Using the idea of electronegativity difference for predicting polarity of bonds and considering the shape of the following molecules, match the compounds in the order of strongest to weakest expected polarity. (1 = strongest; 5 = weakest) hbr i2 hf hcl hi

See electronegativity is the tendency of an atom to gain an electron and flourine with a valecy of one and a vey small size is the most electronegetive because its orbitals are quite closed to the nucleus and hence the attraction is quite strong so it can attract an electron.the question that arises is that some smaller atoms should be more electronegetive as they are closer to the nucleus but it need more energy for them as compared to flourine to complete their octet. now polarity increases when two atoms of quite different sizes form a compound ... the more electronegetive atom will always attract the bond pair forming a negetive charge on it and positive on the less electroneg. one and polarity increases with electronegetivity of the anion.now as your question says 
5=I2.. because both the atoms are same there wont be permanent polarity 
4=HI iodine is the least electronegetive of all the halogens due to its large size,electronegetivity decreases down the group 
3=HBr bromine is the 2nd largest halogen 
2=HCl chlorine is the 3rd largest halogen 
1=HF fluorine is the smallest halogen making and hence makes the most polar