Hot answers tagged

46

Yep, it has to do with the orbitals. $\ce{CO2}$ is linear, so even though the $\ce{C-O}$ bonds have individual dipole moments, the overall dipole moment is zero as these cancel out (they point in opposite directions, as shown in the diagram below). On the other hand, $\ce{H2O}$ is "bent", which means that the individual dipole moments of the bond are at an ...


28

You're correct in assuming that the carbon atom in $\ce{CO2}$ has a partial positive charge. This is because the oxygen atoms are much more electronegative, so they pull the electrons away from the carbon atom. However, this molecule is still nonpolar. This is because, when you draw a dipole moment, you have to take all bonds into account. Take water for ...


27

Mini Research Project Time Updates Added CCSD(T) $n_i$ and dipole moments and tweaked discussion (the delay was caused by a system-wide storage upgrade on the machines which took nearly a week to complete). Preamble This response is in no way meant to be contrary to what Geoff has already posted. I happen to enjoy these types of questions and I like to ...


25

At first, I thought, that those 6.2 D of O’Hagan et al. have been measured somehow but it as later described, they calculated it (1): A molecular dipole value of 6.2 D for 1 was calculated at the M11/6-311G(2d,p) theory level using natural bond orbital (NBO) analysis with NBO 6.0. I was using your given database for a little search, starting with ...


23

Both pyrrole and furan have a lone pair of electrons in a p-orbital, this lone pair is extensively delocalized into the conjugated pi framework to create an aromatic 6 pi electron system. Where pyrrole and furan significantly differ is that, in pyrrole there is an $\ce{N-H}$ bond lying in the plane of the ring and directed away from the ring whereas in ...


21

The necessary formal derivation has already been nicely done by AngusTheMan. I'll start from the last equation: $$ \langle \mu_{z} \rangle = \langle \Psi | \hat{\mu}_{z} | \Psi \rangle $$ where $\Psi$ is the variational wavefunction; it can be any molecular state. It's important that it's variational, otherwise the expectation value approach is not exact. ...


20

How does lone pair of a central atom affect the dipole moment? There is no single answer to your question, let me explain. Unlike a typical covalent bond where the electrons are shared between two nuclei and the electron density is spread out over the entire bond, in a lone pair the electrons are not shared and the electron density is more localized around ...


19

we assume for no particular reason that dipole moments must be behaving like vectors Ah, but there is a reason. Consider the interaction of a molecule with the scalar potential $$ E_{\text{int}} = \int \rho(\mathbf{r})\phi(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r} $$ where the integral over all space is turned into an expansion: $$ \int \rho(\mathbf{r})\phi(...


18

The other answers have done a great job explaining why, even though its bonds are polar, $\ce{CO2}$ lacks a permanent dipole: the molecule's symmetry cancels out the polarity of its bonds. But that's not the whole story. I'd like to add to this a very interesting and environmentally important charateristic of $\ce{CO2}$ — namely that, while it lacks a ...


17

I am told that carbon dioxide is IR inactive. You're right, that's not true. Since carbon dioxide is linear it has $3n-5 = 4$ vibrations and they are pictured below. The symmetric stretch does not result in a change (of the initially zero dipole moment), so it is ir-inactive. The asymmetric stretch does result in a change in dipole moment so it is ir-...


16

Preliminaries: I am using the wrong (but still common) notation of the dipole moment. Please see the question about the direction of the dipole moment. The reason why carbon monoxide is often referred to a being practically nonpolar is because of its very small dipole moment. \begin{align} \ce{{}^{\ominus}\!:C#O:^{\oplus}} && \text{Dipole:}~|\...


15

The dipole moment $\mu$ of a molecule is a measure of charge distribution in the molecule and the polarity formed by the nuclei and electron cloud. We can perturb our system with an external electric field $\vec E$ and gauge the response of the electron cloud and nuclei by the polarisability, i.e how much the dipole moment changes. In practice the nuclei ...


15

I (accidentally) stumbled upon the following statement in Atkins' "Elements of Physical Chemistry" (p378): We represent dipole moments by an arrow with a length proportional to $\pmb{\mu}$ and pointing from the negative charge to the positive charge (1). (Be careful with this convention: for historical reasons the opposite convention is still widely used.)...


15

At first I checked common organic compounds and solvents in CRC Handbook [1, 9-59] that have dipole moment above $\pu{4.5 D}$: \begin{array}{llrr} \text{Name} & \text{Formula} & \text{Atom count} & \mu, \pu{D} \\ \hline \text{Thiophenecarbonitrile} & \ce{C5H3NS} & 10 & 4.59(2) \\ \text{3-Methyl-2-butenenitrile} & \ce{C5H7N} & ...


