# Tag Info

56

I think your question implicates another question (which is also mentioned in some comments here), namely: Why are all energy eigenvalues of states with a different angular momentum quantum number $\ell$ but with the same principal quantum number $n$ (e.g., $\mathrm{3s}$, $\mathrm{3p}$, $\mathrm{3d}$) degenerate in the hydrogen atom but non-degenerate in ...

47

That's a good, concise statement of Bent's rule. Of course we could have just as correctly said that p character tends to concentrate in orbitals directed at electronegative elements. We'll use this latter phrasing when we examine methyl fluoride below. But first, let's expand on the definition a bit so that it is clear to all. Bent's rule speaks to the ...

38

General chemistry textbooks tend to explain atomic structure exceedingly poorly using a hodgepodge of obsolete concepts. Your chemistry book provides such a typical example - the notion of penetration only makes sense in the ancient Bohr-Sommerfeld model that has been obsolete since the discovery of quantum mechanics! The idea was that orbits of electrons in ...

37

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 $... 35 I have searched and searched, oh how I have searched. Do you know what I always tell my mom when she asks me to find something in the Internet she was not able to find herself? I ask her: "Are you sure that the thing you are looking for even exists?" I am looking for a 3 dimensional visualization of a whole (moderately complex, hydrogen is just a ball) ... 33 The inert pair effect describes the preference of late p-block elements (elements of the 3rd to 6th main group, starting from the 4th period but getting really important for elements from the 6th period onward) to form ions whose oxidation state is 2 less than the group valency. So much for the phenomenological part. But what's the reason for this ... 30 When people say that Kohn-Sham orbitals bear no physical meaning, they mean it in the sense that nobody has proved mathematically that they mean anything. However, it has been empirically observed that many times, Kohn-Sham orbitals often do look very much like Hartree-Fock orbitals, which do have accepted physical interpretations in molecular orbital theory.... 29 The meaning of covalent bonds being directional is that atoms bonded covalently prefer specific orientations in space relative to one another. As a result, molecules in which atoms are bonded covalently have definite shapes. The reason for this directionality is that covalent bonds are formed by sharing electrons between atoms, or, in other words, as you ... 27 If you have$n$functions (e.g. AOs) you can make a maximum of$n$new linearly independent functions (e.g. MOs). If you try to make$n+1$MOs, then any one of them can be expressed as a linear combination of the other$n$MOs. The most usual way to enforce linear independence is to enforce orthogonality, i.e. all your MOs have to have zero net overlap ... 26 General case There is indeed a mathematical theorem that deals with the number of nodes an eigenfunction corresponding to a certain eigenvalue can possess. It was laid down by Courant$^{[1, 2]}$and it states the following: Given the self-adjoint second order (partial) differential equation \left(\hat{L} + \lambda \rho(\mathbf{x}) \... 26 Unfortunately, the sense in which orbitals are orthogonal is more or less impossible to define rigorously without recourse to functions of some kind. So, I'll give an explanation a shot using some simple, 1-D functions to illustrate the concept, followed by the pictorial orbital example you've asked for. At a basic level, in order to have any two functions ... 26 Omitting j when alphabetically enumerating things has a long tradition. First of all, the alphabet did not always exist in the form we know it today. Quoting Wikipedia: After [...] the 1st century BC, Latin adopted the Greek letters ⟨Y⟩ and ⟨Z⟩ [...] Thus it was during the classical Latin period that the Latin alphabet contained 23 letters: [no J, ... 24 As I understand this, there are basically two effects at work here. When you populate an$\mathrm{s}$-orbital, you add a significant amount of electron density close to the nucleus. This screens the attractive charge of the nucleus from the$\mathrm{d}$-orbitals, making them higher in energy (and more radially diffuse). The difference in energy between ... 24 The answer is... it is not so simple. Some quantum mechanics follow, but the TL;DR version is that while$m_l=0$corresponds to$p_z$, the orbitals for$m_l=+1$and$m_l=-1$lie in the$xy$-plane, but not on the axes. The reason for this outcome is that the wavefunctions are usually formulated in spherical coordinates to make the maths easier, but graphs in ... 23 When combined at a given atomic center, any two atomic orbitals add in a vectorial manner. For example, consider the orbital$\phi$defined by$\ce{p_{x}}$and$\ce{p_{y}}atomic orbitals as \begin{align} \phi = c_1 \ce{p_{x}} + c_2 \ce{p_{y}} \end{align} The orbital addition can be pictured like this for the two casesc_1 = c_2 > 0$and$c_l = -c_2 ...

