Use the chart to determine which is an example of a highly polar covalent bond.

Question

Elements	Electronegativity
aluminum (Al)	1.61
calcium (Ca)	1
carbon (C)	2.55
chlorine (Cl)	3.16
fluorine (F)	4
hydrogen (H)	2.2
iron (Fe)	1.83
nitrogen (N)	3.04
oxygen (O)	3.44
phosphorus (P)	2.19
potassium (K)	0.82
silicon (Si)	1.9
sulfur (S)	2.58

A. 

Ca – Cl

B. 

Fe – O

C. 

H – S

D. 

F – F

Answer 1

The difference in electronegativity is used to define the type of bond. The ranges for these bonds is somewhat arguable (different sources list different ranges) but generally you could say that for electronegativity difference (ΔEN) greater than 1.7 the bond is mostly ionic

For ΔEN of about 0.4 to 1.7 the bond is mostly polar covalent

For ΔEN of approx less than 0.4 the bond is mostly covalent

And for ΔEN of zero, the bond is nonpolar covalent

For a) the ΔEN is 4 - 1 = 3 (mostly ionic)

For b) the ΔEN is 3.44 - 1.83 = 1.61 (mostly polar covalent)

For c) the ΔEN is 2.58 - 2.2 = 0.32 (mostly covalent)

For d) the ΔEN is 4 - 4 = 0 (nonpolar covalent)

So the answer would be b)