Which is polar??

[BaylorDDS]

Which of the following is polar?

a. CBr4
b. SO3
c. XeF4
d. ClF5
e. PBr3

The answer is d. It is from Destroyer (Q7). The explanation shows SO3 as all three Os double-bonded to S. The S has no lone pair, so it has only 6 valence electrons. why is this??

Also, how do you know the max number of bonds/lone pairs that S, P, Xe, etc can have??

Thanks mucho!

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[cyrano]

This is essentially a question on electron/molecular geometry. You can just imagine adding the dipoles as vectors.

a. CBr4: tetrahedral, dipole addition = zero
b. SO3: trigonal planar, dipole addition = zero
c. XeF4: four bonds in a plain with two lone pairs, one on top and one on the bottom. Lone pairs cancel each other out as do the four other bonds. dipole addition = zero
d. ClF5: this would have four bonds in a plain like XeF4, and would have one bond downwards and a pair of electrons on top. This pair of electrons is much more polar than and sigma bonds could ever be so there is a net dipole in that direction.
e. PBr3 : WTF?! I think this is polar too. It is trigonal pyramidal with a lone pair sitting on the top. This should cause a net dipole in that direction! Can someone explain why this isn't the case? Thx

As to the S there are pi bonds as well (remember that when you draw lewis structures there is actually a true structure which is a hybrid of all the resonance structures so there would be no dipole towards or away from pi bonds you draw).

 

[BaylorDDS]

my bad, PBr5. so its non-polar.

but in the XeF4, or ClF5 examples: the sigma bonds provide the central atom with more than 8 electrons, so why do you add lone pairs? and how do you know how many lone pairs you should have?

 

[cyrano]

As you move down the periodic table you'll see that atoms can have more than 8 valence electrons. For the first few rows stick to the octet rule, but after that some of these atoms go hog-wild with the number of valence electrons.

If you were to try to draw a lewis structure of Cl bonded to 5F without some extra valence electrons you'd have quite a few electrons left over. You need to place all your electrons. Chlorine doesn't mind taking the extra's b/c it's already a bit positive as a result of the 5F's each taking the electrons in the bond for more time (remember F is more electronegative than Cl) and Cl is also just bigger and more capable of taking on these electrons. The hybridization of these orbitals is dsp3. So things are a bit crazy, but this gives you the five hybridized bonding orbitals you need to bind the five F.

Same kind of thing with XeF4. Make sure to place all your electrons!

Hope this clears things up a bit. Sometimes its nice to go back to the basics and just draw your Lewis structures and make to sure place all your electrons.

 

[jbrenn4]

my bad, PBr5. so its non-polar.

but in the XeF4, or ClF5 examples: the sigma bonds provide the central atom with more than 8 electrons, so why do you add lone pairs? and how do you know how many lone pairs you should have?

there's an easy way to do this just count out the # of total electons draw your lewis structure with your most electorpositive atom as the central with all the electronegative atoms coming out wtvr leftover electrons you have should go as lone pairs.....once you know that you candetermine which is polar or non-polar

 

[jbrenn4]

also you should know as a general rule that S may disobey the octet rule....hope this helps

 

[Ryltar]

Here is the easiest way imo. Forget that valence electron formula, it's confusing. For ClF5, first determine the number of valence electrons for the center element. Because Chlorine is in group 7, the halogens, it will have 7 valence electrons. Each bond counts as 1 valence electron, and double bonds count as 2 and triple bonds as 3. Each nonbonding electron counts as a valence (e.g. a lone pair will count as 2) So with five fluorines we have 7 total valence electons minus 5 bonds, so 2 valence electrons remaining, thus we have a single lone pair on the chlorine that causes polarity.

 

[klutzy1987]

Here is the easiest way imo. Forget that valence electron formula, it's confusing. For ClF5, first determine the number of valence electrons for the center element. Because Chlorine is in group 7, the halogens, it will have 7 valence electrons. Each bond counts as 1 valence electron, and double bonds count as 2 and triple bonds as 3. Each nonbonding electron counts as a valence (e.g. a lone pair will count as 2) So with five fluorines we have 7 total valence electons minus 5 bonds, so 2 valence electrons remaining, thus we have a single lone pair on the chlorine that causes polarity.

I would do it a drop differently. You have 5 Flourine's and since each Flourine has 7 valence electrons you have a total of 35 valence electrons and since chlorine has 7 valence electrons you add those two giving you a final total of 42 electrons. That means that in the resulting molecule there will be a total of 42 valence electrons. So then you begin to draw it out, Draw your ClF5 and give all of the flourines an octet. That makes a total of 40 electrons allocated. However you have 42 valence electrons in the molecule so you need to add 2 more electrons as a lone pair. The lone pair is added to the central chlorine making the molecule polar with a total of 42 valence electrons.

 

[Ryltar]

Bottom of the line:
My method is a quick and dirty way to determine how many lone pairs there are, but will not answer questions correctly such as how many total valence electrons are there?

If you're asked for the total number of valence electrons, then you have to sum all of the valence electrons using Klutzy's method.

 

[BaylorDDS]

Thanks klutzy, that was my problem. I wasn't summing the valence electrons and then finding a place for them.