Quadruple CC-bond

[Function]

Hi everyone

 

Just wondering what's the problem with a quadruple bond between two C-atoms.

Why is this not possible?

I mean, if you promote an electron from the full s-orbit to the empty p-orbit, without hybridization, on each atom, why can't they have one sigma-bond and 3 pi-bonds? Or is hybridization 'obliged' with promotion of an s-electron?

 

Thanks.

 

F.

[Fuzzwood]

It is the nature of these bonds and the angle that they make with each other.

 

In an sp³ hybridized system you have 4 equal bonds to other atoms, ideally making an angle of cos^-1(-1/3) = 109.47°. In an sp² hybridized system you have 3 equal bonds with other atoms and an additional pi-bond with one of the 3 atoms. Ideally, the bonds now make an angle of 120°. In an sp hybridized system you have 2 equal bonds with other atoms and two additionals pi-bond with one or 2 of the other atoms. Ideally, the bonds now make an angle of 180°. The reason behind these angles is that the elektrons making up the bonds are as far away from the other bonds as possible.

 

The orbitals making up the pi bonds are perpendicular to the one making the sigma bond. To engage in a pi bond with another atom, these p orbitals sort of have to bend towards each other. This works for the p_x and p_y orbitals, but not for the p_z orbital as that one is antiparallel to the sigma bond and thus has to wrap around the entire atom for it to reach the other orbital. You can imagine the strain on this bond to exceed whatever bonding force the bond might have.

[Function]

It is the nature of these bonds and the angle that they make with each other.

 

In an sp³ hybridized system you have 4 equal bonds to other atoms, ideally making an angle of cos^-1(-1/3) = 109.47°. In an sp² hybridized system you have 3 equal bonds with other atoms and an additional pi-bond with one of the 3 atoms. Ideally, the bonds now make an angle of 120°. In an sp hybridized system you have 2 equal bonds with other atoms and two additionals pi-bond with one or 2 of the other atoms. Ideally, the bonds now make an angle of 180°. The reason behind these angles is that the elektrons making up the bonds are as far away from the other bonds as possible.

 

The orbitals making up the pi bonds are perpendicular to the one making the sigma bond. To engage in a pi bond with another atom, these p orbitals sort of have to bend towards each other. This works for the p_x and p_y orbitals, but not for the p_z orbital as that one is antiparallel to the sigma bond and thus has to wrap around the entire atom for it to reach the other orbital. You can imagine the strain on this bond to exceed whatever bonding force the bond might have.

 

i'm really convinced that your answer would make sense to me if I could visualize this...

see the image attached here; is the orange line that prevents the C-atom to undergo 3 pi-bonds in casu?

If so, why couldn't the electrons in these 'problematic' p-orbits just move to the part of the p-orbits that lie between the two atoms, between the sigma-bond?

[Fuzzwood]

That is exactly how I wanted you to grasp my explanation. As for your final bit: that won't work because of the nature of the p orbital, which has a negative and positive lobe, which in turn are solutions of quantummechanical functions. Both parts contribute to a pi bond equally. This essentially means that your image is half-complete: you need to draw 2 addiional purple lines between the P_x and _y yet unbound 'half-orbitals'.

 

http://en.wikipedia.org/wiki/Pi_bond

Edited September 27, 2014 by Fuzzwood

[Function]

That is exactly how I wanted you to grasp my explanation.

 

Great, but why won't the electrons just migrate to the parts of the p-orbits between the two atoms, forming a pi-bond lying 'on' the sigma-bond?

Edited September 27, 2014 by Function

[studiot]

I have given fuzzwood +1 for getting you to this stage.

 

However you have not quite drawn the two sp hybrid orbitals quite correctly.

For the situation we are talking about there are two number sp hybrid orbital per atom.

The original p orbitals are symmetric about the origin, both lobes are equal.

The sp hybrids are not. They have one large fat lobe sticking out to one side of the origin and one small tail lobe on the other side.

The fat lobes form the bonds.

