Option A

A charge–charge interaction falls off as

$$U_{qq}\propto\frac1r$$

while a charge–dipole interaction goes as

$$U_{q\mu}\propto\frac1{r^2}\,. $$

So the charge–dipole potential actually decays faster.

⇒ A is false.

Option B

A fixed dipole–dipole energy is

$$U_{\mu\mu}\propto\frac1{r^3}\,, $$

but once both molecules tumble freely, the thermal (Keesom) average gives

$$\langle U\rangle_{\rm Keesom}\propto -\frac{\mu^4}{kT\,(4\pi\varepsilon_0)^2\,r^6}\,. $$

That falls off as $1/r^6$, not $1/r^3$.

⇒ B is false.

Option C

The permanent–induced dipole (Debye) interaction is

$$U_{\rm Debye}=-\tfrac12\,\alpha\,E^2\propto -\frac{\alpha\,\mu^2}{(4\pi\varepsilon_0)^2\,r^6}\,, $$

and when you average over random orientations (assuming the interaction is weak compared with $kT$), no $T$ appears in the result.

⇒ C is true.

Option D

Even totally nonpolar species experience instantaneous‐induced (London) dispersion forces that attract as

$$U_{\rm disp}\propto -\frac1{r^6}\,. $$

⇒ D is true.

Answer: C and D.