Question

Arrange the bonds in each of the following sets in order of increasing polarity:

1. C-F, O-F, Be-F

2. Be-Cl, Br-Cl, C-Cl

3. C-Cl, C-O, C-H

Answer 1

increasing polarity: the more electronegative difference, the more polar

therefore

1. C-F, O-F, Be-F ---> F is fixed, therefore, choose the least to most electronegative

Be < C < O ... then the "most polar" should be Be-F then C-F; finally O-F

2. Be-Cl, Br-Cl, C-Cl --> Cl is fixed so

Be < C < Br ... then the most polar shoul be: Be-Cl then C-Cl; finally Br-Cl

3. C-Cl, C-O, C-H ..  C is fixed, therefore:

special note; H has 2.1 in electronegativity therefore

C-Cl = 3-2.5 = 0.5

C-O = 3.5-2.5 = 1

C-H = 2.1-2.5 = -0.4

Then

most "polar" is C-O; then C-Cl and finally C-H;