Diatomic molecules bond energies

[sludger13]

The bond enthalpies for these bonds are:
C-C 347 kJ/mol
C-N 276 kJ/mol
C-O 351 kJ/mol

Why is carbon-nitrogen bond enthalpy lower? What factor makes that?

And also the halogens bond enthalpies:
F-F 157 kJ/mol
Cl-Cl 243 kJ/mol
Br-Br 193 kJ/mol
I-I 151 kJ/mol

Why has fluorine lower enthalpy? I suppose a polarizability plays a role - iodine is highly polarizable, so when approaching two atoms of iodine, they make large dipoles so they do great work... or something like that. Is it true?

 

[Adithyan]

C-N bond is less stable as compared to C-O due to various factors. An interplay of electronegativities, bond length etc may be responsible for this. One can't really explain the exact cause of the order of bond enthalpies, so we have to rely upon the experimental values.

Speaking of F-F bond, flourine has a very high charge/radius ratio which results in high repulsion between the two flourine atoms. Hence it has greater bond length and lower bond enthalpy.

 

[sludger13]

C-N bond is less stable as compared to C-O due to various factors. An interplay of electronegativities, bond length etc may be responsible for this. One can't really explain the exact cause of the order of bond enthalpies, so we have to rely upon the experimental values.

As I'm aware of, bond length (as well as bond angle) is a consequence of two factors - atom radius and hybridised orbital character. Bond lengths are:
N-N 0.146 nm
O-O 0.148 nm
despite smaller oxygen atom. That is another mystery.

Speaking of F-F bond, flourine has a very high charge/radius ratio which results in high repulsion between the two flourine atoms.

That might make a sense - electrostatic repulsion of valence electrons. I'd need to check it deeper - as I'm adding the protons in period and not significantly moving with valence electron orbitals (I'm also adding electrons, but their repulsion is obviously minor), the electrons are more attracted so they are closer to the nucleus. Their charge density increases, but - what about nucleus electrostatic force? As I'm closer to nucleus in smaller atom with more protons, should I feel greater positive electrostatic force as well? That is a vital thing. Perhalps the nucleus is becoming more shielded in period as the electron density increases (I'm adding electron as well as proton plus I'm increasing ELECTRON density).
Also the electrostatic repulsion of atoms in period could be another factor for bond enthalpies, lengths.

Hence it has greater bond length and lower bond enthalpy.

Greater bond length than what?

 

[DrClaude]

N-N 0.146 nm
O-O 0.148 nm
despite smaller oxygen atom. That is another mystery.

The N-N is a triple bond, while O-O a double bond.

 

[DrDu]

Yes, for fluorine (not flourine) molecule would not even be stable according to MO theory. Only when electron correlation is taken into account, a stable bond results. Valence bond gives a stable bond directly. So due to the high electron density, much care has to be taken to allow the electrons to avoid each other to get stabe (though weak) bonding.

 

[sludger13]

The N-N is a triple bond, while O-O a double bond.

I considered it as an analogy to C-C and C=C, where smaller bond length is PRIMARY due to more (s) characted hybridized orbitals, not PRIMARY because of another (∏) bond.

Unless in nitrogen (and oxygen) molecule are sp (and sp2) hybridized orbitals, too! THAT WOULD MAKE A SENSE :)

 

[sludger13]

I just checked the tabulated values (http://jdem.cz/bageg7), and the bonds are indeed simple, not multiple (below them there are their multiple bond lengths).

Edit:

N-N (sp3) 0.146 nm
N=N (sp2) 0.12 nm
N≡N (sp) 0.11 nm

O-O (sp3) 0.148 nm
O=O (sp2) 0.121 nm

1) bond length N-N(sp2) = bond length N=N(sp2)

2) The oxygen has greater energy gap between (2s) and (2p) orbitals, so (sp3) hybridisation has relatively bigger (p) character, so the orbitals may be even larger on the contrary of smaller size of oxygen!

