# 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 ...

48

I'm not sure about the existence of molecules with bridges through rings. However, there are several publications of synthesis of molecules mimicking wheels and axles ([2]rotaxanes; The “[2]” refers to the number of interlocked components) as one shown below (Ref. 1): (The diagram is from Reference 1) This specific molecule (8; an “impossible” [2]rotaxane)...

47

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 ...

44

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 ...

42

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 ...

30

A variation on this theme is Ice VII, in which two cubic ice structures are intertwined with hydrogen bonds from each component structure passing through the hydrogen-bonded rings formed by the other component. Known to occur naturally on Earth as a high-pressure phase trapped in diamonds, Ice VII is a stepping-stone to the macromolecular and superionic ...

27

Inflate balloons, and tie them «at their stem» like a bouquet of flowers. If you take four of them, not too much inflated, you well demonstrate a situation close to $sp^3$ hybridization. These models equally work well in larger lecture halls by the way, and intentionally using different colors allows many options. (source) (screen photo You need some ...

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 (...

25

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 ...

24

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 ...

24

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

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 ...

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

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 ...

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.

21

The name of the compound is 1-chloromethyl-4-fluoro-1,4-diazoniabicyclo[2.2.2]octane bis(tetrafluoroborate) (CAS #: 140681-55-6), which is commonly known as Selectfluor, a trademark of Air Products and Chemicals (see Waylander's comment elsewhere). Different view of the compound is given below (to you to understand the zig-zag feature): Introduced in 1992, ...

20

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 ...

20

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 $\ce{C(sp^2)-C(sp^3)}$ single ...

20

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.

19

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 ...

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

Take home message: Just because you have an ortho substituent does not mean you will have an ortho effect. To have an ortho effect two conditions must be met: you need an ortho substituent and, importantly, it must be of sufficient steric bulk to disturb the adjacent substituent. One measure of steric bulk is the cyclohexane $A$-Value, a table of which can ...

19

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 $\... 19 In 2005, a field ion microscope captured an image of a very sharp tungsten needle. The small round features are individual atoms. At even smaller scale than molecules, a quantum microscope has visualized the electron orbital of a hydrogen atom: 19 Your book was correct that a five coordinate metal complex is able to adopt both square pyramidal and trigonal bipyramidal geometries, and in both cases the sp3d hybridisation scheme applies (if you believe in hybridisation...). Which geometry is adopted depends upon a combination of steric and electronic factors, and isn't necessarily trivial to predict, ... 19 No, this is not possible. Actually, if I would have to think of the most unlikely chemical conceivable, that would be it. Let's see why: Krypton is a noble gas that doesn't bond to anything. All of the known krypton compounds can be counted on one hand, and most of them contain fluorine. Putting krypton in a large molecule like this just can't be. This is ... 19 Perhaps not a complete answer, but this might put you on the right track. It looks as though a lot of the system we have today grew out of recommendations made by August Wilhelm von Hofmann in a paper titled On the Action of Trichloride of Phosphorus on the Salts of the Aromatic Monamines He actually suggested using the Latin numerals as a prefix for the ... 18 The bonding situation in the compound$\ce{(ICl3)2}$is by far more complex than what is depicted in this book. The molecule itself has very high symmetry, i.e.$D_\mathrm{2h}$, that needs to be satisfied in the molecular orbital/ valence bond description.[1,2] That is why one cannot distinguish between the bridging chlorine bonds. Analysing this bonding ... 17 It is very convenient to use crystal field theory to discuss this. It is usually assumed that in octahedral coordination the energy levels of the five d-orbitals are split, with two orbitals ($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 ...

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