Latent heat of Vaporization Vs Latent heat of fusion Why is less energy required to change ice into water than to change the same mass of water into steam? (I do understand its required to separate the molecules and do work against the atmosphere to accommodate the volume expansion. I am looking for an answer in terms of the attractive forces).
 A: Ice has a lattice structure which is relatively "open plan" which results in its density being less than that of liquid water.  The molecules are held totter with strong bonds due the fact the the water molecule has a permanent dipole moment.
When you change solid ice into liquid water the liquid water still has within it regions where there is the open structure of the regular lattice of ice.  So the water molecules form clusters which exhibit a regular less dense crystalline structure.  As the temperature these structures break down which will tend to increase the density of water even though the increase in temperature has the effect of increasing the spacing between the water molecules.  The explains the fact that the density of liquid water is a maximum around $4^\circ \rm C$.
Above that temperature there is still strong bonding between the water molecules particularly as they are polarised.
So at the melting point not all the lattice bonds are broken and so less than the "expected" amount of energy is required to turn solid ice into liquid water. 
At the boiling point there is still very strong bonding between the water molecules in the liquid phase and so a lot of energy is required to break these bonds completely.
Please note that it is generally true that the latent heat of fusion is smaller than the latent heat of vapourisation because the bonds between molecules have to be completely broken in the transition from liquid to vapour whereas that is not so when going from solid to liquid as there is still strong bonding between molecules in the liquid phase.
There is a table here of values of latent heat which illustrates this point.
