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There is this one question that akss for the sum of bond dipoles. I don't understand that at all. What is the question asking? I thought I was figuring out if the difference between the dipoles and if I got a 0 then the molecule would be non-polar and if I got anything greater than 0 then the molecule would be polar. Could someone please clarify this? Thanks :)

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?