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2017-10-27 Update [NOTE: My earlier notation-focused answer, unchanged, is below this update.] Yes. While having an octet of valence electrons creates an exceptionally deep energy minimum for most atoms, it is only a minimum, not a fundamental requirement. If there are sufficiently strong compensating energy factors, even atoms that strongly prefer octets ...


87

Yes, it can. We have molecules which contain "superoctet atoms". Examples: $\ce{PBr5, XeF6, SF6, HClO4, Cl2O7, I3- , K4[Fe(CN)6], O=PPh3 }$ Almost all coordination compounds have a superoctet central atom. Non-metals from Period 3 onwards are prone to this as well. The halogens, sulfur, and phosphorus are repeat offenders, while all noble gas compounds ...


50

In chemistry, and in science in general, there are many ways of explaining the same empirical rule. Here, I am giving an overview that is very light on quantum chemistry: it should be fairly readable at a novice level, but will not explain in its deepest way the reasons for the existence of electronic shells. The “rule” you are citing is known as the octet ...


45

Something worth adding to this discussion that I'm surprised hasn't been mentioned about such "hypervalent" molecules like $\ce{SF6}$. One of my professors at university informed me that the common explanation (that the empty d-orbitals are empty and are thus accessible) is actually most likely incorrect. This is an old-model explanation that is out-of-date,...


44

Shells and orbitals are not the same. In terms of quantum numbers, electrons in different shells will have different values of principal quantum number n. To answer your question... In the first shell (n=1), we have: The 1s orbital In the second shell (n=2), we have: The 2s orbital The 2p orbitals In the third shell (n=3), we have: The 3s orbital The ...


39

This question may be difficult to answer because there are a couple of definitions of valence electrons. Some books and dictionaries define valence electrons as "outer shell electrons that participate in chemical bonding" and by this definition, elements can have more than 8 valence electrons as explained by F'x. Some books and dictionaries define valence ...


39

Yes, there are coordination complexes of large elements which have coordination numbers greater than eight. Some examples are: $\ce{[ReH9]^2-}$ with a tricapped trigonal prismatic structure. The nine hydride ligands are small enough to fit around the relatively large rhenium atom fairly easily. This ion can be isolated as a potassium salt $\ce{K2ReH9}$. $\...


35

There is an explanation to this that can be generalized, which dips a little into quantum chemistry, which is known as the idea of pairing energy. I'm sure you can look up the specifics, but basically in comparing the possible configurations of $\ce{Nb}$, we see the choice of either pairing electrons at a lower energy, or of separating them at higher energy, ...


35

In addition to the general rules of how electronic configurations of atoms and ions are calculated, the elements from the d-block (aka the transition metals) obey one special rule: In general, electrons are removed from the valence-shell s-orbitals before they are removed from valence d-orbitals when transition metals are ionized. (I took this ...


34

Here's a graphic I use to explain the difference in my general chemistry courses: All electrons that have the same value for $n$ (the principle quantum number) are in the same shell Within a shell (same $n$), all electrons that share the same $l$ (the angular momentum quantum number, or orbital shape) are in the same sub-shell When electrons share the same $...


30

There is a big difference between a "rule" and a law of nature. The "octet rule" is a turn-of-the-last-century concept that somehow managed to get into introductory chemistry books and never got kicked out with the advent of modern quantum mechanics. (Circumstantial proof: it is impossible to identify individual electrons to label them "valence" or "not ...


30

14 coordination is claimed in $\ce{U(BH4)4}$ (ref_1, p. 268). The molecule exists as a polymer in the solid state. Six hydrogens from two of the $\ce{BH4}$ groups bond between the boron and uranium (a bridge bond). Two hydrogens from each of the two remaining $\ce{BH4}$ groups also bridge bond to uranium; the other two hydrogens bond to an adjacent ...


27

Klaus Warzecha's answer pretty much answers your question. But I know that this subject is easier to understand if supported by some pictures. That's why I will take the same route as Klaus at explaining the concept behind why the absorption in conjugated systems is shifted to higher wavelengths but I will provide some pictures on the way. In a conjugated ...


24

As I understand this, there are basically two effects at work here. When you populate an s orbital, you add a significant amount of electron density close to the nucleus. This screens the attractive charge of the nucleus from the d orbitals, making them higher in energy (and more radially diffuse). The difference in energy between putting all the electrons ...


24

I will try to describe what happens when two hydrogen atoms approach each other from infinity. At infinite separation the hydrogen atoms don't feel their mutual presence and each atom has one electron localized in its atomic 1s orbital. In the absence of magnetic fields it will not matter whether the spins of the electrons are parallel or antiparallel and ...


23

Disclaimer My following answer is the "traditional" explanation of Hund's first rule, which is based on a smaller value of $V_\mathrm{ee}$ (electron-electron repulsions) in the triplet state arising from Fermi holes. According to Levine's Quantum Chemistry 7th ed.: This traditional explanation turns out to be wrong in most cases. It is true that the ...