22

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

22

For the azimuthal quantum number (l) of an atom, there is no "j" because some languages do not distinguish between the letters "i" and "j". L is the total orbital quantum number in spectroscopic notation and uses capital letters. The nomenclature just follows suit with the suborbital notation and skips J since there is no corresponding j.

21

The wavefunction of a particle actually has no physical interpretation to it until an operator is applied to it such as the Hamiltonian operator, or if you square it which gives its probability of being at a certain place. So having a negative wavefunction doesn't mean anything physically. However, let's say for a particle in a box, if you solve the momentum ...

21

NOTE: In the below, I'm implicitly discussing a ground-state, closed-shell wavefunction, where all occupied orbitals are doubly occupied. The discussion would be similar for open-shell wavefunctions, but there are complexities that I won't address here. Also, once one starts noodling at excited states, things get complicated pretty quickly, and (AFAIK) some ...

20

It seems like it should be the average distance that matters No. It is the average energy that matters. Note that this stuff about "spends so much time here and so much there..." is really just a (not particularly good) way of describing a quantum-mechanical wave function's absolute square. The electron is actually never at any particular place in the ...

20

In fact I find a more simple reasoning with resonance structure. When phenol loses the $\ce{H+}$ the phenolate ion is stabilized due to the resonance effect, as shown below: The energy of the dissociated form is lower and so phenol has more chance to be in the solution dissociated with the phenolate ion. Aliphatic alcohols are not stabilized by resonance so ...

20

Let me approach this another way than the others: orbitals are NOT physical objects! They do not exist in physical sense, they are theoretical constructs, chemical concepts that help understand / visualize / etc. mathematical solutions of Schrodinger / Dirac / Kohn–Sham / etc. equations. Orbitals are not unique: given linear combinations are ...

20

First, you have to know the geometry of your compound. The complex $\ce{[PtCl4]^2-}$, for example, is square planar. The next step is to determine the point group of the compound from this geometry by identifying the symmetry elements that it posesses (a procedure on how to do this can be found here). The point group of the complex $\ce{[PtCl4]^2-}$, for ...

20

As orthocresol mentioned, this is all about relativity, so let's talk about it. I am hardly an expert myself, but I'll try to give an answer to the best of my limited knowledge. For an interesting and accessible overview of the incorporation of relativistic effects into chemistry, I recommend the 2012 review article "Relativistic Effects in Chemistry: More ...

19

tl;dr The next in the series is called φ bond. There is even a tiny Wikipedia article about it. Nicolau pointed me to the Wikipedia article, that had at the time a tiny section about the φ symmetry of the bond. Ben also kindly agreed with my naming proposition. I'd like to back up just a little bit an quote one sentence of this article: The type ...

18

First, this isn't quite true. It is true for the first row of the periodic chart (from lithium to neon). It is almost true for the second row (from sodium to argon. But there are exceptions there. Beyond that it really isn't true at all for the elements beyond the first two columns. The reason for the increased stability for the first two rows lies in ...

18

This is a complex issue, particularly because people often like to think in terms of an indepedent-particle picture (i.e. the aufbau filling up orbitals), even though the exact many-body wavefunction has strong electron-electron correlations. So let me rephrase your question: What is the relationship between the KS eigenfunctions and the exact many-body ...

18

In the definition of an s shell, you will find that its $\ell$ number is zero. In classical terms, that corresponds to an orbit with zero orbital or angular momentum — which for a large object is a clear impossibility. For an electron, it gives the peculiar result that any electron in any s shell is, classically speaking, moving back and forth through the ...

18

1. How to get the number and type of nodes for an orbital As you said, nodes are points of zero electron density. From the principal quantum number $n$ and the azimuthal quantum number $\ell$, you can derive the number of nodes, and how many of them are radial and angular. $$\text{number of nodes}=n-1$$ $$\text{angular nodes}=\ell$$ \text{radial nodes}=...

18

The effect is indeed amazing, if you compare the $\mathrm{p}K_\mathrm{a}$ of tert-butanol (17.0) with that of phenol (9.95). Deprotonation is facilitated when the reaction goes downhill (energywise). In order to obtain stabilization of the anion, the negative charge needs to be distributed over a larger number of centres. This distribution (only) is ...

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