# Diatomic molecules bond energies

by sludger13
Tags: bond, diatomic, energies, molecules
P: 75
 Quote by DrDu 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?
P: 3,596
 Quote by sludger13 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.
P: 75
 Quote by DrDu 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.

 Quote by DrDu 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°) (reference). 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?
P: 3,596
 Quote by sludger13 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.
P: 3,596
 Quote by sludger13 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?
P: 75
 Quote by DrDu 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.
 Sci Advisor P: 3,596 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.
P: 75
 Quote by 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.
 Quote by DrDu ...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.

 Related Discussions Advanced Physics Homework 1 Chemistry 12 Quantum Physics 1 Advanced Physics Homework 9 Quantum Physics 0