Key Idea:- Molecules having the same number of hybrid orbitals, have same hybridisation and number of hybrid orbitals, H=21[ V+X−C+A] where, V= no. of valence electrons of central atom X= no. of monovalent atoms C= charge on cation A= Charge on anion. (a) In SF4 H=21[6+4−0+0)=5 (b) In I3− H=21[7+2+1]=5 (c) In SbCl52−, H=21[5+5+2)=6 (d) In PCl5, H=21[5+5+0−0]=5 Since, only SbCl52− has different number of hybrid orbitals (ie, 6) from the other given species, its hybridisation is different from the others, ie, sp3d2. (The hybridisation of other species is sp3 d).