# Tag Info

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 ...

21

It depends on what you mean by "spin". If you mean "have intrinsic internal angular momentum, independent of its trajectory through space", then yes, electrons spin, and that's what the quantum number is measuring. Though if by "spin" you mean "undergoes rotation" ("there's a little billiard ball, and if I were to put a mark on it and watch it, the mark ...

18

Technical Note: This page makes heavy use of MathJax, give it time to load. $%some shortcuts \newcommand{\op}[1]{\mathbf{#1}} \newcommand{\ve}[1]{\mathbf{#1}} \newcommand{\id}[1]{\mathrm{#1}} \newcommand{\bra}[1]{\left\langle#1\right|} \newcommand{\ket}[1]{\left|#1\right\rangle} \newcommand{\bracket}[2]{\left\langle#1\middle|#2\right\rangle} \newcommand{... 17 The lowest energy state has parallel spins to maximize the exchange energy. As you say, there's a Coulomb repulsion between two electrons to put them in the same orbital. There's also a quantum mechanical effect. The exchange energy (which is favorable) increases with the number of possible exchanges between electrons with the same spin and energy. Going ... 12 The effect is due to the symmetry properties of the rotational energy levels and those of the nuclear spin. The change with temperature is, as usual, traced back to the Boltzmann distribution. In homonuclear diatomic molecules such as$\ce{H2}$it is necessary to consider the spin of the overall wavefunction to exchange of nuclei. If this is$\Psi = \psi_t\...

11

I have seen multiplicity 13 referred to as simply a 13-let. This #-let nomenclature is sometimes used for high multiplicities (e.g. higher than 10).

10

Despite the success of the Schrödinger Equation in predicting energy levels of the hydrogen atom, experimental observations suggest that it doesn't tell the whole story of electron behavior in atoms. Firstly, spectroscopists observed "anomalous" splitting in the previously studied "normal" Zeeman effect, sometimes with unequal spacing. Secondly, some ...

10

Some understanding can be gained by looking at the symmetry of the orbital parts of the wavefunctions involved. The total wavefunction for electrons must be anti-symmetric with respect to exchanging electrons, the Pauli Principle. The total wavefunction has spin and spatial parts, $\psi =\psi _{space}\psi _{spin}$. When electrons are paired with total ...

10

As Martin has mentioned, carbenes are a good starting point if you are looking for organic compounds with a triplet ground state. In these carbenes, the HOMO is not twofold degenerate, as your question suggests. Instead, there are two singly occupied orbitals with different energy. The energy gap, though, is typically small, and since there is an energy ...

10

Atoms Atomic carbon with its $\mathrm{1s^2 2s^2 2p^2}$ configuration has a triplet ground state ($S = 1$), precisely because of Hund's first rule. However, in the context of the Stern–Gerlach experiment, you might run into a problem with orbital angular momentum, as carbon's ground state also has nonzero orbital angular momentum ($^3\mathrm{P}$ ground ...

10

I'm not aware of the Russell–Saunders effect, but the Russell–Saunders coupling scheme is definitely a thing. As you noted, the Wikipedia page on "spin-orbit interaction" doesn't talk about it, but a different Wikipedia page does, and basically tells you the same thing as I will. The answer is... yes and no. The word "coupling" refers to ...

10

It is very tempting (and often also very useful!) to picture electron spin as an angular momentum vector, similar to a spinning top. Using this analogy, there are two properties (or numbers) of this angular momentum vector that we need in order to describe the electron spin. The first one is the spin itself and this is often designated the symbol $s$. The ...

8

You assumed that coupling to the three chlorides would yield some type of quartet. This is correct in principle. However, chlorine is one of the many quadrupole nuclei that are basically unobservable by NMR due to their rapid relaxation. I hope another answer is around explaining that as I am not good at it. I can, however, help you interpret the $1:1:1$ ...

8

There are various examples of molecules in triplet ground state. The smallest (organic) is possibly methylene, $\ce{:\!CH2}$. Carbenes in general may adopt triplet state as their ground states. Other possibilities are biradicals (From the IUPAC gold book): An even-electron molecular entity with two (possibly delocalized) radical centres which act nearly ...

