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Hot answers tagged molecular-structure

63

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 p character tends to concentrate in orbitals directed at electronegative elements. Why does F replace an axial bond in $\ce{PCl5}$? In order to answer this question, we need to start by understanding ...

45

Yep, it has to do with the orbitals. $\ce{CO2}$ is linear, so even though the $\ce{C-O}$ bonds have individual dipole moments, the overall dipole moment is zero as these cancel out (they point in opposite directions, as shown in the diagram below). On the other hand, $\ce{H2O}$ is "bent", which means that the individual dipole moments of the bond are at an ...

43

Yes. Researchers have been using atomic force microscopy (AFM) and scanning tunneling microscopy (STM) for some time for this purpose. Do note that these images are not photographs in the sense that we usually think of "pictures" and are indirect measurements of constituents of the molecule. However, they do yield "pictures" that show the geometry of the ...

41

TL;DR Xenon hexafluoride has a fluxional structure in the gas phase, with multiple rapidly interconverting conformers. The three most important conformers have $C_\mathrm{3v}$, $O_\mathrm{h}$, and $C_\mathrm{2v}$ symmetries. The minimum energy conformer is probably $C_\mathrm{3v}$. Experimental evidence The structure of xenon hexafluoride ($\ce{XeF6}$) has ...

38

Here are the $\ce{H-X-H}$ bond angles and the $\ce{H-X}$ bond lengths: \begin{array}{lcc} \text{molecule} & \text{bond angle}/^\circ & \text{bond length}/\pu{pm}\\ \hline \ce{H2O} & 104.5 & 96 \\ \ce{H2S} & 92.3 & 134 \\ \ce{H2Se}& 91.0 & 146 \\ \hline \end{array} The traditional textbook explanation would argue that the ...

26

There's an interesting article here - Illustrating Atoms and Molecules - that discusses atom colors: In 1865, the chemist August Hoffman gave a Friday Evening Discourse at London’s Royal Institution on the “Combining Power of Atoms.” In order to demonstrate chemical bonding of atoms, he drilled holes in croquet balls and connected them with metal pipes (...

23

The answer is you are referring to neither of them. That is because resonance structures don't actually exist in reality. We only use them to give us a rough idea what the actually molecule and bonds look like. A common way to explain resonance structures is this: An explorer from a far distant land travels to a new continent and sees a strange animal that ...

23

In addition to the answer provided by Todd, there is also the very established technique of single crystal X-ray diffraction. The basic principle is sending X-rays through a single crystal of a compound. These X-rays interact with the electrons of the molecules or ions in question in a certain way that gives a diffraction pattern (basically, a series of ...

23

Interestingly, nobody addressed the reason why diamonds are hard in the first place. The pressure (and temperature) are not the reason why they're hard, only the reason why they are formed. The diamonds are hard because the carbon atoms are bonded together by sigma ($sp^3$) bonds, which are the strongest chemical bonds. Other materials exhibiting the same ...

23

The first thing to say is that I'm not sure where that image is taken from; it's neither in the original article nor in the supporting information to the article. Therefore, it appears to be more of an "artist's impression" rather than an actual atomic force microscopy (AFM) image, which is what was reported in the paper. Nevertheless, the actual AFM images ...

22

It's all about the 3D structure of double bonds. If we look at the tub form we see that all dihedral C-C=C-C angles are 0°. the C=C-C angles are 125°, also pretty close to the optimal value. Everything works out fine, there's basically no strain on the whole molecule. I tried to build the chair form of all cis, but every optimization ended up in either the ...

21

Here are some compounds that have other structures, followed by their hardest structure (based on Moh's Scale). Titanium dioxide: Rutile structure or Cotunnite structure Aluminum oxide: Corundum Silicon Oxide: Stishovite Boron Nitride: Wurtzite Boron Nitride There are many, many more.

20

The carbon is not hexavalent, it is hexacoordinated. A covalent bond does not equal to a total of two electrons between the bonding partners and the nature of the chemical bond may lie somewhere between totally covalent and totally ionic. Examples of this include boranes, with their three-centre-two-electron bonds. But we don't need to stop there; any π ...

19

You can look up the molecule on chemspider, where you have a little applet for the 3D structure. Or you can download a coordinate file from NIST and view it in a molecular viewer, like Avogadro. Or keep on reading for some deeper insight. As Klaus already pointed out, if VSEPR is a valid concept, one would arrive at the conclusion, that the molecule is ...

19

No, this is not possible. Actually, if I would have to think of the most unlikely chemical conceivable, that would be it. Let's see why: Krypton is a noble gas that doesn't bond to anything. All of the known krypton compounds can be counted on one hand, and most of them contain fluorine. Putting krypton in a large molecule like this just can't be. This is ...

18

I approach this question from the opposite direction. Benzene is commonly drawn as a ‘cyclohexatriene’ corresponding to the Kekulé structure, i.e. with three single bonds and three double bonds, despite the fact that the six bonds of benzene are actually indistinguishable from each other. This graphical representation of benzene is in accordance with ...

