I heard on a lecture of an online site that, let us say, we have two orbitals (1 and 2). Now, it will be like a spherical cloud. Now, what they say is that

London Forces

Now, we can notice that for orbital number 1. There is a dark region and a light region (right side). Now, what they say is that the darker region means the electron finding probability is more there. Similarly, for the number 2, electron finging probability is more on left side.

Therefore, we say that 1 develops $\delta +$ charge and 2 develops $\delta -$ charge. Therefore, the two will attract each other.

Now, my teacher on the other hand told it a bit differently:

H-bonding in HF

The more electronegativity means more $\delta -$ charge. But, I have a confusion in it. Between $\ce{H}$ and $\ce{F}$. I have drawn a line. Fluorine has 7 electrons. My teacher says that $\ce{F}$ and $\ce{H}$ share that one electron where $\ce{H}$ has one and fluorine 7th electron. Now, since $\ce{F}$ is more electronegative. The electron (by hydrogen) is more pushed towards fluorine. My confusion is that if $\ce{F}$ is more electronegative than $\ce{H}$. Then, the electron should be present more towards the $\ce{H}$ side.

  • 1
    $\begingroup$ Please do not use (+)ve or (-)ve notations in your text body. $\endgroup$ Commented Mar 10, 2021 at 22:06

2 Answers 2


Although, OP indicated the confusion regarding $\delta^\pm $ charge compared with electronegativity, the text body indicated whole lot of confutations over interatomic/intermolecular forces. Thus, I start with the electronegativity.

According to IUPAC Gold Book, electronegativity is the concept introduced by Linus Pauling as the power of an atom to attract electrons to itself. There are several definitions of this quantity. However, Wikipedia gives a simple to understand version better suited for OP:

Electronegativity is the measurement of the tendency of an atom to attract a shared pair of electrons (or electron density).

The relative electronegativity of $\ce{H}$ atom has given a value of $\pu{2.1 eV^{-\frac{1}{2}}}$ and the electronegativity for each of other atoms in the periodic table have been calculated relative to this value for $\ce{H}$ atom (IUPAC Gold Book). These calculations are beyond OP’s field of study so should be disregarded. The important point for OP to consider is the higher the associated electronegativity, the more an atom or a substituent group attracts electrons. For instance, the electronegativity of $\ce{F}$ is $\pu{4.0 eV^{-\frac{1}{2}}}$ and hence, shared electron fair in $\ce{H-F}$ bond stays more closed to $\ce{F}$ than to $\ce{H}$. As a result, $\ce{F}$ carries permeant $\delta^- $ charge while $\ce{H}$ carries permeant $\delta^+ $ charge in the $\ce{HF}$ molecule, creating a permanent dipole. Hence, $\ce{HF}$ is a polar molecule. Therefore, among $\ce{HF}$ molecules, there are always dipole-dipole interactions (the image 2 of OP's text). Since the positive end of these dipole is $\ce{H}$ atom, the relevant dipole-dipole interaction is called $\ce{H}$-bonding, the strongest intermolecular force among dipole-dipole interactions. These dipole-dipole interactions occur only in molecules.

Now, I think I have addressed the OP's main confusion:

My confusion is that if $\ce{F}$ is more electronegative than $\ce{H}$, then, the electron should be present more towards the $\ce{H}$ side.

According to the above explanation, PO's understanding of electronegativity is not correct. In reality, if an atom is relatively more electronegative than that of the other atom it attaches to, that means it attracts the shared pair of electrons, increasing its electron density compared to its protons (gaining $\delta^- $ charge). Meanwhile, the other atom feels decreased electron density compared to its protons (gaining $\delta^+ $ charge). This is completely opposite of what OP's indicated understanding.

On the other hand, the London dispersion force is also one of the intermolecular forces, but is the weakest among them and occurs between atoms as well. For example, why do you think gaseous $\ce{He}$ become a liquid when the temperature is lowered sufficiently? You can find the answer in chem.purdue.edu:

The London dispersion force is a temporary attractive force that results when the electrons in two adjacent atoms/molecules occupy positions that make the atoms/molecules form temporary dipoles. Thus, this force is sometimes called an induced dipole-induced dipole attraction. London forces are the attractive forces that cause nonpolar substances to condense to liquids and to freeze into solids when the temperature is lowered sufficiently.

An atom or a molecule can develop a temporary (instantaneous) dipole when its electrons are distributed unsymmetrically about the nucleus because of the constant motion of the electrons (e.g., the image 1 of OP's text). The explanation can be depicted in an image:

London dispersion force

For instance, assume two helium atoms approaching each other. When they get closed to each other, the electrostatic attraction force between the nucleus of first $\ce{He}$ atom with two positive charges and the two electrons in the second atom causes each atom to polarize (creating an induced dipole in each atom). Note that at the same time, there is an electrostatic repulsion acing as well between electrons of each other as depicted in the image above.

