Match the following columns :

	List-I (Pair of species)		List-II (Same characteristics of pair)
P)	$\left({\mathrm{NF}}_3,{\mathrm{OF}}_2\right)$	1)	d-orbit.al participation in hybridisation of central atom is observed
Q)	$\left({\mathrm{PCl}}_5,{\mathrm{ICl}}_3\right)$	2)	Tetrahedral electron geometry of central atom
R)	$\left({\mathrm{BeF}}_2,{\mathrm{SO}}_3\right)$	3)	Total number of lone pair
S)	$\left({\mathrm{CO}}_2,{\mathrm{SO}}_2\right)$	4)	$p_{\pi }\mathit{.}{p}_{\pi }$bond is present

 

Because nitrogen has Sp³ hybridization and forms three bond pairs, one with each fluorine atom, the molecular geometry of NF₃ is a trigonal pyramid, and its electron geometry is tetrahedral.
Hybridization is hence sp³d. Due to its correct orientation, the d-orbital involved in the sp³d hybridization is 3d_z2.
Eight lone pairs of electrons make up sulphur trioxide (SO₃), also known as sulfuric oxide and sulfuric anhydride.
When there are more bonds than there are unhybridized p orbitals, p-d bonds will develop. P-P bonds are created first.

Thus the option B is correct.