NF3 Dipole: Understanding the Electron Distribution and Polarization

[Stalker23]

Does NF3 have a dipole?

my teacher says that teh electronegativity attraction fo the electorns of the Flourines cancel each other out, and teh lone pairs of electrons cause a dipole towards the Nitrogen...is that right...shouldnt it be toward the flourines since there electronegativitis is higher, and they have more electrons.

thanks

 

[bomba923]

Lone pairs of electrons will almost always contain a larger partial negative charge than an atom. However, the flourines cancel out much of the lone pair's negative charge in the calculating the net dipole. But, the effect of the flourines is not enough to change the direction of the net dipole vector.

And so, your teacher is correct. The greater partial negative is held by the lone pair of electrons, and thus the net dipole moment vector will point towards the lone pair of electrons.

Click here, http://srdata.nist.gov/cccbdb/exp2.asp?casno=7783542 , or even http://srdata.nist.gov/cccbdb/dipole2.asp ! (for more information)

 

[Stalker23]

i gues that makes sence

at least i was rig tin some senses...although if she had told me that i wuld aagreed with he

thanks again

 

[Stalker23]

wait what am i saying

what about all the lone pairs around the flourines...each flourine has 3 lone pairs...thats 9 al together agaisn nitogens one lone pair? if lone pairs are so significant, then it would most deifinatly go towars teh flourines right?

dont mean to argues...jsut tryingo t understand a concept not even my tutor could show me

 

[bomba923]

Hmm...

Looking at the structure, you would see that the lone pairs of the fluorines do NOT extend out towards a single direction.

Rather, they are dispersed about the fluorine atoms and extend into mutliple directions, which severly reduces the gross dipole moment of that group of flourines.

~The lone pair on the nitrogens does extend into a single direction,
**where both electrons are within an sp3-hybridized orbital of the nitrogen.
-In comparison, the flourines are sp-hybridized, and their electrons will not generally extend comparatively farther from the nucleus.

Thus, with the multiple dipole moments of each lone pair on the fluorines reducing the gross dipole moment provided by the fluorines --, and the comparison of hydridizations of nitrogen and the fluorines (when referring to their particular lone-pairs)

will point the net dipole vector in the direction of the nitrogen's lone pairs.

 

[Stalker23]

well in our book, it never tells us to be conserned with the lone pairs, so excluding them, NF3, would have a dipole towards the flourines.

 

[Gokul43201]

wait what am i saying
what about all the lone pairs around the flourines...each flourine has 3 lone pairs...thats 9 al together agaisn nitogens one lone pair? if lone pairs are so significant, then it would most deifinatly go towars teh flourines right?
dont mean to argues...jsut tryingo t understand a concept not even my tutor could show me

These are all good questions. And the only way to answer them is to tell you that textbooks often ignore the details you are probing, in order to present the basic concept involved as concisely as possible.

Picture the NF3 molecule with the 3 F-atoms below the N-atom, and a lone pair above the N-atom. Now keep in mind that F being much more electronegative than N will have a greater share of the bonding electrons. For every pair of bonding electrons, let's make the crude assumption that F gets 1.5 of them and N nets 0.5 (I'm too lazy to try and dig up the actual numbers). Armed with this "data", let's look at each of the atoms.

First the N-atom : It has 2 electrons (lone pair) above but 3 regions below, each containing only 0.5 electrons. So, there's clearly a net upward moment.

Now let's concentrate on the F-atoms : I'm going to pick anyone of the 3 F-atoms, and refer to the direction towards the N-atom as upwards. So, each F atom has 1.5 electrons above it and 3 lone pairs (2 electrons each) below. This gives each of the F-atoms a small downward (to mean away from the N-atom) dipole moment.

So, looking at the big picture, we have an upward dipole moment on the N-atom and a downward moment from the 3 F-atoms.

It takes a little more work to tell which of these two dipoles is stronger (very briefly, the dipole moment on N will be stronger - even though it's one against three - because it's a bigger atom and the charge imbalance is greater**), but at the very least, you can see why the two contributions act against each other, resulting in a small net dipole moment (upwards).

Your teacher was wrong to say that the dipoles from the 3 F-atoms cancel each other. That can happen only in the molecule was planar (which it clearly is not). No, they do contribute a net downward dipole moment which causes a reduction in the total moment. Hence (and assuming I've not made any careless error here), NF3 should have a much smaller dipole moment than NH3 (where the moments from the H-atoms point upwards - towards the N-atom - and hence, increase the total moment).

(** Ask, if you want a description of how to do this calculation. I might not respond in a hurry - I'm going out of town - but I'll throw in a neat way to figure this out, when I find some time. It's something you won't find in most any text. )

 

[Stalker23]

hey man. That would be awsome if you have the time. maybe i could show that to my teacher. i finally convinced her that NF3 has a dipole towards the F's (which according to you is wrong), so now i need to know how to explian this to her. And when i say a dipole towards the F's, i mean the arrow goes towards the F's, showing that it is more negative...isnt that right?

thanks for all the help.