# Tag Info

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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 ...

55

A lithium atom has one valence electron, easily lost (shared), so it is connected to other atoms by a metallic bond. This is a bit like the shell game where a pea (electron) is hidden under a walnut shell... the uncertainty of where it is at any instant implies, in a quantum sense, that all the atoms share it, and are held together. This bond is so strong ...

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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 ...

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TL;DR Fluorine is electronegative and can support the extra negative charge that is dispersed on the six X atoms in $\ce{SX6}$, whereas hydrogen cannot. First, let's debunk a commonly taught myth, which is that the bonding in $\ce{SF6}$ involves promotion of electrons to the 3d orbitals with a resulting $\mathrm{sp^3d^2}$ hybridisation. This is not true. ...

39

The following ring opening reaction will occour: You are quite right about the angle strain. Because orbital interactions are not optimal in this geometry. Consider p-orbitals, then a natural bond angle would be $\theta\in [90^\circ; 180^\circ]$. A mixing of s- and p-type orbitals allows a wide range of angles $\theta\in (90^\circ,\dots, 180^\circ)$. In ...

36

In the original 1771 experiment, Scheele used a very simple setup consisting of a glass retort with a glass receiver (round-bottom flask). Yes, the glass was etched to some degree by the fumes, but it was not drastic enough to destroy the apparatus. From Anders Lennartson's The Chemical Works of Carl Wilhelm Scheele [1, p. 22]: 3.1 Publication 1. ...

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XKCD's source material is an article from the blog of one of the funnier chemists I've read, Dr. Derek Lowe. The chemical in question made his "Things I Won't Work With" list, and the article is found here. Dioxygen difluoride, $\ce{O_2F_2}$, sometimes evocatively labeled according to its atomic arrangement $\ce{FOOF}$, is first and foremost a vigorous ...

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In alcoholic solution, the $\ce{KOH}$ is basic enough ($\mathrm{p}K_{\mathrm{a}} =15.74$) to deprotonate a small amount of the alcohol molecules ($\mathrm{p}K_{\mathrm{a}}= 16–17$), thus forming alkoxide salts ($\ce{ROK}$). The alkoxide anions $\ce{RO-}$ are not only more basic than pure $\ce{OH-}$ but they are also bulkier (how much bulkier depends on the ...

22

$\ce {O_2F_2}$ doesn't spontaneously combust. It is a supporter of combustion, which means that it's basically a better version of oxygen when it comes to supporting fires. Basically, when placed in contact with $\ce{O_2F_2}$, other materials spontaneously combust. Here's an analogy: Substances such as nitroglycerin and TNT are like a person with a short ...