 

[sludger13]

But again, from this image:

the thesis just doesn't make sense, because energy gap "goes" down the energy, oxygen (p) orbital has of course lower energy than nitrogen (p) orbital.
Any opinion?

 

[DrDu]

I considered it as an analogy to C-C and C=C, where smaller bond length is PRIMARY due to more (s) characted hybridized orbitals, not PRIMARY because of another (∏) bond.

You have a reference for this statement?

 

[DrDu]

There are multiple factors which determine bond length.
In comparing oxygen and nitrogen, you also have to consider that p-orbitals in oxygen are smaller than in nitrogen as the p electrons are not very efficient in screening each other from the nuclear charge which is higher in oxygen than in nitrogen.

 

[sludger13]

You have a reference for this statement?

No, I take it back, the (π) bonding is the second factor shortening the bond.

Also these bonds are shortening because of more (s) character hybridisation AND more bonding electrons:

N-N (sp3) 0.146 nm
N=N (sp2) 0.12 nm
N≡N (sp) 0.11 nm

O-O (sp3) 0.148 nm
O=O (sp2) 0.121 nm

I don't understand one thing that could vindicate the (2s)-(2p) energy gap effect on hybridisation:

because the 3s-3p energy gap is large, hybridization of the orbitals costs too much energy, so bonding occurs primarily between the hydrogens and "pure" sulfur p-orbitals

N has a small energy difference between the 2s and 2p orbitals, so it's not a great energy penalty to hybridize and form sp3 hybrids to both hold the lone pair and form the bonds -- indeed, the overlap is stronger with the hybrids, the bonds are stronger, so you get a big benefit by doing so. But not tue with P. P-H is weaker than N-H, so the benefit to the bonds you can get from hybridization is reduced. P is bigger, Hs farther apart, steric benefit by avoiding steric repulsion is also reduced. The s-p gap is much much greater in P, so the energy penalty to hybridize is greater -- moving the lone pair out of a pure s into an ap3 hybrid isn't worth it. So you don't hybridize. PH3 uses three atomic p orbitals for bonds, keeps a (nearly) 90º angle, and puts the LP in a pure s orbital. As and Sb same argument but moreso.

In my point of view, the hybridization doesn't change the overall energy. Also what energy is needed for forming hybrid orbitals?

 

[Yanick]

In my point of view, the hybridization doesn't change the overall energy. Also what energy is needed for forming hybrid orbitals?

Just like MO's, hybrid orbitals are 'formed' when atomic orbitals fuse together to form other types of orbitals. In my limited understanding of all of these concepts, the difference between hybrid orbitals and MO's is that MO's are formed when two atoms fuse their respective orbitals into one molecular orbital whereas hybridization just looks at a sole atom fusing s and p orbitals into sp^x orbitals. The energy difference comes from taking, say, one lower energy s orbital and three higher energy p orbitals and combining them to make 4 medium energy sp³ orbitals. Obviously from an energy standpoint, the resulting orbitals should be in an overall lower energy than the separate orbitals we started with. I don't think these energies are necessarily calculated directly but more inferred from observations and calculations showing, for example, that the each C-H bond in methane is equivalent therefore the orbitals should be drawn as degenerate in whatever bonding/orbital model you decide to use.

This link explains it better than I can: Orbital Hybridization

 

[DrDu]

Hybridization is a useful concept in valence bond theory, however, it is not a property of the atoms in a molecule.
At the end of the day, you can get quite similar results starting out with different hybridizations. E.g. the double bond in ethene can equally well be described in terms of sp2+p or sp3 hybrid orbitals. In the latter case you get so called "banana bonds". Such a description makes a close connection to similar molecules like B2H6.
A description in terms of banana bonds lowers the energy as the bonding electron pairs better avoid each other, reducing electronic correlation energy.
Also for molecules like NH3 or H2O there is little difference between a description in terms of s+p or sp3 hybrids.

