Chemblog!

I wrote most of this explanation last year so I want to pass it on to this year's classes, which are also overly enamored with the octet rule:
Though most of you can express the connection between attraction and potential energy or repulsion and potential energy, others have not expressed that connection and, instead, have developed a completely disproportionate sense of the relevance and/or importance of the octet "rule". This "rule" has about as much contribution/relevance to potential energy lowering or stabilization as an eyelash contributes to the mass of an ELEPHANT!
Before I reiterate what I said in class about "Attraction, Repulsion and Potential Energy", let me reiterate why the octet "rule" exists. Take a fluorine atom, for example. A fluorine atom tends to bond to ONE and ONLY ONE other fluorine atom to form a diatomic fluorine molecule (draw the Lewis structure for fluorine). Why is there no stable triatomic fluorine molecule? Once a fluorine atom shares its one unpaired electron with another fluorine atom, each F atom has eight valence electrons. ..very nice but no big deal. There is actually greater NET repulsion (which is ALWAYS energy raising) among electrons when the eighth electron enters the valence shell BUT the bond forms due to the competing and MORE SIGNIFICANT factor of the SIMULTANEOUS attraction of the TWO fluorine nuclei (Zeff = +7) for the shared PAIR of valence electrons. ALSO, this extra repulsion among electrons is compensated for A LITTLE BIT due to the (on average) symmetric distribution of the electrons about each nucleus when the s and p sublevels are completely filled (due to the symmetric orientation of s and p orbitals). This symmetric distribution of electrons causes a bit LESS repulsion compared to the amount of repulsion when there is an asymmetric electron distribution (i.e. when at atom has 6 or 7 valence electrons). If a third F atom bonded to the molecule, the central F atom would have 9 valence electrons! There is no "room" (due to too much electron-electron repulsion...i.e. Pauli Exclusion Principle, Aufbau Principle) for a 9th electron in the 2nd principal energy level (2s2 2p6) so that next bonded electron would have to go to the much higher potential energy 3rd principal energy level. This would have a destabilizing effect (higher potential energy = destabilizing by definition!) that outweighs the attraction that would occur from a third covalent bond. So, that is why most (BUT NOT even ALL!) non-metal atoms stop bonding once they acquire an octet.
To really blow your mind, let me tell you what you actually already know: a sodium atom, BY ITSELF, is MORE STABLE than a sodium ION (BY ITSELF in the gas phase, NOT in a lattice of cations and anions) even though the sodium ION has an octet in its outermost occupied PEL (principal energy level). Didn't think that you knew that? I bet that you do. Here's why: Recall the definition of ionization energy. Look at the table of ionization energies. Locate sodium. Aha, so energy is actually REQUIRED to remove the valence electron from sodium? YES! Why? Simple: the positive sodium nucleus attracts (with a Zeff of +1) the negative valence electron! Energy is required to overcome that attraction. Why doesn't sodium lose two electrons then? The octet rule? NOPE! The second ionization energy of Na is so high because, to remove the second electron, the attraction from the nucleus on that electron is +9 AND that electrons is in a closer (to the nucleus) principal energy level (which, by Coulomb’s Law means that there is greater electrostatic attraction); therefore, too much energy is required for Na to lose a second electron. See this by drawing a Bohr model of a sodium atom and a sodium +1 ion and calculating the Zeff's on each shell of electrons.
You object: Compared to sodium atoms, why is the sodium +1 ion so stable and unreactive then? ahaaa! Sodium ions are NEVER by themselves! They are ionically BONDED (VERY STABILIZING!!!) to any and all surrounding anions in the salt’s crystal lattice! (draw lattice of ions in a typical salt)...or, as you will learn, in any aqueous solution or mixture (saliva, blood, the ocean), cations are surrounded by the partially negatively charged oxygen end of a water molecule (very stabilizing).

Now, the main reason for potential energy lowering WHENEVER a bond is formed: (write this out and MEMORIZE it if you have to)

COVALENT BONDS: the (negative) electrons that are SHARED between the (positive) nuclei will ALWAYS have a potential energy LOWERING effect (whether OR NOT there is an octet of electrons i.e. in Hydrogen) because:
1. positive particles (protons) attract negative particles (electrons) and vice versa
2. attraction is BY DEFINITION potential energy LOWERING!
3. relative to having a valence electron around one nucleus (an atom), there is MORE ATTRACTION when a valence electron is shared/located BETWEEN TWO NUCLEI (a COVALENT BOND BETWEEN two atoms’ nuclei) because there is more positive charge (protons from TWO nuclei) surrounding negative charge (shared valence electrons) and vice versa!

IONIC BONDS: again, the octet rule is practically IRRELEVANT to ionic bonding:
the actual potential energy lowering effect is due almost SOLELY to:
1. here, the CATIONS are the positive particles and the ANIONS are the negative particles in all ionic bonds.
2. positive particles attract negative particles and vice versa
3. attraction is BY DEFINITION potential energy LOWERING!
4. since electrons are NOT shared in ionic bonds, mentioning them as a significant final factor is meaningless. Each cation can have (depending on the geometric arrangement of ions in the lattice) four, six, or maybe even eight ionic BONDS to surrounding anions and vice versa! Each INDIVIDUAL ionic bond is between TWO fully charged IONS of opposite charge. The fact that the cations and the anions each have (usually) an octet of outermost PEL electrons is a MINOR almost irrelevant factor.

That is the relationship among Attraction, Repulsion, and Potential Energy. Pass it on.