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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 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 ...


47

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 ...


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Water, as simple as it might appear, has quite a few extraordinary things to offer. Most does not seem to be as it appears. Before diving deeper, a few cautionary words about hybridisation. Hybridisation is an often misconceived concept. It only is a mathematical interpretation, which explains a certain bonding situation (in an intuitive fashion). In a ...


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If you can assign the total electron geometry (geometry of all electron domains, not just bonding domains) on the central atom using VSEPR, then you can always automatically assign hybridization. Hybridization was invented to make quantum mechanical bonding theories work better with known empirical geometries. If you know one, then you always know the other. ...


42

Let's consider, for example, a tetrahedral Ni(II) complex ($\mathrm{d^8}$), like $\ce{[NiCl4]^2-}$. According to hybridisation theory, the central nickel ion has $\mathrm{sp^3}$ hybridisation, the four $\mathrm{sp^3}$-type orbitals are filled by electrons from the chloride ligands, and the 3d orbitals are not involved in bonding. Already there are several ...


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 ...


35

Atoms at the edge of a crystal that have an unsatisfied valence are said to have "dangling bonds." Many elements, in addition to carbon, can have dangling bonds. Dangling bonds is a subject of current interest because of the impact these structures can have on semiconductor properties. These dangling bonds are very similar to free radicals, except since ...


34

Look carefully, it's (distorted) tetrahedral--four groups at nearly symmetrically positions in 3D space{*}. So the hybridization is $sp^3$. As you can see, the shape is distorted, but it's tetrahedral. Technically, the banana bonds can be said to be made up of orbitals similar to $sp^3$ but not exactly (like two $sp^{3.1}$ and two $sp^{2.9}$ orbitals--since ...


31

I'll present a LCAO-MO argument. But first let's debunk a myth. $\ce{SF6}$ has "hypervalent" sulfur and the 3d orbitals on sulfur participate in bonding No. This is not true. I would close one eye if it is taught in high school but really, this cannot be further from the truth. Here's one reference: J. Am. Chem. Soc. 1986, 108, 3586; there are many more. ...


23

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 ...


21

First of all, are they correct? ChemBioDraw had some complaints, but as far as I can see there's the same amount of electrons, and no valence orbitals exceeding capacity. Yes, these are the six most important resonance structures for this compound. The reason ChemDraw complains is that it is trying to act smarter than you, and it most certainly is not. It ...


21

Here is a plot of the Quantum Theory of Atoms in Molecules answer to you question. I have shown the bond paths of $\ce{B2H6}$. Indeed, they are "banana-like" but interestingly they are curved inward, unlike the case of cyclopropane which are curved outward. (Hybridization does not exist. Also, I'm not sure if there is a point of ascribing "number of ...


19

You can find the hybridization of an atom by finding its steric number: The steric number = the number of atoms bonded to the atom + the number of lone pairs the atom has. If the steric number is 4, the atom is $\mathrm{sp^3}$ hybridized. If the steric number is 3, the atom is $\mathrm{sp^2}$ hybridized. If the steric number is 2, the atom is $\mathrm{...


18

Classical hybridization theory does not allow for noninteger hybridizations. However, ab initio calculations can be interpreted using a bond order analysis method such as NBO, where the MO coefficients are used to provide the closest analogue possible to a classical hybridization picture. For example, one of the pure $sp$ orbitals in Pauling's valence bond ...


18

Hybridization is given by the following formula: $$H= \frac{1}{2} (V + X - C + A)$$ Where: $V$ = number of valence electrons in central atom $X$ = number of monovalent atoms around the central atom $C$ = positive charge on cation $A$ = negative charge on anion $$H=4 \to \ce{sp^3},\;2\to \ce{sp,\;3}\to \ce{sp^2}...$$ e.g.: in $\ce{NH3}$, the ...


18

You asked a question, belonging to surface chemistry. It is a relatively new area of research, as it relies heavily on atomic-resolved microscopy and computational methods. Generally, the answer depends on prehistory of the surface and its environment. In case you crack a diamond, making new surface, two processes happens. so-known reconstruction, or ...


17

You're right in that bond length, and therefore bond strength does affect acidity (see: $\ce{H2S}$, $\mathrm{p}K_\mathrm{a} = 7$ and $\ce{H2O}$, $\mathrm{p}K_\mathrm{a} = 15.7$). If we defined acidity with the following equation $$\ce{HX -> H + X}$$ then the bond strength would indeed be the only deciding factor in the acidity of $\ce{HX}$, since the ...


16

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, ...


16

I already pointed it out in the comments, but I believe it is time to give it some more thought and explanation. Let's deal with some conceptual issues first. Hybridisation is a model, that can be used to describe a bonding situation. It is never cause for a certain geometric arrangement, it is always the result of a bonding arrangement. Strictly speaking ...


16

The nitrogen in aniline is somewhere between $\ce{sp^3}$ and $\ce{sp^2}$ hybridized, probably closer to the $\ce{sp^2}$ side. We are correctly taught that the nitrogen in simple aliphatic amines is pyramidal ($\ce{sp^3}$ hybridized). However in aniline, due to the resonance interaction between the aromatic ring and the nitrogen lone pair, considerable ...


16

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 ...


16

You got it backwards. The $\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$, $\mathrm{f}$ orbitals stand for sharp, principal, diffuse, and fundamental. Wikipedia: Electron Configuration § Notation: The choice of letters originates from a now-obsolete system of categorizing spectral lines as "sharp", "principal", "diffuse" and "fundamental" (or "fine"), based on ...


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

Until this post, $\mathrm{sp^1,~sp^2~\text{and}~sp^3}$ meant for me, that the hybrid orbitals would consist from one s orbital and one, two or three p orbitals. I thought of it being so, as this seems to be what the most textbooks suggest, as can be seen in the following image: One 2s orbital and three 2p orbitals hybridize to four $\mathrm{sp^3}$ orbitals. ...


15

Background Amine basicity correlates with, among other things, the hybridization of the nitrogen orbital that is holding the lone pair of electrons. The less s-character in this orbital, the more basic the amine. So, as the following figure indicates, a trialkylamine (the lone pair is in an $\ce{sp^3}$ orbital) is more basic (less acidic or higher $pK_{a}$)...


15

Don't put too much trust in the absolute signs of wavefunction, for they all are arbitrary anyway. Look at it this way: an s orbital has one sign (*). One lobe of p orbital has the same sign, so when they add up, it grows huge. Another lobe inevitably has the opposite sign, so when they interfere, it is reduced to a tiny pig tail. (*) This is not quite true ...


14

(I strongly disagree with the comprehensiveness of the accepted answer, but here I go...) 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 ...


14

Whether sulfur or phosphorous actually expand their octet is contested within the chemistry community. Another term for this octet expansion is "hypervalency." You can find many works of research regarding hypervalency. The consensus, according to Wikipedia, is that both can expand their octets, but not to a significant extent. In other words, d-orbital ...


14

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 $\mathrm{sp}^3$ ...


14

Yes, but. Yes, in the conventional low-level models, one would consider the two carbons in benzyne you mentioned as sp-hybridised. And that does mean that their orbitals seem to be pointing the wrong way. However, a better picture would be to use an orbital which is a lot closer to sp²-hybridisation, and an even better picture would calculate orbitals ...


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