Of the two in each atom

One fat lobe sticks out to the left and one to the right (along the x axis)

 

So the two overlapping fat lobes form a single bond along the x axis and the other bonding lobe for each atom sticks out on the other side of each atom (and therefore can easily bond to a hydrogen).

 

The other two p orbitals per atom overlap sideways on and form 2 weaker pi bonds

 

 

 

http://www.chemtube3d.com/orbitalsacetylene.htm

 

also

 

http://science.uvu.edu/ochem/index.php/alphabetical/g-h/hybridization/

Edited September 27, 2014 by studiot

[Function]

I have given fuzzwood +1 for getting you to this stage.

 

However you have not quite drawn the two sp hybrid orbitals quite correctly.

For the situation we are talking about there are two number sp hybrid orbital per atom.

The original p orbitals are symmetric about the origin, both lobes are equal.

The sp hybrids are not. They have one large fat lobe sticking out to one side of the origin and one small tail lobe on the other side.

The fat lobes form the bonds.

Of the two in each atom

On fat lobe sticks out to the left and one to the right (along the x axis)

 

So the two overlapping fat lobes form a single bond along the x axis and the other bonding lobe for each atom sticks out on the other side of each atom (and therefore can easily bond to a hydrogen).

 

The other two p orbitals per atom overlap sideways on and form 2 weaker pi bonds

 

 

 

http://www.chemtube3d.com/orbitalsacetylene.htm

 

But that's what I've drawn: 1 s-orbital per atom and 3 p-orbitals. That was my intention. I could've drawn sp-orbitals, but then my whole question wouldn't be necessary anymore.

Let me reform my question: is a hybridization of s- and p-orbital into spⁿ-orbitals obliged when promoting an s-electron to a p-orbital?

If not, then why can't just the two sole s-orbits form a sigma-bond, and the three p-orbits each a pi-bond, forming a quadruple bond? (Which was my original question)

Edited September 27, 2014 by Function

[studiot]

 

If not, then why can't just the two sole s-orbits form a sigma-bond, and the three p-orbits each a pi-bond, forming a quadruple bond? (Which was my original question)

 

 

 

The Geometry of three dimensions.

 

Two of the sets of p orbitals are sideways on and can form pi bonds (say p_y and p_z)

 

But that would leave the p_x orbitals facing end on as the bond is along the x axis.

[Function]

Very well, question answered clearly. Thank you both for the scientifical enlightment. +1 each.

 

Additional, secondary question: so as long as hybridization hasn't found place, one of the P-orbitals will stay 'useless'?

Edited September 27, 2014 by Function

[studiot]

Well that was why there is no common carbon to carbon quadruple bond and how the carbon to carbon triple bond works.

 

But there are also carbon to carbon double and single bonds which employ a different form of hybridisation.

These are discussed in my links.

[Function]

Yes, I understand hybridisation, thanks

[studiot]

I don't know if it was clear but my post#10 was meant to be an answer to your question about spare p orbitals in post#9

[Function]

I don't know if it was clear but my post#10 was meant to be an answer to your question about spare p orbitals in post#9

 

Sort of; just a small yes or no question: as long as no hybridisation has found place, one of the p-orbitals of an atom is useless?

[studiot]

You cannot have hybrid orbitals in an isolated free atom.

 

The standard s, p etc orbitals are the lowest energy solution.

 

But as soon as the atoms bond the combined orbitals (molecular orbitals) exist in (slightly) lower energy states by hybridisation, resonance or other mechanisms.

 

So the rearrangement for bonding from an s, p d ............ structure to a hybrid structure is a bit like clearing the decks for action in a warship.

It only happens in certain circumstance ie when there is a suitable dancing (bonding) partner or partners available.

 

The alternative approach to bonding is to go straight for solving quantum mechanics for the molecule rather than the atoms.

This would be tough postgraduate and research work.

 

The normal approach is called the Linear Combination of Atomic Orbitals (LCAO) method.

[Function]

Thanks. Think I get it now.