[Winga]

C in alkanes is sp3 hybridized.
C in alkenes is sp2 hybridized.
C in alkynes is sp hybridized.

sp hybridized C has more s character than that of the others.

s-orbital holds the e- density more firmly than that of p-orbital.
But the reason which my lecturer give me is that the electronegativity of s-orbital is higher that that of p-orbital, so it holds the e- density more firmly.

My question is that can we use electronegativity to apply to orbitals? (My lecturer said yes) Was he wrong?

I learnt it from my perious lecturer which she told me the reason is due to the penetrating power of orbitals, s > p.

Is there any other terms to describe it besides these two?

[Mitch]

Yes there is an other way it is called Molecular Orbital Theory. The way you are being taught is the VESPR way. As an undergraduate just try to soak in as much as you can of either theory.

[movies]

The second explanation that you gave is the better one.  Since s-orbitals penetrate closer to the nucleus of the atom, they are lower in energy.  The p-orbitals are "shielded" from the nucleus by the electrons in the s-orbital, so they don't get to "see" as much of the positive charge at the nucleus.

Strictly speaking, electronegativity is a property of atoms and atoms only.  You can calculate similar values for different functional groups (e.g. nitro) which are like electronegativity, but are really not quite the same.

[Donaldson Tan]

when compare sp3, sp2 and sp, you find increasing s character in the the hybrid orbital involved. since the s orbital is more stable than the p orbitals, an increasing s character suggest more stable hybrid orbitals. This leads to stronger bond, and thus shorter bond length.

[Winga]

The diagram below shows the rdf curves for the 1s, 2s and 2p orbitals shown above plotted on the same figure.



The most probable place to find an electron in a 2s or 2p orbital is further out from the nucleus than that for an electron in a 1s orbital, as expected from Bohr's atom. However, an electron in a 2s orbital has quite a large probability of being found closer to the nucleus than a 1s electron, as indicated by the shaded region on the diagram. Because of this time spend close to the nucleus, the 2s electron experiences a larger Zeff. The probability of finding a 2p electron closer to the nucleus is much smaller and so,

Zeff (Li 2s) > Zeff (Li 2p)


(Quoted from a website )

[Winga]

an electron in a 2s orbital has quite a large probability of being found closer to the nucleus than a 1s electron

Is it correct?

[Mitch]

Look at your own chart, it doesn't look like it.

[Winga]

The 4s orbital, however, is more penetrating than the 3d orbital. (The 4s orbital has 3 radial nodes and a 3d orbital has none). An electron in a 4s orbital thus has a fairly high probability of being closer to the nucleus than the core electrons. The core electrons shield the nuclear charge from a 4s electron less than they do for a 3d electron. A 4s electron experiences a higher Zeff than a 3d electron and so it is better to occupy the 4s orbital in K. (When there are no core electrons to penetrate - as in the hydrogen atom - the penetrating power of an orbital is irrelevant).

(from the same website)

The 4s orbital has 3 radial nodes and a 3d orbital has none

What is the relationship between the nodes and penetrating power?

(I think I have mixed-up)
4s is more penetrating than 3d orbitals, what's that mean?
It means the e- in 4s orbital has more chances to penetrate the core electrons to the nucleus?

[Mitch]

Doesn't quite sound like this is appropriate in the Organic Chemistry forum. Look up the graph of a 4s versus 3d electron and you'll see why.

[Winga]

An electron in a 4s orbital thus has a fairly high probability of being closer to the nucleus than the core electrons.

The ascending order of penetrating power of of s-orbitals:
1s < 2s < 3s < 4s ..

Is it correct?

[Mitch]

looks good.

[Mitch]

Actually these are better.

http://www.shef.ac.uk/chemistry/orbitron/AOs/4s/radial-dist.html

http://www.shef.ac.uk/chemistry/orbitron/AOs/3d/radial-dist.html

[Winga]

I have found the relationship between the electronegativity and the s character of carbon from a textbook.

It said that the higher the s character of carbon , the higher the carbon's electronegativity.

I understood now, but the term, electronegativity is only applied to atoms but not orbitals, right?

[Mitch]

right

[dexangeles]

as far as i remember, orbitals are only there to show the probability of finding electrons....is this correct?  

if so, then electron density in orbitals tells you right there that it cannot be electronegative because it's negative on it's own right, and the charge law states that they should repulse, correct?

aren't they also known as wave functions?

just questions to brush up on