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14

The issue stems from the fact that you may not have understood what each bond notation means. Here, the solid wedge indicates that the bond is toward us and so a 3-D view of the same molecule would look as follows: So as you can see the dipole moments – instead of cancelling, add up. So simply by elimination option (d) is the right answer For trans-1,3-...


10

The electrons are not shared equally among the oxygen atoms. The central oxygen atom alone donates three electrons to the covalent bonding whereas the other two oxygen atoms combined donate the other three (there is a 3-center, 4-electron pi bond in which two of the rlectrons are shared between just the end atoms). Since the central oxygen atom is ...


8

In my opinion, instead of trying to cancel out dipoles, you should fall back to first principles and just do what you are effectively doing: using symmetry. The key concept is that the symmetries of the nuclei translate to symmetries of orbitals and electrons, and these together translate to symmetry of any properties derived from the positions of nuclei and ...


4

Symmetry is very useful in cases like this. The rules are that (a) any molecule with a centre of inversion ($(x,y,z)\to(-z,-y,-z)$ which makes the molecule indistinguishable) cannot have a permanent dipole. In addition (b) a dipole cannot exist perpendicular to a mirror plane ($\sigma$) and (c) a dipole cannot exist perpendicular to a rotation axis ($C_n$). ...


4

PubChem gives dipole moment ($\mu$) of 1,4-dibromobenzene (p-dibromobenzene) as $\pu{1.43 D}$ in gas phase and $\pu{1.87 D}$ in liquid phase at $\pu{20 ^\circ C}$. It has given 1987 version of Handbook of Organic Chemistry (J. A. Dean, Ed.) as the reference. PubChem also gave $7.77$ as dielectric constant at $\pu{10 ^\circ C}$ and $6.7$ at $\pu{40 ^\circ C}$ ...


3

Basically what is the meaning of a permanent dipole? I understand how it is different from induced dipole and instantaneous dipole, but these examples feel like instantaneous dipole examples to me. Addressing your second question, permanent dipole moment experimentally means that a molecule will be deflected by an electric field. Consider water as an ...


3

Indeed, $\ce{XeF6}$ has a distorted tetrahedral structure in all three phases. Two of solid state crystal structure modifications ($\bf{A}$ and $\bf{B}$) are shown in following diagram (Ref.1): The abstract of Ref.1 tells them all: According to single crystal X-ray diffraction, neutron powder diffraction, solid state MAS NMR data, and differential ...


3

Distance In vacuum, the calculated (classical) interaction between to aligned dipoles decreases with the square of the distance. If the dipoles can't come very close to each other, the interaction will be weak. If you treat hydrogen bonds as a special case of dipole-dipole interaction, you will find that a N-H ... O=C interaction is much stronger than a C-H ....


2

There are two reasons why dichloromethane is polar and the tetrahedral shape is only one The reasons why any molecule is polar–which is usual chemistry talk means that the molecule has a dipole moment–is that the individual dipole moments of its constituent bonds don't balance out. That means that we need to know both the polarity of the constituent bonds ...


2

However, we were taught in my class that the dipole moment of a molecule is equal to the vector sum of the dipole moment of each bond in the molecule. But Ozone only has bonds between two Oxygens, so why is the dipole moment of each bond not equal to zero? Chemical education is in such a sorry state. Why do they teach chemistry like chicken or eggs story? ...


2

In aqueous medium, the proton ($\ce{H+}$) is thought to exist as hydronium, $\ce{H3O+}$. However, in reality, probably bigger complexes of water may also be associated with one proton. In liquid water, everything is interconnected with strong hydrogen bonds anyways, and it is difficult to distinguish the free species. All of this means that it would be ...


1

In H2O, as O is more electronegative than hydrogen, the resultant bond dipole is towards O, which means both the lone pair and bond pair dipole are acting in the same direction and the dipole moment of H2O is high. (1.84 D) On the other hand, In the case of F2O, the bond dipole is acting towards fluorine, so in F2O the lone pair and bond pair dipole are ...


1

When a $\ce{H}$ atom is bound to an electronegative atom like oxygen, its only electron is mostly placed between the two nuclei. Outside of this $\ce{H}$ atom, the $\ce{H}$ nucleus is nearly "naked". Speaking naively, there is no other electron to occupy the space around the nucleus towards outside. It looks as if the nucleus $\ce{H}$ is not ...


1

It is common practice at a high school or first year undergraduate level to determine whether bonds are in any way polar (and, by extension, whether a molecule can be a dipole) just by looking at the atoms on either side of the bond. If you do that for ozone, you will obviously come across two $\ce{O-O}$ bonds and would have to assume that these bonds are ...


1

Note that polarity can be considered for the whole molecule, functional group or particular bonds.E.g. $\ce{CO2}$ has zero permanent dipole moment, as bond dipoles cancel each other. But the bonds themselves are polar enough due their dipole moment, being able to involve the molecule in polar intermolecular interactions. Bonds of both $\ce{CH4}$ and $\ce{...


1

I read up in my text book that dipole moment direction of phenol is towards OH group so I guess O being highly electronegative pulls the electrons more and participates less in resonance hence it has deta negative charge on it thus acc to convention dipole moment direction is from positive to negative hence direction is towards OH group.


1

I would say that your analysis of first, second and fourth one is correct, as H-bonds are stronger than dipole-dipole interactions. However, you made a mistake while filtering compounds having dipole-dipole interaction. The $1$ and $3$, both have dipoles! The $\ce{C=O}$ bonds in $\ce{CO2}$ are polar and hence they possess dipoles, but unlike a water molecule,...


1

EDIT (after title edited): $\ce{HBr}$ doesn't have stronger interaction than $\ce{CH2NH2}$, but it has dipole-dipole interaction as the strongest forces between it's molecules, which is obviously weaker than H-bonding. Hydrogen bonding is the strongest intermolecular attraction. It is a type of dipole-dipole interaction1, but it is specific to Hydrogen. In ...


1

At some time the electrons will be at the fluorine end more often as it has higher electronegativity and thus making the hydrogen end partly positive. How does this work? This is the charge distribution on the "surface" of the HF molecule, estimated by the program molcalc.org: In a thought experiment, if you placed a negative sample charge (...


1

I guess for homogeneous read homo-nuclear. He is trying to explain how the electric field of the radiation induces a dipole by shoving the electrons around, hence reference to polarisability: this relates to the standard description. This induced oscillating field then radiates, hence scattering. The electron and proton centre is when the field is zero, i.e. ...


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