# Tag Info

7

The configuration for the free atoms is: Ba - $\mathrm{[Xe]\ 6s^2}$ Br - $\mathrm{[Ar]\ 3d^{10} 4s^2 4p^5}$ S - $\mathrm{[Ne]\ 3s^2 3p^4}$ Si - $\mathrm{[Ne]\ 3s^2 3p^2}$ Thus: Barium has no unpaired electrons. Bromine has one unpaired electron in 4p subshell. Sulfur has two unpaired electrons in the 3p subshell. Silicon has two unpaired electrons in the ...

7

For p-orbitals $i$ and $j$ which are orthogonal to each other (i.e. different colour in your diagram), the off-diagonal matrix elements are simply zero: $$\langle i | H | j \rangle = 0$$ If we ignore the overlap matrix,* the eigenvalue equation to solve therefore has the form: $$\mathbf{Hc} = E\mathbf{c},$$ or in explicit form, using the fact that the matrix ...

6

Based on available theoretical considerations and a available literature, a tetrahedral geometry appears to be a good assumption for $\ce{Cr(NO)4}$. The NO ligand can be rendered as a three-electron donor, if we take it as uncharged. It is actually similar to CO, really, in that it interacts through its $\pi^*$ orbitals as well as its $\sigma$ orbital; but ...

6

I agree in part with Mithoron and MaxW's comments: these are different molecules, not even sharing a mutual atom (e.g. C-O vs C-C), so direct comparison is restricted. However, Pauling's concept of electronegativity does explain why, in general, the heteronuclear A-B bond is stronger than the average of the homonuclear A-A and B-B bonds. the difference in ...

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 ...

5

Not a full answer, but some general thoughts: In general, what we can learn from individual orbitals (and their energies) is very limited, since orbitals are just an abstract tool to represent the many particle electronic wave function. This is manifested by the need for methods beyond Hartree-Fock (DFT, post-HF, etc) and the fact that the orbitals in these ...

5

The answer technically depends on the method. In most cases, meaning HF or DFT with standard basis sets, it is infinite separation of all bare nuclei and all electrons. One exception known to me is VASP (a periodic, solid-state code), which uses pseudopotentials to avoid having to model the core singularity and core electrons. There, the zero-point is in ...

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

Powerful electron withdrawing substituents, like $\ce{-NO2}$ can be interpreted as having a large effect on the $\pu{\pi}$-system of an aromatic ring. Resonance structures can be drawn that put a positive charge at the ortho and para positions and put the electron out in the nitro group, on an oxygen. This must be adjusted for $\ce{-N(CH3)3+}$, because it ...

4

Hückel theory would still work if there would be overlap. In fact, it is possible to create a completely new basis by applying a transformation to the basis set by which a new set of orthogonal basis functions is created. The advantage of setting $\hat{S}$ (the overlap matrix) to $\hat{I}$ (the identity matrix) is that the eigenvectors of the matrix ...

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

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 ...

3

Choices B and C are two different ways of saying the same thing. Say $E_{MO}(A)$ and $E_{MO}(AB)$ are the energies of electrons in bonding and anti-bonding orbitals, respectively, relative to the electrons in the original atomic orbitals from which the molecular orbitals are formed, or, to make matters simple, set the energy of electrons in the original ...

3

There is a significant portion of s-p mixing in dinitrogen. Please see this great answer by Wildcat on Molecular orbital (MO) diagram for $\ce{N2}$ and $\ce{N2^-}$. For the impatient, I'll quote an image from myself from How to rationalise with MO theory that CO is a two-electron donor through carbon? I have marked these interactions with the dotted lines. ...

3

The question is so good that maybe it deserves another correct answer. While my answer above attempts to answer using the criterion mentioned by user57048 (hyperconjugation), hyperconjugation is generally a small effect, but the $-I$ effect of $-N(CH3)3+$ is so large that maybe an entirely different approach could be appealing - not that you could ...

3

In the linear $C_{\infty \mathrm{v}}$ and $D_{\infty \mathrm{h}}$ point groups there are two notations for the irreducible representations which are equivalent, in that \begin{align} \mathrm{A_1} &\equiv \Sigma^+ \\ \mathrm{A_2} &\equiv \Sigma^- \\ \mathrm{E_1} &\equiv \Pi \\ \mathrm{E_2} &\equiv \Delta \\ \mathrm{E_3} &\equiv \Phi \\ &... 3 I think you are close to the right idea, but let me try to clarify some points. The full Schrodinger equation for a molecule should depend on the 3N nuclear coordinates, as well as the 3n electronic coordinates, where N and n are the number of nuclei and electrons respectively. This is challenging both due to the number of coordinates and, in ... 3 First, these Coulombic interaction integrals can be simplified to  \langle \mu\nu|\frac{1}{r_{12}}|\lambda \sigma \rangle , which is the common two electron integral in the AO basis (your atom-centered gaussian basis functions). You are correct that energies are only well defined for eigenfunctions of the Hamiltonian, and thus basis set functions (the ... 3 Bond order is the number of electrons shared between two atoms divided by two. There are a few things that limit how high a bond order can possibly go, however. First, atoms can usually only form bonds until their valence electron shells are filled (any more electron would be unstable). The heaviest elements, those of periods 6 and 7, have at most 32 valence ... 3 As correctly noted in the question, \ce{CO} and \ce{N2} are iso-electronic, so they should be comparable. But a difference remains: \ce{CO} is hetero-nuclear, \ce{N2} is homo-nuclear. The latter case typically means that there is better overlap between the orbitals (because they are of similar extent/energy). This translates into a greater gap ... 3 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 ... 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: ... 2 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 ... 2 The number 6 relates to a central atom in a molecule with octahedral geometry such as sulfur hexafluoride. \ce{SF6} (image from wikipedia) A central atom with 8 other atoms does occur in ionic crystals (body centered cubic), but not in any molecules to my knowledge (memory...). (image from website location) 2 To show that your two functions do represent the \mathrm{E} irreducible representation, we can approach the problem algebraically. We start by representing the operators of the C_{3v} point group in the basis of the AOs. The identity element is easy in any basis:E=\pmatrix{1 &0 &0\\0 &1 &0\\0 &0 &1\\} We can see that ...

2

A cousin of mine asked me to share her answer here: In the molecular orbital diagrams shown in textbooks, the highest occupied molecular orbitals of $\ce{O2}$ and of $\ce{F2}$ are shown as antibonding. If the molecules and their cations had the same bond length, and the electrons in a species didn't "talk to each other", and the orbital energies ...

2

Such questions troubled physicists in the early 20th century. Their inability to add more than one electron to the planetary Bohr-model was one of the reasons that quantum mechanics was developed. Nevertheless, continuing to ask this type of questions may help us sharpen our thinking and develop our intuition, as far as it goes (you might also take a look at ...

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

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 ...

1

When a test question is so vague that you can't figure out what answer is best, it's difficult to formulate a question about the answers to the question. The test question specifies bonding electrons equal to antibonding electrons. No energy statements. There needs to be one more answer: E. All of the above, although some answers are a little better than ...

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