[Ahmed Abdullah]

In chemistry van der Waals strain is strain resulting from van der Waals repulsion when two substituents in a molecule approach each other with a distance  less than the sum of their van der Waals radii.
I understand van der Waals force should always be attractive unless they are closer than the sum of their van der Waals radii. But when I look to an eclipsed conformation of an ethane molecule,  it doesn't seem that hydrogen atoms are very close to each other.  At least in the way they are drawn, even 3d model cannot diminish my confusion.
It implies that when ethane gas condense to form solid , the distance between ethane molecule should be larger than the distance in hydrogen atoms in it.
It's a large gap! I have seen that van der Waals radii is comparable to that of bond length, so it seems odd that eclipsed Hydrogens in C2H6 are closer than C-H or C-C distance.
Could anyone help me?

[Rabn]

Are you taking into consideration the atomic size of the hydrogen? Center on center compared to carbon hydrogen is much smaller.  Is it possible that the missing distance is made up for, in part, by differing atomic sizes?  You also need to keep in mind that the covalent bond is much stronger than repulsion caused by van der Waals interactions.  It is likely that the C-H bonds become skewed when ethane is in the eclipsed conformation.  You should also keep in mind the dynamics that a molecule has.  Unfortunately our models cannot replicate what a molecule is really like.  The bonds aren't rigid and the atoms don't sit like pegs in a hole. Every part of the molecule is dynamic, the data we have are just averages of what has been observed.