 

[cgk]

DrDu, I don't want to unnecessarily add to the confusion, but I think your statement that MO theory cannot properly describe F2, while VB theory can, is wrong. F2 most definitely has a minimum on the restricted Hartree-Fock potential energy surface (the prototypical MO method), and it is not far from the experimental equilibrium distance. The issue is simply that the energy of this minimum is lower than the energy of two isolated F atoms. This is a problem of electron correlation, as you pointed out: Any other MO method alternative to Hartree-Fock (e.g. Kohn-Sham) or beyond Hartree-Fock (e.g., all kinds of single-reference and multi-reference correlation methods) do get a stable F2 molecule, not only a meta-stable one. The reason this comes out of, say, a generalized valence bond treatment directly could be regarded simply as a consequence of GVB de facto being a more costly and more-difficult-to-implement approximation to CASSCF/CASCI and thus capturing a bunch of the correlation energy. It is not necessarily a consequence of this being a chemically more meaningful description.

Similarly, the description of ethene with banana bonds or sigma/pi bonds is mathematically identical in almost all MO theories, since those are generally unitarily invariant with respect to occupied orbital rotations, and both the pi/sigma and two-banana-bond descriptions can produce identical Hartree-Fock or CASSCF reference functions.

 

[DrDu]

F2 most definitely has a minimum on the restricted Hartree-Fock potential energy surface (the prototypical MO method), and it is not far from the experimental equilibrium distance.

Thank you, for pointing that out, I didn't know of this local minimum. However, I wouldn't have much confidence in properties calculated for F2 with Hartree Fock, only.



Concerning the banana bonds, all I was saying is that it yields an almost equivalent description of bonding in ethene, so that it makes no difference to use sp2 or sp3 hybrids.
The banana bonds are interesting as they are in a closer relation to other beginners ideas like VSEPR theory, but are not very present in teaching, that's why I gave some arguments what makes them worth to study.

 

[cgk]

Concerning the banana bonds, all I was saying is that it yields an almost equivalent description of bonding in ethene, so that it makes no difference to use sp2 or sp3 hybrids.
The banana bonds are interesting as they are in a closer relation to other beginners ideas like VSEPR theory, but are not very present in teaching, that's why I gave some arguments what makes them worth to study.

That is an excellent point. I misunderstood this before, but you are absolutely right.

 

[sludger13]

When I think about it, the reason of different hybridisation for this effect seems to me most probable. As I found out, (π) bonds are shortening the overall bond. Also multiple bonds with the same hybridisation are shorter.

When examinating other bond lengths, it is (qualitatively) obediently changing with different atomic radii. Just in case of nitrogen-oxygen this happens. And just in case of nitrogen-oxygen a triple bond occurs before double bond (not already in case of P-S).
I realized oxygens in (∙O-O∙) are (sp2) hybridised, the unpaired electrons are unhybridized. That is the best expression of single bond length of oxygen (when a substituent binds oxygen forming (sp3), the O-O length obviously always differs, so there must be marked what substituent it is; O2{2-} bond length is 0.162 nm).
Analogously, (∙∙N-N∙∙) is (sp) hybridized.

Also the chart should look like:

N-N (sp) 0.146 nm
N=N (sp) 0.12 nm
N≡N (sp) 0.11 nm

O-O (sp2) 0.148 nm
O=O (sp2) 0.121 nm

 

[DrDu]

I realized oxygens in (∙O-O∙) are (sp2) hybridised, the unpaired electrons are unhybridized.

No, that isn't possible. With oxygen, O2, the lone pair has high s character, something between s and sp. This does not leave enough s to form sp2.

 

[sludger13]

No, that isn't possible. With oxygen, O2, the lone pair has high s character, something between s and sp. This does not leave enough s to form sp2.

Also what is the hybridization of (∙O-O∙)? What is the hybridization of (∙∙N-N∙∙)? Why is (N-N) shorter than (O-O)?
Are you suggesting oxygen's lone pairs posseses more (s) character than in nitrogen -> oxygen's bonding orbitals are longer?