Dispersion forces are present between all atoms or molecules. In molecules, these forces are present regardless of whether they are polar or nonpolar. London dispersion forces depends on few factors:

  • Larger and heavier atoms and molecules exhibit stronger dispersion forces than smaller and lighter ones. In a larger atom or molecule, the valence electrons are farther from the nuclei than in a smaller atom or molecule. They are less tightly held and can more easily form temporary dipoles (e.g., $\ce{Cl2}$ is a gas at room temperature and $\pu{1 atm}$ pressure, while $\ce{Br2}$ is a liquid and $\ce{I2}$ is a solid under the same conditions).
  • The strength of London dispersion forces varies according to molecular shapes. For example, $n$-$\ce{C5H12}$ and $neo$-$\ce{C5H12}$ are both pentanes, but former is more cylindrical shape while the latter is more spherical shape. Although, both have the same molar mass, $n$-pentane is a liquid (average boiling point of pentanes: $\pu{36.1 ^\circ C}$) at room temperature and $\pu{1 atm}$ pressure, while $neo$-pentane is a gas (boiling point: $\pu{9.5 ^\circ C}$) under the identical conditions.

Also, London dispersion forces are depend of the molecule's polarizability (The ease with which the electron distribution around an atom or molecule can be distorted is called the polarizability). London dispersion forces tend to be stronger between molecules that are easily polarized, but weaker between molecules that are not easily polarized.

  • $\begingroup$ Ohk. Thank you very very much for the efforts in solving my confusion . I would just like to give a short summary. 1 London dispersion forces can be between atoms to atoms or molecules with molecules. 2 If F atom is more electronegative, that means it has a stronger electric field to attract the H atom. Therefore , the Electron of H atom shifts towards F. It shifts at such a point where the forces by nucleus to F electrons and Forces by F’s electric field to the electron of H atom cancels out. So ,electronegativity does not mean that the electron is present more on the right side or left side $\endgroup$
    – Rider
    Commented Mar 11, 2021 at 4:13
  • $\begingroup$ Now , since the F atom shifted towards electron . It carries permeant $\delta$ + ve charge and F carries negative charge. Why ? Electron is still present more on the right side for H atom but it still is far away so it carries $\delta$ +ve charge. $\endgroup$
    – Rider
    Commented Mar 11, 2021 at 4:17
  • $\begingroup$ Also , as you pointed that it is a temporary dipole attraction. Does it mean it has a probability of it to happen or not ? Moreover , you took two He atoms which have same electronegativity difference. So , they won’t form LDF right. It would be a non polar covalent bond? $\endgroup$
    – Rider
    Commented Mar 11, 2021 at 4:53
  • $\begingroup$ How did induced dipole in each formed. Just when a Cl gets closer to H atom . It forms it automatically? $\endgroup$
    – Rider
    Commented Mar 11, 2021 at 5:00
  • $\begingroup$ @Rider: You still did not resolve your confusion. Your understanding is some what different from that before. Yet, get tangled with two different concepts. Please read my post carefully from beginning to finish. You may get some clues. Electronegativity applies in specific bond within the molecule. My example is $\ce{HF}$ molecule, and the bond in concern is $\ce{H-F}$, which is a covalent bond in gas phase. The $\ce{H-F}$ is made by $\ce{H^.}$ and $\ce{F^.}$ sharing the each valence electron. $\endgroup$ Commented Mar 11, 2021 at 5:46

Now , since F is more electronegative . The electron(by hydrogen ) is more pushed towards Fluorine. My confusion is that if F is more electronegative than H. Then , the electron should be present more towards the H side.

Electronegativity is a measure of an atom or an ion to attract electrons in a covalent bond and by definition both electrons of the covalent bond will be closer to F.

Now , we can notice that for orbital number 1. There is a dark region and a light region(right side). Now , what they say is that the darker region means the electron fining probability is more there. Similarly , for the number 2. Electron fining probability is more on left side.

This is not a London Dispersion Force.We use London Dispersion Forces to understand intermolecular forces of molecules which are non-polar.

  • $\begingroup$ You are right that what OP needs is a definition of Electronegativity. For the second part, the sketch could be well one to speak about London forces. It might represent an instant, we weren't present where and when it was sketched at what was the accompanying speech/text. This is just to no confuse even further the OP. $\endgroup$
    – Alchimista
    Commented Mar 10, 2021 at 14:45
  • $\begingroup$ London forces are between molecules not atoms.The OP clearly states he believes it is between atoms so I added it to get rid of the confusion. $\endgroup$
    – Miss Mulan
    Commented Mar 10, 2021 at 14:47
  • 1
    $\begingroup$ well there is no reference if not electron clouds influencing each other. We can't really reteach from scratch. It seems that OP is indeed confused and mixing too much things. For instance the first sketch is unrelated to the question about El Neg. Note that I have nothing against your answer. $\endgroup$
    – Alchimista
    Commented Mar 10, 2021 at 14:52
  • 2
    $\begingroup$ Op: keep things easy. Electronegativity is the tendency of an atom to attract towards itself the electrons of a covalent bond. Look at chapter about Electronegativity. If you are limiting the scenario to biatomic molecule X-Y, the difference between El Neg of X and Y dictates the electrical dipole moment. Molecule having a relevant one, means the molecules can interact each others via electrostatic interactions. Besides, the electron clouds can mutually influence each other leading to dynamical induced dipoles. Also leading to electrical interaction. Those are relevant in molecule without.. $\endgroup$
    – Alchimista
    Commented Mar 10, 2021 at 16:53
  • 1
    $\begingroup$ @user883022..... permanent dipole. $\endgroup$
    – Alchimista
    Commented Mar 10, 2021 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.