Problem: Which of the following is correctly ranked in order of increasing bond polarity?a) B-F < C-Fb) Cl-S < Cl-Brc) F-O < O-Od) Al-O < P-Oe) Li-F < C-Of) H-H < N-Ng) O-O < O-F

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Electronegativity Difference        

              (ΔEN)                                     Bond Classification
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Zero (0.0)                                          Pure Covalent
Small (0.1 – 0.4)                                Nonpolar Covalent
Intermediate (0.4  – 1.7)                    Polar Covalent
Large (Greater than 1.7)                  Ionic               


↑ higher electronegativity difference → more ionic
↓ lower electronegativity difference → more covalent


Calculate the electronegativity difference and determine their bond classification:   

*the electronegativity value for each element can be found in books or the internet
*negative differences can be ignored since we are only looking at the value


a) B-F < C-F

 B–F

ΔEN = 2.0 – 4.0
ΔEN = 2.0


 C–F

ΔEN = 2.5 – 4.0
ΔEN = 1.5


 C-F < B-F


b) Cl-S < Cl-Br

 Cl–S

ΔEN = 3.0 – 2.5
ΔEN = 0.5


 Cl–Br

ΔEN = 3.0 – 2.8
ΔEN = 0.2


 Cl-Br < Cl-S


c) F-O < O-O

 F–O

ΔEN = 4.0 – 3.5
ΔEN = 0.5


 O–O

ΔEN = 3.5 – 3.5
ΔEN = 0.0


 O-O < F-O


d) Al-O < P-O

 Al–O

ΔEN = 1.5 – 3.5
ΔEN = 2.0


 P–O

ΔEN = 2.1 – 3.5
ΔEN = 1.4


 P-O < Al-O

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Problem Details

Which of the following is correctly ranked in order of increasing bond polarity?

a) B-F < C-F

b) Cl-S < Cl-Br

c) F-O < O-O

d) Al-O < P-O

e) Li-F < C-O

f) H-H < N-N

g) O-O < O-F

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