18

The electron didn't go anywhere. It's in an unhybridized p orbital on the central bromine, and yes, $\ce{Br3O8}$ is a free radical. That is why it decomposes above -80ºC.$^{[1]}$ $^{[1]}$ Cotton, F. A. Progress in Inorganic Chemistry - Volume 2; Interscience Publishers: New York, NY, 1960.

17

It is sometimes challenging to determine if a molecule is going to be acidic or basic if the system in which it is reacting is not considered. An important point to consider when dealing with acids and bases is that acid/base strength is inherently tied to the solvent. For this answer, I'm going to limit the discussion to acids and bases in an aqueous ...

17

General Rule #1: Most elements use only s and p orbitals to form bonds, only transition elements and heavier elements use d, f, etc. orbitals in bonding. General Rule #2: The more s-character in a bond the shorter the bond (reference). For example a $\ce{C(sp^3)-C(sp^3)}$ single bond length is ~ 1.54 $\mathrm{\mathring{A}}$ a $\ce{C(sp^2)-C(sp^3)}$ ...

17

TL;DR: there have been many theoretical investigations of the relative energies of these two forms for the parent vinyl cation with more recent work indicating that the bridged form is slightly more stable (by about 1–3 kcal/mol)4,5. This prediction has received support from recent experimental work as well6. Taking such relatively small energy differences ...

17

Your book was correct that a five coordinate metal complex is able to adopt both square pyramidal and trigonal bipyramidal geometries, and in both cases the sp3d hybridisation scheme applies (if you believe in hybridisation...). Which geometry is adopted depends upon a combination of steric and electronic factors, and isn't necessarily trivial to predict, ...

16

In 2005, a field ion microscope captured an image of a very sharp tungsten needle. The small round features are individual atoms. At even smaller scale than molecules, a quantum microscope has visualized the electron orbital of a hydrogen atom:

16

Ah, a (fairly) common conundrum that assails us Chemistry students when we start Biochem. ;-) At first glance, the fructose molecule in your first picture, and that in the second picture appear to have different conformations: Believe me, they're the same. "How?" you ask? Stay with me here (it requires a little imagination): Draw a vertical line/...

15

The question asks why water has a larger angle than other hydrides of the form $\ce{XH2}$ in particular $\ce{H2S}$ and $\ce{H2Se}$. There have been other similar questions, so an attempt at a general answer is given below. There are, of course, many other triatomic hydrides, $\ce{LiH2}$, $\ce{BeH2}$, $\ce{BeH2}$, $\ce{NH2}$, etc.. It turns out that some are ...

15

Well, let's reconstruct that starting from the very right side, where it says a $\ce{C_6H_5}$. The ring and the $\ce{CO}$ group would be a benzaldehyde if it had an $\ce{H}$ instead of an $\ce{N}$, right? Or a benzoic acid if it was $\ce{OH}$ instead of $\ce{N}$. So what would it be if it had an $\ce{NH_2}$-group? It would be a benzamide. If the $\ce{N}$ is ...

14

CCDC doesn't contain any molecules of such geometry. The closest topology observed is among pnictogens such as $\ce{P, As, Sb}$, which form tetrahedron units $[\ce{P4}]$, $[\ce{As4}]$, $[\ce{Sb4}]$ which are readily coordinating with Au, Rh, Ag and Cu by edge-sharing: The first publication (1) tetraphosphido-ligand presents the following structure of $\ce{... 14 It is very convenient to use crystal field theory to discuss this. It is usually assumed that in octahedral coordination the energy levels of the five d-orbitals are split, with two orbitals ($d_{z^2}$and$d_{x^2-y^2}$) well above the other three. The splitting is assumed to be large enough to overcome electron pairing energy. The first six electrons ... 14 No, VSEPR theory does not allow for a coplanar arrangement. Let us consider allene and its orbitals: Source. Green indicates$\mathrm{p}$orbitals; blue$\mathrm{sp^2}$; and red$\mathrm{s}$. Hybridization The terminal carbons are$\mathrm{sp^2}$hybridized, and form three$\sigma$-bonds each. This means that each terminal carbon has one unhybridized$\...

13

In a rigorous geometrical sense, there is no difference between tetrahedron and trigonal pyramid--the terms both mean the same thing. In colloquial and chemical use, however, 'tetrahedron' typically implies the 'regular tetrahedron', where all four faces are equilateral triangles. Chemically speaking, when referring to these two shapes as descriptors of ...

13

First off, it was very astute of you to recognize that if the radical formed at carbon 3 (the methylene carbon in the starting compound) is pyramidal, then the radical would be chiral ( the lone radical electron serving as the 4th different substituent on that carbon) and 4 different radicals would be possible. In a typical carbocation, the carbon bearing ...

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