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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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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 ...

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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 ...

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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 $\... 19 Forget about applying hybridization outside the second row, especially in 'hypervalent' compounds. I know, that it is common to use and sometimes works, but it is incorrect. The$\ce{XeF6}$molecule is a hard spot. While, indeed, experimental data suggest that it adopts distorted octahedral geometry in the gas phase, there is evidence that the minimum is ... 18 Short Answer The structure on the left is "preferred" because the structure on the right cannot exist. You cannot put 4 electrons in a p-orbital. Detailed Explanation In$\ce{ClF3}$the central chlorine atom is roughly$\ce{sp^2}$hybridized. This means that we will have 3$\ce{sp^2}$orbitals emanating from the chlorine; they will form an equatorial ... 15$\ce{XeO4}$has a tetrahedral structure as you noted, but$d$-orbitals are not involved in explaining its geometry or hybridization. The central xenon atom in$\ce{XeO4}$is best described as simply$\ce{sp^3}$hybridized. We have 8 xenon electrons so we need 4 orbitals to put them in. The 4 equivalent orbitals are formed by mixing the$s$- and$p$-... 15 No 6-centre-10-electron bond. The bonding situation in$\ce{IF7}$is explained by Christe et. al. (see ssavec's answer) as a 6-centre-10-electron bond. In the article itself I found no real evidence that supports this statement and I believe it to be completely wrong. The following section is taken from Karl O. Christe, E. C. Curtis, David A. Dixon, J. ... 14 So, from Wikipedia article on VSEPR theory we read: The overall geometry is further refined by distinguishing between bonding and nonbonding electron pairs. The bonding electron pair shared in a sigma bond with an adjacent atom lies further from the central atom than a nonbonding (lone) pair of that atom, which is held close to its positively ... 13 The simplest way to look at this trend is through VSEPR theory, which produces a very good qualitative understanding of molecular geometry even if it is not compatible with modern molecular orbital theory. However, VSEPR is not the whole story for the heavier Group V and Group VI hydrides. According to VSEPR, electron domains (i.e. bonds and lone pairs) are ... 12 In bonded pairs of electrons, the repulsion of the negative charges is somewhat reduced by the positive charge of the bonded atom's nuclei. Since lone pairs don't have to deal with this positive charge, naturally their repulsion is stronger. 12 Generally, these terms are used to refer to lone pairs rather than s-orbitals. In the context of VSEPR theory, lone pairs usually affect molecular geometry. For example$\ce{NH3}$adopts a trigonal pyramidal structure, because of the presence of the lone pair on nitrogen; if there was no lone pair, then it would be trigonal planar, cf.$\ce{BF3}$. These are ... 11 The following disilyne has been prepared and found to be stable to ~100 C. An X-ray crystal structure found that the two silicon atoms in the triple bond, Si(1) and Si(1'), along with the two attached silicon atoms, Si(2) and Si(2') are coplanar and the Si(1)-Si(1')-Si(2') angle is 137.44 degrees ("trans-bent"). The full text of the article describing the ... 11 There are only five Platonic solids: tetrahedron, octahedron, cube, dodecahedron, and isocahedron. Therefore five atoms cannot be arranged in five equivalent positions, unless all five lie in a plane. Trigonal bipyramidal is the minimum energy arrangement of five charges on a sphere. See Kevin Brown's MathPages Min-Energy Configurations of Electrons ... 11 how would people give estimates of the actual bonding angle of water What physics would be involved in the calculation Background That's a very good question. In many cases Coulson's Theorem can be used to relate bond angles to the hybridization indices of the bonds involved. $$\ce{1+\lambda_{i} \lambda_{j} cos(\theta_{ij})=0}$$ where$\ce{\lambda_{i}}$... 10 I took an interest in this question because it's something I recently wondered myself. First of all, I should clarify that while you mention hypervalency, what you seem interested in is hypercoordination, or even more generally, just compounds with high coordination numbers (hypercoordination is used specifically when the number of ligands in a compound is ... 10 The hybridisation of any atom is determined by the total count of its sigma bonds and localised lone pairs. This is because, while localised lone pairs occupy hybrid orbitals (as in ammonia), delocalised lone pairs occupy pure$\ce{p}$orbitals instead. Hence, the latter is not counted in determining hybridisation. The$\ce{N}$atom labelled "B" has its ... 10 Xenon is a noble gas and thus does not need to form any bonds to complete its octet. When such a bond is formed, this is known as a hypercoordinate structure: it contains more bonding partners than the most simple of simple Lewis structure theories would allow. Thanks to Linus Pauling, for many years chemists have been trying to invoke d orbitals to explain ... 9 That structure is incorrect. Several online chemical distributors have similar structures posted in their entries for aluminum carbide; so does PubChem! Some sources show the structure as linear; some show it as bent. That structure basically assumes that aluminum forms primarily covalent bonds with an "incomplete octet" in VSEPR structures, rather like ... 9 Simply put: a lone-pair orbital is "fatter" than a bonding orbital as I have illustrated here. This can be rationalized as follows: in a bonding orbital the electron pair is mainly attracted to 2 different nuclei which helps localize the electron pair. In a lone-pair orbital the electron pair is mainly attracted to a single nucleus and can therefore spread ... 9 Using the usual "AXE method" all molecules of$\ce{AX3E2}$type, where$\ce{A}$represents the central atom;$\ce{X}$represents the number of ligands (atoms) bonded to A;$\ce{E}$represents the number of lone electron pairs surrounding the central atom; are assumed to be T-shaped. I think it does not matter what the central atom$\ce{A}$and ligands$\...