22

The relative energies of the electronic subshells have been calculated for atoms in the vicinity of $Z=20$ (J. Chem. Educ., 1994, 71 (6), 469), and the result is surprising: Looking at this graph, by all means the electronic configuration of scandium should in fact be $\ce{1s^2 2s^2 2p^6 3s^2 3p^6}$ $\color{blue}{\ce{3d^3}}$ in order to minimize orbital ...


22

The pattern of maximum possible electrons = $2n^2$ is correct. Also, note that Brian's answer is good and takes a different approach. Have you learned about quantum numbers yet? If not... Each shell (or energy level) has some number of subshells, which describe the types of atomic orbitals available to electrons in that subshell. For example, the $s$ ...


22

Disclaimer: I now believe this answer to be fully incorrect. Please consider un-upvoting it and/or downvoting it. I do not like seeing incorrect answers at +22. However, I will leave it up for now. It is a reflection of what is taught in many undergraduate-level textbooks or courses. However, there have been criticisms of this particular graph in ...


21

This is just a confirmation to Aesin's answer... Say, we take copper. The expected electronic configuration (as we blindly fill the d-orbitals along the period) is $\ce{[Ar] 3d^9 4s^2}$, whereas the real configuration is $\ce{[Ar] 3d^{10} 4s^1}$. There is a famous interpretation for this, that d-orbitals are more stable when half-filled and completely-...


21

Atomic copper has the electron configuration $\ce{[Ar] 3d^{10} 4s^1}$. By removing one electron and producing $\ce{Cu^{+1}}$, an inert gas configuration $\ce{[Ar] 3d^{10} 4s^0}$ is produced. While it does take a lot more energy to remove the second electron from copper (first IP=745 kJ/mol, second IP=1,958 kJ/mol), if this energy can be offset by the ...


21

Yes and no. Elements are defined by the number of protons only. It does not matter if (say) a carbon nucleus has six or seven (or eight) neutrons, they will all react the same.* With that, to create new elements, you would need to get up to some 115 or so protons fused together. But there is a reason for neutrons: all the positively charged protons in the ...


20

First off, gold does react. You can form stable gold alloys and gold compounds. It's just hard, mostly for reasons explained by the other answer The reason bulk gold solid is largely unreactive is because the electrons in gold fall at energies which few molecules or chemicals match (i.e., due to relativistic effects). A nice summary of some work by Jens K. ...


20

s, p, d, f and so on are the names given to the orbitals that hold the electrons in atoms. These orbitals have different shapes (e.g. electron density distributions in space) and energies (e.g. 1s is lower energy than 2s which is lower energy than 3s; 2s is lower energy than 2p). (image source) So for example, a hydrogen atom with one electron would be ...


20

Full Coupled Cluster (FCC) vs. Full Configuration Interaction (FCI) The main theoretical difference is the way excitations are used. "Excitation" refers to putting one or more electrons in higher orbitals than the "reference" calculation, which is often a Hartree-Fock calculation, would. CI uses a "simple", linear excitation operator, whereas CC uses a more ...


19

I have been taught that the MO diagram is different for molecules with 14 or less electrons than the one used for molecules with 15 or more electrons. This is (partly) wrong because the change in the order of $\mathrm{\sigma_{2p_{z}}}$ and $\mathrm{\pi_{2p_{xy}}}$ MOs to the left of $\ce{N2}$ is not directly related to the number of electrons. Rather, ...


19

Jan's answer is correct. I will try to fill in a few details about why neutrons are essential to creating stable nuclei. All stable isotopes excepting Hydrogen-1 have neutrons in their nuclei. Hydrogen, for example has two stable isotopes: The first simply has a proton with no neutrons in the nucleus, while the second, often called Deuterium has a proton ...


18

This exception rule is actually orbital filling rule. For two electrons to be in same orbital they need to have different spins (Pauli exclusion principal). This electron pairing requires additional energy and thus it is easier to add electrons if there are free orbitals. When element has a half-filled p sublevel all 3 orbitals have one electron and pairing ...


18

There are 3 types of octet rule "violations" or exceptions molecules with an odd number of electrons, such as nitric oxide (image source) molecules with less than 8 electrons around an atom, $\ce{BeCl2}$ and $\ce{BH3}$ serve as examples (image source) molecules with more than 8 electrons around an atom, such as $\ce{PCl5}$ or $\ce{SF6}$ Take a look at ...


17

Alas, the problem of course is that you need both position q and momentum p to get an accurate picture of the overall electron state. This is not true in quantum mechanics; it is sufficient to characterize the wavefunction $\left\langle x |\psi\right\rangle$ in position space or $\left\langle k|\psi\right\rangle$ in momentum space. Which is chosen is ...


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