[Pranav Jha]

I get that despite having polar bonds CF4 is an overall non-polar molecule. However, what i cannot understand is why CF4 has so low solubility in water (around 0.038% at 20 C)? Shouldn't the polar C-F bond mean that the C and F atoms can have dipole dipole interaction with O and H respectively? How does the overall polarity of the molecule play a role in this?

[Grundalizer]

Because the four C-F bonds dipoles "cancel eachother out" seeing that the molecule has symmetry.  Water on the other hand has two free pairs of electrons which are not "balanced" by the O-H dipoles, giving it a net dipole of whatever it's value is.  Remove one of the F and replace with anything else, and you will see an increase in water solubility is that "cancelling" effect is reduced/gone

[Schrödinger]

Because the four C-F bonds dipoles "cancel eachother out" seeing that the molecule has symmetry.  Water on the other hand has two free pairs of electrons which are not "balanced" by the O-H dipoles, giving it a net dipole of whatever it's value is.  Remove one of the F and replace with anything else, and you will see an increase in water solubility is that "cancelling" effect is reduced/gone

The OP has told us that he understands why CF4 is non-polar.

What he's asking is why the electronegativity of F doesn't make the molecule soluble in water by forming Hydrogen bonds...

[Pranav Jha]

yeah, i already understand why it is an overall non polar molecule and why the net dipole moment is zero . What i don't understand is shouldn't the polarity of the C-F bond mean that it should have a dipole-dipole interaction with the water molecule?

[Grundalizer]

is an overall non polar molecule

the net dipole moment is zero

What i don't understand is shouldn't the polarity of the C-F bond mean that it should have a dipole-dipole interaction with the water molecule?

No, because as you just said above, there can be no dipole-dipole interaction if there is no net dipole.  The bond length and strength of the C-F bond in CF₄ is also very short and extremely strong, so you aren't going to get any Van-der Walls bonding, and even less if any dipole interaction because the bond length if so short.

[Schrödinger]

What i don't understand is shouldn't the polarity of the C-F bond mean that it should have a dipole-dipole interaction with the water molecule?

@Pranav Jha : This part isn't correct... I thought you were referring to F's electronegativity all along.. The C-F bond isn't polar. The only reason I can see the molecule being soluble in water is the formation of Hydrogen bonds, which is again because of F's electronegative nature, not the bond's polarity