Order the following bonds according to their degrees of polarity, based on electronegativities of the element?

Question

P-H
H-F
O-Br
Al-Cl
O-N

Answer 1

1) First we must know the electronegativities of each participating element in the bond:

P - H (electroneg. of P = 2.19, electroneg of H = 2.2)
H - F (electroneg. of F = 3.98, electroneg. of H = 2.2.)
O - Br (electroneg. of Br = 2.96, electroneg of O = 3.44)
Al - Cl (electroneg. of Al = 1.61, electroneg. of Cl = 3.16)
O - N (electroneg. of O = 3.44, electroneg. of N = 3.04)

2.- We can take the difference of electronegativities in each bond:

P-H diff=0.01
H-F diff=1.78
O-Br diff=0.48
Al-Cl diff=1.55
O-N diff=0.4

3) Finally we can order this differences from higher to lower. The higher the difference the more polar the bond:

H-F > Al-Cl > O-Br > O-N > P-H

H-F is the more polar bond and P-H the less polar bond.

That´s it!

Good luck!