22

The $\ce{H-C-H}$ angle in cyclopropane has been measured to be 114°. From this, and using Coulson's theorem $$1 + \lambda^2 \cos(114^\circ) = 0$$ where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\mathrm{sp^{2.46}}$ hybridized. Now, using the equation $$\frac{2}{1 + \lambda_{\... 21 \ce{SF6} is extremely stable for purely steric reasons, because S is completely blocked by fluorine atoms from all directions, so the reactions starting with an attack on S that otherwise would readily occur (hydrolysis, etc.) never have the chance to occur. This has nothing to do with electronegativity. For another similar example, look at \ce{CCl4}. ... 20 The answer simply has to do with the accessibility of the high +6 oxidation state. In Cr, the 3d electrons drop in energy extremely rapidly as you remove electrons. So, it is much harder to remove multiple electrons one after another; the only Cr(VI) compounds that we know of are paired with extremely hard bases like the oxide ion, viz. CrO3, CrO42−, Cr2O72−... 18 \ce{XeF8} is not known to exist though O.N is +8. Why is this so? At least 2 compounds have been reported that contain the \ce{XeF8^{2-}} unit. See, for example: \ce{(NO^+)2[XeF8]^{2-}} (reference) Metal salts of the form \ce{(M^{+})_2[XeF8]^{2-}} where M is a metal salt such as \ce{Cs, Rb} (see the above reference) or \ce{Na} (see p. 62 in ... 18 Part of the reason why there has been very little research done on organoastatine chemistry is the extreme radioactivity of astatine; the half-life of \ce{^210At}, the longest-lived isotope, is about 8 hours. However, the element can be incorporated into organic compounds (e.g., via a destannylation reaction).1 In organic chemistry, the prefix is "astato-... 18 As @Waylander pointed out, it appears this reaction has not been performed and/or recorded in any literature, so it is quite dangerous to speculate. But keeping that aside, A 3D perspective reveals that abstraction of protons from the methyl group in quite unhindered. Hence, the triiodo intermediate is well anticipated. However, a quick glance at spatial ... 17 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, ... 17 The halogens, particularly in their diatomic free states and within various oxoacids, are strong oxidizing agents by virtue of their high electronegativities, electron affinities, and reduction potentials. The polarizability of the heavier halogens also makes them almost uniquely versatile as both good leaving groups and strong nucleophiles, depending on ... 17 It doesn't stop, it just gets a lot slower the more chlorine atoms you add. From the abstract of an article titled Photochemical side-chain chlorination of toluene[1]: Where k_1, k_2, and k_3 are the rate consts. for the successive reactions: \ce{PhMe} → \ce{PhCH2Cl} → \ce{PhCHCl2} → \ce{PhCCl3}, resp.; r = \frac{k_1}{k_2} and s = \frac{... 16 Why can't fluorine be the central atom in inter-halogen compounds? First off, fluorine can be at the "center of things." Examples would include the strongly hydrogen bonded hydrofluoric acid and the very relevant example of the trifluoride anion$$\ce{[F-F-F]^-}$$The trifluoride anion example is critical as it demonstrates that fluorine can be ... 16 First of all as @chipbuster says \ce{HF} in diluted solutions in water is nearly completely dissociated and therefore shouldn't be called weak. Wikipedia describes this nicely and cites several sources for this claim. It was rather difficult to prove (spectroscopic methods were used), because hydronium ions created in dissociation are mostly bound to ... 16 Forget about applying hybridization outside the second row, especially in 'hypervalent' compounds. I know, that it is common to use and sometimes works, but it is incorrect. The \ce{XeF6} molecule is a hard spot. While, indeed, experimental data suggest that it adopts distorted octahedral geometry in the gas phase, there is evidence that the minimum is ... 16 Thionyl chloride is preferred for preparing alkyl chlorides from alcohols because the by-products formed in the reaction are \ce{SO2} and \ce{HCl} which are in gaseous form and escape into the atmosphere leaving behind pure alkyl chlorides.$$\ce{CH3CH2-OH + SOCl2 → CH3CH2-Cl + SO2 ↑ + HCl ↑}$$16 Preparation of alkyl chlorides from alcohols by thionyl chloride is preferred over other methods mentioned because of the following reasons :$$\ce{R-OH + PCl5 → R-Cl + POCl3 + HCl ↑}\ce{3R-OH + PCl3 → 3R-Cl + H3PO3}\ce{R-OH + SOCl2 → R-Cl + SO2 ↑ + HCl ↑}Both the byproducts formed in the last reaction i.e., \ce{SO2} and ... 15 The reason that they do not exist (or at least are not the most stable form) is because the decomposition reaction is exothermic. \begin{aligned}\ce{ (1) && SF6 &-> SF4 + F2\\ (2) && SH6 &-> SH2 + 2 H2 }\end{aligned} Reaction (2) is much more exothermic than (1), and it can be argued in two ways: either there is something ... 15 The equation for the formation of \ce{MX(s)} from its constituent elements is:\ce{M(s) + 1/2 X2(g) -> MX (s)}$$In general, \ce{X2} could be liquid (\ce{X} = \ce{Br}) or solid (\ce{X} = \ce{I}). However, since your question is asking about the fluorides and chlorides, we can just limit it to the gaseous state. The principles that will be ... 15 Dipole moment is not just about charges, it also has L term. Bond length of \ce{C-Cl} is greater than \ce{C-F} and in this case, that is more dominating factor. The dipole moment is in order$$\ce{CH3Cl} \gt \ce{CH3F} \gt \ce{CH3Br} \gt \ce{CH3I} You can see that electronegativity plays a more dominating role in $\ce{CH3X}$ when $\ce{X}$ is $\ce{... 14 Based on research inspired by andselisk's answer, chemists stored it in glass vessels coated in wax (similar to the receiver setup Scheele used to prove the silicon dioxide precipitate was from the glassware itself. The fourth paragraph down in this blog post on The Chronicle Flask touches on it (emphasis mine): Where do you put something that eats ... 13 Secondary alkyl halides are on the borderline of$\ce{S_{N}2}$and$\ce{S_{N}1}$, so either could be operating. Branching at the position beta to the halide will further hinder the backside attack necessary for$\ce{S_{N}2}$. Furthermore, the solvent is polar protic, favoring$\ce{S_{N}1}$. In this case, the secondary carbocation can rearrange to a more ... 13 Fluorine reacts with ammonia to give dinitrogen and hydrogen fluoride as the major product. The Formation of Dinitrogen Tetrafluoride in the Reaction of Fluorine and Ammonia - J. Am. Chem. Soc. 1959, 81 (23), 6338-6339 - states that dinitrogentetrafluoride accounts for 6% of the yield while the major reaction is:$\ce{3F2 + 2NH3 → N2 + 6HF}$The same ... 13 TL;DR Initially published crystal structure of$\ce{[NEt4]2[InCl5]}$[1] according to the further investigations [3], is not valid. The$\ce{InCl5^2-}$ion does not have$C_\mathrm{4v}$symmetry, and VSEPR theory pretty much explains formation of numerous slightly distorted trigonal bipyramidal$\ce{InCl5^2-}\$-containing complexes according to the most ...

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