Which of the following molecules shows sp^3d hybridisation?

The hybridisation of a central atom can be determined using the formula for steric number ($H$):

$H = \dfrac{1}{2}(V + M - C + A)$

where $V$ is the number of valence electrons on the central atom, $M$ is the number of monovalent atoms attached, $C$ is the cationic charge, and $A$ is the anionic charge.

For $\text{PCl}_5$:
Central atom is P ($V = 5$).
Number of monovalent Cl atoms ($M = 5$).
$H = \dfrac{1}{2}(5 + 5) = 5$
A steric number of $5$ corresponds to $\text{sp}^3\text{d}$ hybridisation.

For $\text{SO}_2$:
Central atom is S ($V = 6$).
Number of monovalent atoms ($M = 0$).
$H = \dfrac{1}{2}(6 + 0) = 3$ ($\text{sp}^2$ hybridisation).

For $\text{BF}_3$:
Central atom is B ($V = 3$).
Number of monovalent F atoms ($M = 3$).
$H = \dfrac{1}{2}(3 + 3) = 3$ ($\text{sp}^2$ hybridisation).

For $\text{CO}_2$:
Central atom is C ($V = 4$).
Number of monovalent atoms ($M = 0$).
$H = \dfrac{1}{2}(4 + 0) = 2$ ($\text{sp}$ hybridisation).

Thus, $\text{PCl}_5$ shows $\text{sp}^3\text{d}$ hybridisation.

Answer: $\text{PCl}_5$