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VSEPR works by accident in the cases of $\ce{NH3}$ and $\ce{NF3}$. In reality, there are much more things to consider as shown in this answer. All four compounds should have bond angles of $90^\circ$ if there were no other effects present. For both nitrogen compounds, the effects are the short $\ce{N-X}$ bonds which lead to steric clash of the three ...

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You can check out the calculated molecular orbitals of water on Professor Zipse’s (LMU Munich) page. The surrounding ‘text’ in German need not interest you, just click on mo-number to access an image of the corresponding MO. The lowest orbital with the lowest energy is, of course, oxygen’s core $\mathrm{1s}$-orbital. Next up we have an all-bonding orbital ...

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This can be argued on the basis of Bent's rule; concisely stated Atomic s character concentrates in orbitals directed toward electropositive substituents What follows below is a crude explanation. Before that, I'll note that we concern ourselves with the hybridisation of the orbitals at the central atom. Since s orbitals are lower in energy than p ...

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The answer that you are supposed to give is presumably something along the lines of "dπ–pπ" bonding between silicon and nitrogen. This essentially means that the lone pair of nitrogen is involved with backbonding into a silicon 3d orbital, leading to a linear geometry at nitrogen (i.e. the bond angle is 180°). This can be represented with the following ...

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With VSEPR theory you can predict the structures of $\ce{BX3}$. Since boron only has three valence electrons, all of which are used for σ-bonding to the halogen atoms, they arrange in a trigonal planar fashion. This means that all molecules $\ce{BF3}$, $\ce{BCl3}$, and $\ce{BBr3}$ have the same point group $D_\mathrm{3h}$. From this symmetry you can ...

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Why? Because there is no other choice. Starting from the AB6 octahedral configuration, all six vertices of the octahedron are symmetric, so it doesn't matter whichever one you “choose” to replace by the lone pair. All will yield the same final configuration.

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I find orbital hybridization of very limited use. Orbital hybridization is a mathematical manipulation of atomic wavefunctions, but that that is the only relationship to a quantum mechanical description of an atom. Presently, we have no evidence that supports the claim that orbital hybridization happens. The reason why I am not in favor of using orbital ...

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Here are the bond angles for each molecule (data from wikipedia): \begin{array}{|c|c|}\hline \mathrm{Molecule} & \mathrm{Bond \space Angle \space (^\circ)} \\ \hline \ce{H2S} & 92.1 \\ \hline \ce{H2O} & 104.5 \\ \hline \ce{NH3} & 107.8 \\ \hline \ce{SO2} & 119 \\ \hline \end{array} So $L \propto \frac{1}{BA}$ where $L$ is the number ...

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I actually have a (or many) big issue(s) with the quote: The central C-atom is in an sp2 hybridized state, for which the carbocations have planar geometry. The p$z$-AO remains empty. The authors here have clearly scrambled up their reasoning, making carbocations seem as something they are definitely not. Suffice to say (tl;dr) the above statement cannot ...

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