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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 ...
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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. ...
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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 ...
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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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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 ...
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The $\ce{H-C-H}$ angle in cyclopropane has been measured to be $114^\circ$. 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 + \...
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$\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}$.
...
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The answer has to do with two things. Note that HSAB theory is dubious at best and doesn't have very useful predictive power, so I am going to avoid talking about it.
(1) 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 ...
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As noted in the referenced question, fluorine is not in the +1 oxidation state in hypofluorous acid (which, incidentally, is the only hypohalous acid that has been isolated) nor in any other compound where its bonds give it a complete octet. Even where the fluorine has a positive formal charge, attaching it to less electronegative atoms will lead to an ...
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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 ...
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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-...
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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 ...
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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]^-}$$
The trifluoride anion example is critical as it demonstrates that fluorine can be ...
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First of all as @chipbuster says $\ce{HF}$ in diluted solutions in water is nearly completely dissociated and therefore shouldn't be called weak. Wikipedia describes this nicely and cites several sources for this claim.
It was rather difficult to prove (spectroscopic methods were used), because hydronium ions created in dissociation are mostly bound to ...
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The bonding situation in the compound $\ce{(ICl3)2}$ is by far more complex than what is depicted in this book. The molecule itself has very high symmetry, i.e. $D_\mathrm{2h}$, that needs to be satisfied in the molecular orbital/ valence bond description.[1,2] That is why one cannot distinguish between the bridging chlorine bonds.
Analysing this bonding ...
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The equation for the formation of $\ce{MX(s)}$ from its constituent elements is:
$$\ce{M(s) + 1/2 X2(g) -> MX (s)}$$
In general, $\ce{X2}$ could be liquid ($\ce{X} = \ce{Br}$) or solid ($\ce{X} = \ce{I}$). However, since your question is asking about the fluorides and chlorides, we can just limit it to the gaseous state. The principles that will be ...
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$\ce{XeF8}$ is not known to exist though O.N is +8. Why is this so?
At least 2 compounds have been reported that contain the $\ce{XeF8^{2-}}$ unit. See, for example:
$\ce{(NO^+)2[XeF8]^{2-}}$ (reference)
Metal salts of the form $\ce{(M^{+})_2[XeF8]^{2-}}$ where M is a metal salt such as $\ce{Cs, Rb}$ (see the above reference) or $\ce{Na}$ (see p. 62 in ...
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It doesn't stop, it just gets a lot slower the more chlorine atoms you add. From the abstract of an article titled Photochemical side-chain chlorination of toluene[1]:
Where $k_1$, $k_2$, and $k_3$ are the rate consts. for the successive reactions: $\ce{PhMe}$ → $\ce{PhCH2Cl}$ → $\ce{PhCHCl2}$ → $\ce{PhCCl3}$, resp.; $r = \frac{k_1}{k_2}$ and $s = \frac{...
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Preparation of alkyl chlorides from alcohols by thionyl chloride is preferred over other methods mentioned because of the following reasons :
$$\ce{R-OH + PCl5 → R-Cl + POCl3 + HCl ↑}$$
$$\ce{3R-OH + PCl3 → 3R-Cl + H3PO3}$$
$$\ce{R-OH + SOCl2 → R-Cl + SO2 ↑ + HCl ↑}$$
Both the byproducts formed in the last reaction i.e., $\ce{SO2}$ and ...
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Thionyl chloride is preferred for preparing alkyl chlorides from alcohols because the by-products formed in the reaction are $\ce{SO2}$ and $\ce{HCl}$ which are in gaseous form and escape into the atmosphere leaving behind pure alkyl chlorides.
$$\ce{CH3CH2-OH + SOCl2 → CH3CH2-Cl + SO2 ↑ + HCl ↑}$$
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Dipole moment is not just about charges, it also has $L$ term. Bond length of $\ce{C-Cl}$ is greater than $\ce{C-F}$ and in this case, that is more dominating factor.
The dipole moment is in order
$$\ce{CH3Cl} \gt \ce{CH3F} \gt \ce{CH3Br} \gt \ce{CH3I}$$
You can see that electronegativity plays a more dominating role in $\ce{CH3X}$ when $\ce{X}$ is $\ce{...
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Aqueous $\ce{KOH}$ is alkaline in nature i.e. it dissociates to produce a hydroxide ion. These hydroxide ions act as a strong nucleophile and replace the halogen atom in an alkyl halide.
$$\ce{RCl + KOH (aq) -> ROH + KCl}$$
This results in the formation of alcohol molecules and the reaction is known as nucleophilic substitution reaction.
Alcoholic, $\...
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Based on research inspired by andselisk's answer, chemists stored it in glass vessels coated in wax (similar to the receiver setup Scheele used to prove the silicon dioxide precipitate was from the glassware itself.
The fourth paragraph down in this blog post on The Chronicle Flask touches on it (emphasis mine):
Where do you put something that eats ...
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The key point is driving the equilibrium where you want it. You could do it by relative reaction rates. The Finkelstein reaction, however, uses a better method: precipitation of one of the products. Acetone is specifically chosen because sodium chloride, bromide and fluoride are insoluble in it while sodium iodide is soluble. So the equation becomes:
$$\ce{...
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Gold(III) chloride does not exist as $\ce{AuCl3}$. According to Wikipedia, the name gold trichloride is a simplification of the name, which is referring to the empirical formula, $\ce{AuCl3}$. The X-ray defraction studies revealed that the chemical exists as a chloride-bridged dimer, $\ce{Au2Cl6}$ (Ref.1):
In $\ce{Au2Cl6}$, each gold center is square planar,...
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Yes, that reaction occurs, but generation of the phenyl carbocation was much more difficult than anyone might have guessed. Here is a drawing of the phenyl carbocation.
First note that the 6 p orbitals making up the aromatic pi system are all still intact and overlapping - the aromatic nature of the benzene ring has not been tampered with. Then notice that ...
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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 ...
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At Wikipedia you find this structure (drawn by Benjah-bmm27 on Wikimedia Commons):
So, the answer (c) is correct. Yomen Atassi correctly stated that in such a hydrogen bond the two electronegative partners and the hydrogen prefer a linear arrangement, as this maximizes the orbital overlap for the hydrogen bond. That the configuration (c) is preferred over (...
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