8

Triplet oxygen has two unpaired electrons with the same spin, and a total spin value of 1. In fact, by Hund's rule, the triplet states are preferred over the singlet states which have two electrons with opposite spins. Further reading: https://en.wikipedia.org/wiki/Triplet_oxygen https://en.wikipedia.org/wiki/Singlet_oxygen

8

The first question is: What is the mechanism by which spin isomers of hydrogen switch between the ortho and para forms? There is some explanation in ChemPhysChem 2006, 7 (3), 551–554 (non-paywall version here): One can define three situations. In the first, a magnetic conversion occurs without bond elongation or breaking. For example, in solid ...

8

Throughout your question it is clear that you are considering electrons as particles. This is not entirely incorrect; for example, the cathode ray tube is best explained by electrons being considered negatively charged particles that can travel through vacuum and be deflected by (electro)magnetic fields. However, particles as small as electrons can also be ...

7

If volume is understood classically, i.e. as the quantity of three-dimensional space enclosed by a closed surface, then quantum systems in general do not have a certain volume. One could, in principle, redefine the notion of volume for a quantum system to be, say, an area chosen so that there is a certain probability (for example, 90%) of finding the system ...

7

In the most common case you can only observe 3J-couplings, any spins further away than three bonds do not cause a visible splitting in your spectra. The exception are couplings via double-bonds or aromatic systems, there you can often see small couplings over four bonds. In this case the spins are too far away from each other to cause a visible splitting, ...

7

It is impossible. Having two electrons with the same spin in the same orbital is a violation of the Pauli exclusion principle.

7

I will go into the theoretical details first and come to the practical answer for your exact case later; separated from the theory by a horizontal rule. Approaching the complex theoretically, our first observation is that we are going to have six atoms coordinating our $\ce{Fe^3+}$ centre. Six ligands means that we should consider an octahedric environment ...

7

Maybe OP has already contacted the authors of this paper by now, but this was interesting. I'm no expert but I can Google things and this was too long for a comment. Antisymmetric exchange: At first I thought it was simply an exchange interaction where the wave function's sign is changed during exchange, now I don't think it's so simple. Antisymmetric ...

7

No it does not matter whether you start with 'spin up' or 'spin down' electrons as long as you are consistent: If you start filling some orbitals with 'spin up' electrons you will have to keep doing this for all the orbitals you want to fill. But: The convention is that you start with 'spin up' as this usually is taken to represent the $\frac{1}{2} \! \hbar$...

7

The "n+1" rule refers to a situation where you have a proton of type A with $n$ protons of type B next to it. Proton A's signal will be split $n+1$ times by the B's. However, none of the B's will split each other because they are equivalent.

6

Firstly I recommend you to understand the concept of "Magnetic Susceptibility". My source for this is Miessler,Tarr Inorganic Chemistry Book 2nd edition. Consider Ni2+, its electronic structure is 4s03d8. This is a d8 ion. Then it has 8 electrons in d orbitals. d orbitals have 5 degenerate level of orbitals. Also you should know gyromagnetic ratio which ...

6

Commuting operators indeed admit a set of simultaneous eigenfunctions, and since the non-relativistic electronic Hamiltonian commutes with total spin operators $\hat{S}_{z}$ and $\hat{S}^{2}$, the exact non-relativistic electronic wave function (which is an eigenfunction of the electronic Hamiltonian) is also an eigenfunction of the total spin operators. And ...

6

There are two things going on here. You have conflated with "forbidden" with unfavorable. In many systems, the first excited triplet is lower in energy than the first excited singlet. Hence, the order of energies excited singlet > triplet > ground state singlet is correct and sensible. If your question is "why is this the order?" that should be a separate ...

6

Here's what I believe they're trying to say. First, note that the discussion is mostly limited to high spin complexes (3 unpaired electrons out of the 7 d electrons), so the figure you reproduced is only representative of high spin states. Second, recall that when D > 0 (ie left side of figure), the five d orbital energy levels split so that we have three ...

5

Some peer review journal articles and an India high school textbook and various study guides do refer to the spin isomers as allotropes. There are physical differences such as different heat capacities. See Orthohydrogen, Parahydrogen and Heavy Hydrogen for comprehensive information. However, aside from the spin isomers, there are genuine allotropes ...

5

For starters, it is incorrect to think of the ground state being substantially more populated in NMR spectroscopy. Because the energy difference between aligned and misaligned nuclear spins in the external magnetic field is very low, we can say that both states are almost equally populated due to thermal excitation alone. For a $500~\mathrm{MHz}$ experiment, ...

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