why an atom with a full valence shell such as a noble gas or covalently bonded atom is incapable of bonding or being attracted to other atoms. Are the electrons not attracted to protons in the same way? Is it because the repulsion between the electrons wins out over the electron-proton attractive forces. Furthermore, why do valence shell arrangements with few electrons bring about attractive forces?

## closed as unclear what you're asking by Jan, jerepierre, M.A.R., bon, Klaus-Dieter WarzechaJan 7 '16 at 22:57

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• it is not clear why an atom with a full valence shell such as a noble gas or covalently bonded atom is incapable of bonding or being attracted to other atoms - It is, in some cases. – bon Jan 7 '16 at 8:46
• Possible duplicate of Why are noble gases stable – Corundum Jan 7 '16 at 18:05

## 2 Answers

Firstly, why should covalently bonded atoms be incapable of bonding? Bond changes are the very basis of quite a huge amount of chemical reactions. Oxygen occurs in the $\ce{O2}$ form, which is covalently bonded, but readily breaks its O–O bond in order to form other bonds.

Now, to your question. Why are elements in group 18 (generally: think of the xenon fluorides) inert? Why is sodium so willing to give up its electron, while neon doesn’t form any compounds at all, when they only differ by one proton in the nucleus?

The answer lies in the orbital energies. You can find a table listing the outer shell orbital energies page 16 of Appendix B of Miessler and Tarr's Inorganic Chemistry (5th ed.). Let us compare the outer orbital energies of Ne and Na. Neon has two 2s electrons with an energy of $-48.47\ \mathrm{eV}$ and four 2p electrons with an energy of $-21.59\ \mathrm{eV}$, while the 3s electron in the sodium atom lies at $-5.14\ \mathrm{eV}$. The difference is overwhelming. When a ionic compound is formed, an electron is removed from the less electronegative atom and added to the more electronegative atom, forming a cation and an anion. Now, the Na electron is much easier to remove than the Ne electron – you just need $-5.14\ \mathrm{eV}$ to do the task, while eliminating the Ne electron is much more difficult. The electron is then placed into an atom such as fluorine, whose outer 2p subshell lies at $-18.65\ \mathrm{eV}$: you needed energy to remove the Na electron, but energy is released when you place it in the F atom (since it is put into a low energy orbital). Placing an electron from F to Na wouldn’t release energy because it is much more stable in the fluorine atom, due to their orbital energies (if you want to simplify things my anthropomorphizing the electrons, you could say they’re “happier” in the fluorine atom). Removing an electron from the neon valence shell is very difficult because it has very low energy; adding one doesn’t bring much because since the second shell is full, you would have to add it to the third shell – its energy is similar to the Na 3s shell, so it’s not convenient.

I have to note here that while the first ionization energy corresponds more or less to the orbital energy (Na: $495.8\ \mathrm{kJ/mol}$ which corresponds to $5.139\ \mathrm{eV/atom}$), the electron affinity doesn’t (F: $-3.399\ \mathrm{eV/atom}$), because of electron-electron repulsion in the fluorine p subshell. That’s why an isolated $\ce{Na+ - F-}$ compound cannot exist: the additional energy released when forming the lattice is needed to form a stable entity.

Having said this, we can move on. You may ask: Yes, but why are the F electrons more stable than the Na ones, when Na has more protons? The reason is that the outer Na electrons are in the third shell, and thus they have higher energies. In a hydrogen-like atom (an atom with only one electron, which is clearly an oversimplification, since it doesn’t involve screening) the orbitals have the following energies:

$$E \approx -13.6 \left(\frac{Z}{n}\right)^2 \mathrm{~eV}$$

You can see that raising the $Z$ (atomic number) value from 9 (F) to 11 (Na) isn’t enough for lowering the energy, since $n$ (the principal quantum number, i.e. the shell) changes from 2 to 3.

So, this is why sodium forms positive ions, fluorine forms negative ions, and neon doesn’t form any ions at all (chemically). Nota bene: as I said, the formula above doesn’t take screening into account – when going down a group, it (usually: see d-block contraction) prevails over the growing nucleus. That’s why Ar is less inert than Ne, even though the formula above would suggest the contrary.

But what about covalent compounds? According to VB theory, bonding occurs through orbital overlap. Lower energy orbitals mean smaller overlaps, which is inconvenient if you want a stable bond: Carbon forms bonds while neon doesn’t because its orbitals are less stable (see table) and can overlap with other orbitals. Its orbitals are less stable because neon has more protons with the same principal quantum number – see the equation above to understand the concept.

In short: noble gases, especially those positioned higher in the periodic table, don’t bond because their outer orbitals have the lowest energies. Ultimately, this is the result of having the largest number of protons possible for a given outer shell in the ground state.

If you consider Hydrogen bonding to itself, the bond is stabilized both by the electric attraction between proton and electron. As well the 1s orbital of both atoms merge into one bonding orbital, this single orbial is more energetically stable because electrons like to be paired off (spin 1/2 and -1/2) in orbitals.

The electron configurations of the noble gas feature valence shells with completely filled orbital pairs. According to the Pauli Exclusion Principle (PEP), no two electrons may inhabit the same quantum state in the same system. If all available electronic states for the atom are filled then there is no stabilizing phenomena for either the loss or gain of a new electron, and the formation of bonds is not favored energetically.

An analogy commonly used for the PEP is that there is an additional repulsive "force" between elections beyond the electric repulsion, it is precisely this "force" that prevents the noble gasses from bonding. Though it's not so much a force as it's simply the inherent fermionic behavior of the electrons at play. However, exotic noble gas compounds do exist, but are not terribly stable.

• Spin pairing has a net energy cost associated with it, not an stability gain. The lower energy of bonding orbitals has nothing to do with spin pairing. – bon Jan 7 '16 at 8:47
• I don't understand "If all available electronic states for the atom are filled then there is no stabilizing phenomena for either the loss or gain of a new electron" - there is no such thing as running out of electronic states – orthocresol Jan 7 '16 at 9:27
• There are noble gas compounds that are almost stable as a rock. – Martin - マーチン Jan 7 '16 at 10:51