# Tag Info

72

Recently, there has been a lot of discussion of Bent's rule (see for example "What is Bent's rule?") here in SE Chem. Simply stated, the rule suggests that $\mathrm{p}$-character tends to concentrate in orbitals directed at electronegative elements. Why does $\ce{F}$ replace an axial bond in $\ce{PCl5}$? In order to answer this question, we need ...

60

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 ...

40

This answer is intended to clear up some misconceptions about resonance which have come up many times on this site. Resonance is a part of valence bond theory which is used to describe delocalised electron systems in terms of contributing structures, each only involving 2-centre-2-electron bonds. It is a concept that is very often taught badly and ...

26

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 cases $c_1 = c_2 > 0$ and $c_l = -c_2 ... 18 TL;DR VB theory treats atomic orbitals (including hybridized orbitals) as providing a good mathematical/physical description of the true form of the molecular wavefunction. MO theory uses atomic orbitals (with Gaussian radial functions) as a tool of computational convenience in an effort to define a molecular wavefunction that in its final form often bears ... 17 Geoff has summarized well the mathematical, VB wavefunction-based description of charge-shift bonding. This answer aims to provide further exposition, from the perspective of the distribution of the electron density and of the electron localization in charge-shift bonded as compared to traditional covalent systems. Much of my understanding and argumentation ... 17 OK, this was a new one to me too, but I found a few resources which might be useful: A Computational Organic Chemistry blog post Michael Scott Long blog post Nature Chemistry review The basic idea is this. Most bonds are ionic (i.e., both atoms are charged, but electrostatically attracted) or covalent (i.e., the electron pair is shared). (This is largely ... 17 Hybridisation is a purely mathematical concept, which makes it possible to explain experimentally found structures. The most prominent example for this is methane, where you can consider the central carbon atom to be$\mathrm{sp^3}$hybridised. Formally, the$\mathrm{s}$orbital and the three$\mathrm{p}$orbitals can be linearly combined to form four ... 17 Yes, they do exist and were characterised spectroscopically, see here (and there is a note on similar clusters for sodium): Blanc, J.; Bonačić‐Koutecký, V.; Broyer, M.; Chevaleyre, J.; Dugourd, P.; Koutecký, J.; Scheuch, C.; Wolf, J. P.; Wöste, L. Evolution of the electronic structure of lithium clusters between four and eight atoms. J. Chem. Phys. 1992, 96 ... 16 Unfortunately the key to understanding the delocalisation of electrons lies in understanding rudimentary molecular orbital theory. It is also necessary to understand that hybridisation is a mathematical concept which can be used to describe bonding. It is no necessity to have bonding. Your explanation of benzene is the most common description, but it is ... 15 Have you read the Wikipedia article to Bent's rule (especially the Justification paragraph). I think it explains the things rather well. In the example of$\ce{H3CF}$the$\ce{H}$is more electropositive than$\ce{C}$and the$\ce{F}$is more electronegative than$\ce{C}$. So, using the assumption that like in$\ce{CH4}$the$\ce{C}$atom is$\mathrm{sp}^3$... 15 Bon has a very nice example of motion potentially restricted because of a barrier, and although it turns out that this is not the case for the norbornyl cation, there are simpler examples that do show inversion through a barrier, such as ammonia inversion or cyclopentene ring puckering. The two structures either side of the barrier would be not be called ... 15 Firstly, note that hybridisation theory as applied to transition metal complexes is an incorrect, flawed theory. It is an attempt to rationalise experimental observation (e.g. geometries), but in the process it invokes models of bonding which are highly unrealistic and demonstrably untrue. See: Why is it wrong to use the concept of hybridization for ... 14 In non-nuclear chemistry, everything is electrostatic interactions. This is why you can learn and predict so much just by "following the electrons" Covalent bonds are also formed because of electrostatic interactions - they are just more complicated conceptually than ionic (actually, ionic bonds are more accurately described by wavefunctions, we just try to ... 14 Why 8? has not really been addressed by the above answers, and while tangential to the question, it is somewhat important to be considered. In general, but not always, atoms react to form complete quantum 'shells', with electrons interacting with all their orbitals. The principal quantum number ($n$) determines the maximum azimuthal quantum number ($l$), in ... 