Among all three PCl₃, NO₃⁻ , I₃⁻, H₂Se only I₃⁻ will involve participation of d-orbitals in hybridization as bonding in I₃⁻ will include, s, p and d orbital as shown in the image attached. there will be sp³d hybridization, there will be presence of three lone pairs and 2 bonds as
I⁻ has 8 valence electrons and 2 neighbours atoms which will need one electron each to satisfy their valency.
so the number of electrons on central atom will be:
8-1-1=6
That 6 electrons will make 6/2 =3 lone pairs.
