Posted by Lena on Wednesday, July 8, 2009 at 9:58pm.
You need to type the entire question. I'm not sure what it means.
Basically it gives me a formula like Ch3Cl and it says to draw the lewis structure and the VSEPR name. The it asks for the number of bonded atoms and number of lone pairs the molecule geometry, whether or not there are polar bonds, the sum of bond dipoles and whether the molecule is polr or not. So this is what I wrote for CH3Cl.
VSEPR Name: AX4
Bonded atoms: 4
Lone Pairs: 0
Molecular Shape: tetrahedral
Polar Bonds: YES
Sume of Bond dipoles: the options are =0, not = 0 or N/A
Polar Molecule: YES
All of your answers are correct. The sum of bond dipoles is not zero BECAUSE the molecule is not symmetrical. A molecule like CH4 would be AX4 also, it would have individual dipoles between each H and C but the sum would be zero and the CH4 molecule is tetrahedral and symmetrical; therefore, it is a non-polar molecule.
Where would I use N/A ? If the molecule is polar then the sum would be not = 0 and if it isnt polr then the sum would be 0. Correct>?
Yes. I suspect, but I don't know, that the N/A was there to make it an answer from a choice of three instead of a choice of two; otherwise, its just a true/false type question.
IF5 would be not equal to 0 right?
You are right. The IF5 molecule is AX5U type, coordination number of 6, which makes it an octahedral molecule but the U (the lone pair) the molecule is not symmetrical; therefore, it is polar (the sum of dipoles is not zero).
where is the lone pair? isn't iodine completely bonded to the fluorines?
I may not have made it clear that its ELECTRONIC geometry is octahedral.
the molecular geometry would be a trigonal bipyramidal though, correct?
Isn't it square pyramidal?
Square base and four edges with F on each corner and the top.
oops, you`re right. Thanks :)
Would the I have a lone pair?
Yes, I had a lone pair. Put I in the middle, and place 5 F atoms around itmore or less to form a pentagon. Now start adding electrons so you have 8 around each F. That will take 40. But you have 7 x 6 = 42 total so the extra 2 electrons must go to the central atom or I. That's the lone pair and the U in AX5U.
should I have 6 around every F? :S
You should have 8 electrons around each F. That uses 40 electrons (8*5=40) and if you count the electrons around I there are 10. But you have 42 electrons total (6 atoms of 7 electrons each = 42); therefore, you must place 2 more electrons on I to make 12 electrons on I and 8 on each F.