# Tag Info

17

What does MO formation entail? A very common misconception is that the formation of MOs involve addition, or subtraction, of two physical objects. And the confusion arises because, how can the same physical object have two different phases when added or subtracted? However, this is not an accurate picture. MO construction is not a physical process which ...

15

Unfortunately, in quantum mechanics there is rarely an explanation "in simple terms." Quantum mechanics is a mathematical construct that so far seems to predict the results of all experiments that have been done to test it, but explaining those maths in terms of anything we have learned from the larger-than-quantum scale "classical" ...

6

You have to consider the system as a whole - you can't directly compare $\ce{O_2}$ and $\ce{O_2^2+}$ because they have different numbers of particles. To put it another way when you consider the relative stability of two interconvertible specifies you really have to write down a chemical reaction that connects them, and then consider which direction is ...

6

What is the endo-selectivity of Diels-Alder reactions? If the diene used in the Diels-Alder reaction has asymmetric substituents at the end carbons, and if the dienophile is unsymmetric, then two different isomers of the final adduct can form. The isomer, where the functional group(s) (usually carbonyl) on the alkene end up on the same side as the newly ...

6

The very question you pose is addressed thoroughly in this open access work: https://www.nature.com/articles/ncomms9287 . The short answer to your question is that the electron density can be mapped using a technique akin to diffraction as described in their work. You mentioned a distinction between the electron density and the orbitals, and the orbital is ...

6

What you have calculated is the energy required to make two separated ions out of two separated $\ce{Na}$ ad $\ce{Cl}$ atoms. When they are separated, these ions do not make a molecule or a crystal of salt. Separated ions attract one another and are able to release a huge amount of Coulomb energy when approaching each other. Anyhow your calculation should ...

5

Anti-aromaticity is often not taught very clearly. Let me start, then, by emphasising that this anti-aromatic diradical state should not be taken as a real thing. It is a purely hypothetical state that may arise if the molecule adopted the shape of a planar regular polygon (i.e. square for $\ce{C4H4}$, octagon for $\ce{C8H8}$). Because of various reasons, ...

4

I assume you mean from the classical electrodynamics side only, not from quantum electrodynamics side. By the former, even H atom cannot exist, as the electron would fall along a spiral curve on the nucleus, continually emitting radiation being radially accelerated. For a hydrogen molecule, both electrons move around both protons, they are not dedicated to ...

4

Why are hydrogen bonds in an antiparallel beta sheet stronger than those in parallel beta sheets? They are probably not. The difference is small, and depends on sequence context. Also, the diagrams do not reflect the typical conformation of the backbone in beta sheets. To complicate matters, most sheets are mixed rather than purely parallel or anti-parallel,...

4

First of all, how can a "half" sigma bond exist? Usually, you expect double bonds to be shorter and stronger than the corresponding single bonds, and triple bonds even shorter and stronger. The OP already mentioned bond-orders of 1.5 that occur for conjugate double bond systems, and those have properties in between single and double bonds. Perhaps ...

4

Yes, they result from the HF equations just like the occupied orbitals. The diagram should also be show them as present in the initial set of equations, unless the basis (used for the LCAO) was changed between the two states shown. The number of occupied orbitals plus the number of virtual orbitals is equal to the basis set size because they are the ...

4

In multielectronic atoms we have a relatively large difference between $s$ and $d$ orbitals when they have the same $n$ quantum number, or in terms of the actual quantum mechanics when they have the same total number of nodes in the wavefunction (this being what we label as $n-1$). But in the case of transition metals the $s$ orbital that mixes in with the $... 4 You can visualize isosurfaces of the molecular orbitals in gview. This allows you to identify the orbitals visually. That is a way to determine that the HOMO is a lone p-orbital of the oxygen atom. I think that you have to load the *.chk file with gview and not the *.log file, if you want to plot isosurfaces since the required information is saved there. ... 3 The way to tackle this is to look at the lone pair repulsions between the 2 molecules. It is known that$\ce{N-N};\ce{O-O};\ce{F-F}$single bonds are quite unstable due to lone pair (lp) repulsions. An evidence of this is that$\ce{N}$exists in molecular state as$\ce{N2}$using multiple bonds, so that it's lone pairs do not repel each other, however$...

3

It's all about the Pauli exclusion principle. Assume the available MOs in each of two approaching molecules are fixed, and that each molecule is in a closed shell configuration (all electrons are paired). The electrons in closed shell systems cannot accept additional electrons that share principal, azimuthal and magnetic quantum numbers, since all the spin ...

3

I disagree with your statement that "The dxz and dyz don't seem to have the correct symmetry to sigma bond with any p orbitals of the ligands. They are nonbonding in a perfect octahedron" Consider the case of a halide ligand like Cl-, which has filled p orbitals perpendicular to the axis of the M-Cl sigma bond. If these are aligned with the ...