 

[DrDu]

Edit:

N-N (sp3) 0.146 nm
N=N (sp2) 0.12 nm
N≡N (sp) 0.11 nm

O-O (sp3) 0.148 nm
O=O (sp2) 0.121 nm

I just checked the source of this table you gave. There is nothing about hybridization in that table! It would be much easier to help you if you were to stop presenting us a mixture of facts and fiction.

Let's take some definite molecules: Hydrogen peroxide and hydrazine. In H2O2, the bond length in gas phase is 147.4 pm and the OOH bond angle is 94.8 degrees, see
http://en.wikipedia.org/wiki/Hydrogen_peroxide

In hydrazine, the bond length is 145 pm and the NNH angle 112 degrees.
These values correspond very well to the ones from your table.
If you want so, you could take the bond angles as an indication that oxygen is not appreciably hybridized while nitrogen is sp3 hybridized. Hence in hydrazine both the bonding overlap might be higher and the sterical repulsion between the hydrogens and lone pairs be lower than in hydrogen peroxide.

 

[sludger13]

I just checked the source of this table you gave. There is nothing about hybridization in that table! It would be much easier to help you if you were to stop presenting us a mixture of facts and fiction.

Sorry, the purpose was not to confuse someone, I thought everybody understand I only suggested some explanation.

In hydrazine, the bond length is 145 pm and the NNH angle 112 degrees.
These values correspond very well to the ones from your table.
If you want so, you could take the bond angles as an indication that oxygen is not appreciably hybridized while nitrogen is sp3 hybridized. Hence in hydrazine both the bonding overlap might be higher and the sterical repulsion between the hydrogens and lone pairs be lower than in hydrogen peroxide.

That is false, NNH angle in hydrazine is (103.1°) (http://www.chemeddl.org/resources/models360/models.php?pubchem=9321). The only lone pair still repels some more.
Oxygen should be (primary, of course) also (sp3) hybridised. Because of its two lone pairs, the (OOH) angle is smaller than in hydrazine.

ALSO, WHAT IS WRONG ABOUT THAT TABLE?

Thus, it seems to me like oxygen (σ) bonding orbitals are slightly bigger than nitrogen (σ) bonding orbitals despite smaller oxygen atomic radius, because of its higher (p) character in oxygen. Anyone possesses better explanation?
In fluorine (σ) bonding orbitals are fully (p), yet are smaller due to smaller fluorine radius.

Anyone possesses better explanation?

 

[DrDu]

That is false, NNH angle in hydrazine is (103.1°)

There seem to be different values around. Quite a reliable source is
www.nist.gov/data/PDFfiles/jpcrd146.pdf‎
which quotes a value of 109.2 degrees.

 

[DrDu]

Thus, it seems to me like oxygen (σ) bonding orbitals are slightly bigger than nitrogen (σ) bonding orbitals despite smaller oxygen atomic radius, because of its higher (p) character in oxygen.

If you assume both to be sp3 hybridized as in your table, how do you think that the p character can be higher in oxygen?

 

[sludger13]

If you assume both to be sp3 hybridized as in your table, how do you think that the p character can be higher in oxygen?

Because of its TWO lone pairs, compared to ONE nitrogen's lone pair. The lone pair possesses some (s) character; the more lone pairs, the greater is their overall (s) character.

 

[DrDu]

That's clear, but if the lone pairs are more as and the bonding orbital more p, that means that they are no longer sp3 orbitals.

 

[sludger13]

That's clear, but if the lone pairs are more as and the bonding orbital more p, that means that they are no longer sp3 orbitals.

...while nitrogen is sp3 hybridized.

As well as nitrogen hybridisation. Its lone pair is more (s) and bonding orbitals are more (p), also the real hybridisation is not the ideal (sp3) (+ the second factor of its shape is different atoms binding). But the shape is still most similar to tetrahedral. That's what I mean by PRIMARY hybridisation.