a polar bond is created when there is unequal sharing of electrons between two atoms. This unequal sharing in an assymmetrical compound creates an
electric dipole.
The dipole modifies the strength of the charges and this is what creates the difference between polar and non-polar molecules(that have partial
charges) in solutions?
That last paragraph is what i'm unsure of. Water is a polar compound but if it was symmetrical like this H-O-H it would not be. the partial charges
would be there but would the Hydrogen and Oxygen charges still have the same strength?
[Edited on 4-9-2008 by omario4]sonogashira - 4-9-2008 at 12:13
The more electronegative atom attracts the bonding pair of electrons towards itself more than the less electronegative atom (by definition).
Electronegativity increases from left to right across the periodic table and
Electronegativity increases from bottom to top of the periodic table, towards fluorine - the most electronegative atom.
Since electrons are negatively charged, the atom which attracts them the most will have a partial negative charge, which is written d- (where "d"
should be a greek delta), leaving the other atom with a slight positive charge, d+.
eg.
d+....d-
C-----N (nitrogen to the right of C)
d+.....d-
C------F (F to the right of C)
d+....d-
Br----F (F above Br)
In water there are 2 O-H bonds, and since O attracts the bonding pair of electrons more than H (it is more electronegative) it will have a partial
negative charge, d-, since there are negatively charged electrons closer to it than the hydrogen atoms, which will therefore have a slight positive
charge, d+.
d+...d-....d+
H----O----H
Since the charges are partial and not precisely defined there is no need to balance the charges.
"Dipole" refers to the presence of polarity within the molecule, so water is polar, as shown above.
A fluorine molecule, F----F is non-polar since both atoms are pulling at the electrons with equal force and neither are winning. Similarly H---H is
non-polar.
[Edited on 4-9-2008 by sonogashira]omario4 - 4-9-2008 at 14:15
err i was kind of wondering what difference thepolarity made to the partial charges on the atoms. Is waters hydrogen bond less effective in the linear
H-O-H structure?Blind Angel - 4-9-2008 at 14:20
Also, you may consider that the water molecule is not straight, it has a 105deg bend, it form some kind of triangle. You can imagine that the negative
part would be the top of the triangle and the base of the triangle would be more positive.
In case where the molecule is perfectly symmetric like in the case of C-Cl4 where it form a small triangular based pyramid with the carbon atom in the
middle, all the charge cancel each other. Imagine this same pyramid with 4 people with the same strength each pulling in their own direction, the
pyramid in the middle won't move since they all pull with the same force (if you know what vectorial mathematic is, when you add all vector you end up
with a null vector). If you take a very similar molecule, let say H-C-Cl3 (which is chloroform), this would be like replacing one of the person
pulling with a weaker one, the weakest person will be pulled slightly in the direction of the three other thus creating a slightly more positive
charge this way. The chloroform molecule is slightly polar.omario4 - 4-9-2008 at 15:07
Thanks for the help guys but i wanted to clarify that i know water has a bend in its structure, i was just wondering if the electrostatic bonds would
be weaker if the structure itself were no longer polar and was linearly shaped.Blind Angel - 4-9-2008 at 15:54
yes, if it was linear, the polarity vector would be null thus it would loose it's polarity. See CO2 for an example of such effect.omario4 - 4-9-2008 at 16:17
I''m curious why? the partial charges are still on there so shouldnt they attract other oppositely charged particles?Blind Angel - 4-9-2008 at 16:38
The partial charge are still there, but the resulting charge is 0 thus it's non-polar. If you take only the left part of O=C, this would be
effectively polar, but since the other side has exactly the same charge they cancel each other.
See this link under the Addition and Scalar Multiplication section for more info on how vector work:
http://en.wikipedia.org/wiki/Vector_(geometric)
You can consider charge as vector, the have an orientation and a magnitude which is bond strength or electro negativity. Let consider that the left
O=C bond has a magnitude of 2 and an orientation of -90deg, now the C=O bond has also a magnitude of 2 and an orientation of 90deg which is exactly
the opposite. To calculate it you can use X,Y coordinate.
for the magnitude: ((2sin90 + 2sin-90)2 + (2cos90 + 2cos-90)2)1/2 which =0
and for the orientation: tan(magnitude/(2sin90 + 2sin-90)) which also equal 0
(please correct this if I'm wrong, this is old in my memory and I'm doing it quickly)
You can try to calculate it also using a theoretical molecule using random value in O=C=S, the O=C bond has a greater magnitude than the C=S bond. If
you calculate the resulting vector you'll end-up with a vector pushing to the left.omario4 - 4-9-2008 at 16:51
thank you very much for your help Blind Angel, i will read your links and think it over.kmno4 - 5-9-2008 at 02:19
Quote:
(...) The partial charge are still there, but the resulting charge is 0 thus it's non-polar. If you take only the left part of O=C, this would be
effectively polar, but since the other side has exactly the same charge they cancel each other.
Your wrong - compensation of internal dipole moments does not mean non-polarity.
Something about 1,4 dioxane for example:
The low dipole moment of dioxane (i.e., 0.45 Debye) is an example of internal dipole compensation. Compared with diethyl ether, which has a dipole
moment of 1.15 Debyes, the dipole moment would be expected to be about half that of dioxane.However, there is strong intramolecular compensation
between the dipoles from each ether group within the molecule of dioxane. This reduces the field, as measured by external techniques, and gives a
value for dioxane of only about one-third of that of a diethyl ether molecule alone. However, another molecule approaching one ether group of the
dioxane molecule will be subject to the uncompensated field of a single dipole and interact accordingly. Consequently, although dioxane has a
very low dipole moment of 0.45, it is still a very polar substance that is completely miscible with water.
[Edited on 5-9-2008 by kmno4]Blind Angel - 5-9-2008 at 05:32
Yes thanks for pointing that out. You're right. My example was with very simple molecule though, with more complex molecule you must usually split
them to estimate the polartiy, as you have pointed. Usually for chain of longer than 3 carbon, each of you can consider that the active point are far
enough to be unafected by each other. It's simply a matter of spatial representation to evaluate the polarity of each chunk.
