Question

How can you determine which bond in a structure is more polar without using an electronegativity table?

Answer 1

How can you determine which bond in a structure is more polar without using an electronegativity table?

Step-by-step explanation:

When we say something is polar, it has a very specific meaning. It means that there is a spatial separation of charges, or a dipole moment. In the simplest case, this is given by:

μ=Qr

Where μ is the dipole moment, Q is the charge on each pole (equal and opposite) and r is the distance between the charges.

Anything with charges can have a dipole moment, and therefore be polar. In chemistry, we typically worry about whether or not two things are polar:

Bonds

Molecules

Bond Polarity

Chemists are concerned with the polarity of bonds because polarity affects the character of the bond - the more polar the bond is, the more it behaves like an ionic bond. The less polar, the more it behaves like a covalent bond. This has implications for everything from naming rules to reactivity, but all you need to know for this problem is:

All bonds between two atoms are polar, unless the atoms are identical.

(This is including whatever else the atoms may be bonded to). The only question after that is "how polar is the bond," and that is where tables of electronegativity come in. Using an arbitrarily defined scale, we can measure the relative polarity of bonds by comparing the difference in electronegativity between the two elements.

Molecule Polarity

For molecules, polarity isn't quite so simple. We know that all bonds between dissimilar atoms are polar, but in a molecule, sometimes the dipole moments add up to form a net dipole moment of zero. This is a little bit difficult to explain if you don't have a mathematics and physics background that includes vectors and summations of forces/moments.

If you do, then when I say something like:

The net dipole moment for a molecule is equal to the sum of dipole moments over each bond.

It will make sense to you. If you don't, then you have to get a little more creative. I tell my students to imagine that there are ropes connecting each outer molecule to the central molecule, along each bond. Then imagine pulling each rope towards the side that is more electronegative (has a higher electron density). The more polar the bond, the harder you pull on the rope. Then imagine whether the molecule moves. If the forces balance, it stays put, so the net dipole moment is zero and it is not polar. If it does move, there was a net dipole moment, so it is polar.

This works pretty well - as long as you can visualize the molecular geometry. That's the hard part. To know how the bonds are oriented in space, you have to have a strong grasp of Lewis structures and VSEPR theory. Assuming you do, you can look at the structure of each one and decide if it is polar or not - whether or not you know the individual atom electronegativity. This is because you know that all bonds between dissimilar elements are polar, and in these particular examples, it doesn't matter which direction the dipole moment vectors are pointing (out or in).

Let's look at each one, using wikipedia's geometric pictures.

BrF5

BrF5

As you can see, there is a lone pair at the bottom (this is a square pyramidal geometry). The net dipole moment will be pointing "up", which makes this a polar molecule.

PBr5

PBr5

This one has a trigonal bipyramidal geometry, with each bond symmetrically opposed to each of the others. If you add up the vectors, they result in a net dipole moment of zero. Therefore, it is non-polar.

EDIT: As ron points out in the comments below, this is not really a binary molecular compound, and in reality forms an ionic crystal structure with PBr4+ and Br−. I don't think the author of the question intended you to worry about that, but if they did, the answer is still "non polar," although we wouldn't really call it a "molecule" any more.

CCl4

CCl4

Carbon tetrachloride has a tetrahedral geometry, and all the dipole moment vectors cancel. Therefore, it is non-polar.

XeF2

XeF2

This one is linear - no net dipole moment, so it is non-polar.

XeF4

enter image description here

I could only find the ball-and-stick model for this one. It is square planar - there is a lone pair of electrons (not shown) on each "face" of the square on Xe. There is no net dipole moment, so it is also non-polar.

Out of all of these, the only polar molecule is BrF5 (the first one).

If the net dipole moment is zero, it is non-polar. Otherwise, it is polar.

Answer 2

To know this you pretty much do have to kind of memorize a few electronegativities. I don't recall ever getting a table of electronegativities on an exam.
From the structure, you have:

I remember the following electronegativities most because they are fairly patterned:
EN
H
=
2.1
EN
C
=
2.5
EN
N
=
3.0
EN
O
=
3.5
EN
F
=
4.0
EN
Cl
=
3.5
Notice how carbon through fluorine go in increments of
~
0.5
. I believe Pauling made it that way when he determined electronegativities in the '30s.
Δ
EN
C
−
Cl
=
1.0
Δ
EN
C
−
H
=
0.4
Δ
EN
C
−
C
=
0.0
Δ
EN
C
−
O
=
1.0
Δ
EN
O
−
H
=
1.4
So naturally, with the greatest electronegativity difference of
4.0
−
2.5
=
1.5
, the
C
−
F
bond is most polar, i.e. that bond's electron distribution is the most drawn towards the more electronegative compound as compared to the rest.
When the electron distribution is polarized and drawn towards a more electronegative atom, the less electronegative atom has to move inwards because its nucleus was previously favorably attracted to the electrons from the other atom.
That means generally, the greater the electronegativity difference between two atoms is, the shorter you can expect the bond to be, insofar as the electronegative atom is the same size as another comparable electronegative atom.
However, examining actual data, we would see that on average, in conditions without other bond polarizations occuring:
r
C
−
Cl
≈
177 pm
r
C
−
C
≈
154 pm
r
C
−
O
≈
143 pm
r
C
−
F
≈
135 pm
r
C
−
H
≈
109 pm
r
O
−
H
≈
96 pm
So it is not necessarily the least electronegativity difference that gives the longest bond.
Therefore, you cannot simply consider electronegativity. Examining the radii of the atoms, you should notice that chlorine is the biggest atom in the compound.
r
Cl
≈
79 pm
r
C
≈
70 pm
r
H
≈
53 pm
r
O
≈
60 pm
So assuming the answer is truly
C
−
C
, what would have to hold true is that:
The
C
−
F
bond polarization makes the carbon more electropositive (which is true).
The now more electropositive carbon wishes to attract bonding pairs from chlorine closer, thereby shortening the
C
−
Cl
bond, and potentially the
C
−
H
bond (which is probably true).
The shortening of the
C
−
Cl
bond is somehow enough to be shorter than the
C
−
C
bond (this is debatable).