3

There are many ways in which bond lengths can be changed, and these will be accompanied by changes in bond energy. I will give examples for four different classes of substances, though they may not all count as "forcing" the bonds to be shorter in the way in which you mean. You can look these over, and perhaps clarify your question: Glasses: ...

3

This question requires idea of the Linear Combination of Atomic Orbitals (LCAO). It involves the following equation: $\psi_n = \sum\limits_{i} c_{ni}\phi_{i} = c_{n1}\phi_{1} + c_{n2}\phi_{2} + ... c_{ni}\phi_{i}$ Here $\psi_n$ represents the wavefunction for the resulting molecular orbital, $\phi_{i}$ represents the atomic orbitals that contribute to the ...

3

The discussion surrounding SF6 is often centered around two opposing hypotheses: (1) Hypothesis # 1: SF6 actually obeys the octet rule, because the sulfur atom has a net positive charge and some of the S-F bonds are ionic in nature. For example, four single S-F bonds and two ionic S - F bonds. The six equal S-F bond lengths are then explained as a resonance ...

3

You can actually calculate orbital "energies" with MO theory. The resulting "energies" for $\ce{N2}$ are the following (calculated with molpro and CASSCF(10,8)/aug-cc-pVTZ): Orbital "energies" in atomic units orbital energy/a.u. 1σg -1.11 2σg -0.99 1σu -0.77 πu -0.59 πg 0.29 2σu 1.22 As you can see, the "energy"...

3

The short answer is that $\sum |c|^2$ is basically equal to "the number of electrons in the orbital". You might think that this should be 2, not 1. But note that each molecular orbital actually comprises two different "spin orbitals". Each "spin orbital" contains one electron of a particular spin - so there is a spin-up orbital ...

3

As $$\gamma = \dfrac{c_A}{c_B}$$ when $c_A=0$, $\gamma$ must also be zero. This follows as $c_B$ can not be also be zero as the wavefunction is normalised to unity. Thus under such conditions it is invalid to divide through by $\gamma$. Instead simply use the normalisation condition and the fact that $c_A$ is zero to show that $|c_B|=1$ .

3

Using the normalisation condition $$c^2_A + c_B^2 + 2c_A c_B S = 1$$ while $c_A = 0$ $$0 + c^2_B + 2 \cdot 0 \cdot c_B S = 1$$ $$c^2_B = 1$$ $$c_B = \sqrt{1} = \pm 1$$

3

Your problem seems to stem from confusing the number of linear independent basis functions(which is the same number as the size of the atomic orbital basis with which we started) and the number of all possible functions that can be built using this basis. Just as in a two dimensional vector space, where you have a maximum of two independent basis functions, ...

2

The confusion seems to arise from the incorrect statement that the S-O bond has pi character and that there exist $\pi$ and $\pi^*$ molecular orbitals in DMSO. Instead, DMSO (despite the way it is often drawn) is most accurately depicted as having three single bonds and one lone pair on the the S atom (clearly visible as a dominant element in the HOMO shown ...

2

In the strictest sense, the orbital energies are simply eigenvalues of the multi-electron Schrödinger equation; unfortunately, the multi-electron Hamiltonian takes a nasty form and is impossible for us to solve exactly, so we can't get any insight on the energies from that approach... Instead we inadvertently resort to LCAO-based methods, whereby exploiting ...

2

Atomic orbitals are not the same as molecular orbitals You seem to be assuming that the electron orbitals of isolated atoms are the same as the orbitals that exist in stable molecules. So the issue is which orbitals get filled to create the lowest energy configuration. But the quantum mechanics of orbitals is a lot more complex than that. The orbitals that ...

2

To recognize the more stable molecule you may extend the description with bond orders is the one considering how the mathematical concept to mix atomic orbitals by LCAO and visualizing the results in Molecular orbital diagrams and compute the overall energy of such a molecule. Eventually, you compare the total energy of $\ce{O^-_2}$ with $\ce{C^+_2}$ using ...

2

We are not trying to maximize overlap in this diagram, quite the opposite. You can imagine it rather as pushing negative charges, the negative ligands ions, unto the electrons on the central atom while looking at the orbital energies during this process. That is a repulsive interaction between two negative charges, that raises the energy of orbitals that ...

2

The atomic orbital is a mathematical model that tries to explain the behaviour of electrons around the nucleus of an atom. Like any other theory explaining the behaviours of sub-atomic particles, this is purely based on experimental data and mathematics. No one has actually seen how an electron looks like because it is simply not possible to 'see' sub-atomic ...

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