If the electronegativity difference between A and B is 0.8, what type of bond is formed?

Question

If the electronegativity difference between A and B is 0.8, what type of bond is formed?

Answer 1

well, it really depends on the two elements.  

for example, Si and O are more than 1.0 apart, but SiO2 is considered pretty covalent.

but Fr and Rh are more than .8 apart, but they're both metals so they can't bond as far as I know.

and Hg and I are about .8 apart, but I think HgI exists as an ionic compound. not sure if it exists, though.

I would think polar covalent, though.  

Answer 2

If B is fluorine, electronegativity 4.0, then A could be oxygen, electronegativity 3.5, or nitrogen or chlorine, electronegativities 3.0. NF3 or ClF would be covalent. If B is Cs, electronegativity 0.7, then A is titanium, electronegativity 1.5. A cesium-titanium compound would be intermetallic, bonded by metallic bonding in a crystalline lattice.

Which just goes to show that you can't dictate ionic bonding by saying the electronegativity difference is 0.8.

Answer 3

it would be most likely a polar covalent bond.