# Tag Info

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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 ...

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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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There are 3 types of octet rule "violations" or exceptions molecules with an odd number of electrons, such as nitric oxide (image source) molecules with less than 8 electrons around an atom, $\ce{BeCl2}$ and $\ce{BH3}$ serve as examples (image source) molecules with more than 8 electrons around an atom, such as $\ce{PCl5}$ or $\ce{SF6}$ Take a look at ...

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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 ...

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In terms of a simple bonding concept the answer of ron is correct, however due to symmetry constraints it is not possible to distinguish the two $\pi$ systems. The reason for this is, that these $\pi$ orbitals are degenerated and perpendicular to each other. Carbon dioxide The molecule is very symmetric and one of only a few with the point group $D_{\infty ... 14 Technically, both the structures$\ce{H2N-OH}$and$\ce{H3N\bond{->}O}$may exist. However, in reality hydrogen atom is rather prone to migration and the second structure is not favorable. So, for a compound with composition$\ce{NH3O}$, the correct structure would be$\ce{H2N-OH}$. The structure of second type is stable for compound$\ce{NOF3}$and may ... 12 According to Electronic Structure of NO2 Studied by Photoelectron and Vacuum-uv Spectroscopy and Gaussian Orbital Calculations J. Chem. Phys. 53, 705 (1970) : The highest molecular orbital$4a_1$is occupied by the 1 unpaired electron. The electron population of this orbital is (see table VI): 0.53 on the N atom (0.16 2s, 0.37 2pz) 0.24 on each O atom (... 12 Those are all valid resonance structures for ozone (except the second from the left in the "four-form structure" row, that one has two negative charges, maybe it's a typo). In fact you can draw even more structures like this one, where this is no single bond between the middle and terminal-right oxygens. All of these structures contribute to the "real" ... 11 Nitrogen is in the second row with no$d$orbital in the valence shell. It obeys the octet rule and cannot have more than 8 electron. There are exceptions to the octet rule. Having less than 8 electron is less preferable but still possible, and is commonly seen in free radicals and cations. On the other hand, having more than 8 electron is extremely ... 10 We can draw the 3 Lewis structures (or the corresponding resonance structures) pictured below for$\ce{O_2}$Since an oxygen atom has 6 electrons, A would correspond to a structure with a single bond between the oxygen atoms, 2 lone pairs on each oxygen and an unpaired electron on each oxygen; however A does not have an octet around each oxygen, in fact, ... 10 There is resonance between various Lewis structures as: $$\ce{:\!\overset{\ominus}{C}#\overset{\oplus}{O}\!: ~<->~ :\!C=O\!:: ~<->~:\!\overset{\oplus}{C}-\overset{\ominus}{O}\!:::}$$ The term "valence" is not used much nowadays, and "oxidation state" is used but has a different meaning. The IUPAC definition of oxidation state requires the ... 10 I've seen the following drawing used instead of the three double bonds. The circle signifies that the ring is aromatic. It's an alternative that is used to show that all bonds are the same. 10 Both of your Lewis structures are wrong. In both of the shown structures, the solid line from sodium to oxygen/ sulfur implies that there is a discrete bond between these atoms. This is not the case. The sodium ions and the thiosulfate ions form ion pairs (or ion triples) and are attracted to each other via diffuse electrostatic interactions. Usually that ... 10 Lewis dot structures predate the advent of modern quantum theory and quantum chemistry. There is no Schroedinger equation associated with them -- in fact placing a dot on an atom (seems to me that it) violates the concept that electrons must be indistinguishable. There are no slots, there are no "octet rules". There are only particles, and kinetic and ... 10 As Jan mentions in the other answer, in general, a Lewis structure is preferred if it retains the octet rule. There can be more complicated bonding, and sulfur and phosphorous are two common examples (e.g.,$\ce{SF6}$). I find that many students can come up with "non-traditional" Lewis structures that, like yours, satisfy the number of valence electrons and ... 9 With familiarity you will recognize that$\ce{COOH}$in a formula generally refers to a carboxylic acid group. Sometimes you will also see$\ce{CO_2H}$. Either is acceptable. Vinegar, or acetic acid, is a carboxylic acid. Its formula is$\ce{CH3COOH}$or$\ce{H3CCOOH}$. It is important to realize that a carboxylic acid group is NOT a peroxide. Peroxides ... 