Question

arrange tbe following in oreer of increasing boiling point

CH4

H2Se

H20

also could you please explain how to find answer.

Answer 1

Since H2O can form strong H-BONDING, thus it has high boiling point whereas CH4 is non polar molecule and thus weak vanderwaal interactions are present in it. Hence CH4 has low boiling point. In H2Se, dipole-dipole interaction is present which is stronger than vanderwaal interaction but weaker than H-BONDING. Hence, order is :

CH4 < H2Se < H2O.

Please give feedback!