14

Draw the structures in 3D and then you will see why one is polar and the other not. $\ce{CF4}$: As you can see this molecules adopts a tetrahedral geometry which is perfectly symmetrical in every direction and so the dipoles of the four $\ce{C-F}$ bonds cancel out, leaving no overall dipole. $\ce{CHF3}$: Although the molecule has some symmetries, it is ...


14

Very simply, you explain the reason for this solubility rule by taking in consideration the energy requirements for the breaking of intermolecular forces between the molecules in the solute and the solvent. Note: this is only a simplified explanation as it also depends on other factors such as change in entropy Here is some background information on ...


14

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

According to Wikipedia, bond dipole moment depends on: Distance between atoms and Overall charge difference, not just electronegativity difference. Resonance tells us that there is some amount of charge separation in $\ce{C=O}$ bonds because of the $\ce{C+-O-}$ contributor. This difference in charge, in addition to the electronegativity difference, is ...


14

You are correct, the carbon does have a positive charge. We cannot measure a dipole, but that doesn't prove anything. However, $\ce{CO2}$ does have a quadrupole moment. Imagine a $\ce{CO2}$ molecule oriented along the $x$-axis, and a bit further along the $x$-axis there's also a $\ce{H2O}$ molecule with its dipole oriented along the $x$-axis. Its dipole ...


13

Well, it turns out that this is a very active area of research. I will only summarize what I understand to be true about the covalent nature of the hydrogen bond, so I'm sure the explanation could be more detailed and potentially more accurate in some places (I hope someone gives a more detailed answer), but here's what I've got. As you said, it has been ...


13

Because it takes two to tango. Dipoles interact with each other. A Lone dipole has nothing to interact with (other than an electric field, but if we ignore some externally applied macro field, there is nothing for a lone dipole to interact with). So molecules with an inherent dipole (like water or chloroform) interact with each other. One molecule's dipole ...


13

It is safe to say that there will always be intermolecular forces at play. At the time where you will consider these you should already have a good idea about the molecules involved in your system. Based on the composition and molecular structures you can make certain assumptions. In a molecule it is straight forward to estimate (bond) polarities based on ...


12

According to this paper: Planar configurations for resorcinol and hydroquinone are more acceptable on comparison of experiment and calculation than are the structures assuming either free or partially restricted rotation of the hydroxyl groups. It states that the dipole is present because there is restricted rotation with the hydroxyl groups. The ...


12

This was actually an interesting problem. Well, I ran a quick calculation using Avogadro and GAMESS, although other packages would work. This is a CCSD/aug-cc-pVTZ calculation, pretty much the gold standard in quantum chemistry. (CCSD(T) and a larger basis set might be better, but are unlikely to differ much here and I can't run them on my laptop.) The ...


12

Summary A systematic examination of the $\ce{CH3F}$ and $\ce{CH3Cl}$ complexes with density functional techniques and high-level wave function theory indicates that the charge-separation within these complexes influences the relative corresponding dipole moments for these compounds. Though fluorine is indeed more electronegative than chlorine, $\ce{CH3Cl}$ ...


12

An ionic bond could maybe be described as an inter-ionic force. All electron interactions are most accurately described by wavefunctions and quantum mechanics, but in practice we use successively more detailed approximations for convenience, stopping at the lowest level of detail that suits our needs at the time. At the lowest-detail end of the spectrum, ...


12

In non-neutral species the dipole moment in calculations is dependent on the choice of the co-ordinate system. For this reason the origin is chosen to be the centre of mass. With $r=107.8~\pu{pm}$ on RMP2/def2-QZVPP previously determined by a geometry optimisation. The program used for the computations is Gaussian 09 Rev. D.01 and used input files are ...


12

The question itself is void, as all methods with the exception of Hartree-Fock predict the direction of the dipole moment correctly. Deathbreath found a Full CI calculation in Jeremy P. Coe, Daniel J. Taylor, and Martin J. Paterson. J. Comp. Chem. 2013, 34 (13), 1083-1093 of which luckily an open version exists (arXiv:1301.4904 [physics.chem-ph]). The ...


12

The said lesson tends to oversimplify things to the point of getting them downright wrong. Atom dipoles do not line up with each other. They are way too weak for that. Should it be according to your description, they would indeed line up all in one direction and stay locked forever, much like magnetic particles in a magnet. Now, I brought up the magnetic ...


12

The previous answers by mpprogram6771 and MSalters nailed it. I'd like to add that, as $\ce{CO2}$ is a very small molecule, you can, with a bit of effort, set up a little numeric experiment to answer your own question, and even get approximate partial charges in each atom, and dipole moment of the whole molecule, using just free/open source software. First, ...


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