11 First off, hybridization is a concept chemists developed to help explain reality (their observations). Just like resonance theory and Huckel MO theory, it is often (but not always) a useful way to explain the world around us. A "rule" on hybridization: hybridization occurs in response to a bonding interaction. Further, hybridization involves the mixing of ... 11 Comparing modern valence bond and electronic structure theories one can argue that the generalized valence bond (GVB) wave function can be regarded as a special form of the multi-configurational self-consistent field (MCSCF) wave function.1 Thus, for instance, for the hydrogen molecule, the GVB wave functions has the following form (ignoring normalization ... 11 If you limit consideration to hybridization of atomic orbitals, a good reference to see is On the role of d orbitals in sulfur hexafluoride J. Am. Chem. Soc., 1986, 108 (13), pp 3586–3593. First, it is found that of the 6 valence electrons that atomic sulfur has (two 3s and four 3p), in SF6 a total of 3.1 electrons worth of electron probability density ... 11 Surely1. At the end of the day, it is just the matter of supplying enough energy: the homolysis requires approximately 4.5 eV per hydrogen molecule, while for heterolysis more than 17 eV per molecule must be provided. 1) G. Busca, A. Vaccari, Heterolytic dissociation of hydrogen on high-temperature methanol synthesis catalysts, Journal of Catalysis, 108(2), ... 10 I took an interest in this question because it's something I recently wondered myself. First of all, I should clarify that while you mention hypervalency, what you seem interested in is hypercoordination, or even more generally, just compounds with high coordination numbers (hypercoordination is used specifically when the number of ligands in a compound is ... 10 I am guessing that you have an understanding of the atomic orbitals of an atom such as 1s and 3p atomic orbitals. Atomic orbitals are the probability distribution of where an electron is going to be$90$% of the time. Technically, electrons don't revolve around the nucleus but are rather quantum particles that are superimposed in several positions at the ... 10 It is still LCAO-MO theory, but just dumbed down a lot. The difference is that, instead of feeding the "pure" atomic orbitals into the LCAO mechanism, you carry out an additional mathematical step in order to get orbitals that have nice directional properties, and you feed those into the LCAO mechanism instead. Let's talk about a simple example, methane. ... 10 Hybridisation is determined by geometry! This is the number one thing that you should learn from this answer. If you know the geometry of a molecule then you can work out the hybridisation of the atoms within it because the hybrid orbitals must have the correct geometry to account for the molecular structure. The second important thing that you should ... 9 A$\pi$bond has a plane of symmetry along the bond axis. It cannot be formed by s-orbitals; it needs at least p-orbitals to be created.$90\,\%$of all bonds described some time or another are somehow involving carbon, nitrogen or oxygen. (In fact, I probably underestimated). But these elements can only use p-orbitals to create$\pi$bonds. To do that, one ... 9 First, don't forget that the phosphorus is sitting on a diamond cubic silicon lattice site, which pretty much defines the symmetry of the potential around it - the P cannot change that, but must respond to it. So, you should not think in terms of atomic orbitals. Second, the fifth electron is not initially a free electron - it occupies a state that ... 8 The so called gold standard of quantum chemical calculations of ground state energies and properties is usually considered to be Coupled Cluster Singles Doubles (with perturbative) Triples as a method. This method is based on ab initio molecular orbital theory and it is size consistent, but not variational. A basis set of at least a triple zeta quality with ... 8$\sigma$-bonding MOs tend to have lower energy than$\pi$-bonding MOs so they will be formed first. One explanation is that the$\sigma$-bonding MOs have a lot of$s$-AO character and$s$-AOs have lower energy than$p$-AOs. I don't know of any double bond that is purely$\pi$-bonding. Basic MO theory suggest that$\ce{C2}$should have a double bond made of ... 8 It's just a matter of definition. If a "layer" is understood as an infinitely thin plane passing through the centers of carbon atoms, then the electron density of the huge delocalized$\pi$orbital is located above and below that plane, i.e., technically between layers. If a layer is defined so as to include whole atoms, then the said orbital is within layer.... 8 The bond between$\ce {Pt} $and ethene is not purely a$\sigma $bond and also not a pure$\pi \$ bond. The bonding actually consists of a combination of both bonding and back-bonding effect. If you look into the Molecular Orbital Diagram of Ethene, the picture will be more clear. For the formation of bond between the metal and the ligand, electrons from ...

Only top voted, non community-wiki answers of a minimum length are eligible