48

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 ...


44

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 $...


40

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 $\ce{K2ReH9}$. $\...


33

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 ...


31

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 ...


26

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 $s$ ...


26

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 ...


25

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 ...


24

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 ...


23

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. ...


23

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 ...


23

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, ...


23

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 ...


22

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

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 ...


21

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 ...


20

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 ...


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 ...


20

This is because we live in a world dominated by oxygen and water. In other words, it is an oxidized world. Most metals occur naturally in the form of oxides, silicates, halides, or other derivatives. Hydrogen occurs as $\ce{H+}$. In a hypothetical world dominated by metals, all that could have turned out otherwise. Oxygen would be a scarcity, and would come ...


19

I think it is important to understand that for hydrogen atom (or any other one-electron system) all orbitals from the same shell have same energy. For instance, $E_\mathrm{2s} = E_\mathrm{2p}$, $E_\mathrm{3s} = E_\mathrm{3p} = E_\mathrm{3d}$, etc. Thus, The first excited state of hydrogen atom would be one in which either $\mathrm{2s}$ or one of the three $\...


18

The lowest energy state has parallel spins to maximize the exchange energy. As you say, there's a Coulomb repulsion between two electrons to put them in the same orbital. There's also a quantum mechanical effect. The exchange energy (which is favorable) increases with the number of possible exchanges between electrons with the same spin and energy. Going ...


17

This is yet another interesting effect of the anomalous compactness of orbitals in the first appearance of each type of subshell ($1s$, $2p$, $3d$, $4f$, $5g$, etc). The solutions to the Schrödinger equation for electron wavefunctions in hydrogen-like atoms are such that these subshells are composed of orbitals with no radial nodes. This means the electrons ...


17

Relativistic effects account for gold's lack of reactivity. Gold has a heavy enough nucleus that its electrons must travel at speeds nearing the speed of light to prevent them from falling into the nucleus. This relativistic effect applies to those orbitals that have appreciable density at the nucleus, such as s and p orbitals. These relativistic ...


17

Surprisingly, I learned that there are also usages for orbitals g,h,i and even j. Actually, the letter "j" is not used, so it is s, p, d, f, g, h, i, k, l, etc. The higher angular momentum orbitals do enter the domain of science, due to excited states of atoms. Transitions to and from excited states are observable through atomic spectroscopy. For ...


17

Out there in the real world, university students and school pupils alike favour strict rules that are true as often as possible (or at least have clear, easily remembered exceptions), while their professors and teachers on the other hand try to tell them that these rules are merely to be considered ‘guidelines’ rather than actual rules (and attempt to ...


17

The normal distinction between "steric" and "electronic" is based on whether the effect is transmitted through space or through bonds All the normal physical interactions we experience are arguably electronic. When you touch your desk, you feel force because of interactions between the molecules of the desk and the molecules of your hand ...


16

There has been quite some interesting work by my former co-worker and my supervisor on metal-rich molecule with the co-ordination higher than eight. For more see Timo Bollermann, Thomas Cadenbach, Christian Gemel, Moritz von Hopffgarten, Gernot Frenking, Roland A. Fischer, Chem. Eur. J., 2010, 16 (45), 13372-13384. Also available through researchgate.com. ...


16

As stated in this answer, these are irrep (irreducible representation) labels for molecular symmetry point groups. In the context of chemistry, point groups are usually introduced when learning about structural symmetry (atoms and bonds). This is a broad topic with many technical points, too many for a single answer, so I won't cover the basics but hopefully ...


15

What you say is correct. The [Ar] configuration we are left with does have 8 valence electrons. But I think it is just semantics. Elemental potassium has an [Ar] 4s1 electron configuration. One would say it has one valence electron. If we take that one valence electron away, it makes sense to say that it now has zero valence electrons because "1 - 1 = ...


15

Usually when adding electrons based on the Aufbau principle, you go from one element to the next highest one, e.g. from $\ce{Ti}: \ce{[Ar] 4s^2 3d^2}$ to $\ce{V: [Ar] 4s^2 3d^3}$. Thus you add not only an electron but also a proton to your atom. When you remove electrons to get to a cation, you only remove electrons. Thus it is a different situation, with ...


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