9 As mentioned in the comments,$\ce{SbF5}$in the gas phase has a rather high symmetry, the point group looks like$D_\mathrm{3h}$to me, containing a principal three-fold rotational symmetry axis (along the axial bonds) and three two-fold axis (along each of the equatorial bonds), among other operations. As mentioned in this text, a dipole moment can only ... 9 Oxyphosphorus compounds, all of which contain phosphorus-oxygen linkages, are the most dominated subset in Phosphorus Chemistry. You may find good review of oxyphosphorus compounds in Ref.1. In particular, most of these commonly known as phosphates are described in Chapter 3 of Ref. 1 (Pages 169-305) which states that: Oxyphosphorus compounds may be ... 8 Fluorine is more electronegative than phosphorus. The phosphorus is at the apex of a pyramid, the base of the pyramid being an equilateral triangle with a fluorine atom at each vertex. The F-P-F angles are 96 degrees. Each F-P bond contributes to the net dipole moment, as a vector from the P to the F. The net dipole moment is the vector sum of the vectors ... 8 I completely agree with Gonçalo Justino's answer, but would like to add another shade to it. In principle there are quite a few possible arrangements thinkable for this molecule - at least in the gas phase. In this thought it simply refers to the fact that they are local minima on the potential energy surface. The most stable of all is the one that has been ... 8 I highly doubt that it could exist. The co-linear arrangement,$C_{\infty\mathrm{v}}$, of$\ce{Cl-O-S-Cl}$is a second order saddle point on the DF-BP86+D3(BJ)/def2-SVP level of theory on the$S=0$potential energy surface. It has two imaginary bending modes,$\nu_1=371.39\mathrm{i}$,$\nu_2=108.23\mathrm{i}$. (Please note that they are actually ... 8 Which is a better Lewis structure? There is no such thing as a better or a worse Lewis structure. They should be properly referred to as major or minor resonance contributors/forms. I looked up a textbook to check the rules of determining which resonance structure contributes more. The order is: Structures that satisfy the octet rule are greater ... 8 Consider that the carbon in carbon dioxide is bonded to two electronegative oxygen atoms. This will cause shift the electrons towards the oxygens and away from the carbon, making it electron deficient:$\hspace{7.5cm}$This makes the central carbon electrophilic, and explains the reactivity of the carbonyl moiety towards nucleophiles. Similarly,$\ce{CO2}$... 7 Nitrogen dioxide has 17 valence electrons, and is bent with an angle of 134 degrees and bond lengths 0.119 nm. Electron spin resonance experiments place the odd electron on the nitrogen in a$\sigma$rather than a$\pi$orbital. Thus formally we can consider the odd electron to be in a sigma bonding orbital (of$a_1$symmetry species in$\ce{C_2v}$) of$sp^2$... 7 As I mentioned in my comment, the Wikipedia article on resonance provides a rather nice description of the subject where the Resonance in quantum mechanics section is of particular interest for the question asked. However, I decided to somewhat clarify the matter emphasising few important points. And I will start by quoting the OP. We know that in ... 7 If you count electrons and determine the formal charge on each atom, you find that in structure #1, the negative charge is on the oxygen. Do the same exercise for structure #2 and you find that the negative charge is on nitrogen. Since oxygen is more electronegative then nitrogen, the negative charge is more stable when its on the oxygen atom. Therefore, ... 7$\ce{O2NCl}\$ does not have two or three resonance structures but rather almost infinitely many that have different probabilities. Resonance structures will never bring you close to the actual picture; for the true picture to evolve you will need to do molecular orbital calculations. That said, resonance structures are often a somewhat good approximation if ...

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In your drawn example, the red-boxed structure requires the carbon atom to have five bonds, which, to channel my secondary school teacher, says "yuck, I don't like that". This is a case of breaching the octet rule which has to with how the orbitals hybridise. Five full bonds on a carbon is not "allowed". However three full bonds and two partial bonds is ...

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The principle of full octet is that most reactivity patterns are described well if there is a driving force / energetic gain for full octets in main-group atoms. However, the only facet of an atom that dictates anything about the number of valence electrons is its nuclear charge. Boron has a charge of 5. This is balanced by 5 electrons. Two of them are ...

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