# Covalent bonding

1. Nov 2, 2016

### pinochet

Hi, I read in a textbook that covalent bonds are formed when atoms share an electron. The book gave an example of the H2 molecule.

I'm having trouble understanding how H2 can still bond when the nuclei and electrons are repelling. Shouldn't that cancel the attraction force between the atoms?

Last edited by a moderator: Nov 2, 2016
2. Nov 2, 2016

### Steven Hanna

This is an excellent question! Unfortunately, the explanation given in most gen/org chem classes is a lie. This explanation says that when electrons get to be in between the two nuclei, they are stabilized by attraction to the nuclei. However, you're right, the electrons being next to each other more than outweighs attraction to the two nuclei. You should draw a free body diagram to convince yourself of this.

So, based on electrostatics alone, a chemical bond should be a destabilizing interaction, not a stabilizing one. Therefore, we need to look to dynamics.

Quantum mechanics tells us that when electrons spread out into larger regions they are stabilized. This is because the kinetic energy of the bonding electrons is lowered. The potential energy of the electrons is actually increased due to repulsion between the bonding electrons.

Delocalization of a quantum mechanical entity lowers its kinetic energy because the kinetic operator is proportional to the second derivative of its wavefunction with respect to space. When the wavefunction gets spread out, the second derivative goes down.

Note that this is a very simplified explanation of what's going on, and this answer doesn't answer the question: "why does the kinetic operator depend on the 2nd derivative of the wavefunction?." This answer requires physics beyond what we covered in my undergrad.

Last edited: Nov 2, 2016
3. Nov 2, 2016

### James Pelezo

Yes, definitely good question. To understand the process of covalent bonding, you will first need to understand a few basic factoids...
1. Covalent bonding occurs between electrons in the valence shell (outer most electron energy level) of an element because of the inherent tendency of an unstable electronic system (unpaired electrons) to form a more stable electronic system (paired electrons).
2. A single covalent bond is, at it's simplest configuration, an electron pair.
3. The driving force for covalent bonding (again, at it's most fundamental level) is the pairing of an unstable system of unpaired electrons at the valence level of elements having the capacity to form an energetically more favorable pairing of electrons to form covalent bonds. However, in some cases, pairing of electrons occur forming a 'non-bonded electron pair' to fill an octet valence requirement.
4. An electron - as an individual entity - is a concentrated spinning mass having a negative charge. A spinning charged particle will have a dipole character; i.e., a weak negative end and weak positive end. The term used to describe an unpaired electron is 'Paramagnetic' and a paired set of electrons is 'Diamagnetic'.

Now, two Hydrogen atoms in close proximity will have a tendency to bond by covalent electron pairing of valence electrons because the dipole character of the two electrons will form a more energetically favorable configuration than if the electrons remained unpaired. As two electrons approach, one electron will 'flip' to give a +/- dipole that aligns with the -/+ dipole of another electron creating a stabilized electron pair. A commonly used image of this is as follows...

The 'North Pole - South Pole' attraction forms an electron pairing that is more energetically stable than for unpaired electrons. Such is the most fundamental driving force in all of covalent bonding. Beyond this, the nature of electron-electron interaction is described in many good texts on Quantum Mechanics and gives a more comprehensive overview of the behavior of electrons before and after forming the covalent bond consisting of 'Diamagnetic' paired electrons.

4. Nov 2, 2016

### Staff: Mentor

Eh. Unless the textbook's writers are deliberately trying to deceive you then it's not a lie as much as it is a simplification/over-simplification. Chemistry is inherently quantum-based but lower level textbooks can't teach at that level without completely losing their intended audience.

5. Nov 3, 2016

### Staff: Mentor

I have to agree with @Drakkith that you are overstating some things here. While the gist of what you said is correct, that the delocalization of electrons over more than one atom is the main contributor to covalent bonding, the screening effect of the elctrons in-between the nuclei cannot be neglected. It allows to understand, for instance, why the bond length of H2+ is much longer than that of H2.

That's too much of a classical picture to be instructive here.

6. Nov 4, 2016

### DrDu

You are right, atoms are overall neutral and therefore, classically, we wouldn't expect a strong attraction based on electrostatics which could give rise to a bond. As
Steven Hanna already pointed out, the mechanism behind bond formation is lowering of kinetic energy. This is fundamentally a quantum mechanical effect: According to Heisenbergs uncertainty relation, the mimimal momentum a particle can have is inversely proportional to the space which its wavefunction can occupy. In a bond this space becomes larger as compared to the single atom and momentum and kinetic energy become smaller.

7. Nov 4, 2016

### DrDu

Sorry, but the dipole dipole interaction is a very, very weak effect and does not explain bond formation. This should already been clear from the fact that even a single electron like in H2+ can form a very strong bond. The reason electronic spins pair up upon formation of two-electron bonds has nothing to do with spin spin interaction but with the Pauli principle which in its simplest form states that no two electrons with equal spin can be at the same position. This effectively leads to a strong repulsion of the electrons with equal spin whence they can't form a bond. For anti-parallel spins, no such repulsion arises.

8. Nov 4, 2016

### TeethWhitener

Just to be clear, Coulomb repulsion between these electrons still exists.

9. Nov 8, 2016

### Steven Hanna

Sorry about the exaggeration. At any rate, this explanation is taken from Anslyn & Dougherty's Modern Physical Organic Chemistry (p 812-813).

10. Nov 8, 2016

### Steven Hanna

So I was actually wondering about that for H2+. Isn't the bond length of H2+ longer than that of H2 simply because a second electron is not delocalized into the internuclear region? Attached is a figure from Anslyn & Dougherty that shows how the kinetic and potential energies of the first two solutions to the schrodinger equation for H2+ change with r. The potential energy of the gerade combo (V_g in the diagram) of 2 1s orbitals is always positive (relative to infinite internuclear separation), presumably due to electron/electron repulsions. To me this indicates that the e-/e- repulsions outweigh the attraction to the protons at any r. Why do you need to invoke attraction to protons to explain the longer bond in H2+?

11. Nov 9, 2016

### Staff: Mentor

This for H2+: there is no e-/e- repulsion.

12. Nov 9, 2016

### Steven Hanna

that's true I didn't consider that. but then why is V_g always positive?

13. Nov 10, 2016

### Staff: Mentor

Labeling error? The potential energy should have a minimum. Here is a figure from Demtröder, Atoms, Molecules and Photons

14. Nov 10, 2016

### DrDu

I don't think so. Rather, in the figure the energy curves resulting from bonding $1s_a+1s_b$ and anti-bonding $1s_a-1s_b$ are shown.

15. Nov 10, 2016

### Staff: Mentor

I would have expected the curve labeled $V_u$ to be $V_g$. What am I missing?

16. Nov 10, 2016

### DrDu

Sorry, I didn't see this. I thought you were talking about excited states in your diagrams. I agree that the u and g labels have been interchanged.
Edit: To be more specific: I think the labels Tg and Vg have been exchanged.

17. Sep 11, 2017

### Heisenburgundy Gold

I'm in my third week of Bio 101 class and I have to say this thread is somewhat helpful!