38

Okay, this is not so much of an answer as it is a summary of my own progress on this topic after giving it some thought. I don't think it's a settled debate in the community yet, so I don't feel so much ashamed about it :) A few of the things worthy of note are: The bond energy found by the authors for this fourth bond is 13.2 kcal/mol, i.e. about 55 kJ/...


35

Starting point: 2s orbitals are lower in energy than 2p orbitals. The $\ce{H-N-H}$ bond angle in ammonia is around 107 degrees. Therefore, the nitrogen atom in ammonia is roughly $\ce{sp^3}$ hybridized and the 4 orbitals emanating from nitrogen (the orbitals used for the 3 bonds to hydrogen and for the lone pair of electrons to reside in) point generally ...


32

I'll present a LCAO-MO argument. But first let's debunk a myth. $\ce{SF6}$ has "hypervalent" sulfur and the 3d orbitals on sulfur participate in bonding No. This is not true. I would close one eye if it is taught in high school but really, this cannot be further from the truth. Here's one reference: J. Am. Chem. Soc. 1986, 108, 3586; there are many more. ...


28

Mathematical Explanation When examining the linear combination of atomic orbitals (LCAO) for the $\ce{H2+}$ molecular ion, we get two different energy levels, $E_+$ and $E_-$ depending on the coefficients of the atomic orbitals. The energies of the two different MO's are: $$\begin{align} E_+ &= E_\text{1s} + \frac{j_0}{R} - \frac{j' + k'}{1+S} \\ E_- &...


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 \begin{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 ...


25

Carbon cannot have more then 4 double-electron bonds in reasonable conditions. However, in can form a bond with 5 or 6 atoms, like $\ce{Fe6C}$ fragment, where iron atoms form octahedron around the carbon atom. However, the sum of orders of 6 $\ce{C-Fe}$ bonds will be still 4. The situation is different if we consider exited states. Indeed, it is possible ...


24

Here is a picture of a "classical" carbocation, there is an electron deficient carbon bearing a positive charge. There are many examples of "non-classical" carbocations, but the 2-norbornyl carbocation is among the best known. Labeling experiments have shown that the positive charge resides on more than one carbon in the 2-norbornyl ion. Early on, the ...


23

As stated in both the links Geoff and Philipp have kindly commented (1, 2) they are to do with symmetry labels we chemists like to assign to orbitals. Knowing an orbitals symmetry class can lead to a lot of simplifications down the road when you use quantum mechanical calculations and even dictate the reactivity of which orbitals are "allowed" to interact ...


23

The geometry of the complex changes going from $\ce{[NiCl4]^2-}$ to $\ce{[PdCl4]^2-}$. Clearly this cannot be due to any change in the ligand since it is the same in both cases. It is the other factor, the metal, that leads to the difference. Consider the splitting of the $\mathrm{d}$ orbitals in a generic $\mathrm{d^8}$ complex. If it were to adopt a ...


22

Unfortunately, nothing in the bonding situation in carbon monoxide is easily explained, especially not the dipole moment. According to the electronegativities of the elements, you would expect the partial positive charge to be at the carbon and a partial negative charge at oxygen. However, this is not the case, which can only be explained by molecular ...


21

First of all, are they correct? ChemBioDraw had some complaints, but as far as I can see there's the same amount of electrons, and no valence orbitals exceeding capacity. Yes, these are the six most important resonance structures for this compound. The reason ChemDraw complains is that it is trying to act smarter than you, and it most certainly is not. It ...


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 ...


19

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, ...


19

Very interesting question, and it kept me up despite daylight saving time cheating me of one hour of sleep last night... A good reference is Albright, Burdett and Whangbo, Orbital Interactions in Chemistry 2nd ed. pp 282ff. which explains this in much greater detail than I can. (In general, that book is fantastic.) I will try my best to summarise what they ...


18

The hypervalent carbon do exist. Look at this article: Kin-ya Akiba, et. al., J. Am. Chem. Soc., 2005, 127 (16), 5893–5901. X-ray measurements confirmed the 10-C-5 structure.


18

TL;DR VB theory treats atomic orbitals (including hybridized orbitals) as providing a good mathematical/physical description of the true form of the molecular wavefunction. MO theory uses atomic orbitals (with Gaussian radial functions) as a tool of computational convenience in an effort to define a molecular wavefunction that in its final form often bears ...


18

TL;DR "The destabilisation of an antibonding MO is equal to the stabilisation of a bonding MO, relative to the constituent AOs." Several key assumptions are inherent in this statement. Born–Oppenheimer approximation (not discussed further) "Orbital approximation": wavefunction of $\ce{H2, He2, ...}$ can be expressed as an antisymmetrised product of one-...


17

Carbon can make 5 bonds on methanium, $\ce{CH5+}$. It is a kind of carbonium ion and superacid. Methanium can be produced in the laboratory as a dilute and low-temperature gas. Reaction: $$\ce{CH4 + H+ -> CH5+}$$


17

Alas, the problem of course is that you need both position q and momentum p to get an accurate picture of the overall electron state. This is not true in quantum mechanics; it is sufficient to characterize the wavefunction $\left\langle x |\psi\right\rangle$ in position space or $\left\langle k|\psi\right\rangle$ in momentum space. Which is chosen is ...


17

The real issue is that no one has ever taken a picture (i.e. electron density) of genuine, unambigious, cases of a single, double, triple, quadruple??? bonds. And they never will, because these concepts are not based on quantum mechanics. Two atoms reside next to each other, and if they have a favorable electrostatic interaction, then a certain type of ...


17

Unfortunately, the arguments presented by buckminst and Uncle Al aren't completely right. The MO schemes are correct but the HOMO-$\sigma$ orbital ($s_{\sigma}^{*}(5\sigma)$ in buckminst's diagramm, $\sigma_{3}$ in Uncle Al's diagram) is not antibonding but slightly bonding in character because there is some mixing with the $\ce{p}$ atomic orbitals of the ...


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

One atomic (or molecular) orbital is said to be orthogonal to another atomic (or molecular) orbital if there is no interaction between the electrons in one orbital with the electrons (wavefunction) in the other orthogonal orbital. Acetylene is an example of a molecule in which the two pi bonds are orthogonal to each other (see the light-blue pi bond and the ...


16

This is caused by the molecule $\ce{SF6}$ being hypervalent, which means that the main element (in this case sulfur) has more then 8 valence electrons. The reason why this can happen is extremely complex and, to be honest, I am not even sure whether it is a fully solved issue. I do know that the effect is related to the electronegativity of the ligands, ...


16

Hybridisation is a purely mathematical concept, which makes it possible to explain experimentally found structures. The most prominent example for this is methane, where you can consider the central carbon atom to be $\mathrm{sp^3}$ hybridised. Formally, the $\mathrm{s}$ orbital and the three $\mathrm{p}$ orbitals can be linearly combined to form four ...


16

Let’s take a look at a qualitative MO scheme for a tetrahedric transition metal complex whose ligands have three p-type orbitals each. On the left of figure 1 you have the metal orbitals ($\mathrm{3d}$, $\mathrm{4s}$ and $\mathrm{4p}$) and on the right the twelve degenerate ligand p-orbitals (transform as $\mathrm{a_1 + e + t_1 + 2t_2}$). Only orbitals of ...


16

No. The reason why a gas particle in a large volume has a large entropy is not because it has a lot of space to move around per se. A better explanation is that for a given energy, there are many accessible translational states (these states can be derived from the particle in a box model). If we assume that all of these translational states are equally ...


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