To begin, drag the Na (sodium) and Cl (chlorine) atoms into the simulation area. Turn on Show valence electrons . A valence electron is found in the outermost energy level of the atom.

Let's take a compound NaCl ,

The outer shell electronic configuration of Na = [Ne] 2S2 2P0

On releasing one electron Na gets a stable noble gas like electronic configuration

But Cl on accepting one electron gets a noble gas electronic configuration, like Kr.

As the outer shell electronic configuration of Cl = [Ar]2S2 2P5

On accepting one electron it's electronic configuration is like Kr.

Hence, Na has low tendency to pull on the electrons they share during the bond formation between Na–Cl, compared to the Cl.

The tendency of Cl to pull on the electron , makes the Cl (-)ve end as the electrons intensifies here and Na become (+)ve end on lack of electrons.

Due to very much different tendency to pull on the bonding electrons NaCl become a ionic compound, the force acting between two ions is electrostatic in nature and the force in between two molicules also.

When the tendency to pull on the shared electron is different but not too much, then the the molicules are polar. Then the bond between two atoms are polar bond and the force between two molicules is dipole- dipole interaction.

And when the tendency to pull on the shared electron is almost equal, then the the molicules are covalent. Then the bond between two atoms are covalent bond and the force between two molicules is Van Dar Walls interaction.