Question

The electronegativities for the elements vary from 0.7 for cesium to 4.0 for fluorine. The electronegativity for iodine is 2.5. Based entirely on the general guidelines for electronegativities and bond character,

binary compounds with iodine should all be polar covalent with a δ+ on I.
no binary compounds with iodine should be substantially ionic.
binary compounds with iodine should all be polar covalent with a δ- on I.
compounds with iodine may be ionic, polar covalent, or nonpolar covalent.

Answer 1

i know that they should all be polar covalent, but I don't know whether you can say that they should all be polar covalent

Answer 2

@whpalmer4 @SolomonZelman can someone please help me with this? Thank you ^_^

Answer 3

@robtobey @AccessDenied

Answer 4

Looking at the general rules for bond polarity:
If the electronegativities are equal (i.e. if the electronegativity difference is 0), the bond is non-polar covalent
If the difference in electronegativities between the two atoms is greater than 0, but less than 2.0, the bond is polar covalent
If the difference in electronegativities between the two atoms is 2.0, or greater, the bond is ionic

If we look at the difference in electronegativities between I and Cs, then I and F, we could not get >2.0 difference for an ionic. But Iodine would be the positive end with respect for fluorine and negative with respect to cesium. So the only truth is that it is never quite ionic / 2nd option...

Answer 5

Oh, I kept thinking that it had to be one of the polar covalent answers

Answer 6

Iodine is one of the diatomic elements, so it does form a nonpolar covalent bond with itself in the form \( \rm I_2 \). :)

Answer 7

ok thank you!!

Answer 8

Glad to help!